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l^ouuL |g 0^.8 (.3 




SCIENCE CENTER LIBRARY 




h---i 



'7< 



ELEMENTS 



OF 



QUATERNIONS 



BY 

A. 8. HARDY, Ph.D., 

PROFESSOR OF MATHEMATICS, DARTMOUTH COLLEGE. 



BOSTON: 

PUBLISHED BY GINN, HEATH, & CO. 
1881. 



JbtL.k\^ allege Library 
x/ Aug. 28, 19V *, 

FffOxn .l';«t9te of 
F-Qf e Oh^rlos J. W l -t- . 



Entered according to Act of Congress, in the year 1881, by 

A. S. HARDY, 
in the office of the Librarian of Congress, at Washington. 



Gink, Heath, & Co. 

J. S. Cushing, Printer, i6 Hawley Street, 

Boston. 



PREFACE. 



rpHE object of the following treatise is to exhibit the 
elementary principles and notation of the Quaternion 
Calculus, so as to meet the wants of beginners in the 
class-room. The Elements and Lectures of Sir William 
Rowan Hamilton, while they may be said to contain the 
suggestion of all that will be done in the way of Quater- 
nion research and application, are not, for this reason, as 
also on account of their diffuseness of style, suitable for 
the purposes of elementary instruction. Tait's work on 
Quaternions is also, in its originality and conciseness, 
beyond the time and needs of the beginner. In addition 
to the above, the following works have been consulted: 

Calcolo dei Quaternione. Bellavitis ; Modena, 1858. 

Exposition de la Mithode des Equipollences. Traduit 
de l'ltalien de Giusto Bellavitis, par C.-A. Laisant ; Paris, 
1874. (Original memoir in the Memoirs of the Italian 
Society. 1854.) 

Theorie Elementaire des Quantites Complexes. J. 
Hoiiel; Paris, 1874. 

Essai sur une Maniere de Representer les Quantites 
Imaginaires dans les Construction Greometriques. Par 
R. Argand ; Paris, 1806. Second edition, with preface 



IV PREFACE. 

by J. Hoiiel ; Paris, 1874. Translated, with notes, from 
the French, by A. S. Hardy. Van Nostrand's Science 
Series, No. 52; 1881. 

Kurze Anleitung zum Rechnen mit den {Hamilton 'scheri) 
Quatemionen. J. Odstrcil; Halle, 1879. 

Applications Mecaniques du Calcul des Quaternions. 
Laisant; Paris, 1877. 

Introduction to Quaternions. Kelland and Tait; Lon- 
don, 1873. 

A free use has been made of the examples and exercises 
of the last work ; and, in Article 87, is given, by permis- 
sion, the substance of a paper from Volume I., page 379, 
American Journal of Mathematics, illustrating admirably 
the simplicity and brevity 6i the Quaternion method. 

If this presentation of the principles shall afford the 
undergraduate student a glimpse of this elegant and pow- 
erful instrument of analytical research, or lead him to 
follow their more extended application in the works above 
cited, the aim of this treatise will have been accomplished. 

The author expresses his obligation to Mr. T. W. D. 
Worthen for valuable assistance in the preparation of 
this work, and to Mr. J. S. Cushing for whatever of 
typographical excellence it possesses. 

A. S. HARDY. 
Hanover, N.H., June 21, 1881. 



CONTENTS. 



CHAPTER I. 

Addition and Subtraction of Vectors; or, Geometric 
Addition and Subtraction. 

Article. Page. 

1. Definition of a vector. Effect of the minus sign before a 

vector 1 

2. Equal vectors 2 

8. Unequal vectors. Vector addition 2 

4. Vector addition, commutative 3 

5. Vector addition, associative 3 

6* Transposition of terms in a vector equation 4 

7. Definition of a tensor 4 

8. Definition of a scalar 5 

9. Distributive law in the multiplication of vector by scalar 

quantities 6 

10. If 2a -f 2/3 = 0, then 2a = and 2/3 = 7 

11. Examples 8 

12. Complanar vectors. Condition of complanarity 15 

13. Co-initial vectors. Condition of collinearity 16 

14. Examples 17 

16* Expression for a medial vector 24 

16. Expression for an angle-bisector 25 

17. Examples 26 

18. Mean point 28 

19. Examples 28 

20. Exercises 30 

CHAPTER II. 

Multiplication and Division of Vectors; or, Geometric 
Multiplication and Division. 

21. Elements of a quaternion 32 

22. Equal quaternions 34 

23. Positive rotation 35 



VI CONTENTS. 

Article. Page. 

24. Analytical expression for a quaternion. Product and quo- 
tient of rectangular unit-vectors. Tensor and versor of a 

quaternion 36 

25* Symbolic notation q = TqVq 39 

26. Reciprocal of a quaternion 39 

27. Quadrantal versors, i, j, k 40 

28. Whole powers of unit vectors. Square of a unit vector is — 1, 41 

29. Associative law in the multiplication of rectangular unit- 

vectors 42 

30. Negative sign commutative with i, j, k 43 

31. Commutative law not true for the products of i, j, k . . . . 43 

32. Reciprocal of a unit vector 44 

33. A unit vector commutative with its reciprocal. Reciprocal 

of any vector . . . . ; 45 

34. Product and quotient of any two rectangular vectors .... 47 

35. Square of any vector 47 

36. Distributive law true of the products of i. j % k 48 

37. Exercises 48 

38. Symbolic notation, q = Sq + Vq 49 

39. De Moure's theorem 50 

40. Products of two vectors. Symbolic notation 56 

41. General principles and formulae 58 

42. Powers of vectors and quaternions 61 

43. Relation between the vector and Cartesian determination of 

a point 04 

44. Right, complanar, diplanar and collinear quaternions. Any 

two quaternions reducible to the forms i, Z and 2, Z . . 65 

45. Reciprocal of a vector, scalar and quaternion 66 

46. Conjugate of a vector, scalar and quaternion 68 

47. Opposite quaternions 70 

48. UKg=ui = — = KVq 71 

49. Representation of versors by spherical arcs 71 

50. Addition and subtraction of quaternions. K, S and V dis- 

tributive symbols. K commutative with S and V. T and 

U not distributive symbols 72 

51. Multiplication of quaternions ; not commutative. Vllq = TlUq. 

Tng=nTg. Kgr=KrKg. Product or quotient of com- 
planar quaternions 74 

52. Distributive and associative laws in quaternion and vector 

multiplication 71) 

53. General formulae 82 



CONTENTS. VU 

Article. Page. 

54. Applications 82 

55* Formulae relating to the products of two or more vectors . . 99 

66. Exercises 106 

57. Examples. Applications to Spherical Trigonometry .... 108 

68. General Formulae 119 

59. Applications to Plane Trigonometry 125 



CHAPTER in. 
Applications to Loci. 

60. General equations of a line and surface 131 

61. Use of Cartesian forms in conjunction with quaternion rea- 

soning 132 

62. Non-commutative law in quaternion differentiation. Differ- 

entiation of scalar functions 132 

63. Quaternion differentiation 134 

64. Illustration 136 

65. Distributive principle 138 

66. Quadrinomial form 139 

67. Examples 140 

The Right Line. 

68. Right line through the origin 145 

69. Parallel lines 145 

70. Right line through two given points 146 

71. Perpendicular to a given line ; its length 147 

72. Equations of a right line are linear and involve one indepen- 

dent variable scalar 150 

The Plane. 

73. Equation of a plane 151 

74. Plane making equal angles with three given lines 152 

76. Plane through three given points 153 

76. Equations of a plane arc linear and involve two independent 

variable scalars 154 

77. Exercises and problems on the right line and plane 154 

The Circle and Sphere. 

78. Equations of the circle 164 

79. Equations of the sphere 165 



Vlll CONTENTS. 

Article. Page. 

80. Tangent line and plane 166 

81. Chords of contact 167 

82. Exercises and problems on the circle and the sphere .... 167 

83. Exercises in the transformation and interpretation of ele- 

mentary symbolic forms 176 

The Conic Sections. Cartesian Forms. 

84. The parabola * , 178 

86. Tangent to the parabola , 178 

86. Examples on the parabola 180 

87. Relations between three intersecting tangents to the para- 

bola 185 

88. The ellipse 191 

89. Examples' on the ellipse . 192 

90. The hyperbola ' 195 

91. Examples on the hyperbola 195 



92. Linear equation in quaternions. Conjugate and self-conju- 

gate functions 199 

93. General equations of the conic sections 201 

94. The ellipse 204 

95. Examples 206 

96. The parabola 214 

97. Examples 216 

98. The cycloid 222 

99. Elementary applications to mechanics 223 

100. Miscellaneous Examples 231 



ELEMENTS OF QUATERNIONS. 



QUATERNIONS. 



CHAPTER I. 

Addition and Subtraction of Vectors, or Geometric Addition and 

Subtraction. 

1. A Vector is the representative of transference through a 
given distance in a given direction. 

Thus, if a, b are any two points, vector ab implies a trans- 
lation from a to b. 

A vector may be represented geometricallj- bj' a right line, 
whose length denotes the distance over which transference takes 
place, and whose direction denotes the direction of the trans- 
ference. In thus designating a vector, the direction is indicated 
by the order of tiie letters. 

Thus, ab (Fig. 1) denotes transference FIg ' lm 

from a to b, and ba from b to a. 

Retaining the algebraic signification of the signs -f- and — , if 
ab denotes motion from a to b, then — ab will denote motion 
from b to a, and 

AB=— BA, — AB=BA . . . . (1). 

Hence* the effect of a minus sign before a vector is to reverse 
its direction. 

The conception of a vector, therefore, implies that of its two 
elements, distance and direction; it was first defined as a directed 
right line. It is now applied more generallj r to all quantities 
determined by magnitude and direction. Thus, force, the path 




2 QUATERNIONS. 

of a moving bod} r , velocit}', an electric current, etc. , are vector 
quantities. 

Analytically, vectors are represented by the letters of the 
Greek alphabet, a, /?, y, etc. 

2. It follows, from the definition of a vector, that all lines 
which are equal and parallel may be represented by the same vec- 
tor symbol with like or unlike signs. 
If equal and drawn in the same 
direction, they will have the same 
sign. Hence an equality between 
two vectors implies equalit}' in dis- 
tance with the same direction. 

Thus, if ab (Fig. 2) , cd, be, ep 
and hg are equal and drawn in the same direction, the} r may be 
represented by the same vector symbol, and 

AB = CD = BE = EF = HG = a . . . . (2) . 

3. It follows also from the definition of a vector that, if vec- 
tors are not parallel, they cannot be represented by the same 
vector symbol. 

Thus, if the point a (Fig. 3) move over the right line ab, 
from a to b, and then over the right line bc, from b to c, and 

ab = a, bc must be denoted by 
some other S3'mbol, as /?. 

The result of these two succes- 
sive translations of the point a is 
the same as that of the single and 
direct translation AC=y, from a to 
c ; in either case a is found at the 
extremity of the diagonal of the 
parallelogram of which ab and bc are the sides. This combina- 
tion of successive translations is called addition, and is written 
in the ordinary way, a + /3 = y (3). 

This expression would be absurd if the symbols denoted mag- 
nitudes only. It means that transference from a to b, followed 




GEOMETRIC ADDITION AND SUBTRACTION. 



3 



by transference from b to c, is equivalent to transference from 
a to c. The sign + doss not therefore denote a numerical ad- 
dition, or the sign = an equality between magnitudes. It is, 
however, called an equation, and read, as usual, "a plus /? is 
equal to y." This kind of addition is called geometric addition. 

4. If the point a (Fig. 3) , instead of moving over the sides 
ab, bc of the parallelogram abcd, had moved in succession over 
the other two sides, ad and dc, the result would still have been 
the same as that of the single translation over the diagonal ac. 
But since ab and bc are equal in length to dc and ad respect- 
ively, and are drawn in the same direction, we have (Art. 2) 

ab = dc and bc = ad, 

and if the first two translations are represented by ab and bc, 
the second two may be represented by bc and ab, or 

a + /? = /? + a = y (4). 

Hence the operation of vector addition is commutative, or the 
sum of any number of given vectors is independent of their order. 



5. If the point a (Fig. 4) move in succession over the three 
edges ab, bc, cg of a parallelopiped, 

Fig. 4. 



we have 
and 


AB + BC 
AC + CG : 


= AC, 
= AG, 




or 


(AB -f BC) + 


CG = 


AG. 


In like 


manner 








BC + CG : 


= BG, 






AB -f- BG : 


= AG, 




or 


AB + (BC + 


cg) = 


AG. 




Hence 



(AB + BC) -f CG = AB + (BC + CG) 



(*), 



and the operation of vector addition is associative, or the sum 
of any number of given vectors is independent of the mode of 
grouping them. 



4 QUATERNIONS. 

6. Since, if ac = y (Fig. 3) , then ca = — y, we have 

a + /?-y=0, 

or, comparing with equation (3) , 

a + /? = y, 

a term may be transposed from one member to another in a vector 
equation by changing its sign. 

Also, in every triangle, any side may be considered as the 
sum or difference of the other two, depending upon their direc- 
tions as vectors. Thus (Fig. 3) 

y-/? = a, 
y-a = /3. 

It is to be observed that no one direction is assumed as posi- 
tive, as in Cartesian Geometry. The only assumption is that 
opposite directions shall have opposite signs. The results must, 
of course, be interpreted in accordance with the primitive as- 
sumptions. Thus, had we assumed ba = a (Fig. 3) , y and /? 

being as before, then 

/?-a==y, 

a — f3= — y. 

7. If two vectors having the same direction be added together, 
the sum will be a vector in the same direction. If the vectors 
be also equal in length, the length of the vector sum will be the 
sum of their lengths. If n vectors, of equal length and drawn 
in the same direction, be added together, the sum will be the 
product of one of these vectors by n, or a vector having the same 
direction and whose length is n times the common length. If 
then (Fig. 2) 

v AF = SCAB = SCCD = Xa, 

where a, b and f are in the same straight line, cd = ab, and x 
is a positive whole number, x expresses the ratio of the lengths 
of af and a. From the case in which x is an integer we pass, 
by the usual reasoning, to that in which it is fractional or in- 
commensurable. Vectors, then, in the same direction, have the 
same ratio as the corresponding lengths. 



$M 



GEOMETEIC ADDITION AND SUBTRACTION. 5 j 

i 
If ab = a be assumed as the unit vector, then 

af = ma, 

in which m is a positive numerical quantity and is called the 
Tensor. It is the ratio of the length of the vector ma to that 
of the unit vector a, or the numerical factor by which the unit / 
vector is multiplied to produce the given vector. 

Any vector, as /?, may be written in general notation 

In this notation, T/3 (read "tensor of /?") is the numerical 
factor which stretches the unit vector so that it shall have the 
proper length ; hence its name, tensor. It is, strictly speaking, 
an abstract number without sign, but, to distinguish between it 
and the negative of algebra, it may be said to be always posi- 
tive. U/3 (read " versor of /?") is the unit vector having the 
direction of /? ; the reason for the name versor will appear later. 

T and U are also general symbols of operation. Written be- 
fore an expression, they denote the operations of taking the 
tensor and versor, respectively. Thus, if the length of ft is n 
times that of the unit vector, 

T(/?) = n, 

where T denotes the operation of taking the stretching factor, 
i.e. the tensor. While 

U(/?)=U/? 

indicates the operation of taking the unit vector, that is, of 
reducing a vector /? to its unit of length without changing its 
direction. 

& If bc (Fig. 5) be any vector, and ba = #bc, then 

-BA = AB=-yBc; Fig5 

and, in general, if ba and bc be b c a 

any two real vectors, parallel and 

of unequal length, we ma} r always conceive of a coefficient y 

which shall satisfy the equation 

ba = ^BC, 



6 QUATERNIONS. 

where y is plus or minus, according as the vectors have the same 
or opposite directions, y may be called the geometric quotient, 
and is a real number, plus or minus, expressing numerically the 
ratio of the vector lengths. This quotient of parallel vectors, 
which may be positive or negative, whole, fractional or incom- 
mensurable, but which is always real, is called a Scalar, because 
it may be always found by the actual comparison of the parallel 
vectors with a parallel right line as a scale. 

It is to be observed that tensors are pure numbers, or signless 
numbers, operating only metrically on the lengths of the vectors 
of which they are coefficients: while scalars are sign-bearing 
numbers, or the reals of Algebra, and are combined with each 
other by the ordinary rules of Algebra ; they may be regarded 
as the product of tensors and the signs of direction. 

Thus, let 

a= aUa. 

Then Ta = a. If we increase the length of a by the factor 6, 
b is a tensor, but the tensor of the resulting vector is ba. If we 
operate with — &, — b is not a tensor, for a is not only stretched 
but also reversed ; the tensor of the resulting vector is as before 
ba ; in other words, direction does not enter into the conception 
of a tensor. As the product of a sign and a tensor, — b is a 
scalar. The operation of taking the scalar terms of an expres- 
sion is indicated by the symbol S. Thus, if c be any real alge- 
braic quantity, 

S(-&al T a + c) = c, 

for — ba Ua is a vector, and the only scalar term in the expres- 
sion is c. 

9. It is evident from Art. 7 that if a, 6, c are scalar coeffi- 
cients, and a any vector, we have 

(a + b + c) a = aa -f 6a + ca . . . . (6) . 

Furthermore, if (Fig. 6) 

OA = a, AB = /3, BC = y, OA' = ma, 



GEOMETRIC ADDITION AND SUBTRACTION, 
then, a'b' being drawn parallel to ab and b'c' to bc, 
a'b' s= m/3, b'c' = my. 



Now 



OC = a + /? + y, 



Fig. 6. 



and 



oc' = moc = m (a + fi 4- y) . 
Bat we have also 




Hence 



OC f = OA f -f A f B f -f B f C' 

= ma + mfi + my. 

m (a + ft -f- y) = ma + m/J + my 



(7), 



or the distributive law holds good for the multiplication of scalar 
and vector quantities. 



10. It is clear that while 

a — a= 0, 

a ± 8 cannot be zero, since no amount of transference in a direc- 

* r 

tion not parallel to a can affect a. 
Hence, if 

na + m/J = 0, 

since a and ft are entirely independent of each other, we must 
have 

na = and m/3 = 0, 
or 

n = and m = 0. 
Or, if 

ma + nf$ = m'a + w'/J, 
then 

m = m' and w = w r . 

And, in general, if 

2a-i-2/? = 0, 
then 

2a=0 and 2/3: 



;..l 



(8). 



Pig. 7. 



QUATERNIONS. 

Three or more vectors may, however, neutralize each other. 
Thus (Fig. 7) 

€-/?-a=0, 

' P and this whether abcd be plane or 
gauche. In any closed figure, there- 
J fore, we have 

A a B 

* + P + y + S + =0, 

where a, /?, y, 8, , are the vector sides in order. 




U. Examples. 

1. The right lines joining the extremities of equal and parallel 

right lines are equal and parallel. 

Fig. 8. 

o ^ v Let oa and bd (Fig. 8) be 

the given lines, and oa = a, 
bo = ft, da = y. Then, by 
condition, BD = a. 
B ^ ^> Now, 

BA = BO + OA = p -f- a ; 

BA = BD + DA = a -f- y ; 

or, equating the values of ba, 

P + a = a + y. 
Hence (Art. 2) , y = /?, and bo is parallel and equal to da. 




also, 



2. The diagonals of a parallelogram bisect each other. 

In Fig. 8 we have 

bd = oa = op -f- pa ; 

also 

bd = bp -f pd ; 

.' . OP -f PA = BP -f PD. 

But, op and pd being in the same right line, 

op = mPD. 
Similarly 

J PA = WBP. 



GEOMETRIC ADDITION AND SUBTRACTION. 



Hence 
and 



mPD -f- ?ibp = pd + BP, 
m= 1, w= 1, 

op = PD, bp = PA. 



3. If two triangles, having two sides of the one proportional 
to two sides of the other, be joined at one angle so as to have 
their homologous sides parallel, the remaining sides shall be in a 
straight line. 

Let (Fig. 9) ab = a, ae = p. Then, n Fi * 9 - 

by condition, dc = xa, db = xft. 
Now 

/ \ /<* 

CB = CD -|- DB = X (/? — a) . 




But 



BE = ft — a. 



Hence (Art. 2), b being a common point, cb and be are one 
and the same right line. 

4. If two right lines join the alternate extremities of two 
parallels, the line joining their centers is half tlie difference of 
the parallels. 



Pig. 10. 



We have (Fig. 10) 

AB = AD + DC + CB, 

and, also, 

AB = AE -f- EP -f- FB. 

Adding 

O ° 

2 AB = (AD -f- AE) + (DC + EP) + (CB + Ffi) 
= EF — CD ; 

or, as lines, 

ab = i (ef — cd) . 




10 



QUATERNIONS. 



5. The medials of a triangle meet in a point and trisect each 
other. 

Let (Fig. 11) bo = o, cd = /3. Then 

OC = a, DA = /?. 

Now 

BA = 2a+2/?=2(a + /3), 

and, since od= (a + /J), ba and od are 
parallel. 
Again 




^ O A> 



BP -f- PA = BA = 20D = 2 (OP + PD) . 

But bp and pd, as also op and pa, lie in the same direction, 

and therefore _ 

bp = 2 pd and pa = 2 op. 

Hence the medials oa and db trisect each other. 
Draw cp and pe. Then 



and 






BP=2PD = $BD = $(2a + 0), 

'cP=CB + BP = f (2a-f /3)-2a = $ (0-a), 
PE = PB + BE = a + /?--f (2a + 0)=i(/?-a). 



Hence pe and cp are in the same straight line, or the medials 
meet in a point. 

6. In any quadrilateral, plane or gauche, the bisectors of 
opposite sides bisect each other. 

We will first find a value for op (Fig. 12) under the supposi- 
tion that p is the middle point of 
ge. We shall then find a value for 
op, under the supposition that p is 
the middle point of fh. If these 
expressions prove to be identical, 
these middle points must coincide. 
In this, as in many other problems, 
the solution depends upon reaching 
the same point by different routes and comparing the results. 



Fig. 12. 




GEOMETRIC ADDITION AND SUBTRACTION. 
Let OA = a, OB = /?, OC = y. 

OC + CG = OE + EG. 



1st. 
But 



which, in (a), gives 



y + *(j8-y)=*a + EG. 
.-. EP = ^EG = i(y-f /? — a), 
OP = OE + EP = ba + i (y + /? — a) 



or 



11 



(a) 



<») 



2d. FH — £AB = FO + OA, 

FH-i(/3~a) = -iy+a. 
.-. FP = iPH = i(a + ^-y), 
OP = OF + FP = iy + i (a + /? — y) 

= ±(a + /? + y), 



which is identical with (b) . Hence, the middle points of fh 
and ge coincide. 



7. If abcd (Fig. 13) be any parallelogram, and op any line 
parallel to dc, and the indicated lines be drawn, then will mn 
be parallel to ad. 



Fig. 13. 



Let AM = a, 


BM = /?. 




Then 










A0 = 


ra, 






AD = 


8a + q/3, 






OD = 


— ra + sa + q/3 


We have 


NM = 


NO + 


in which 










no = x (— ra + sa + qf$) , 

OM=(l — r) a, 

NP=y (-rp + 8a + qf3)> 

pm = (1 - r) 0. 



D 



12 QUATERNIONS. 

Substituting in the above equation, we obtain, by Art. 10, 

rig. 13. — xr -f- xs + 1 — r = ys, 

xq=—yr + yq + l—r. 

Eliminating y 

„ I-' 
x== . 

r 
Substituting this value in 
nm = no + OM 

= — — ( - ra + sa + qfi) + ( 1 - r ) a 

T 

1-r 




r (s« + qP) 
1-r 



AD. 



r 
Hence ad and nm are parallel. 

8. If through any point in a parallelogram, lines be drawn 
parallel to the sides, the diagonals of two of the parallelograms 
so formed will intersect on the diagonal of the original parallelo- 
gram. 

«*•". Let (Fig. 14) oa = a, ob = /?. 

c Then or = ma, oe = nfi. 
We have 




7» 

, RD=RO+OE-f ED = n/J+(l— m)a, 

B B ES=EO-j-OR-j-RS=ma-j-(l— n)/3. 

Also 

FO = FR + RO = ORD + RO = tt[> j 3 + (l-m)al-ma, (a) 
and ' v / 

fo = fe + eo = yss + eo = y [ma + (1 - n) /?] - n/3. (b) 

From (a) and (b) 

nx = y(l—n) — n and x (1 — m)- m = ym.. 
Eliminating y 

1 — m — n 



GEOMETRIC ADDITION AND SUBTRACTION. 13 

Substituting this value of x in (a) 

fo = ™ [nB + (1 — m) a] — ma 

1 — m — n 

mn //?, \ 



1—m — n 
or, fo and oc = (/? -f a) are in the same straight line. 

9. If, in any triangle oab (Fig. L5), a line od be drawn to 
the middle point o/ab, and be produced to any point, as f, and 
the sides of the triangle be produced to meet af and bf in h and 
r, tfiew tw# hr be parallel to ab. 

Let oa = a, ob = /?. Then or = a?a, 
oh = yfi, AB = /? — a. 
Now 




OD = OA + £AB = i (a + /?). 

Also, of = 2 (a-f /?), that is, some 
multiple of od. 
Then, 1st. 

BR=pBF, 
— P + Xa=p (— /3-f-OF) 

.•. #=#3 and — l=pz — p. (a) 

Eliminating z 

p = x + l. 

And, 2d. 

ah = gAF, 

-a + yp = q (-a + OF) 

= g[-a + *(a + /?)]; 
.-. y=qz and — 1 = qz — g. (6) 

Eliminating z 

q = y + l. 
From (a) and (b) 

x y 
p q 



14 



QUATEBKIONS. 



and, since p = x + 1 and g = y + 1, 



r,g - M - x = y and p = q. 

B ^ .•. rh = ro + oh = y/3 — xa = x(/3 — a) 
= o?ab, 




Pig. 16. 



or, rh and ab are parallel. 

10. If any line pr (Fig. 16) be drawn, cutting the two sides 
of any triangle abc, and be produced to meet the third side in q, 
then 

PC . BQ . RA = CR . AQ . BP. 

Let bp = a, cr = /?. Then pc =i>a, 
ra = r/8 and ba = bc -f ca = (1 -f p) a 
+ (l + r)/8. 

We have 

aq = x*a = x [(1 +p) a + (1 + r) 0], 
as also 

aq = ar + rq = - r/3 + ypR = - rfi + y (p a + 0). 
.\ a?(l+p) =yp and a? (1 -f- r) = — r + y. 

Eliminating y * = (X + *) pr . 

whence 

AQ_BQ PC RA 

ba~~ba bp * cr' 




or 



PC . BQ . RA = CR . AQ • BP. 



11. If triangles are equiangular, the sides about the equal 
angles are proportional. 

Let (Fig. 17) Bc = a, ca = 0. Then be = ma, ed = n& 
bd = ma + n/3 and ba = a + f$. 
Now 

BD = 2>BA, 

ma + n0 = i >(a + 0). 
Whence 

m = jj, n = p and m = n. 

.'. be: bc: : ed : ca. 



*Z. 




GEOMETRIC ADDITION AND SUBTRACTION. 15 

12. If, through any point o (Fig. 17), within a triangle abc, 
lines be drawn parallel to the sides, then will 

ED gf , hi 

CA CB AB 

Let ca = /?, cb = a. Then ab = 
a — p, ed = m/?, hi = p (a — p) and 

GF = fia. 

We have 

CO = CG + GO = CH -f- HO. (d) 

Now, as lines, 

GF GA .^ \ n 

— = — = n, .\ cg = ca — ga= (1 — n) B. 

CB CA \ / r- 

EB ED , H v 

— = = m, .\ GO= CE = CB — EB =(1— m) a. 

CB CA V ' 

DB DE /i \ / fl \ 

= = m, .'. HO = AD = AB — DB = (1 — m) (a — /?). 

AB AC 

Substituting in (a) 

(1 - n) £ + (1- m) a = .p/3 + (l - m)(a- 0), 
or (Art. 10) » + „*+* = 2. 



12. Complanar vectors are those which lie in, or parallel to, 
the same plane. If a, p, y are any vectors in space, they are 
complanar when equal vectors, drawn from a common origin, 
lie in the same plane. 

If a, p, y are complanar, but not parallel, a triangle can al- 
ways be constructed, having its sides parallel to and some mul- 
tiple of a, p, y, as aa, bp, cy. If we go round the sides of the 
triangle in order, we have 

aa + bp + cy = 0. 

If a, p, y are not complanar, conceive a plane parallel to 
two of them, as a and p. In this plane two lines may be drawn 
parallel to and some multiple of a and /?, as aa and bp ; and 
these two vectors may be represented by p8 (Art. 3) . 



16 QUATEENIONS. 

Now pS, being in the same plane with aa and bp, cannot 
therefore be equal to y, or to any multiple of it ; pS and y can- 
not therefore (Art. 10) neutralize each other. Hence 

p8 -|- cy = aa + bp + cy cannot be zero. 

If, then, we have the relation 

aa + bp + cy = 

between non-parallel vectors, they are complanar; or, if a, p, y 
be not complanar, and the above relation be true, then, also, 

a = 0, 5 = 0, c = 0. 

13. Co-initial vectors are those which denote transference 
from the same point. 

(a). If three co-initial vectors are complanar, and give the 

relations, 

(a) aa+bp + cy = 0\ m 

(6) a + b + c=:0 J V ; ' 

they will terminate in a straight line. 

For, let oa = a (Fig. 15) . ob = p, od = y. Then da == a — y, 

BA = a — p. 

From Equation (9), (b) 

(a -f b + c) a = 0, 
from which, subtracting (a) of Equation (9), 

b (a - (3) -h C (a - y) = 0, 

6ba -f- cda = ; 

and, since these two vectors neutralize each other, and have a 
common point, they are on the same straight line. Hence, 
a, d and b are in the same straight line. 

(b) . Conversely, if a, p, y are co-initial, complanar and ter- 
minate in the same straight line, and a, b, c have such values 

as to render „ . -. , „ A 

aa + bp -f cy = 0, 

then wiU „ , * , „ A 

a 4- b -\- c = 0. 

0r da = a — y and BA = a — p. 



GEOMETRIC ADDITION AND SUBTRACTION. 
But, by condition, 
or 
in which 



17 



a-P = X (a-y), 

(1 _ x) a - $ + Xy = 0, 

(1 — x) — 1 + x = 0. 



14. Examples. 

1. The extremities of the adjacent sides of a parallelogram 
and the middle point of the diagonal between them lie in the same 
straight line. 

Fig. 18. 
Let OA = a, OB = /?, OC = y. A__. D 

Then 

OD = OB -f- BD, / _^ — -q 

2 y — P — a = °- ° l 




But, also, 



2-1-1 = 0, 



hence, b, c and a are in the same straight line (Art. 13). 

2. If two triangles, abc and smn (Fig. 19) , are so situated 
that lines joining corresponding angles meet in a point, as o, 
then the pairs of corresponding sides produced will meet in three 
points, p, q, r, which lie in the same straight line. 



Let OA = a, OB = /?, OC = y. 

Then os = ma, om = n/2, 

ON = py, BA = a — /?, 

ms = ma — nft, 
• br = x (a — ft) and 
mr = y (ma — nf$) . 



Fig. 19. 



1st. 



BM = BR — MR, 



or 




tt/? — p = x (a — p) —y (ma — nfi) , 
r. n — 1 = — x + yn, x — my = 0. 



Eliminating y 



m(n — \) 
m — n 



18 QUATERNIONS. 

Also 

m (n — 1) 

OR = OB + BR = p + X (a - fi) = p - ^ __ % (a - 0), 

whence n (m - 1) - m (n - 1) a , * 

OR = — - * ' 

m — n 

2d. cn = cp — NP, 

or 

P7 - 7 = v (P - y) - w {np - py) . 
.•. p — 1=— v + uop y v — wn = 0. 

Eliminating w n (p _ i) 

v = • 

Also n ~P 

op = oc + cp = y + ^(/3-y) = y- W /_ j> (ft-y), 

WheDCe qr _ j>(— Dy— (P-D/I (6) 

n — p 

3d. In the same manner, we obtain 

= m(j>-l)a-jp(m-l)y .* 

Q p — m 

From (a), (6) and (c) we observe that, clearing of fractions, 
and multiplying (a) by p — 1, (6) by m — 1, (c) by n — 1, and 
adding the three resulting equations, member by member, the 
collected coefficients of a, /?, y, in the second member of the 
final equation, are separately^ equal to zero. Hence the first 
member 

or (m—n) (p — 1) + op (n— p) (m — 1) + oq (p— m) (n — 1) = 0. 

But 
(m - n) (p - 1) + (n -p) (m - 1) + (p - m) (n - 1) = 0. 

Hence, r, p and q are in the same straight line. 



GEOMETRIC ADDITION AND SUBTRACTION. 



19 



3. Given the relation 



aa + ty3 + cy=0. 



Fig. 20. 



Then a, 0, y are complanar ; but, if co-initial (as they may 
be made to be, since a vector is not changed by motion parallel 
to itself, i.e. by translation 
without rotation) , and a -f 
b •+- c is not zero, they do 
not terminate in a straight 
line. Hence, if o is the ori- 
gin, and a, b, c, their ter- 
minal points, a, b and c 
are not collinear. Let these 
points be joined, forming 
the triangle abc (Fig. 20), 
and oa, ob, oc prolonged to 

meet the sides in a,' bJ c! To find the relation between the 
segments of the sides, let 




whence 



OA'=a'=8a, OB'=/?'=y/?, oc'=y'=Zy, 






Substituting these in succession in the given relation, 



%.'+ bfi + cy-0, 
aa + -B'+ cy — 0, 

z 



whence, since a,' c, b are to be collinear, 



X 



20 QUATERNIONS. 

and, for a like reason, 

a + - + c = 0, 

y 

a + b + - = 0. 

z 

Whence 

a b c 

» = ~r-— , y = — i « = TT' 

o + c a + c a-f6 

and 

a'= — a, /?'= fl, y'= -y, 

6 + c ^ a + <T ' a+6 n 

or, from the given relation, 

, bfr + cy , cy + aa , aa + bp 

6 + c ' ^ c+a ' y a-f & ' 

Whence 

b(a'-f))=c(y- a '), 

c(y3'-y) = a(a- ) S'), 
a (/-a) =6 (/*-/), 
and 

ba' __c cb' __ a ac' _b 

a'c ~~ V b'a ~" c' c'b ~~ a 

or, multiplying, 

ba' . cb' . ac' . = a'c . b'a . c'b. 

4. If o (Fig. 20) 6e any point, and abc any triangle, the 
transversals through o and the vertices divide the sides into seg- 
ments having the relation 

ba' • cb' • ac' • = a'c . b'a . c'b. 

Let a'c = a, bc = aa, cb'= /?, ca = bfi. Then ba = aa + bfi. 

Also let 

BO = flJBBj OA = ^a'a, BC'= mBA, CC'= 2CO. 



GEOMETKIC ADDITION AND SUBTBACTION. 



21 



Then 

bo = «bb' = x (bc -f cb') = x (aa +/?) , 

oa = yx'A = y (a'c + ca) = y (a + ty3), 

bc'= mBA = m (aa -f ty@) , 

cc'=zco = 2 (cb -f- bo) = z [— aa-f-a(aa-f /?)]. 

From the triangle boa we have 

bo -f oa + ab = 0, 
x (aa + fi) + y (a + ty8) - fy3 — aa == 0. 
.*. »a + y — a = 0, #-f-y& — 6 = 0. 

Eliminating # ^(1 — a) 

- 1 - ba ' 
From the triangle bcc' 

bc + cc'+ c'b = 0, 
aa + z [— aa -f- x (aa -f- /?)] — m (aa + tyS) = 0, 

whence, as usual, and substituting the above value of a, 



1— m = 2 — z 



or 



6(1- a) 
l-6a ' 

1 — m _ 1— 6 

m ~~ 1 — a 



m = z 



1-a 
l-5a 



Substituting m, & and a, 



c'a _ AB' CA f 
bc'"" b'c " a'b' 

which is the required relation. 



5. If (Fig. 20) lines be 
drawn through a J b,' cj and 
produced to meet the opposite 
sides rf the triangle in p, q, 
r, then are p, q and r col- 
Unear. 



Fig. 20. 




22 QUATERNIONS. 

With the notation of the last example, 

BC - mBA = ^f^2 (aa + ^- 
1st. From the triangle c'ba' 

c'a'= c'b + ba' 

a " 1 (aa + ty3) + (a-l)a 



a + b — 2 
a-1 



Also 



a + 6-2 



[(& -2) a -6)8]. 



a'r = ac'A'= a'c + cr = a'c — y/3, 



6-2' 

and . 5 •» xv 

br = bc -f- cr = aa p. (a) 

2d. From the triangle c'ab' 

c'b'= c'a + as' 

= (l-m)(aa + 6/8) + (l-6)j8 

- 6 ~ 1 ;[aa-(a-2)£|. 



Also 



a + 6 — 2' 
^'b'= b'c 
[aa-(a-2)/B] = -jB + #i, 



b'q = #c'b'= b'c + cq = b'c + ya, 
6-1 



and 



3d. 



a + 6-2 

- -7=? 

BQ = BC + CQ = (a + «/)a = 2i£L=lll a . (b) 

a — 2 

a'p = <ca'b'= x (a + p) , 
a'p = a'b -f bp = (1 — a) a 4- y (aa + 6)8) , 
a-1 
a — o 



GEOMETRIC ADDITION AND SUBTRACTION. 23 



and 



BP = yBA = (aa -f- b($) . 

a — 6 



(«) 



Multiplying the second members of (a), (&), (c), by (a— 1) 
(b — 2), — (a — 2) (6 — 1), (a — 6) respectively, their sum is 
zero. Hence 

(a -1) (6 - 2) br - (a - 2) (6-1) bq + (a - 6) bp = 0. 

But 

( a _l)(&_2)-(a-2)(&-l) + (a-&) = 0. 

Hence r, q and p are collinear. 



6. If pc (Fig. 20) and po 6e produced to meet aa' and bc, 
£ftew t and s are collinear with c! A similar proposition would 
obtain for q and r. 

With the following notation, 



we have 



ba = a, ba'= /3, bb'= aa + 6/3, 



bo = ba + ab'+ b'o = ba'+ a'o, 
a + b/3-(l-a)a + x(aa + bp) = p + y(a-p). 

a + & 

__ bfi -f- aa # 



bo 



also 



a + b ' 



Fig. 20. 



BP = BA'-f A ; P = BA + AP ; 
P+z[aa+ (&-l)£]= a + wa, 

aa B 



.*. W: 



BP = 



and 



1^6' 



BC = BA'-f A'C = BA + AC, 

£ + v)8 = a + u[(l-a)a-&£), 




24 



QUATERNIONS. 
_ g+ft— 1 



.\ V 

BC 



1-a 



Now to find bs, bc' and bt, we have 



Fig. 20. 




1st. 

bs = b'ba' = bp 4- y'po, 



BS = - 



l_2&_a' 



1-26-a 
2d. 

B</ = V'BA = BC + MCO, 

a 



.'. V': 



2a + 6-l' 



bc'= 



aa 



2a + 6 — 1 



3d. 



BT = BA'+ a't = Ba'+ z'a'o = BP + «/PC, 



a — 6 
o — a 
Clearing of fractions and adding 

(l-26-a)BS + (2a + &-l) bc'+ (b - a) bt = 0, 

as also 

(l_26-a) + (2a + &-l) + (&-a) = 0. 

Hence s, c' and t are collinear. 

15. A medial vector is one drawn from the origin of two co- 
initial vectors to the middle point of the line joining their 
extremities. 



GEOMETRIC ADDITION AND SUBTRACTION. 



25 



Thus (Fig. 21), if p is the middle point of ab, op is a medial 
vector. To find an expression for it, let oa = a, ob = /J, then 



or, adding, 



OP = OA + AP = a -f- AP, 
OP = OB -f- BP =/? — AP, 



OP 



_«+0 



(10). 



The signs in this expression will, of course, depend upon the 
original assumptions. Thus, if ao = a, 

op = — a -f- ap = p — AP, 



OP 



_£-* 



16. An Angle-Bisector is a line which bisects an angle. 

To find an expression for an angle-bi- 
sector as a vector, let OE = a (Fig. 21) 
and of = p be unit vectors along oa and 
ob. Complete the rhombus oedf. Since 
the diagonal of a rhombus bisects the 
angle, od is a multiple of op. Now od 
= a •+- /?, hence 



OP = «(a+/3) 



(11). 




In this expression op is of any length and x is indeterminate. 
If op is limited, as by the line ab, then 



AP = <r(a-r-£) — Oa, 

ap = yAB = y (pp — aa) , 
x{a + P) - aa = y (pp - aa) , 



(a) 



or 



Eliminating x 
Substituting in (a) 



x — a = — ya and x = yb. 
a 



y= 



a + b 
a 



ap = — ^— ab 

a + b 



(12). 



26 



QUATERNIONS. 



17. Examples. 

1. If parallelograms, whose sides are parallel to two given 
lines, be described upon each of the sides of a triangle as diago- 
nals, the other diagonals will intersect in a point. 



rig. 22. 

A 



V* 1 


y^ 


//5^ 





Let abc (Fig. 22) be the given tri- 
angle. Let the diagonals b'f and a'd 
intersect in p, and suppose oe to meet 
a'd in some point as p! 

Let OA = a, ob'=/3, whence oa'= 
ma, ob = nft. 

Now 

b'p — dp = a. 

But 



And 



b'p = ijb'q = y . £ (b'c -f- b'b) 

= 4y[ina+-(n-l)/B]. 

DP= 2DH = 2 . £ (DC -f- CA f ) 

= i 2 [(m-l)a-)8]. 



(a) 
(Art. 15) 



Substituting in (a) , we obtain, as usual, 
._ 2(l-n) 





1 -f m>n — n 


Again 






OP— dp'= a -f p 


But 





(») 



OP f = XOQ = X . $ (OA -f OB) 

= |z (a -f n/3) . 



Substituting in (6) this value of op' and dp'= vdh, we obtain 

as before, 

2 (1 — n) 



v = 



1 -f wiw -- w 



Or, vdh = zdh = dp'= dp. Hence, p and p' coincide, and 
the three diagonals meet in a point. 

2. A triangle can always be constructed whose sides are equal 
and parallel to the medials of any triangle. 



GEOMETRIC ADDITION AND SUBTRACTION. 



27 



In Fig. 23 we have 

aa'= ab -f ba'= ab + £bc. 
bb'= bc + £ca. 

CC' = CA -f ^AB. 

.-. aa'+ bb'+ cc'= f (ab + bc + ca) = 0. (Art. 10). 
3. The angle-bisectors of a triangle meet in a point. 

Let a, /?, y be unit vectors along bc, 
ac, ab (Fig. 23). 
Then (Art. 16) ' 

AP = a(y + £), 
BP = y(a-y). 



Now 



BC = AC — AB, 

aa = bfi — cy 



(a) 



(6) 




where a, &, c are the lengths of the sides. 
Substituting a from (b) in (a) 



BP 



We have also 



-^-r> 



CP = AP — AC = X (y + ft) — 60, 

fbB-Cy \ 
CP = BP + CB = y ( - y ) + cy — 6/3. 



Eliminating y 
Substituting in (c) 

CP = - 



a 



•y + c, a; — 6 = -- — 6. 



x = - 



cb 



a + b + c 



cb 



a-f 6 + c 
_ 6 

a + 6 + c 
= __fc__ 

a-f 6-f-c 

= />(a + /3). 



(y + fl-V 

[cy-(a + &)£| 
(— aa — a/?) 



(o) 



Hence (Art. 16) cp is an angle-bisector. 



28 QUATERNIONS. 

IB. The Mean Point of any polygon is that to which the 
vector is the mean of the vectors to the angles. 

Hence, to find the mean point, add the vectors to the angles 
and divide by the number of the angles. Thus, if a l9 aj, a 8 .... 
be the vectors to the angles, the vector to the mean point is 

<*! + ojj + oa-f ....+0* 
a = .... (13), 

where n is the number of the angles. 

The mean point of a polyedron is similarly defined. It co- 
incides in either case, as will appear later, with the center of 
gravity of a system of equal particles situated at the vertices 
of the polygon or polyedron. 

19. Examples. 

1 . The mean point of a tetraedron is the mean point of the 
tetraedron formed by joining the mean points of the faces. 

Let (Fig. 24) oa = a, ob = /?, oc = 
y. The vectors from o to the mean 
points of the faces are 

l(« + + y)> 
*(« + ?)> 
*<« + «. 
i(y+Ph 

and that to the mean point of the tetraedron formed by joining 
them is 

which is the vector to the mean point of oabc. 

The same is true of the tetraedron formed by joining the mean 
points of the edges ab, bc and ca with o, since 

i [l+M+2 + l+/]_j ( . + , + ,,. 





GEOMETRIC ADDITION AND SUBTRACTION. 29 

The above is, of course, independent of the origin, and would 
be true were o not taken at one of the vertices. 

2. The intersection of the bisectors of the sides of a quadri- 
lateral is the mean point. 



Let (Fig. 25) oa = o, ob =s /?, oc = y, 
o = 8, or = p. Then (Art. 15) 

p = i (op + oe) 

-i [*(« + «) +*(y + »] 
= i(a + £ + y + 8). 

If o is at a, then oa = a = 0, and 

3. If the sides (in order) of a quadrilateral be divided propor- 
tionately, and a new quadrilateral formed by joining the points 
of division, then will both quadrilaterals have the same mean 
point. 

Let a, ft, y, 8 be the vectors to the vertices of the given 
quadrilateral, from any initial point o. 

Then, for the vector to the mean point, we have 

i(a + + y + 8). 

If m be the given ratio, and a\ /?( y{ 8' the vectors to the ver- 
tices of the second quadrilateral, then 

a'= a -f- m (/? — a) = (1 — m) a -f- mfi, 
)8'=(l-m)0 + my, 
y = (1 — m) y + m8, 

8' = a + (1- m) (8 - a) = 8 - m (8 - a) ; 
whence 

1 08'+ y'+ 8'+ a') = H« + £ + y + 8). 



30 



QUATERNIONS. 




4. In any quadrilateral, plane or gauche, the middle point 
of the bisector of the diagonals is the mean point. 

Let (Fig. 26) oa = a, ob = /?, oc = y, os = £y. 
Fig. 26. Then (Art. 15) 



op = i (oq + os) 

«*»(• + » +iy] 
= i("+0 + y). 

5. If the two apposite sides of a quadrilateral be divided pro- 
portionately, and the. paints of division joined, the mean points 
of the three quadrilaterals will lie in the same straight line. 

Let cj A f (Fig. 27) be the points 
of division, and m the given ratio. 
Then, if oa = a, bc = y, oa'= ma, 
c'c = my, ab = £ and o is the in- 
itial point, the vectors to the mean 
points p, p,' p" are 

op = |(3a+20 + y), 

op' =i[(m + 2)a + 20 + (2-m)y], 

O p"=i[(m + 3)a + 20+(l-m)y]; 




pp' = 



\ — m 



(r— ). 



ih. 



P"p=^(y-a). 
Therefore, pj p," p are in the same straight line. 



20. Exercises. 

1. The diagonals of a parallelopiped bisect each other. 

2. In Fig. 58, show that bg and ch are parallel. 

3. If the adjacent sides of a quadrilateral be divided propoi-s 
tionately, the line joining the points of division is parallel to the 
diagonal joining their extremities. 



GEOMETRIC ADDITION AND SUBTRACTION. 31 

4'. The medial to the base of an isosceles triangle is an angle- 
bisector. 

5. In any right-angled triangle abc (Fig. 58), the lines bk, 
cf, al meet in a point. 

6. Any angle-bisector of a triangle divides the opposite side 
into segments proportional to the other two sides. 

7. The line joining the middle point of the side of an}' paral- 
lelogram with one of its opposite angles, and the diagonal which 
it intersects, trisect each other. 

8. If the middle points of the sides of any quadrilateral be 
joined in succession, the resulting figure will be a parallelogram 
with the same mean point. 

9. The intersections of the bisectors of the exterior angles 
of any triangle with the opposite sides are in the same straight 
line. 

10. If ab be the common base of two triangles whose vertices 
are c and d, and lines be drawn from any point e of the base 
parallel to ad and ac intersecting bd and bc in f and g, then is 
fg parallel to dc. 




CHAPTER II. 

Multiplication and Division of Yectors, or Geometric Multipli- 
cation and Division. 

21. Elements of a Quaternion. 
The quotient of two vectors is called a Quaternion. 
We are now to see what is meant by the quotient of two 
vectors, and what are its elements. 

Let a and /?' (Fig. 28) be two vec- 
Fig. 28. ^ torg fl rawn f rom Q an( j » respectively 

and not lying in the same plane ; and 
let their quotient be designated in the 

usual way by ±. 

Whatever their relative positions, we 
o ,y B jg7 fc ; may alwa}'s conceive that one of these 

vectors, as fi\ may be moved parallel 
to itself so that the point o' shall move over the line o'o to o. 
The vectors will then lie in the same plane. Since neither the 
length or direction of /?' has been changed during this parallel 
motion, we have ft = /?! and the quotient of any two vectors, a, 
p\ will be the same as that of two equal co-initial vectors, as a 

and B. We are then to determine the ratio -, in which a and B 

P 
lie in the same plane and have a common origin o. 

Whatever the nature of this quotient, we are to regard it as 
some factor which operating on the divisor produces the dividend, 
i.e. causes /3 to coincide with a in direction and length, so that 
if this quotient be g, we shall have, by definition, 

qB = a when ~ = g. . • . . (14). 
32 * M j8 



GEOMETRIC MULTIPLICATION AND DIVISION. 33 

If at the point o' we suppose a vector o'c = y to be drawn, 
not parallel to the plane aob, "and that this vector be moved as 
before, so that o' falls at o, the plane which, after this motion, 
y will determine with a, will differ from the plane of a and /?, so 
that if the quotient 

H 

q and q' will differ because their planes differ. Hence we con- 
clude that the quotients q and q' cannot be the same if a, fi and 
y are not parallel to one plane, and therefore that the position 
of the plane of a and /? must enter into our conception of the 
quotient q. 

Again, if y be a vector o'c, parallel to the plane aob, but 
differing as a vector from $ then when moved, as before, into 
the plane aob, it will make with a an angle other than boa. 
Hence the angle between a and /? must also enter into our con- 
ception of q. This is not only true as regards the magnitude of 
the angle, but also its direction. If, for example, y have such a 
direction that, when moved into the plane aob, it lies on the 
other side of a, so that aoc on the left of a is equal to aob, then 

the quotient q\ of -, in operating on y to produce a must turn y 

in a direction opposite to that in which q = - turns /? to produce 

a. Therefore q and q* will differ unless the angles between the 
vector dividend and divisor are in each the same, both as regards 
magnitude and direction of rotation. Of the two angles through 
which one vector may be turned so as to coincide with the other 
is meant the lesser, and it will therefore, generally, be < 180? 

Finally, if the lengths of /? and y differ, then - = q will still 
differ from - = q'. Therefore the ratio of the lengths of the vec- 

r 

tors must also enter into the conception of q. 

We have thus found the quotient q, regarded as an operator 
which changes ft into a, to depend upon the plane of the vectors, 
the angle between them and the ratio of their lengths. Since 



34 QUATERNIONS. 

two angles are requisite to fix a plane, it is evident that q 
depends upon four elements, and performs two distinct opera- 
tions : 

1st. A stretching (or shortening) of /?, so as to make it of 
the same length as a ; 

2d. A turning of /?, so as to cause it to coincide with a in 
direction, 

the order of these two operations being a matter of indiffer- 
ence. 

Of the four elements, the turning operation depends upon 
three ; two angles to fix the plane of rotation, and one angle to 
fix the amount of rotation in that 
Fig. 28. plane. The stretching operation de- 

pends only upon the remaining one, 
i.e., upon the ratio of the vector 
lengths. As depending upon four 
elements we observe one reason for 
calling q a quaternion. The two ope- 

c / *5? — b' rations of which q is the s3*mbol being 

entirely independent of each other, a 
quaternion is a complex quantity, decomposable, as will be 
seen, into two factors, one of which stretches or shortens the 
vector divisor so that its length shall equal that of the vector 
dividend, and is a signless number called the Tensor of the 
quaternion ; the other turns the vector divisor so that it shall 
coincide with the vector dividend, and is therefore called the 
Versor of the quaternion. These factors are symbolically repre- 
sented by Tq and U</, read " tensor of g" and " versor of q" 
and q may be written 

q = Tq .Uq. 

22. An equality between two quaternions may be defined di- 
rectly from the foregoing considerations. 

If the plane of a and ft be moved parallel to itself; or if the 
angle aob (Fig. 28) , remaining constant in magnitude and esti- 
mated in the same direction, be rotated about an axis through o 
perpendicular to the plane ; or the absolute lengths of a and ft 




GEOMETBIC MULTIPLICATION AND DIVISION. 35 

vary so that their ratio remains constant, q will remain the same. 

Hence if , 

- = q and ~= q\ 

then will 

when 

1st. Tlie vector length* are in the same ratio, and 

2d. The vectors are in the same or parallel planes, and 

3d. TJie vectors make with each other the same angle both as 

to magnitude and direction. 

The plane of the vectors and the angle between them are 

called, respectively, the plane and angle of the quaternion, and 

the expression -, a geometric fraction or quotient. It is to be 

P 
observed that q has been regarded as the operator on /?, produc- 
ing a. This must be constantly borne in mind, for it will sub- 
sequently appear that if we write qp = ato express the operation 
by which q converts /? into a, qfi and /3q will not in general be 
equal. 

23. Since g, in operating upon /3 to produce a, must not only 
turn p through a definite angle but also in a definite direction, 
some convention defining positive and negative rotation with 
reference to an axis is necessary. 

By positive rotation with reference to an axis is meant left- 
handed rotation when the direction of the axis is from the plane 
of rotation towards the eye of a person who stands on the axis 
facing the plane of rotation. 

[If the direction of the axis is regarded as from the eye 
towards the plane of rotation, positive rotation is righthanded. 
Thus, in facing the dial of a watch, the motion of the hands is 
positive rotation relatively to an axis from the eye towards the 
dial. For an axis pointing from the dial to the eye, the motion 
of the hands is negative rotation. Or again, the rotation of the 
earth from west to east is negative relative to an axis from north 
to south, but positive relative to an axis from south to north.] 

On the above assumption, if a person stand on the axis, fac- 
ing the positive direction of rotation, the positive direction of 



36 



QUATERNIONS. 




the axis will always be from the place where he stands towards 

the left. 

If t, &, j (Fig. 31) be three axes at right angles to each other, 
with directions as indicated in the 
figure, then positive rotation is from i 
to j, from j to &, and from k to i, rela- 
tively to the axes A;, i, j respectively. 
A precisely opposite assumption would 
be equally proper. The above is in 
accordance with the usual method of 
estimating positive angles in Trigo- 
nometry and Mechanics. 

24. Let oa and ob (Fig. 29) be any 

two co-initial vectors whose lengths are a and &, a and ft being 

unit vectors along oa and ob, so that 

oa = Oa, 

Fig. 29. OB = b(3. 

Let the angle aob between the 
vectors be represented by <£ ; also 
draw ad perpendicular to ob, and 
let the unit vector along da be B. 
The tensor of od is evidently 
a cos <f> and that of da a sin <£. If 
we assume that, as in Algebra, geometrical quotients which 
have a common divisor are added and subtracted by adding and 
subtracting the numerators over the common denominator, so 
that 




then, since 
we have 



OA 



a y _ a ± y 

OA = OD + DA, 
OD-f-DA 



OB OB 

__ a cos <£ . /? 
" b./3 
___ a /cos </> . /? 

"H1~ 



OD 
OB 



DA 
OB 



a sin <£ . 8 



b./3 

sin <£ • & 



p 



-) 



GEOMETRIC MULTIPLICATION AND DIVISION. 37 

We have already defined (Art. 8) the quotient of two parallel 
vectors as a scalar, and in the first term of the parenthesis, /? 

being a unit vector, E = 1, and 

— = ~(cos<f> + sin<£ . |Y (a) 

ob b \ pj 

8 
The last term contains the quotient - of two unit vectors at 

right angles to each other. This quotient is to be regarded, as 
before, as a factor which, operating on the divisor /J, produces 
8, i.e., turns /? left-handed through an angle of 90° ; and this 
quotient must designate the plane of rotation and the direction 
of rotation. If we define the effect of any unit vector, operating 
as a multiplier upon another at right angles to it, to be the turn- 
ing of the latter in a positive direction through an angle of 90° 
in a plane perpendicular to the operator, then the unit vector c, 
drawn from o perpendicular to the plane of 8 and /J, and in the 
direction indicated in the figure, will be the factor which oper- 
ating on fi produces 8, and 

€0 = 8 or |=e. 
P 

The unit vector c, as an axis, indicates the plane of rotation ; 
its direction determines the direction of rotation, and by defini- 
tion its rotating effect extends through an angle of 90° ; as a 
quotient, therefore, it completely determines the operator which 
changes /? into 8. Equation (a) thus becomes 

<* = 2(cob* + esin<£), 
ob b 

or, if oa and ob be themselves denoted by a and /?, and the ten- 
sors of a and /? by Ta and T/3, 

g = ^(cos^ + c sin<£) . • • . (15), 



38 QUATEBNIONS. 

To. 
in which — is the tensor of g, being the ratio of the vector 

lengths, and cos <j> + c sin <j> is the versor of g, its plane, deter- 
mined by the axis €, and angle <£ being the. plane and angle of 

the quaternion. 

To. 
When a and £ are of the same length, or Ta=T/2, Tq= — = 1, 

■*" 
and the effect of q as a factor, or operator, is simply one of 
version. 

Like T, the symbol U is one of operation, indicating the oper- 
ation of taking the versor, so that 

Uq = cos <£ -f- c sin <f>. 

This operation takes into account but one of the two distinct 
acts which we have seen the quotient q must perform, as an 
agent converting /? into a, namely, the act of version ; it thus 
eliminates the quantitative element of length. In this respect it 

is similar to the reduction of a Tec- 
Flg ' **' tor to its unit of length, an opera- 

tion which also eliminates this same 
element of length, and has been 
designated by the same symbol U. 

When a and ft are at right angles 
to each other, <j> = 90? and the ver- 
sor cos <f> + € sin <f> reduces to the 
unit vector €, which has been de- 
fined, as an operator, to be a versor turning a line at right 
angles to it through an angle 90? Any vector, therefore, as a, 
contains, in its unit vector in the same direction, a versor 
element or factor of which Ua is the symbol, U indicating the 
reduction of a to its unit of length or the taking of its versor 
factor. Hence the appellation versor of a (Art. 7) . 

If in Equation (15) the vectors be reduced to the unit of 
length, 




GEOMETRIC MULTIPLICATION ANI> DIVISION. 39 

25. We may now express the relation 

J = 5J(cob + + csm*)=g (Eq. 15) 



in the symbolic notation 



* = T^.U^ 



or 




(16), 



and say that the quotient of two vectors is the product of a tensor 
and a versor; and that 

1st. Tlie tensor of the quotient, [—\ is the ratio of their 
tensors; ^ P' 

2d. The versor of the quotient, (cos $ -f- c sin <f>) , is the cosine 
of the contained angle plus the product of its sine and a unit 
vector, at right angles to their plane and such tliat the rotation 
which causes the divisor to coincide in direction ivith the dividend 
shall be positive. 

26. If, for - = q, we write 6 = q[ it is evident that q' differs 
P a 

from q both in the act of tension and ver- 
sion ; the tensor of q' being the reciprocal 
of the tensor of q, and the unit vector c, 
while still parallel to its former position, 
is reversed in direction (Art. 23) since 
the direction of rotation is reversed (Fig. 
30). Hence 

^ = ^(cos<f»- € sin6) .... (17). 
a Ta r ry v ' 

C is called the reciprocal of -. As already remarked, the 

a P 

positive direction of € is a matter of choice. It is only neces- 
sary that if we have + c in U-, we must have — € in Uil, or 

P a 

conversely. 




40 



QUATERNIONS. 



Fig. 31. 



27. Let t, jj k (Fig. 31) represent unit vectors at right angles 
to each other. The effect of any unit vector acting as a multi- 
plier upon another at right angles to it, 
has been defined (Art. 24) to be the 
turning of the latter in a positive direc- 
tion in a plane perpendicular to the ope- 
rator or multiplier through an angle of 
90? Thus, i operating on j produces k. 
>r _. This operation is called multiplica- 
tion, and the result the product, and is 
expressed as usual 




V = 



(18). 



The quotient of two vectors being a factor which converts 
the divisor into the dividend, we have also 



- = i 



.(19). 



either the product or quotient of two unit vectors at right angles to 
each other being a unit vector perpendicular to their plane. 

This multiplication is evidently not that of algebra ; it is a 
revolution, which for rectangular vectors extends through 90? 
Nor is k in Equation (18) a numerical product, nor i in Equa- 
tion (19) a numerical quotient. This kind of multiplication and 
division is called geometric. 

In accordance with the above definition we may write the fol- 
lowing equations : , 

ij = k - = i 

j 
i 



jk = i 



k 



=J 



ki=j 


l=k 

i 


ji = — k 


-k . 


kj = — i 


=i=k 



(20). 



GEOMETRIC MULTIPLICATION AND DIVISION. 



41 



ik = -j 
i{-j)=-t 
'<(-*) = / 
k(-i) = -j 
k(-j) = i 
j(-k) = -i 
j(-i) = k 



~l = i 



k 
-k 



hr< 



-J 



= k 



• = k 



=J 



-J 

— i 
^k 

— i 



(20). 



Since the effect of i, &, j as operators is to turn a line from one 
direction into another which differs $rom it by 90? they are 
called quadrantal versors. 



28. Since 
we have 
or 



iXj=:Jc and i x & = — J = — 1 Xi, 

iX ix J = — lxj, 

IX i = — 1. 



We may denote the continued use of i as an operator by an 
exponent which indicates the number of times it is so used. 
This is consistent with the meaning of an exponent in algebraic 
notation. In both cases it denotes the number of times the 
operator is used, in one instance as a numerical factor, in the 
other as a versor. Thus 

•« •« III I A 

jjai = j 2 i s , — =-, etc. 
JJ J 2 

In conformity to this notation the above equation becomes 

* 2 =-l (21), 



42 



QUATERNIONS. 



and in a similar manner, 



>. — ii 



(22). 



Hence the square of a unit vector is — 1. 
The meaning of the word " square " is more general than that 
which it possesses in Algebra, as was that of the word " product" 
in Art. 27. The propriety of this ex- 
tension of meaning lies in the fact that 
for certain special cases, the processes 
above defined reduce to the usual alge- 
braic processes to which these terms 
were originally restricted. The conclu- 
sion i 2 = — 1 is seen to follow directly 
from the definition, since if i operates 
twice in succession on either ± j or ± &, 
it turns the vector, in either case suc- 
cessively through two right angles, so 
that after the operation it points in the opposite direction. A 
similar reversal would have resulted if the minus sign had been 
written before the vector. Thus — ( ± j) = T j* Hence % X i, 
or £ 2 , as an operator, has the effect of the minus sign in revers- 
ing the direction of a line. 




29. It is to be observed that so long as the cyclical order i, J, 
&, t, J, &, i, .... is maintained, the product of any two of these 
three vectors gives the third ; thus 



and therefore 



as also 



ij=k, jk = i, ki = j; 

(ij)k = kk = Jc*=-l, 
(jk) i = ii = i 2 = — 1 , 

i(jk)= ii = i 2 =— 1, 



GEOMETRIC MULTIPLICATION AND DIVISION. 43 

hence 

i'0'fc) = (#)&> 

Jc(ij) = (Jci)j\ 

which involves the Associative law. 

We may therefore omit the parentheses and write 

ijk = jki=Jcij=z—l (23), 

or, tJie continued product of three rectangular unit vectors is the 

same so long as the cyclical order is maintained. 

But 

k(ji)=k(-k) = -7c i =l .... (24), 

or, a change in the cyclical order reverses the sign of tJie product. 

30. In Equation (24) we have assumed that 

jc(^]c) = -kk. 

That this is the case appears from the fact that i operating on 
—j produces — A:, or 

*(-,/) = -&, 

and that the same result would be obtained by operating with i 
on J, producing &, and then reversing k. That is, to turn the 
negative, or reverse, of a vector through a right angle, is the 
same as turning the vector through a right angle and then re- 
versing it. TJie negative sign to, therefore, commutative with i, 

h *j or 

t(-J) = -0 = -* (25). 

3L It follows directly from the definition of multiplication, 
as applied to rectangular unit vectors, that the commutative prop- 
erty of algebraic factors does not hold good. For 

!/ = &, 
but 

ji = -k. 



44 



QUATEBNIONS. 



Hence, to change the order of the factors is to reverse the sign 

of the product. The operator is always written first ; and, since 

the order cannot be changed without affecting the result, in 

reading such an expression as ij = &, this sequence of the factors 

must be indicated by saying "t into j 

equals k" and not " i multiplied by j 

equals k" the latter not being true. 

Hence also the conception of a quo- 
tient as a factor requires a similar dis- 
tinction, which in Algebra is unneces- 

sary. In the latter, from - = a we 

have, indifferently, ab = c and ba = c. 

k 
But from - = i', while ij = k is true, 

j 

ji = k is not true. In expressing therefore the relations be- 
tween i, ,; and k by multiplication instead of division, care must 
be taken to conform to the definition, the quotient being used 
as the multiplier or operator on the divisor. This non-com- 
mutative property of rectangular unit vectors, which results 
directl}' from the primary definition of the operation of multipli- 
cation, will be seen hereafter to extend to vectors in general 
and to quaternions, whose multiplication is not commutative 
except in special cases. 

The quotient then being a factor which operates on the divisor 
to produce the dividend, we have 




_/ = &, that is, 



\r=k 



y- 



(26), 



the cancelling being performed by an upward right-handed stroke. 

k 
But^ = ^ is not true, for this would involve ji = ij. 



32. It follows also that the directions of rotation of a fraction, 

k K 

as -, and its reciprocal are opposite. Thus 



'GEOMETRIC MULTIPLICATION AND DIVISION. 45 

and therefore that the reciprocal of the quotient i is — i, or 

t = -* (28); 

that is, the reciprocal of a unit vector is the vector reversed. This 
may be written 

l = i-i = -i (29), 

% 

the exponent denoting that, as a factor or versor, t is used once, 
while the minus sign before the exponent indicates a reversal in 
the direction of rotation. 

33. If a be any unit vector, we obtain from the preceding 

Article 

a- = a(-7~ a) 
a 



= — aa = — a 2 = 1 . 



But 
hence 



y-u 



-a=a- (30), 

a a 



or, a unit vector and its reciprocal are commutative and their 
product plus unity. 

If a is not a unit vector, 

a = TaUa, 

i = _L_ = _i- Ua (31), 

a TaUa Ta V " 

the tensor of the reciprocal of a vector being the reciprocal of its 

tensor. 

k 
It must be carefully observed that a fraction, as -, cannot be 

11 • l 

written indifferently A;- or -fc, for this would involve hi" 1 = i" 1 ^, 
i % 

which is not true. 



46 



QUATERNIONS. 



1 k 

By definition k ( — i) = —J, or ki~ l = k- = — J = -. Hence, 

A; 1 i * 

-zsA;-: or ki" 1 . From the meaning attached to the ordinary 

i i 

notation of algebra, , 

i- = -i (a) 



would appear to be correct; for, cancelling, we have k = k. 

k 1 

Whereas, since - must be written k -, we should have 

i i 



or 



iki" 1 [ = — iki] = ki" 1 i [ = fc] 
^[=-fc]=fc, 



which is not true. Of course that equation (a) is false is 
directly evident from the fact that 

k 

-= —J, and (a) involves i ( — /) = ( —j) i 

or *;'=,/*. The above, however, shows 
that, as cancelling must be performed 
by an upward right-handed stroke 
when the expression is in the form of 
a quotient or fraction, so when ex- 
pressed in the form of multiplication, 
the cancelled factors must be adjacent. 
In such an expression as 




— l = - ji" 1 ij" l =jj = - 1 



<») 



it might be supposed permissible to write also 






(«) 



since in either case the correct result is obtained. This arises, 
however, from the fact that both the fractions in the first mem- 
ber of (6) are equal to Jc, and therefore may be permuted so as 

~~ l ^ "" — -!.. The process of (c) is, how- 



GEOMETRIC MULTIPLICATION AND DIVISION. 47 

ever, illegitimate, and the result is correct, not because the 
process is so, but because the factors are in this case commu- 
tative. 

34. Since the act of tension is independent of that of version, 
and their order is immaterial, 

.xi .yj = xy . ij = yx .ij = zk . . . (32), 

where x and y are any two scalars and xy = z. Hence the com- 
mutative principle applies to tensors. If then a, /?, y are in the 
direction of i, j and k respectively, and a, 6, c are their tensors, 

a/3 = TaT/3 . ij = ab . &, 

ay = TaTy • ik = — ae • j, etc., 

or, the product of any two rectangular vectors is the product of 
their tensors and a unit vector at right angles to their plane. 
So also 

a __ Ta . i _Ta i __ __ a , 

P~fjrTj~T/3 # j~~ b ' 

? = 5iJL* ?/,etc., 
y Ty.fc C 

or, the quotient of two rectangular vectors is the quotient of their 
tensors times a unit vector at right angles to their plane. 

35. If, as above,- a = ai, then 

aa = ai • ai, 

Ji. „2 ,*2 

a = a i , 



(33). 



Hence, £fo square of any vector is minus the square of its tensor. 
Since Ta = a is the ratio of the lengths of a and Ua, the square 
of any vector is the square of the corresponding line, regarded as 
a length or distance only, with its sign changed. 
If ai = a and bi = /?, 

a)8=a6t 2 = — ab. 



48 



QUATERNIONS. 



Fig. 32, 



36. That the multiplication of rectangular vectors is a dis- 
tributive operation may be seen 
directly from Fig. 32 by ob- 
serving that 

*'(J + *) = (/ + * (34), 



i being perpendicular to and in 
front of the plane of the paper. 



37. Exercises in the transformations of t, j, h : 



i. i(-0 = 


= &. 


2. i(-*) = 


3. k(-j)-. 


= i. 


4. k(-i) = 


5. -k(j)- 


= i. 


6. (-£)* = 


7. (-j)k-. 


= — i. 


8. (-/)(_*) = 


9- (-M- 


= 


10. (-*)(-/) = 


ii. x=_ 


&. 


12. Z^ = 


— * 




i 


13. — = 




14. =1 = 


5 




k 


15 - —,= 




16. £=./. 


— k 




J 


17. L = 




18. ^*= 


i 




J 


19. * = _i. 




20. - = 


J 




y 


21. - = 




22.-^.- = 


ji 




i »' 


23. 0» = 




24. i.i.*= 


kji 




* > i 



25. Is it correct to write, in general, the product of any frac- 
tions, as - . -, in the form — ? 

J J m JJ 

26. State whether ^- . - = ^-J- is correct or not, and why. 

k i ki 

27. i 2 j 2 k 2 = -(ijk) 2 . 



GEOMETRIC MULTIPLICATION AND DIVISION. 49 

38. Resuming Equation (15), 

9=-g = ^(cos^ + €sin^); 

the quaternion q was shown (Art. 25) to be the product of a 
tensor and a versor. It may also be regarded as the sum of two 
parts, the first of which — cos <f> is a scalar, whose sign is 
that of the cosine of the angle (<£) between the vectors, while the 
second — sin <f> • € is a vector at right angles to their plane, 

whose sign depends upon the direction of rotation of the fraction 
-. This may be expressed symbolically in the notation 

so that we have both 

q = TqUq 

and 

q = 8q + Yq. 

The second member of this last equation is read " scalar of q 
plus vector of q" Sg and Jq being respectively symbols for the 
scalar and vector parts of the quaternion. As already explained 
in the case of the S3 r mbol S, V is a symbol of operation, denoting 
the operation of taking the vector terms of the expression before 
which it is written. 

The quotient of two vectors is, therefore, the sum of a scalar 
and a vector. 

The scalar of the quotient Sg = — cos <f> is the ratio of the 

tensors times the cosine of the contained angle. The tensor of 
the vector part TYq = — sin <j>\ is the ratio of the tensors times 

the sine of the contained angle. Tlie versor of the vector part 
[UYg = € ] is a unit vector perpendicular to their plane , having a 



50 



QUATERNIONS. 



direction such that the direction of rotation of the divisor is posi- 
tive or left-handed. 

Letting a and b be the tensors of a and /?, and collecting the 
preceding expressions for facility of reference, we have 



Uq= cos <f> + € sin <£ 

Sg=7 cos<£ 
b 

Y</=- sin<£ • c 
o 

TYg=^sin<£ 
o 

UYg=€ 

SUg=cos^ 
YUg=sin<£ • c 
TVU?=sin<£ 



(36). 



These expressions require no further explanation than that 
derived from a simple inspection of Equation (15) in connection 
with the meaning already assigned to T, U, S and V as symbols 
of operation. 



39. De Moivre's Formula. 

The following considerations will explain why the parenthesis 
(cos<£-|-€sin<£) as a versor turns /? left-handed through an 
angle <f>. They also contain the quaternion interpretation of 
imaginary quantities. 

Let v = sin <f> and z = cos <£. 

Differentiating, 



or 



dv = cos <f> dcf>, dz = — sin <f> d<f>, 

dv = zd<f>, 
dz= — vd<f>. 



(«) 



GEOMETRIC MULTIPLICATION AND DIVISION. 51 

Multiplying (a) by V— 1, and adding the result to (6), 

dz + dv . V— 1 = (— v + z V^T) d<£, 
or 

dz + dv . 7^= (v V^H- *) V^l d<£, 
whence 

(Z(Z + vV^T) , J y . x 

which may be written 

or 

cos <£ + sin <f> . V^T= e ^^ (d) 

whence 

cosm<f> 4- sinra<£ • V— 1 = e ro 0V=i. ( e ) 

But we have from (d) 

(cos<£ + sin<£ . ^)™=e^^, (/) 

and therefore, from (e) and (/), 

(cos<£ + sin<£ . V^l) m =cosm<£ -f sinm^ • V— 1 (37), 

which is the well-known formula of De Moivre. 

This formula may be made the basis of a system of analytical 
trigonometry. Thus, for example, to deduce the formulae for 
the sine and cosine of the sum of two angles, we have from (d) 

COS<£ + SUKftV^l:^^ 

cos + sin V^l = e ^- 
Multiplying member by member, 

cos<£ cos0 + cos<£ sin0 • V— 1 + cos0 sin^ • V— 1 — 

sin<£sin0 = e (0+wvzl - (?) 

But from De Moivre's formula 

cos (<£ + 0) + sin (<£ + 0) V^l= e (0+ )v=I - (h) 



52 



QUATERNIONS. 



Equating the first members of (g) and (h) , since in any equa- 
tion between real and imaginary quantities these are separately 
equal in the two members, we have 

cos (0 + <f>) = cos cos <f> — sin sin ^. 
sin (0 + <j>) = sin cos<£ 4- cos0 sin<£. 



These formulae, while they may be of course demonstrated 
independently of De Moivre's formula, are here deduced from 
imaginary expressions. It would therefore appear that these 
expressions admit of a logical interpretation. 

If any positive quantity m be multiplied b}* (V— l) 2 the re- 
sult is — m. That is, in accordance with the geometrical inter- 
pretation of the minus sign, we may regard the above factor 
( V— l) 2 as having turned the linear representative of m about 
the origin through an angle of 180? If, instead of multiptying 
m by (V — l) 2 , we multiply it by V— 1, we may infer from 
analogy that the line m has been turned through an angle of 90° 

about the origin. If, too, we ob- 
serve that each of the four expires- 



Fig. 33. 

F 



X'\ -t, 



my/ -I \ 
m 



sions 



771. 



— m, -mV-1 



-H. 



iV-L, 



Y f 



is obtained from the preceding by 
multiplying by the factor V^l, they 
ma}' be regarded as denoting in 
order a distance m on the co-ordi- 
nate axes OX, OF, OX,' OY' 
(Fig. 33), V— 1 being, as a factor, 
a versor turning a line left-handed througn a quadrant. These 
expressions therefore locate a point on the axes, both as to dis- 
tance and direction from the origin. 

Since every imaginary expression can be reduced to the form 
± a ± b V --1, we may, in accordance with the' above interpre- 
tation of V— 1, regard such an expression as defining the posi- 
tion of a point out of the axes. Thus oa = a (Fig. 34) and 



GEOMETRIC MULTIPLICATION AND DIVISION. 



53 



Fig. 34. 



r'. 


A' 








t* 







A 



ap = &, laid off at a at right angles to oa since b is multiplied 
by V — 1 ; so that in passing over oa and ap in succession we 
reach the point p. It is also evident that such an expression 
implicitly fixes the position of p b}- 
polar co-ordinates, since Va 2 -f- b 2 = op 

and tanP0A = -. In like manner 
a 

— b -f- a V — 1 would locate a point pj 
oa' having a length = a, but laid off 
perpendicular to oa, since V — 1 is a 
factor, and a'p' = — b. As before, 
we have implicitly op'= Va 2 + b 2 and 

tan p'oa = — -. 
b 

Furthermore, if we operate on the 
first expression, a+ &V — 1, which fixes the point p, with 
V— 1, we obtain the second, — b + a V — 1, or V— 1 as a 
factor turns op through 90° so as to make it coincide with 
op! As an operator, therefore, we may regard V— 1, like i, J, 
&, as a quadrantal versor, turning a line through a quadrant 
in a positive direction. Algebraically it denotes an impossible 
operation. (In Algebra quantities are laid off on the same 
line in two opposite directions, -f- and — . It was because quan- 
tities are so estimated only in Algebra that Sir W. Hamilton 
called it the Science of Pure Time, since time can be estimated 
only into the future or the past.) But it is unreal or imaginary 
only in an algebraic sense. If the restrictions imposed by Al- 
gebra are removed, by enlarging our idea of quantity and at the 
same time modifying the operations to which it is subjected, this 
imaginary character disappears. In applying the old nomen- 
clature to these new modifications, it will be seen that the prin- 
ciple of permanence is observed, i.e., the new meaning of terms 
is an extension of the old ; and when the new complex quantities 
reduce to those of Algebra, the new operations become identical 
with the old. 

If now we operate upon 



54 



QUATERNIONS. 



which, if we regard a = OA (Fig. 35) and 6V — 1= ap as 
vectors, is equivalent to op, with the 
Fig. 35. expression 

cos <£ 4- sin <f> . V— 1 
of De Moivre's formula, we obtain 



a cos <£ — b sin<£ -f- V — 1 (a sin <£ + 
6 cos<£). 

Draw OX' so that X'OX = <£ ; also 
pa" and al perpendicular and as par- 
allel to OX'. Then 

a cos <f> — b sin <£ = ol — a"l = oa" 
a sin <£ -j- 6 cos <f> = la -f- sp = a "p. 




V X' 



Make oa'= oa," and laj' off a'p'= a"p perpendicular to OX, 
since it has V— 1 as a factor ; then 

(a cos <£ — b sin <£) 4- V — 1 (a sin <j> -f- 6 cos <£) = oa'-|- a'p'= op,' 

and p'op = <£. 

But the formulae for passing from a set of rectangular axes 
OX, OF, to another rectangular set OX', OF,' are 

x = a'cos <f> -f- y'sin <£, 
y = y'cos ^ — a'sin 0, 

in which XOX'= <£, x = oa, y = ap, a; '= oa" y'= pa," or 



oa = ok -f- KA, 

AP = NP — A"K, 

a"k being perpendicular and a"n parallel to OX. 

Hence the effect of the operator has been to turn op left- 
handed through an angle <£, which is equivalent to turning the 
axes right-handed through the same angle. 



GEOMETRIC MULTIPLICATION AND DIVISION. 55 

+1,-1 and V— 1 are particular cases of the general versor 

cos«£-f sin<£ • V— 1, 

namely, when ^ is 0? 180° and 90° respectively, -f- 1 preserv- 
ing, —1 reversing and V— 1 semi-inverting the line operated 
upon. 

We may now see the meaning of De Moivre's formula 

(cos<£ + sin<£ . V— l) w = cosm<£ -f-sinm<£ . V — 1. 

As operators, the first member turns a line through an angle ^ 
successively m times, while the second member turns it m times 
this angle once, producing the same result. The expressions 
cos <f> + sin <f> . V— 1 and cos </> + sin <£ . c are identical, except 
that in the latter the plane of rotation is not indeterminate, 
being perpendicular to c, V— 1 being any unit vector with in- 
determinate direction in space. 

Equation (37) ma} r be put under the form 

cosm(27m-f 0)-|-sinm(27rw-|-<£) . V— l = [cos(27m + <£) + 
sin (2im + <f>) . V^T]"\ 

In the second member if <f> = and m = £, we have Vl for all 
integral values of n, while the first member for n = 0, n = 1, 
n = 2 becomes 1, -i + ^V^"l, -^-^V^, the three 
roots of unity. 

In the same way for m = £, 

1, 

-i + ^V^T, 



4th 



the six roots of unity. The real roots lie on the axis, along 
which direction is assumed plus and minus, while the imaginary 



56 



QUATERNIONS. 



roots are vectors in a direction not that of the axis, and are the 
sum of two vectors, one of which is in the direction of the axis 
and the other perpendicular to it. 



40. Let a and ft be unit vectors along oa and ob (Fig. 36) . 
Resolve OA = a into the two vectors 
Fig. 36. od, da. Then 




OA = a = OD -f DA. 



But 



OD = COS <f> . /?, 

da = c (sin <£ . /?) = sin <f> . c/?, 

€ being a unit vector perpendicular to 
the plane aob, as in the figure. Hence 



a = cos <f> • /8 -h sin <f> • c/J. 



(a) 



Now when a and /? are unit vectors, we have by definition 
- . P = (cos <£ -f* c sin <t>)f$ = a ; or, comparing with (a) , 
P 

(cos <f> -I- csin <f>)/3 = cos <£ • /? + sin <f> • eft. 

The distributive law, therefore, applies to the multiplication 
of a vector by the scalar and vector parts of a quaternion; for 
if a and /? are not unit vectors, the tensors, as merely numerical 
factors, can be introduced without affecting the versor conclu- 
sion. Resolve /? into the vectors oc, cb, cb being perpendicular 
to oa. Then 

ob = p = oc 4- cb. 

But 

oc = COS <j> . a, CB = — € (sin ^ . a) . 

Hence 

COS <j> • a — sin <f> • ca = /?, 

or, by the distributive principle, 

(cos <f> — sin (j> • c) a = /?. 



GEOMETRIC MULTIPLICATION AND DIVISION. 57 

Using the two members of this equation as multipliers on the 
corresponding members of (a) 

(cos <£ — sin <f> • c) aa = /? (cos $ • p + sin <f> • *ff) , 

or, since a 2 = — 1 , 

— cos<£-|-€sin<£ = /?a (38). 

If a and p are not unit vectors, 

/?a = T/3Ta(-cos<£ + €sin<£) . . . (39). 

Operating with each member of (a) on /?, 

a/? = (cos ^ . /? + sin <f> • c/?)/? 
= cos </> . /3 2 -f- sin <£ • c/P 
= — cos<£ — csin^ (40), 

or, if a and f3 are not unit vectors, 

a£ = TaT/?(— cos<£ — €8in<£) . . . (41). 

The product of any two vectors is, therefore, a quaternion, 
which, as before, ma}' be regarded either as the sum of a scalar 
and a vector or the product of a tensor and a versor. In gen- 
eral notation 

a/? = Sa/?+Va/?=Sg+Yg .... (42), 
a/? = Tg.Ug (43). 

Hie scalar of the product [Sa/3 = — TaT/2 cos <£] is the product 
of the tensors and the cosine of the supplement of the contained 
angle. 

Tlie vector of the product [Va/? = — TaT/? sin <£ . c] has for its 
tensor [TVa/?= TaT/3 sin <£] the product of the tensors and the 
sine of t7ie contained angle, and for a versor [UVa/? = — c] a 
unit vector at right angles to their plane such that rotation about 
it as an axis is positive or left-handed. 



58 



QUATERNIONS. 



Representing the tensors of a and /3 by a and &, we have, as 
in Art. 38, from Equation (41), 



Fig. 36. 




Tq = ab 

TJq = — cos<£ — csin<£ 
Sg = — abcos<f> 
Yq = — absin<t> • c 

TY# = a6sin<£ 

UY? = -€ 

SUg = — cos</> 

VUg = — sin <£ . € 
TYUg = sin</> 
(TY: S)g = — tan<£ 



(44). 



4L Resuming the expressions for the products and quotients 
of a and /3, 

/3a = T/3Ta ( — cos<£ + csin<£), (a) 

a/?=TaT^(— cos<^ — csin^), (b) 

C = -^ (cos 6 — csin^), (c) 

a Ta ™ v ' 



we observe 



a j.a / . , . v 



(d) 



1st. That if a and /? be interchanged the sign of the vector 
part is changed. It is equivalent to a reversal of the angle <£, 
and consequently a change in the direction of rotation. Hence 



UY/?a=€ = -UYa/?-\ 

uy£ = € =-uy£ f 



(45). 



P 



a J 



Vector multiplication is not therefore in general commutative. 
2d. If the vectors are unit vectors, 



/?„ = -£, a/} a - 

a $ 



(46), 



GEOMETRIC MULTIPLICATION AND DIVISION. 59 

the product being expressed also by a quotient. This is of 
course always, possible, as appears from (a), (6), (c) and (d), 
and the transformation may be effected thus : 

£ = _^ = 2£(cos*-«sin*), [Eq.(31)] 
a la la 

— £a = T/3Ta (cos<£ — €sin<£) ; 

or 

/?a =s T/3Ta (— cos <f> -f csin <f>) . 

3d. If <j> = 0, then in either (a) and (ft) or (c) and (d) 
the vector part of q becomes zero, and the quaternion de- 
grades to a scalar. When <j> = the vectors are parallel, and 
a/3 = -TaT/? = -a&, as in Art. 35; also - = ?". = -, in 

Art. 8. If at the same time a and fl are unit vectors - = - =1 

P a 
[or = a6.~ 1 = — a 2 = l] anda/3 = a 2 = — 1, as in Arts. 33 and 28. 
If then q be any quaternion and Yq = 0, the vectors of which q 
is the quotient or product are parallel. 

4th. If <f> = 90,° then in either (a) and (b) or (c) and (d) 
the scalar part of q becomes zero, and the quaternion degrades 
to a vector ; and either the product or quotient of two rectangu- 
lar vectors is therefore a vector at right angles to their plane, 

ap reducing to — abt and 2 to -€, as in Art. 34. If at the 

p b 

same time a and p are unit vectors, a/? = — € and - = c, as in 
Art. 27. & 

If then q be any quaternion and Sg = 0, the vectors of which q 
is the quotient or product are perpendicular to each other. 

5th. If an equation involves scalars and vectors, the vector 
terms having been so reduced as to contain no scalar parts, then 
since the scalar terms are purely numerical and independent of 
the others, the sums of the scalars and vectors in each member 
are separately equal. Thus if 

x + aa + bp = d + y + a'a + (&'- b")p 
then 



x = d -f- y and aa 



-y + a'a + (b'-b»)p ) 

f. (47), 
+ bp = a'a + (V- b")p ) 



60 QUATERNIONS. 

which might also be written (Art. 38) 

S ( X + aa + b/3) =S [e* + y + a'a + (V- &")/?], 
V(x + aa + 6/3) =V[d + 2/ + a f o + (&'- 6")£]. 

6th. C being the quotient which operates on a to produce /?, 



we have by definition Q 

E*-fi (48) 



7th. TYa/3, or ab sin </>, is the area of a parallelogram whose 
sides are equal in length to a and b and parallel to a and p. 
Sa/?, or — a&cos</>, is numerically the area of a parallelogram 
whose sides are a and 6, and angle ab is the complement of <£. 

8th. Since the scalar sj-mbol S indicates the operation of 
taking the scalar terms, 

Sa = (49), 

and, for a similar reason, 

Ya=a . (50). 

Again, since a 2 is a scalar, 

Y(a 2 )=0 (51), 

S(a 2 ) = -a 2 (52). 

V(a 2 ) may be written V • a 2 , as also S(a 2 ) = S . a 2 , but these forms 
must be distinguished from (Ya) 2 and (Sa) 2 , which latter are 
also sometimes written Y*a and S 2 a. 

9th. Comparing (a) and (6), • 

Sa/?=S/?a (53), 

and 

Va0 = -Yj8a (54). 

Adding and subtracting (a) and (b) , we have also 

a/? + /?a=2Sa£ (55), 

a/?-0a=2Ya/S (56). 



GEOMETRIC MULTIPLICATION AND DIVISION. 61 

10th. a/3 . pa = (Sa0 + Ya0) (Sa/? - Ya/3) [Eqs. (53) and (54)] 

= (Sa/3) 2 - Sa/?Ya/? + Sa/?Va/? - (Ya/?) 2 . 
Hence 

a0./?a = (Sa/3) 2 --(Ya/?) 2 .... (57), 

or, from Equation (44) , 

ap .pa = (Tapy (58). 

42. Powers of Vectors. 

The symbol i w , m being a positive whole number, has been 
seen (Art. 28) to represent a quadrantal versor used m times 
as an operator ; the exponent denoting the number of times i is 
used as a quadrantal versor. By an extension of this meaning 
of the exponent, t a would naturally represent a versor which, 

as a factor, produces the — th part of a quadrantal rotation. 

Thus i* produces a rotation through one-third, and i* through 
three-fifths of a quadrant, respectively. With the additional 
meaning attached to the negative exponent (Art. 32), as indi- 
cating a reversal in the direction of rotation, we may in general 
define i*, where i is any vector-unit and t any scalar exponent, 
as the representative of a versor which would cause any right 
line in a plane perpendicular to i to revolve in that plane through 
an angle t X 90? the direction of rotation depending upon the 
sign of t. Hence every such power of a unit vector is a versor, 
and, conversely, every versor may be represented as such a 
power. 2<f> 

Since the angle (<£) of the versor is t x -, we have t = — , 
and any versor 

may be expressed 

and 



cos <f> + € sin <£ 
cos<£-f€sin <£ = €*" (59), 

20 

cos<£ — esin<£ = €~ 7T .... (60), 



the vector base being the unit vector about which rotation takes 
place, and the exponent the fractional part of a quadrant through 
which rotation occurs. 



62 QUATERNIONS. 

The operation of which »* is the agent is one-half that of 
which i is the agent, and therefore two operations with the 
former is equivalent to one with the latter ; or, as in Algebra, 

i 4 i*=i = i* + * (61), 

or, emplojing the other versor form, if a, /?, y are complanar unit 
vectors so that 

a %± 

— = cos<f> + esin<f> — en , 
P 



then since 



we have 



- = cos -f csin = en » 
a p a 

p y y 



(cos<t> -f csin^) (cos0 -f csinfl) = cos<£cos0 + € 2 sin<£sin0 + 

c (sin <£ cos -f- cos <£ sin 0) 

= cos (<£ + 0)+ csin (<£ + 0). 

a 

The second member is the U— , its angle being (<£+0), and 
may be therefore expressed as the power of a unit vector, and 

... 2(0 + 0) 

written e — w — 
the factors, or 



2(0 + 0) 

written e — w — ; this exponent is the sum of the exponents of 



20 20 2(0 + 0) 

This is evidentty an abridged form of notation to which the 
algebraic law of indices is applicable. 

Since c 2 = — 1 and therefore c 4 =l ; if <•'=— 1, t must be an 
odd multiple of 2, and if € r =+l, t must be an even multiple 
of 2. 

In either case the coefficient of -n- in <f> = -7r is a whole num- 
ber, and cos<£±€ sin<£ degrades, as above, to the scalar ±1, 
since sin rmr = when m is an integer. 

If e* = ± €, t must be an odd number ; in which case also 



GEOMETKIC MULTIPLICATION AND DIVISION. 63 

m =ii h £» etc -> cos ma- = and the versor degrades to the 
vector ±€. 

If the vector is not a unit vector, as xi = p, to interpret the 
exponent, say pi, so as to satisfy the formula 

plpi = p (63), 

which is analogous to Equation (61), we must combine with the 
conception of rotation through half a quadrant an act of tension 
represented by the square root of the tensor of p. Thus, if 
a; = 16, and we write 

p i = (16i) i =16 i i i , 
then 

p i p i = (16 i i i )(16 4 t i )=16* = p, 

or, if a? = V8, 

p^^.iWl.i*, 
p^y = (V2 . i 1 ) (V2 . i*) (V2 . i*)= V8 . i = p. 

And, in general, 

pP = (arf) l = ^,t f (64), 

or the tensor of the power is the power of the tensor, and the 
versor of the power is the power of the versor. Symbolically 

T.p' = (Tp)' (65), 

U.p' = (Up)' ., (66). 

Any such power (p*) , as the representative of the agent of 
both an act of tension and version, is therefore a quaternion, 
whose tensor and versor can be assigned b} r the above rules, and, 
conversely, every quaternion can be expressed as the poiver of a 
vector, which quaternion ma}' degrade to either a scalar or a 
vector as seen in the preceding versor conclusions. Hence it 
follows that the index-law of Algebra is applicable to the powers 
of a quaternion. 



64 QUATERNIONS. 

43. Relation between the Vector and Cartesian deter- 
mination of a point. 

If *, J, k are three unit vectors perpendicular to each other at 
a common point, then the vector from this point to any point p 
may be written 

p = xi + yj+zk (67), 

in which £, #, z are the Cartesian co-ordinates of p. If the vec- 
tors are not mutually perpendicular and are represented by a, /?, 

y, then 

P = xa + y/3 + zy (68), 

in which x, y y z are the Cartesian co-ordinates of p referred to 
the oblique axes. So long as the vectors a, p, y are not com- 
planar, p refers to any point in space. 

Since any quaternion q maj T be expressed as the sum of a sca- 
lar and a vector, if w be any scalar, then 

q = w ■+- xa -f- yft -+- zy (69). 

As composed of four terms, we observe an additional reason 
for calling this complex expression a quaternion. 
Any vector equation 

p = o- = aa -f- bft + cy, 
involves three numerical equations, as 

unless the vectors are.complanar ; in which case we ma}' write 

y = fta -f- m(2, 
and 

p = (x + zn) a + (y + zm)fi, 
o-= (a -h en) a -f (b 4- cm)fi, 

which, for p = cr, involves but two equations 

x + zn = a -f- en, y-|-zm = &-f cm. 



GEOMETKIC MULTIPLICATION AND DIVISION. 



65 



Resuming the quadrinomial form of g, when the component 
vectors are at right angles, we have 



• (70). 



(71). 



q == w -f- xi -|- yj+ zk 
Vq = xi + yj+ zk 



Since (TV?) 2 = - ( Yq) 2 = x 2 + f + z*, we have 
TV? = Va^-f ^H-2 2 

TV? Vrf + ^ + aV 
Also, since (Art. 41, 10th.) 

(Tg) 2 = (S?) 2 - (Yq) 2 = to 2 + a 2 + y 2 + z 2 , 



Tg= Vi^ + ar' + ^ + z 2 

= _^ = w + xi + yj+ zk 
Q Tg vV + ar' + ^ + z 2 

Tg Vwr' + ar' + ^ + s 2 

Tg \ t0 «+aj" + y , + a? 



44. The plane of a quaternion has been already defined as the 
plane of the vectors or a plane parallel to them. The axis of 
a quaternion is the vector perpendicular to its plane, and its 
angle is that included between two co-initial vectors parallel to 
those of the quaternion. If this angle is 90° the quaternion is 
called a Bight Quaternion. Any two quaternions having a 
common plane, or parallel planes, are said to be Complanar. 
If their planes intersect, the}' are Diplanar. If the planes of 
several quaternions intersect in, or are parallel to, a common 
line, they are said to be Collinear. It follows that the axes of 
collinear quaternions are complanar, being perpendicular to the 
common line. Complanar quaternions are always collinear, and 



(72). 



66 QUATERNIONS. 

complanar axes correspond to collinear quaternions, but the lat- 
ter may of course be diplanar. 

o'a o"c 
Let — -- and -r- be any two quaternions. If complanar, they 

OB 0"d 

may be made to have a common plane ; and, if diplanar, their 
planes will intersect. In the former case let oe be any line of 
their common plane, or, in the latter, the line of intersection 
of their planes. Now, without changing the ratios of their vec- 
tor lengths, the planes, or the angles of the given quaternions, 
two lines, of and og, may always be found, one in each plane, 
or in their common plane, such that with oe we shall have 

c/a __ of , o"c _ og . 

O'fi ~~ OE o"D "~ OE ' 

and, therefore, any two quaternions, considered as geometric 
fractions, can be reduced to a common denominator ; or, in the 
above case 

&A o"c _ OF OG __ OF + OG 

o'b o"d~~oe oe"" oe 

Moreover, a line oh, in the plane ao'b, may always be found 
such that 



and therefore 
and 



o a _ OE 
o'b ""oh' 



O f, C O f A __ OG OE OG 
0"D O'b _ OE OH ~~ Oh' 



45. Reciprocal of a Quaternion. 

The reciprocal of a scalar is another scalar with the same 
sign, so that, as in Algebra, if # be any scalar, its reciprocal is 




GEOMETRIC MULTIPLICATION AND DIVISION. 67 

The reciprocal of a vector has been defined (Art. 33) , so that, 

if a be any vector, - = a" 1 = — — - Ua. 
a . Ta 

The reciprocal of a quaternion has also been defined (Art. 

26) ; thus 

being any quaternion, 

a q 

is its reciprocal. The only difference between the quotients 
- and £ (Fig. 37) is that, as opera- 

P a ' Fig. 87. 

tors, one causes P to coincide with a, 
while the other causes a to coincide 
with p. A quaternion and its recipro- 
cal have, therefore, a common plane 
and equal angles as to magnitude, 
but opposite in direction; that is, ^^~a 
their axes are opposite. Or 

Z - = A q and axis - = — axis q. 

q q 

Since 

P P'P P « « a p P if 

the product of two reciprocal quaternions is equal to positive 
unity, and each is equal to the quotient of unity by the other; 

we have, therefore, as in Algebra, -g=l and £ = t, and no 

q 1 * 

new symbol is necessary for the reciprocal. - is, however, 

sometimes written Eg, B being a general symbol of operation, 
namely, that of taking the reciprocal. It follows from the above 
that 

T-=^- (73), 

q Tq 



68 



QUATERNIONS. 



or, tlie tensors of reciprocal quaternions are reciprocals of each 
other ; while the versors differ only in the reversal of the angle. 
If then 

a = - = — (cos d> -f € sin d>) 
* p T/T * V/ 

we shall have 

Rg = i = a- 1 = ^ = ^ (cos 6 - c sin </>) 
^ a H a Ta ^ ' 



(74). 



46. Conjugate of a Quaternion. 

If /?' (Fig. 37) be taken complanar with ft and a, and making 

with a the same angle that ft does, 

Fig. 37. T/3' being also equal to T/?, then, if 

- = g, ^ is called the conjugate of 

q, and is written Kg. The sj-mbol K 
indicates the operation of taking the 
conjugate. A quaternion and its con- 
^*^ a jugate have, therefore, a common 

plane and tensor, as also, in the ordi- 
nary sense, equal angles ; but their axes are opposite ; or 




and 
If then 

we shall have 



ZKq = Zq = Z I 
Q 

TKg = Tg = -i- 

. Tff 1 

axis Kg = — axis g = axis- 

a Ta / . , • i\ 

? = -g = ^(cos<£ + €sm>) 
K 9 = ^g(cos^-£sin^) 



(75). 



(76), 



or, the tensors of conjugate quaternions are equal, and the versors 
differ only in the reversal of the angle. 

Regarding a scalar and a vector as the limits of a quaternion 



GEOMETRIC MULTIPLICATION AND DIVISION. 69 

(Art. 41, 3d and 4th), we see from Equation (76) that the con- 
jugate of a scalar is the scalar itself and that 

Ka = — a = — TaUa (77), 

or, the conjugate of a vector is the vector reversed. In general 
notation we may write 

g = S? + Yg, 

whence it follows from the above that 



or 



Kq = Sq-Yq } 

(Art. 43) [■ .... (78), 

Kq = w — xi — yj— zk ) 



that is, the scalar of the conjugate of a quaternion is the scalar 
of the quaternion, and the vector of the conjugate of a quaternion 
is the vector of the quaternion reversed ; a result which may be 
expressed symbolically 

YKq-. 



' = \ } (79). 



These are Equations (53) and (54). 

If we add and subtract the two conjugate quaternions 

g = Sg+Vg, Kg = Sg--Vg, 

we have 

q + Kq=2Sq) } 

q-Kq = 2Yq) v J 

The sum of two conjugate quaternions is, therefore, always a 
scalar, positive or negative as the Z q is acute or obtuse. If 

Zq = -, this sum is evidently zero. 

it 

Since, if q is a scalar, Kq = q, then, conversely, ifKq = q, q 
is a scalar. 



70 



QUATERNIONS. 



47. Opposite Quaternions. 

If, for -, we write ^ (Fig. 37), the latter is called the 
P P 

Opposite of g, and is evidently — g, for 



■ a 0-a^0_a =() _ g = _ g 



p p p ft 



As appears from the figure, opposite quaternions have a com- 
mon plane and tensor, supplementary angles and opposite axes ; 
or 

T( — q) = Tg, Z— g = 7r — Zg and axis (— g) = — axis q. 



Since 



— a ,a ct — a r% 



P P P -P 

the sum of two opposite quaternions is zero, or 

Fig. 37. ? + (-?) = <>. 




Also, since 



— a . a — a ft — a 1 

T'p~~p"«~~~ 

or, tfieiV quotient is negative unity. 



If then 



we shall have 



q =7l = -4 ( cos <A + € sin 0) 
-? = ^(-cos<£- € sin<£) 



(81). 



If Z g =• -, Kg = — q ; and, conversely, if Kg = — g, q is a 
vector. 



GEOMETRIC MULTIPLICATION AND DIVISION. 



71 



48. Since JJq is independent of the vector lengths, and only 
dependent upon relative direction, versors are equal whose axes 
and angles are the same. Hence 



But (Art. 24) 



UKg = Ul. 



\ PJ a Ua V/3 



and, Equation (82), 



...ul— L 

q Vq 
1 



(82). 



(83), 



VKq = 



Vq 



Again, since the conjugate of a versor is the same as the re- 
ciprocal of that versor, we have, from Equations (82) and (83) , 



VKq=KVq 



(84). 



49. Representation of Versors by spherical arcs. 

If a, ft, y, are co-initial unit vectors, their extremities will 

all lie on the surface of a unit sphere (Fig. 38). - being any 

a P 

quaternion, U^ turns ft from the position 

ob to oa, and this versor ma}' be repre- 
sented bj T the arc ba joining the vector 
extremities ; for this arc determines the 
plane of the versor as also the magnitude 
and direction of its angle, the direction 
of rotation being indicated by the order 
of the letters as in the case of vectors. 
This representation of versors by vector 
arcs is of importance in the theorems re- 
lating to the multiplication and division of quaternions, and 

may be made upon a unit sphere ; for, if a, ft, y, are not unit 

vectors, the quaternions will differ from the versors b}- a nu- 
merical factor only, the introduction of which cannot affect the 



Fig. 
• < 


88. 




o V\ 



72 



QUATERNIONS. 



Fig. 38. 




versor conclusions. Disregarding, then, the tensors, since ver- 
sors are equal whose planes are parallel and angles equal (in- 
cluding direction), equal arcs on the 
same great circle and estimated in the 
same direction represent equal versors, 
for an}' arc may be slid over the great 
circle on which it lies without change of 
length or reversal of direction. On this 
plan b'a = ab will represent the recipro- 
cal or conjugate of ba, and a quadrantal 
versor would have for its representative 
bc, an arc of 90? Also, the versors of 
all complanar quaternions will be repre- 
sented b}' arcs of the same great circle, while arcs of different 
great circles will represent the versors of diplanar quaternions, 
which are always unequal. 

If m, n and p are the vertices of a spherical triangle, the vector 
arcs mn, np and pm will represent versors, and it will be seen 
that by taking the geometric sum of two of these arcs in a cer- 
tain order, the remaining arc will represent the versor of their 
product ; so that if q' be represented by pm and q by np, q'q may 
be constructed by a process of spherical addition represented by 
pm -+- np = nm, nm representing the versor q'q ; but that because 
q'q and qq' are not generally equal, this process of spherical ad- 
dition, as representing versor multiplication, is not commutative 
as was that of vector addition, pm ■+- np and np -f pm representing 
diplanar versors. 



50. Addition and Subtraction of Quaternions. 

Since a quaternion is the sum of a scalar and a vector, in 
finding the sum or difference of several quaternions the sum or 
difference of their scalar and vector parts may be taken sepa- 
rately. The former will be a scalar and the latter a vector; 
consequently, the sum or difference of several quaternions is a 
quaternion. 



GEOMETRIC MULTIPLICATION AND DIVISION. 73 

1. Both the associative and commutative principles being 
applicable to the summation of scalars, as also to that of vectors 
(Arts. 4, 5), the}' also hold good for the addition and subtrac- 
tion of quaternions ; or 



and 
If then 



in which 



[ . . . . (85). 



r = Sr+Vr 

s = q+r + = &?+Vs; 

8s = S(q + r + )=gg + Sr + . 

Ya = Y(? + r + )=Vg + Vr+. 



and, in general, 

V2$ = 2V?J 



(86), 



or, in quaternion addition and subtraction, 8 and V are distribu- 
tive symbols. 

2. If q -f r +/> + = .9, then, Equation (78) , 



Kry+ Kr + Kp + = S?+ Sr + Sp + - Vg~ Vr- Vp - 

= Ss-Vs=Ks. 

.-. 2K9 = K5^ (87), 

K, like S and V, being a distributive symbol. 

3. Again, since the conjugate of a scalar is the scalar itself, 

KS? = Sg. 
But Sg = SKg. Hence 

KS? = S? = SKg (88). 

Also, since the conjugate of a vector is the vector reversed, 

KV? = -Yg. 



74 



QUATERNIONS. 



Bat — Yq = YKq. Hence 

KYg = — Yq=TKq 
hence K is commutative vrith S and Y. 



(89); 



1 

< 


lg. 38. 




o 7?^ 



4. Since any two quaternions may be 
reduced to a common denominator (Art. 
44), so that 



and since 






TV-f Ty'>T(a'+y') 

unless a f = xy' and a? > 0, it follows that 

Tq + Tq'>T(q + q') 

unless q = xq' and a?>0. Hence, in general, TSg is not equal 
to STg. Moreover, since TJSg is a function of the tensors under 
the 2 sign, while SUg is independent of the tensors, U2<? is not 
equal to 2Ug. This also appears from the representation of ver- 
sors by spherical arcs (Fig. 38) . Hence, in the addition and 
subtraction of quaternions, T and U are not, in general, dis- 
tributive symbols. 

SL Multiplication of Quaternions. 

1. Let 

q = S? +Yg, r = Sr +Yr 

be any two quaternions. Then 

p = qr= SgSr + SgYr + SrYq +YqYr. 

The last member, being the sum of a scalar and a vector, is a 
quaternion. Hence, the product of two quaternions is a quater- 
nion, and 

p == Sp +Yp = Sgr +Ygr, 
in which 

Sgr=S?Sr + S.VgYr .... (90), 
and 

Yqr = 8qYr + SrYq+Y.YqYr . . . (91). 



GEOMETRIC MULTIPLICATION AND DIVISION. 



75 



If we multiply q by r, we obtain 

Srg=SrS? + S.VrVg, 

Yrq = SrVg + SqYv + V . VrYq. 

But, Equation (53), 

S.VrYg = S.YgYr. 
.«. Srg = Sgr. . . . 

But, Equation (54), 

V.YgYr = -V.YrYg, 



(92). 



and therefore the products qr and rq are not equal. Hence, 
quaternion multiplication is not in general commutative. If, 
however, q and r are complanar, Yq and Yr are parallel, and 
Y . YqYr = ; in which case qr = rq. Conversely, if qr = rg, (/ 
and r are complanar. 

Since Reciprocal, Conjugate and Opposite quaternions are 
complanar, they are commutative, or 



qKq == Kq • q 
q- = -q = qq- x = q- l q 



. (93). 



2. It has been shown (Art. 44) that any two quaternions 

</, q', can be reduced to the forms c: and 1 having a common 

a a 

denominator, or to the forms ^ and 2. Hence 

o a 



We have then 



a a a p p 



9 /? T0 Ta IJ5 Ta'Ta~ 1?,l? 
% U U/8 Ua U/3 Ua* Ua Vq ' V<1 



■ (94)- 



76 QUATERNIONS. 

In a similar manner 



w *' [a 8 J 8 T8 Ta T$ * * 



V(q'q) = vZ=Vq1]q 



(95). 



Hence the tensor of the product (or quotient) of any two qua- 
temions is the product (or quotient) of their tensors, and the ver- 
sor of the product (or quotient) is the product (or quotient) of 
their versors. 

In fact, tensors being commutative, we have, in general, 

TUq^UTq (96), 

Uq = THq . JJUq = UTq . IlUg, 

.-. VUq = UVq (97). 

3. The multiplication and division of tensors being purely 
arithmetical operations, we proceed to the corresponding opera- 
tions on the versors. It has been shown (Art. 44) that any 
two versors </, q' t may be reduced to the forms 

q = P=™, </'=£=^', (Fig- 39), 

a OA P OB 

a, b, c\ being the vertices of a spherical triangle on a unit 
sphere. Then 

P a a OA 

If we represent the versors q' and q by the vector arcs bc' 

i 

and ab, then the versor "L, the product of q'q, will be repre- 

a i 

sented by the arc ac' ; moreover if q" = - represent an}' divi- 

P a 

dend and q = - any divisor, then 



a 



q a * p ~ p ~~ OB ' 
the versor of the product q'q being 

bc' -f ab = AC f , 



GEOMETRIC MULTIPLICATION AND DIVISION. 



77 



Fig. 39. 



a" 
and the versor of the quotient i_ 

<i 

AC f — AB = BC'j 

and, as in the addition and subtraction of quaternions, the pro- 
cess consisted in an algebraic addition and subtraction of scalars 
but a geometric addition and subtrac- 
tion of vectors, so the multiplication 
and division of quaternions is reduced 
to the corresponding arithmetical ope- 
rations on the tensors and the geome- 
trical multiplication and division of 
the versors, the latter being con- 
structed by means of representative 
arcs and the rules of spherical addition and subtraction. 

4. The representation of a versor by the arc of a great circle 
on a unit sphere illustrates the non-commutative character of 
quaternion multiplication. For, ab and ba' (Fig. 39) being equal 
arcs on the same great circle, as versors 




and similarty 
Now if 

then 



ab = BA', 



CB = BC'. 



g = -=3 and r = ^ = £, 



qr=z — '-z=z— and 

Py 7 



rq 



-££-£, 



the versors qr and rq being represented b}' the arcs ca' and ac' 
respectively. These arcs, though equal in length, are not in the 
same plane, and therefore the versors rq and qr are not equal. 
Constructing these versors, by spherical addition we should have 

bc' -f ab = ac', 

AB -f BC' = BA' -f CB = CA', 

a change in the order giving unequal results. 



78 QUATERNIONS. 

Hence, unless ac' and ca' lie on the same great circle, in 
which case q and r are complanar, quaternion multiplication is 
not commutative. 

5. Other results, hereafter to be obtained symbolically, maj r 
be readily proved bj T means of spherical arcs, as follows : 

R 
If ab (Fig. 39) represents the versor of q = -, a'b = ba repre- 

1 a 

sents the versor of Kg or -. The spherical sum of ab -f ba 

q 1 

being zero, the effect of the versors in the products qKq and q- 

is to annul each other. Hence, if the vectors are not unit 
vectors, qKq = Kq . q = (Tq)* (98), 



Again, from 
we have 



AB + BC' = CA f , 



qr = -, 
7 



and the versor of K (qr) will therefore be represented by A r c. 
But 

A'c = BC+A r B, 

whence 

K(gr) = KrKg (99), 

or, the conjugate of the product of two quaternions is the product 
of their conjugates in inverted order. 

6. The product or quotient of complanar quaternions is readily 
derived from the foregoing explanation of versor products and 
quotients as dependent upon a geometric composition of rota- 
tions. For, disregarding the tensors, the vector arcs which 
represent the versors, since the latter are complanar, will lie on 
the same great circle, and the processes which for diplanar ver- 

a' 
sors were geometric now become algebraic. Thus for g'="£ 

p " 

and (7 = -5 i r% i 

qq >= q > q =t J = t, 

pa a 



GEOMETRIC MULTIPLICATION AND DIVISION. 79 

and, Fig. 39, 

BA 1 -f AB = AB -f BA' = AA f J 

also for g"= - and q f = ^-, 



and 



g'~"a : a"" a "jS"^ 
BA -f AA* = BA*. 



The product or quotient of any two complanar quaternions is 
therefore obtained by multiplying or 
dividing their tensors and adding or Fig. 89. 

subtracting their angles. Thus 

pq = Up . Tq [cos (<£ + 0) + 
€sin(<£ + 0)]. 



g 2 = (Tg) 2 (cos 2 <£ + € sin 2 <£) , 

or, generally, 

q n = (Tg) n (cos n<f> + c sin n<£) 

whence result the following general formulae, 

T(<T) = (Tg)» 
U( 9 *) = (Ug)» 
SU(g n ) = coswZg 
TYU(g n ) = sin nZg. 

which are all involved in Art. 42. 




(100), 



(101), 



52. 1. Distributive and Associative Laws in Vector 
and Quaternion Multiplication. 

Having assumed (Art. 24) 

- + -= > 

a a a 



80 QUATERNIONS. 




whence 




since a is any vector, we have 




0a + ya = (0+y)a. 


(a) 


Taking the conjugate of (/? -f y)a, 




K[(/? + y)a] = KaK(/? + y) 

= Ka(K£ + Ky). 


[Eq. 99] 
[Eq. 87] 



Taking the conjugate of (/3a -f ya) , 

K(/3a + ya) = K/?a + Kya = KaK/3 + KaKy. 
Hence 

Ka(K/3+ Ky) = KaK/3 + KaKy, 
or 

a'(/?'+y') = a'/?'+ay. ^ (*>) 

Hence, from (a) and (6), the multiplication of vectors is a 
doubly distributive operation, and 

(0 + y)(" + 8) = /3a + ya + 08+yS . . (102). 

2. Let q = y, be any quaternion and a any vector ; also ft a 

vector along the line of intersection of a plane perpendicular to 
a with the plane of q. Then another vector, 8, may be found in 

the latter plane, such that q=£i S having the same angle, plane 

of b b 

and axis as ^. Also let y be a vector in the intersecting plane, 

o 

such that ^ = a. If now a be any scalar, 
"• /J "«~ 8 

- a s + 8- a 8 + -p'J 

= aq + aq. 



GEOMETBIC MULTIPLICATION AND DIVISION. 81 

Taking the conjugates as above, 

q\a'+a') = q'a' + q'a'. 
Hence, in general, 

(a + a) (a'+ a') = aa'-f- aa'-f- a 'a -f- aa'; (c) 

or regarding a, a' y and a, a' each as the sum of two scalars and 
two vectors respectively, 

(«i+ <h + <*i + «2) («'i + « 2 + <*'i + a'2) = 

(<*i + a 2 ) (a\ + a' 2 ) + (a x + a 2 ) (a\ + a' 2 ) + (a\ + a',) 

( a i+ ^2) + (<*i + a s) (a'i + a'2) = 

(«i + ai ) (a'i + a\) + (a! + ttl ) (a' 2 + a' 2 ) + (a 2 + a 2 ) 
(a'i+a'i) + (aH-a*) (a' 2 +a' 2 ), 

since, from (c) , the factors in the expression preceding the last 

are distributive. Putting for the parentheses, which are sums 

of a scalar and a vector, the quaternion symbols p, g, r and s, 

we have 

(p + q)(r + s)=pr+ps + qr + qs . . (103), 

or, the multiplication of quaternions is a doubly distributive 
operation. 

3. Assuming any three quaternions under the quadrinomial 
form, Article 43, i", &, j being unit vectors along three mutually 
rectangular axes, we have 

q = w +xi + yj +zk, (a) 

r = w' +x'i+y'j +z% (b) 

s = w"+ x"i+ y"j+ z"k. (c) 

Multiplying first (c) by (6) and the result by (a), and then 
(b) by (a) and (c) by this result, observing the order of the fac- 
tors, it will be found that the scalar and vector parts of these 
two products are respectively equal, and therefore 

q(rs) = (qr)s (104), 

or, the associative law is true in the multiplication of quaternions. 



82 QUATERNIONS. 

53. 1 . If a and ft be any two vectors, then 

(a + /3)(a + (3) = (a+/3y=a*+(aP + Pa) + /?, 

whence, Equation (55), or, comparing Equations (39), (41) 

and (80), , 

(a + 0) 2 = a 2 + 2Sa/? + /3 2 . . . (105). 

2. Similarly 

(a-iB)(a-iB) = (a-i8)« = a«-(a^+i8a) + ^ 

(a-0) 2 =a 2 -2Sa/? + /3 2 . . . (106). 



or 



Fig. 40. 



3. From Equation (57), or by multiplying q = 8(f+Yq b}* 
a/?.0a = (S?) 2 -(Yg) 2 ; 
hence, from Equation (98), the equalities 

ap ./3a = qKq = (Sap) 2 -(Yap)* = (Tqy . (107). 

54. Applications. 

1 . In any right-angled triangle, the square on the hypothenuse 
is equal to the sum of the squares on the sides. 

Let the sides, as vectors, be repre- 
sented by a and p (Fig. 40), and the 
hypothenuse by y. Then 

Squaring, Equation (105), 
-f = €? + 2*afi + p 9 
7 2 = a 2 + /?, 

or, as lengths simply, changing signs [Equation (33)], 
ba 2 = bc 2 -f CA 2 . 

2. In any right-angled triangle, the medial to the hypothenuse 
is one-half the hypothenuse. 




or, Art. 41, 4, 



GEOMETBIC MULTIPLICATION AND DIVISION. 83 

In Fig. 40, for the medial vector cd = 8, we have (Art. 15) 

8 = +08 -a), 
or 

28 = -a. 

Squaring, and since S/3a = 0, 

48 2 = 2 + a*, 
or 

2== CA* + CB* = AB* 

4 4 ' 

3. If the diagonals of a parallelogram are at right angles to 
each other, it is a rhombus. 

Let the vector sides be represented by a and £. Then a -f ft 
and a — p are the vector diagonals. 
By condition 

S(a + p) (a - P) = 0. [Art. 41, 4] 

But, Equation (53), 

S(a + 0)(a-0) = a 2 -/? = O, 

which is true only when Ta = T/?, that is when the sides are 
equal. 

4. The figure formed by joining the middle points of the sides 
of a square is itself a square. 

Let bc and ca (Fig. 40) be the sides of a square, p and q 
their middle points, and o the middle point of the side opposite 
bc. Then, with the same notation, 

PQ = £(a + 0), QO = £(/?-a); 
.'. S(PQ . QO) = 0, 

or pq and qo are at right angles. 

5. In any triangle, the square of a side opposite an acute 
angle is equal to the sum of the squares of the other sides, less 



84 



QUATERNIONS. 




or 



twice the product of the base and the line between the acute angle 
and the foot of a perpendicular from the angle opposite the base. 



Let ca = /?, cb = a, ba = y (Fig. 41). 
Then 

= a + y, 
/3 2 = a 2 -f2Say-fy 2 l 

Now 

2Say = — *2TaTy COS (180 — b) = 

2 ac cos b. 
Hence 



— b 2 = — a 2 — c 2 + 2accosB zzz — at — c 2 + 2 ad, 
P = a s + c l -2ad. 
If b is a right angle, Say = 0, and, as in Example 1, 
V = a* + (*. 



What does this theorem become for a side opposite an obtuse 
angle ? 

6. In any plane triangle, to find a side in terms of the other 
two sides and their opposite angles. 
In Fig. 41, 

0=a + y. 
Multiplying into a 

($a = a 2 + ya» 

Taking the scalars (Art. 41, 5), 

S£a = - a 2 + Sya, 
or 

— ba cose = — a 2 —ca cos (180°— b) ; 
.-. a = b cosc-f c cosb. 

The above operation with a is indicated by saying simpl}', 
" operating with X S • a," meaning that a is first introduced and 
then the scalars taken. The position of the sign x will indicate 



$?£ i '$:* , -*iJiim& 



GEOMETRIC MULTIPLICATION AND DIVISION. 



85 



how a is used. If used as a multiplier, we should write, " oper- 
ating with S • a x ." 



7. The sines of the angles, in any plane triangle, are propor- 
tional to the opposite sides. 

In Fig. 41 

/3 = a-f y. 

Operating with x T . a, that is, as explained in the preceding 
example, multiplying into a and taking the vectors (Art. 41, 5), 

Y/fo = Y(a + y) a = Y . a 2 +Yya. 



But Y . a 2 = ; hence 



or 



Y0a=Yya, 
ba sinc = ca sins, 

sine : sinB : : c: b. 



Notice that Y/?a and Yya involve a unit vector at right angles 
to their plane, and that, owing to the order of the vector factors, 
c has the same sign in both members of the equality, and ma}* 
therefore be cancelled. The period in Y . a 2 may evidentty be 
omitted, as in Y/?a ; it will be used hereafter only to avoid am- 
biguity. Thus Kg?- means the conjugate of qr ; but Kg . r is r 
multiplied by the conjugate of q. 

8. In a right-angled triangle, to find the sine and cosine of the 
acute angles. 

Let ab = y, AC = /?, bc = a (Fig. 42) . Fig. 42. 

Then 

= y + a, 
whence 

Taking the scalars, since S — = 0, 

1 c b 

1 = -cosa, or COSA=— 

b c 




86 



Fig. 42. 



QUATEBNIONS. 
Taking the vectors 




c . a . A 

-suia-- sinc = 0; 

.•. 8inA=- 

..^^.^ • 

9. To find the sine and cosine of the sum of two angles. 
Let a» /?, y be complaimr unit vectors (Fig. 43) , and c a unit 
vector perpendicular to their plane. We have 

i-i £ 



Fig. 43. 




Hence 



= cos(<£ + 0) + csin(<£ + 0), 
1 = cos <l> + c sin #, 

- = cos0-f€sin0. 



cos(# -f 0) + c sin(# -f 0) = (cos <f> -f c sin <f>) (cos0 -^ c sin0) 
= cos # cos + c(sin ^ cos -f cos ^ sin 0) -f e 2 sin sin #. 

Equating the scalar and vector parts in succession, there re- 
sults, since e 2 = — 1 , 

9 cos(<£ + 0)=cos<£ cos0 — sin<£ sin0, 
sin (<£ -f 0) = sin ^ cos0 + cos ^ sin0. 

10. To find the sine and cosine of the difference of two angles. 
Let the angle between y and a (Fig. 43) be \j/. Then 

y a y 



GEOMETRIC MULTIPLICATION AND DIVISION. 87 

in which 



= cos(i^ — 0) — csin(^ — 0) , 



y 
p 

- = cos0-f csin0, 

- = cos^ — csim/r, 



and, as in the preceding example, 

co8(i/r — 0) = cos0 cos i/r + sin sin 1^, 
sin (\j/ — 0) = cos0 sin i// — sin0 cos i/r. 

11. If a straight line intersect two other straight lines so cts to 
make the alternate angles equal, the two lines are parallel. 

Let a and y (Fig. 44) be unit vectors along ab and cd, and 
P a unit vector along ac. Then 



whence 



a/3 = — cos0 + csinfl, 
fty = — eos0 — csin0 ; 

a/?-/?y=2Ya0, 



Fig. 44. 



B 


\ V 


T> 





and therefore, Equation (56), y = a. 
If a = ab,' then 

a£ = cos0 — csinfl, 
Py = — cos0 — csinfl, 

a£-j3y=2Sa£; 
.-. y = -a. 



[Eq. (55)] 



12. If a parallelogram be described on the diagonals of any 
parallelogram, the area of the former is twice that of thv 
latter. 

Let a and p represent the sides as vectors ; then the diagonals 
are a + p and a — /?, and 

T(a + 0) (a - P) = Y(/?a - a£) = 2 Y/?a, 

since Ya 2 =Yp 2 = and - Yap = T/?a. 





88 QUATERNIONS. 

But, from the order of the factors, 

UV(a + 0)(a-jB)=IJV0a f 
hence 

TY(a + /3)(a-0)=2TY0a, 

which is the proposition (Art. 41, 7). 

13. Parallelograms on the same base and between the same 

parallels are equal. 

We have (Fig. 45) 

Fig. 45. 
A E D P BE = BA -f AE 

= BA + SCBC. 



Operating with Y . bc x 

Y(bc . be) = Y(bc . ba), 
since V#bc 2 = 0. 



BC • BE Sin EBC = BC • BA Sin ABC, 

which is also true when the bases are equal, but not co-incident. 

14. If, from any point in the plane of a parallelogram, per- 
pendiculars are let fall on the diag- 
onal and the two sides that contain 
it, the product of the diagonal and 
its perpendicular is equal to the 
sum, or difference, of the products 
of the sides and their respective per- 
pendicidars, as the point lies with- 
out or within the parallelogram. 

Let oa = a, ob = fi, op = p (Fig. 46) . 
Then 

Vap+Vj8p=Y(a + 0)p. 
But 

UYap = UV/fy= UY(a + ftp. 
Hence 

TYap + TV/fy = TY(a + ft) p. 



Fig. 46. 




GEOMETRIC MULTIPLICATION AND DIVISION. 89 

For p' = op', we have 

UYap'= - UY/V = ±UY(a +P) P '; 
.-. TYap'-TV#>'= TY(a + 0)p.' 

15. If on any two sides of a triangle, as ac, ab (Fig. 47), 
any two exterior parallelograms, as acfg, adeb, be constructed, 
and the sides ed, gf, prodiiced to meet in h, then will the sum of 
the areas of the parallelograms be equal to that whose sides are 
equal and parallel to cb and ah. 

Let AE = a, AB = /3, AC = y 

and ag = 8. Then 

AH = AE -f EH 

= a — Xf$. 



Operating with x V . f$ 
V(AH./3)=Ya/3. 
We have also 



(a) 




Operating with x V • y 



AH = AG -f GH 

= 8 - yy. 



(») 



Y(ah . y)=Y8y. 
Hence, from (a) and (6), 

VAH(/3-y)=Ya/?-Y8y, 
Y(ah . CB)=Ya£-V8y=Ya/?+Yy8. 



These vectors have a common versor ; whence the proposition. 
If one of the parallelograms, as ad', be interior, then ae'= — a 
and ah'= — a — xfi = 8 -f yy, and 

V(AH'../?) = -Va/3, 
V(AH'.y)=Y8y; 
.-. VAH , 08-y) = -Ya)8-Y8y=V^a-Y8y. 




90 QUATERNIONS. 

But in this case 

UY(ah' . cb) = - UY/?a = - UYSy, 

and the area of the parallelogram on ah', cb, is the area of af 
minus the area of ad'. 

16. To find the angle between the diagonals of a parallelogram. 

Let AD = BC = a (Fig. 48), 
Fig. 48. and ba = cd = /3, d and d 1 being 

D the tensors of the diagonals. 
Then 

AC . DB = - (a - p) (a + /?) 

= _ a 2_ (a 0_0 a) + / p 

= _a 2 -2Ya/? + 2 . 
Taking the scalars 

cos doc • dd'= a 2 — V. 

Taking the vectors 

sin doc • dd r = 2a6sin0, 

since UT(ac . db) = — UVa0. 

.-. tanD0c = -tan^ = 2 f 8in f. 

17. TJie sum of the squares on the diagonals of a parallelo- 
gram equals the sum of the squares on the sides. 

In Fig. 48 

BD 2 = (a + 0) 2 = a* + 2Sa/?+/P, 
C A 2 = (P - a ) 2 = /? - 2 Ha/3 + a 2 ; 
.\ CA 2 + BD 2 =2a 2 + 2/]i 2 , 

or 

BD 2 -H CA 2 = BA 2 + AD 2 -f DC 2 + CB 2 . 

18. TJie sum of the squares of the diagonals of any quadri- 
lateral is twice the sum of the squares of the lines joining ilie 
middle points of the opposite sides. 



GEOMETBIC MULTIPLICATION AND DIVISION. 



91 



Let AB = a, ad = /J, DC = y (Fig* 49), For the squares of 
the diagonals, we have 



Fig. 49. 
^ E 




(/* + y) 2 + (/*-a)', 
and for the bisecting lines g. 

Whence the proposition readily follows. 



19. The sum of the squares of the sides of any quadrilateral 
exceeds the sum of the squares on tlue diagonals by four times the 
square of the line joining the middle points of the diagonals. 

Let ab = a, AC = /?, ad = y 
(Fig. 50). The squares of the 
sides as vectors are 



or 



a* + (£-a) 2 + (y-£)' + y 8 , 
or 

2(a« + s + y 2 )- 2S/?a-2Sy0. 
The squares of the diagonals are 

/8 s +(r-a) i , 

P + y* + a i -2&ya. 

The former sum exceeds the latter by 

a* + /3 s + y* - 2 S/3o - 2 Sy£ + 2 Syo> 

(a-P + y)', 
which may be put under the form 




or by 



92 



QUATERNIONS. 



But a ~*~y = ao, and — S. — sa. Substituting these values, 

we obtain 

4(ao + sa) 2 , or 4so 2 , 

which is also true of the vector lengths. 

20. In any quadrilateral, if the lines joining the middle points 
of opposite sides are at right angles, the diagonals are equal. 
With th.e notation of Fig. 49, we have 

FE . GH = H(y — a) + 0]£(a + y)- 
But, by condition, 



Whence 



or 



(Y + /*)* = 08-")', 

AC 2 = BD 2 , 
AC = BD. 



or 



or 



21. In any quadrilateral prism, the 
sum of the squares of the edges exceeds 
the sum of the squares of the diagonals 
by eight times the square of tJie line 
joining the points of intersection of the 
two pairs of diagonals. 

Let oa = a, ob = /?, oc = y, od = 8 
(Fig. 51). For the sum of the 
squares of the edges we have 



2[a 2 + )3 2 + (8-a) 2 -h2y 2 + (8- i S) 2 ], 

2[2a 2 4- 2^ + 2/ + 28 2 - 2S8a - 2S8£). (a) 

The sum of the squares of the diagonals is 

(y + 8) s +(y-8) 2 +(y + a-/?) 2 + (y+0-a)*, 

2(a'+0* + 8» + 2y*-2Sa0). (6) 




; u . j u. u... tWh'f.wp'f 



GEOMETRIC MULTIPLICATION AND DIVISION. 93 

The vectors to the intersections of the diagonals are 
*(8 + y) and *( y + a + /3), 
and the vector joining these points is 

Squaring and multiplying by eight, we have 

, 2[a* + /? + 8' + 2Sa0-2Sa8-2S08], 
which added to (6) gives (a) . 

22. In any tetraedron, if two pairs of opposite edges are at 
right angles, respectively, the third pair will be at riglit angles. 

Let oa = a, ob = /?, oc = y (Fig. 52). 

The conditions give rig * w * 

Sa( j 8-y) = 0, A\ 

S / 8(a-y) = 0. // \ 

Subtracting the first of these equa- / \ \ 

tions from the second o£- -/- --^b 

Sy(a-/3) = 0, \i^^^ 

A 

which is the proposition. 

23. To find .the relations between the edges, angles and plane 
areas of a tetraedron. 

With the notation of Fig. 52, we have 

CA.CB = (a-y)(/J-y), 
or 

CA . CB = a/? — ay — y/? + y*. (a) 

Representing the tensors of ca and cb by m and n, and taking 
the scalars of (a) , 

S(CA . CB)=SajS-Say-Sy^4-y 2 , 
whence 

c 2 — ac cos aoc — be cos boc = mn cos acb — ab cos aob, 



94 QUATERNIONS. 

which is the relation between the edges and their included 
angles. 
Taking the vectors of (a), and squaring, 

[V(CA . CB)] 2 = (Yap) 2 - Ya0Yay -YapYy/3 - VayYa/? ) (b) 

But 

- (YapYyp +Yy/3Yap) = - 2 S . YapYyfi (Eq. 55) 
= 2TVa/?TVy0COSB, 

in which b is the angle 'between the planes aob, boc. 

Also 

- (Va/?Vay + VayYa/3) = - 2 S . Va/?Vay = 2 TVa/?TVay COS A, 

and 

VayYy/? + Vy^Vay =28. YayVy0 = - 2 TVayTVy/?COS (180°- c) 

= 2TVayTVy0COSC, 

in which a, b and c are the angles 
opposite the edges bc, ac and ab re- 
spectively. Hence (b) becomes 

- [TV(CA . CB)] 2 = - (TVa/?) 2 - (TVay) 2 

-(TVy/?) 2 
4- 2 T Ya/3 TVay COS A + 2 T Ya/ST Vy0 COS B 
+ 2TVayTVy0COSC. 

But (Art. 41, 7th) 

TV(ca . cb) = 2 area acb, 

and similarly for the others. Hence, dividing by —4, 

(area abc) 2 = (area aob) 2 -f (area aoc) 2 + (area boc) 2 — 

2 area aob area aoc cos a — 2 area aob area boc cos b — 
2 area aoc area boc cose, 

which is the relation between the plane faces and their included 
angles. 




GEOMETBIC MULTIPLICATION AND DIVISION. 

If the angles are right angles, then 

(area abc) 2 = (area aob) 2 + (area aoc) 2 + (area boc) 2 « 

24. To inscribe a circle in a given triangle. 

Let a, /?, y (Fig. 53) be unit vec- 
tors along the sides. Then, Art. 16, 
the angle-bisectors are 

*(£ + y)> 

-y(y + a), 

Now 

*(/? + y) = cy-3Ky + *) 

Operating with V . (y + a) x 

cVay 




# = 



Vy£+Va/?+Vay 
Hence 

or, since a, /?, y are unit vectors, 

csins /Q . v 

ao = - : : — (/> + y). 

sinA + sinB-j-sinc 

Squaring, to find the length of ao, we have, since (/J-hy) 3 — 
— 2(1 + cos a), 

-ao^-L csinB T 2 (1 + cosa), 

L sin a + sin b -f- sin cj 



AO = 



csmB -V2(l + COSA), 



sin a + sinB + sinc 

csinB 0/1Aoll 

ZCOS$A. 



sin a + sin b + sin c 



25. If tangents be draion at the vertices of a triangle inswibed 
in a circle, their intersections with the opposite sides of the triangle 
will lie in a straight line. 



96 QUATERNIONS. 

Let o be the center of the circle (Fig. 54) whose radius is r, 

and OA = a, ob = £, oc = y. Since oa and ap are at right 

angles, 

S(oa . ap) = 0. 

But 

ap = ab 4- bp = ab 4- 2/bc = /? — a 4- 2/(y — /?) ; 



rig- 54. hence, substituting this value above, 



Sa[0-a + y(y-0)] = O, 
Sa 2 = Sa/?4yS(ay-a/?), 

and 

r' + Safl . 




Say - Sa0 



> = OB4BP = i 3 + ^C = i S-£±^(y- i 8) 

= (7* + Say)/?-(r* + Sa/?)y 
Say - Sa0 

Similarly, or, by a cyclic change of vectors, 

(r* + Sag) 7 -(r* + S/?y)a 
Sa0-S/?y 
0K _ (^4Sffy)a-(r 2 + Say)g 
S/?y - Say 
Whence 

(Say - Sa£)OP + (Sap - S0y)OQ + (S0y - Say) OR = 0. 
But also 

(Say - Sa0) 4" (Sa0 - S£y) 4" (S£y - Say) = 0. 
Hence p, Q and r are collinear. 
26. The sum of the angles of a triangle is two right angles. 



GEOMETRIC MULTIPLICATION AND DIVISION. 97 

Let a, /?, y be unit vectors along bc, ca and ab (Fig. 55). 
Then (Art. 42) 

a 

20 



l=e< 



But 




P 



a y 2* 20 2$ 2(0 + + ,/,) 

y a 



Hence -(<£ -|- + i/r) = an even multiple of 2 (Art. 42) , as 2n, 

as we go round the triangle n times. 

In taking the arithmetical sum, or passing once round, we 
take the first even multiple of 2, or 

-(* + + i/O = 4; 
.-. <£ + + ^=2tt, 

and the sum of the interior angles is 3 tt — 2 tt = tt, or two right 
angles. 

27. TJie angles at the base of an isosceles triangle are equal to 
each other. 

Let a and ft (Fig. 56) be the vector sides Fi s- 56 - 

of the triangle, and Ta = T£. Then, if the 

proposition be true, 

&j 
a _ E ft 

or 

«(a - /?)"' = K/3(/J - «)-'= (ft - a)" 1 /?, 
«(|8- a) = («-£)£; 

which is true, since Ta = T/3. 




98 



QUATERNIONS. 



28. To find a point on the base of a triangle such that, if lines 
be drawn through it parallel to and limited by the sides, they will 
be equal. 

Draw de (Fig. 57) and df parallel to 
the sides. From similar triangles, if 
ae = XAC, 

__ AE __ BF _ BA — AF 
~" AC ~" BA "~ BA 

whence 




Now 



BA 
AD = AF + FD, 



or, since fd = ae, 



= (1 — #)ab +<cac. 
But, since fd is to be equal to ed, 

(1 — o;)Tab = oTac == y ; 

.'. (1 — Oj)TABUAB = yUAB, 
JCTacUaC = #UAC, 

and therefore 

ad = 2/(Uab + Uac) , 

and d is on the angle-bisector. 

29. If any line be drawn through the middle- point of a line 
joining two parallels, it is bisected at that 
point. 

30. If the diagonal of a parallelogram 
is an angle-bisector, the parallelogram is a 
rhombus. 

31. In any triangle the sum of the 
squares of the lines gh, ke, df (Fig. 58) 
is three times the sum of the squares of the 
sides of the triangle. 

32. T7ie sum of the angles about two right lines which intersect 
is four right angles. 




GEQMETRIO MULTIPLICATION AND DIVISION. 99 

33. If the sides of any polygon be produced so as to form 
one angle at each vertex, the sum of the angles is four right 
angles. 

34. Find the eiglU roots of unity (Art. 39). 

35. Tlie square of the medial to any side of a triangle is one- 
half the sum of the squares of the sides which contain it, minus 
one-fourth the square of the third side. 

55. Product of two or more Vectors. 
1 . Let q = a/3, r = y . Then, since Scr = 8rq, 

Sa/fy = Sya/3. 
Let q = ya, r = /?. Then 

Sgr = Srq = Syaj3 = S/?ya ; 
.-. Saj8y=S/3ya = Sya£ (108), 

or, the scalar of the product of three vectors is the same if the 
cyclical order is not changed. 

This may also be shown by means of the associative law of 
vector multiplication as follows: 

ajSy = (a/?)y = (Sa/3 + Va/?)y. 

Taking the scalars 

Sa0y = S(Sa/?+Va/?)y 

= 8(Va0,y), since S(Sa0 • y) = 0, 
= S . y Va/? ; 

introducing the term S . ySa/? = 0, 

= 8,yVa0+S,ySa/? 
= S.y(Sa/?+Va0) 
= Sy(a/?)=Sya/3. 



100 QUATERNIONS. 

In a similar manner 

Sa0y = S . a(S/?y +Y/Jy) 
= S . aV0y 
= S(V0y.a) 
= S(V0y + S0y)a 
= S0ya, 



and, as before, 
2. Again 



Sa/?y = S/?ya = Sya/?. 

Sa0y = S . a(S0y + V0y) 
= S . aV0y 
= - S . aVy/? 
= -Sa(Vy/? + Sy/?); 
.-. Sa0y = - Say/3 (109), 

or, a change in the cyclical order of three vectors changes the sign 
of the scalar of their product. 

3. Resuming 

a/?y = a(0y) 

and taking the vectors, 

Va0y = V-a(S£y+Y0y) 
= aS,Gy +V . aYPy. 

Vy0a = V(Sy0+Vy/3)a 

= V. aSy/?-V.aYy/? 
= V,aSy/?+V.aV/?y 
= V.a(Sy/? + V/?y) 
= aS£y + V,aV0y; 
.'. Va0y=Vy/?a (110), 

or, tfie vector of the product of three vectors is the same as the 
vector of their product in inverted order. 

4. Geometrical interpretation of Sa/fy. 

Let a, /?, y be unit vectors along the three adjacent edges oa, 
ob, oc (Fig. 59) of any parallelopiped, being the angle be- 



Also 



GEOMETRIC MULTIPLICATION AND DIVISION. 101 

tween a and /?, and ff the angle made by y with the plane aob. 
Then 

aft = — cos0 4- € sin0, 

c being a vector perpendicular to the plane aob. 
Operating with xS.y 

Sa0y = S(- COS0 + €Sin0)y Fig. 59. 

= S(sin0 . ey). 



But Sey = — cos of the angle 
between c and y = — sin0' ; 

.*. Sa^y = — sin sin ff. 





Now, if a, /?, y represent as vectors the edges oa, oii, 00, 
whose lengths are «, 6, c, 

Sa/?y = — TaT/>Ty sin sin ff 
= — abc sin $ sin 0[ 

But tt& sin = area of the parallelogram whose sides arc ii and 
6, and csin0' = perpendicular from c on the plane aob. Hence 

— Sa/?y = volume of a parallelopiped ichose edges are 
a, h and c, drawn parallel to a, fi and y. 

Cor. 1. Whatever the order of the vectors, the volume is the 
same ; hence, as already shown, 

± Sa0y = ± S0ya = ± Sya0 = if Say£, etc. 

Cor. 2. If Sa/?y = 0, neither a, ft or y being zero, then citlin 
= 0, or ff = 0, or Me vectors are complanar. 

Cor. 3. Conversely, if a, /?, y are complanar, Sa/fy = 0. 

Cor. 4. The volume of the triangular pyramid of which the 
edges are oc, on, oa, is — } So/Sy. 

5. We have seen that when a, /? and y are complanar, Sa/3y — u, 
and therefore a/?y is a vector. To find this vector, sup] (use, a 



102 QUATERNIONS. 

triangle constructed whose sides ab, bc, ca have the directions 
of a, p and y respectively, a vector not being changed by motion 
parallel to itself. Since the tensor of the vector sought is the prod- 
uct of the tensors of a, ft and y, we have to find U(ab . bc . ca) , 
i.e., its direction. Circumscribe on the triangle abc a circle and 
draw a tangent at a, represented by t'at. Since the angles tab 
and bca are equal, we have 



CA AT' L AT J 



whence 

U(bC.CA) = U(aB . AT f )[=U(BA • AT)]. 

Introducing Uab X 

U(AB . BC • CA) = U(AB . AB . AT f ) [ = U(AB . BA . AT)], 

or, since U(ab . ba) = — (U . ab) 2 =1, 

U(ab . bc . ca) = — U . at' = U . at. 

Hence, if a, b, c are any three non-collinear points in a plane, 
or if a, )3, y are the sides of a triangle joining them, in order 
(in either direction, since Va/Jy = Vy/?a) , 

a Py> 07 a > Y a P 

are the vector tangents to the circumscribing circle at the angles 
of the triangle. 

Again, if a, b, c are any three points in a plane, not in a 
straight line, and a and /? are two vectors along the two succes- 
sive sides ab, bc of the triangle which the}' determine, and cd a 
vector drawn from c parallel to y, intersecting the circumscribed 
circle at d, then is da parallel to Va/fy = 8. For 

8 = ajSy - a£j8y - a/P/J- 1 ? = - (T/?)VV = - (T/8)»|y, 

whence U . ^^, which turns (3 parallel to — a, turns y into a 

direction 8 = da, the opposite angles of an inscribed quadrilateral 
being supplementary. 



GEOMETRIC MULTIPLICATION AND DIVISION. 103 



If y have a direction such that cd crosses ab, or the quadri- 
lateral is a crossed one, it is evident on construction, of the 

figure that 
& U8'= Ua/fy = U(ad) = - VS. 

Hence the continued product of the three successive vector 

sides of a quadrilateral inscribed in a circle is parallel to the 

fourth side, its direction being towards or from the initial point 

as the quadrilateral is uncrossed or crossed ; and, conversely, no 

plane quadrilateral can satisf}' the above formula ± US = Ua/3y, 

unless a, b, c and d are con-circular. The continued product 

of the four successive sides of an inscribed quadrilateral is a 

scalar, for v ^ „ 

a/3yS = (a/?y)8 = ± S 2 = * d\ 

Since the product of two vectors is a quaternion whose axis is 
perpendicular to their plane, while the product of a quaternion 
by a vector perpendicular to its axis is another vector perpen- 
dicular to its axis, and so on, it follows that the continued 
product of any even number of complanar vectors is generally a 
quaternion whose axis is perpendicular to their plane, while the 
product of any odd number of complanar vectors is a vector in 
the same plane. Hence the formulae 

Sa = 0, Sa/?y=0, Sa/fySo-=0, etc., 

for complanar vectors. 

If, however, the given vectors are parallel to the sides of a 
polygon abc mn inscribed in a circle, then 



U(AB . BC . CD MN . NA)= U(AB . BC . CA) U(AC . CD . DA) 

X U(AM • MN • NA). 

But each of the products U(ab . bc . ca) is equal to U . at, 
at being the tangent to the circle at a. Hence 

U(AB • BC • CD MN . NA) = (U . AT) n , 

which reduces, according as n is even or odd, to ± 1 or ±U • at. 
Hence the product of the vectors will be a scalar or a vector 



104 QUATERNIONS. 

according as their number is even or odd, and in the latter case 
this vector is parallel to the tangent at a. 

If the vectors are not complanar, but parallel to the successive 
sides of a gauche polygon inscribed in a sphere, the polygon 
may be divided as above into triangles, for each of which the 
product of the three successive sides is a vector tangent to the 
circumscribing circle, all these vectors lying in the tangent plane 
to the sphere at the initial point. If the number of sides is even, 
their product will be a quaternion whose axis is perpendicular to 
the tangent plane, i.e., lies in the direction of the radius of the 
sphere to the initial point ; if odd, the product is a vector in the 
tangent plane. 

Hence, if a, b, c and d are four given points, not in a plane, 
AB = a, bc = /?, CD = y being given vectors, and p any other 
point such that dp = <t, pa = /j, if p lies on the surface of a 
sphere through the four given points, we have the necessary and 
sufficient Condition 

a/3y<rp = p(ry/?a, 

for each member is equal to minus the conjugate of the other, 
and must therefore (Art. 46) be a vector. 

6. From Equation (56) , 
Operating with V . a x 

2V.aY/?y=y.a(/?y-y/J). 
Introducing in the second member /?ay — /fay, 

= V (a/?y - ay/? + /Jay - /Say) 
= V(a0 + 0a)y - V(ay0 + ya/3) 
= V . 2(Sa£)y - V(ay + ya)0 
= 2ySa0-2/3Say. 



Hence 



Y. aV0y = ySa0-0Say .... (HI). 



GEOMETRIC MULTIPLICATION AND DIVISION. 105 

This formula may be extended. Thus, for a write Ya8, and 

Y . VaSY0y = yS(Va8)0 - /8S(Va8)y, 
y.Va8V/?y = ySa80--£Sa8y . . . . (112). 

An inspection of this formula shows that it gives a vector 
complanar with y and /?. Moreover, since 

Y . ya8V0y = y . Vy0Va8 = 8Sy/?a - aSy/?8, 

It is also complanar with a and 8, and is, therefore, parallel to 
the line of intersection of the planes of a, 8, and /?, y. 
Similarly 

V.V^yVa8 = 8S/?ya-aS^y8 = -.V.ya8y^y . (113). 

Adding Equations (112) and (113) 

8S0ya-aS0y8+ySa80-£Sa8y=O . . (1U), 
or 

8Sa/?y = aS/?y8-/3Say8 + ySa0S . . . (115), 

a formula expressing a vector 8 in terms of any three given di- 
planar vectors, a, /?, y ; so that, if 

S0y8 = 6, — Say8 = Sya8 = C, Sa/J8 = a, Sa/3y = m f 
8 = m- 1 (6a + c0 + ay). 

7. Resuming Equation (111), and adding aS/Jy to both mem- 
bers, 

y • oNpy + aS/?y = y Sa/3 — 0Say + aS0y , 
whence 

V.a(S/Jy + y/?y) = 

yaj8y = aS/?y — 0Say + ySa/3 . . . . (116). 

The form of this equation shows that a and y may be inter- 
changed, or that Va/?y = Vy/?a, as already shown. 
Again, replacing a by Va/? in Equation (111), 

y . ya/?V/?y = yS( Va0)/J - 0S(Va/J)y, 
or 

V.Va/?y0y = -0Sa0y (117). 



106 QUATERNIONS. 

8. Writing Yy8a/J first as Y (y . 8a/?) , and then as V(yS . ap) , 
we have 

Y(y • So/3) = V . y (S8a0 + Y8a0) 

= yS8a0+Y.yY8a0 

[Equation (116)] = ySa08 +Yy8Sa0 -YyaSS0 +Vy0S8a. (a) 

V(y8 • a/?) = V(Sy8 +Yy8) (Sa/3 +Va/3) 

= Vy8Sa0 +Ya/3Sy8 + V . VySVa/3 

= Vy8Sa0 + Va0Sy8 -V . Va0Yy8, 
or, Equation (112), 

= Vy8Sa0 +Ya£Sy8 - 8Sa0y + ySa08. (&) 

Equating (a) and (b) , 

8Saj8y = V^ySo8+YyaS^8 4-Ya^Sy8 . . (118), 

a formula expressing a vector 8 in terms of three other vectors 
resulting from their products taken two and two ; so that, if 
Sa0y = m, Sa8 = a, S/?8 = &, Sy8 = c, 

8 = m-^aYjSy + &Vya + cVajS) . 

Operating on Equation (118) with S • p X , we obtain, since 
S • pYya = Spya, » 

Sp8Sa/?y - S£8Spya - Sa8Sp0y - Sy8S/>a0 = 0, 

or 

Sa8Sp/3y - S/38Sypa + Sy8S/>a0 - S/>8Sa/3y = . (119) , 

a formula eliminating 8. 

56. Exercises. 

Prove the following relations : 

1. Sa0y8 = S8a0y. 

2. a0 . 0y = -ay. 

3. a 2 ^=a^./3a. 

4. S.Va0V0y = S.a/3V0y (120). 

5. Sa0y8 = Sa/?Sy8 - SayS/?8 + Sa8S/?y (121), 

from which show that Sa/}y8 = S/?y8a. 



GEOMETRIC MULTIPLICATION AND DIVISION. 107 

6. S. Ya/^y8=SaSS0y-SayS/38 (122). 

7. S(a+/?)(0 + y)(y + a) = 2Sa0y. 

8. a/?y + y/?a = 2Ya/?y. 

9. a£y-y£a=2Sa/?y. 

10. V(aV0y + 0Vya + yVa/3)=O (123). 

11. Va/?y + Yya/?=2ySa0. 

12. aV£y + £Vya + y Va£ = 3 Sa£y . 

13. S . Va/?V/3yYya = - (Sa/?y) 2 . 

14. S.yY0a = y/3a-yS0a + 0Sya-aS0y . . . (124). 

15. S . Y(Ya0V0y) Y(V0yYya) Y(VyaYa0) = - (Sa£y)\ 

16. S[Ya/?Yy$ + YayY80 + Ya8Y/?y]==O . . . . (125), 

17. If Sa£y = m, Sap = 0, S/fy = 0, Syp = 0, showthatp=0. 

Conversely, if p is not zero, then Sa/?y = 0. 

18. Interpret p = a -1 ^a. 

We have first, directly, 

Tp = Tj8, 
Sap0 = Saa- 1 )8a i S= S0a/? = S0 2 a = ; 

.*. p, a and /? are complanar. 

Sap = Saa-'/fa = S/?a, 
— TpTa COS = — TaT/3 COS <£, 

or, since Tp = T/3, cos 6 = cos <£. 

Similarly Yap = Y/?a, and sin0 = sin <f>. Hence 

= <£, 
and a bisects the angle between ft and p. 

19 . Show that p = a/ftT 1 = a" 1 (Sa/3 - Ya/3) . 

20. p being any vector, show that Y • YapYp/3 = xp. 

21 . If Sa£ = —a 2 , show that a is perpendicular to B — a. 

22. What are the relative directions of a and 8, if K- = - : 

IfK° = "? 
a a 



108 



QUATERNIONS. 



57. Examples. 

1 . TJie altitudes of a triangle intersect in a point. 

Let (Fig. 60) ac = /3, cb = <z, AB = y. 
rig. 60. Then vectors along c'c, b'b and a'a are 

c 

cy, — c/3, — ca 

respectivelj*. Now 

AO = AC + CO = AB + BO, 

— a?€y = y + #c/3. 




y c' 



b or 



Operating with x S • /?, we have, since ySc/P = 0, 

— M. 

Having assumed o to be the intersection of the altitudes bb' 
and ccj let o' be the intersection of aa' and cc: Then 



or 



AO f = AC -f COj 
Zca = P — x'ey. 



Operating with X S • a 



Siya Saey 
_ Sfta = Saff 

»(y-»«y -sp«y 



Hence o and o' coincide, and 



«y- 



GEOMETRIC MULTIPLICATION AND DIVISION. 109 



2. To circumscribe a circle about a triangle. 

Let (Fig. 61) ac = /?, cb = a, ab = y. 
Then Fig. a. 

a'o = — ttea, C 

C'O = yey, 
b'o = — Zcft. 

Operating with x S . ft on the expres- 
sion 

we have 

* 28«y0 

Operating with x S . a on 




we have 



BO' = - i y + y f €y = — |a — tf'ca, 



2Scya 2Ssy0 



Therefore y = y' and o and o' coincide. 
The radius ma}' be found b}* squaring 



AO = iy+ycy = iy. 



Sa/? 
28^' 



whence 



— R' 



c 2 c s a t b a QOB 2 c 
4 4& 2 c 2 sin 2 A 



since, if a, 6, c are the tensors of a, 0, y, 



4 4 

46 2 <rsm 2 A 



Hence 



_ Vc 2 sin 2 a 4- a 2 cos 2 c __ a 

2 sin a 2 sin a 



110 



QUATERNIONS. 



3. In any triangle, the centre of the circumscribed circle, the 
intersection of the altitudes and the intersection of the medials lie 
in the same straight line; and the distance between the last two 
points is two-thirds of the distance between the first tico. 

Let m (Fig. 62) be the intersection of 
Fi &- G2- the medials, a' that of the altitudes, and 

c the center of the circle. 

Then, from Ex. 5, Art. 11, where cp 
(Fig. 11) is given in terms of the adjacent 
sides, we have 



AM = i(/? + y). 
From Ex. 1, Art. 57, 




AA' 



= /j+-M 



From Ex. 2, Art. 57, 



Sey/3 



«y- 



AC = i(y 



But 
and 



Sa/3 
"ScyjB 



•y). 



CM = AM — AC = $/3 — £y -f i 



Sa/3 

MA' = AA'-AM=^-iy + Mcy. 

Scy/? 

.'. ma'= 2cm, 



and, since, as vectors, they are multiples of each other, and have 
a common point, they form one and the same straight line. 

4. To find the condition that the perpendiculars from the angles 
of a tetraedron to the opposite faces shall intersect. 

With the notation of Fig. 52, the perpendiculars from a and u 
on the opposite faces are 

Vfiy and Vya. 

If they intersect, at p say, then must a, b, p lie in one plane. 
Hence, Art. 55, 4, Cor. 3, 



S[(/^a)Wya]=0, 



GEOMETRIC MULTIPLICATION AND DIVISION. Ill 

or 

S (/? - a) [S . V0yVya + V . V/fyVya] = 0, 

S(0-a)V. V0yVya = O. 

Fig. 52 (bis). 

But, Equation (117), ° 

V . V/?y Vya = - yS/?ya ; // \ 

.•.-(S/Jy-Say)S/?ya = 0, / / \ 

or / / \ 

S0y = Say. (a) o<(- j ™£fl 

From the figure, we have ^^\s^^ 

BC 2 + OA 2 =(y-0) 2 + a 2 

= 7 2_2Syj3 + j3 2 + a 2 
or, from (a), =y 2 -2Say + fi 2 + a 2 

= (y-a) f + i8 l 

= AC 2 + OB 2 . 

Hence the condition is that the stims of the squares of each pair 
of opposite edges shall be the same. 

5. Interpret Equation (118), 

8Sa/?y = V/?ySa8 + VyaS08 + Va/3Sy8, 

under the condition that a, /?, y be complanar with 8. 

If a, /?, y are complanar, Sa/?y = 0, and therefore, 8 being in 
or out of the plane, 

Sa8V0y + S08Vya + Sy8 Va0 =0. ( f I ) 

If 8 be in the plane, we have for an}' four co-initial lines 
OA, ob, oc, OD, 

sin boc cos aod -f- sin coa cos bod -f sin aob cos cod = 0, 

and, for a line perpendicular to od, 

sin boc sin aod + sin coa sin bod + sin aob sin cod = 0. 

If 8 is perpendicular to the plane, the terms in (a) vanish 
separately. 



112 QUATERNIONS. 

6. If X, F, Z be the angles made by any line op with three 
rectangular axes, then 

cos 2 X + cos 2 F + cos 2 Z=l. 
From Equation (67) 

ip = xi 2 -f yij + zik = — x + yk — *;, 
whence 

(St» 2 = ^. 

Operating in a similar manner with S .j X and S . k x we obtain 

-p*=($i P y+w P y+(Mpy. 

If Tp = r, then p* = — r 2 , St/a = — r cos X, etc. Hence 

op 2 = op 2 (cos 2 X + cos 2 F + cos 2 Z) , 

or 

cos 2 X + eos 2 F + cos 2 Z = l. 

Applications to Spherical Trigonometry. 

Let abc (Fig. 63) be any spherical triangle on the surface of 
a unit sphere whose center is o ; a, /?, y 
being unit vectors from o to the vertices. 
The sides ab, bc, ca represent versors 
whose angles are c, a, &, and axes are 

Bf oc' = y ; 0A' = a ; ob'=/?'; a ; # y 

V a being unit vectors to the vertices of the 
)b polar triangle whose sides are a\ b\ c\ 
the supplements of the opposite angles 
a' ~ a, b, c of the triangle abc. 




7. We have first 



*-£=. ( «) 

y a y 



GEOMETRIC MULTIPLICATION AND DIVISION. 113 
Taking the scalars, we have [Equation (90)], 



But 



and 



8 8a 8 a 

y ay ay 

P 8 a 

S- = cosa, S- = cosc, S- = cos&, 
y a 7 

8 a 
8 . T^T - = S(sin c . y f ) (sin b . p') 

= sine sin bSy'ft 

= — sine sin & cos a' 

= sine sin b cos a. 
Hence, in (a), 

cos a = cos c cos b + sin c sin b cos a. 
By a cyclic permutation of the letters in (a), we obtain 



Whence, as before 



a""J3 a" 
a p a p a 



(b) 



or 



iu which 



cos 6 = cos a cose + sin a sine Sa'y', 
Sa'y' = — cost' = COSB. 
.\ cos& = cosa cosc + since sine cos b. 



(«) 



Similarty, or directl}' b} r cyclic permutation in (c) , 

cos c = cos b cos a + sin b sin a cos c. 

From the relation 

F F a ' 
7= a' 7 

may be deduced in like manner 

— cosa = cose cosb — sine sinB cosa. 



114 QUATERNIONS. 

8. Resuming the equation 

7 a 7 

of the last example, and taking the vectors, we have [Equa- 
tion (91)], 

8 8 a a 8 8 a , \ 

y ay y a ay 

But 

V- = — a'sina, 

y 

8 a 

S- V- = cosc(/3'sin&)=coscsin& . fi\ 

a 8 , 

S-V- = cost (y f sin c) = cost sine • yj 

£ a 
T . V^V- = V(y'sinc)(/?'sin&)=sincsin&Vy7?' 

= sincsin6(— asina f ) = — sincsin&sinA • a. 

Substituting in (a), 

— sin a . a'=coscsin& . /?' -f cos & sinewy'— sincsin&sinA .a. (&) 

Operating with X S • y'" 1 , 

a 1 8' y' a 

— sina . S— l = coscsin6S cl 7+cos&sincS— . — sincsin&sinAS-jv 
7 7 7 7 

in which 

S — = cos b' = — COSB, 

S — = — COS A, 

S — = 0, since a and y' are at right angles. 

Hence 

sin a cos b = cos b sin c — cos c sin 6 cos a, 



GEOMETRIC MULTIPLICATION AND DIVISION. 115 
and in the same manner, or by a cyclic permutation of the letters, 

sin b cos c = cos c sin a — cos a sin c cos b, 
sin c cos a = cos a sin b — cos b sin a cos c. 

9. Operating on Equation (b) of the last example with 
X V . y'" 1 instead of X S . y'" 1 , 

— sinaV — = cose sin6 V — , + cos b sine V— r — sine sin 6 sinA V— ,- 



But 



Fig. 03. 



T- = painb' = /JsinB, 
Y-, | = — a sm a ' = — a sin a, 

y 

r 



Substituting these values 

— sin a sin b . /? = — cos c sin b sin a . a 

a 
— sin c sin b sin a . V~ # 

y 

Operating with X a" 1 , and substituting for 




P 

- = cos c + y' sm c, 



we obtain 



or 



— since sinB cose — sina sinB sine • y f = — cose sin6 skiA 
— sine sin b sin a . y'. 

Equating the scalar or vector parts, we have in either case 

sina sins = sin a sin 6, 

sina : sin 6 : : sin a : sinB. 



The formulae of the preceding examples have all been deduced 

from the equation - = The product as well as the quotient 

ma} r also be employed, as follows : 



116 QUATERNIONS. 

10. Assuming the vector product 
Ya0V0y, 
and taking the vector part, we have [Equation (117)], 

T . Va/?V0y = - 0Sa0y. (a) 

But 

Y • Va/TV/fy = Y(y'sinc) (a' sin a) = sine sin a sinu • /?, 

and, Art. 55, 4, 

Sa/Jy = — sine sin 6' f 

0' being the angle made bj r oc with the plane of c. Substituting 
in (a), 

sine sina sin b • /?= sine sin 0' . /?, 
Fig. ©. 

c' 



or 




sm0 ' = sina sin b. 

Bj r permutation, from (a), 

~- B ' V . VyaVa/3 = - aSya/? = - aSa0y, 

b or 
c sin £> sine sin a . a = sine sin 0' • a, 

.*. sin0' = sin&sinA. 

Equating these values of sin# we have, as in Example 9, 

sina : sin 6 : : sin a: sine. 

11. Let p a , p 6 , p e represent the arcs drawn from the vertices 
of abc perpendicular to the opposite sides. 

Resuming Equation (a) of the preceding example, and taking 
the tensors, 

TV . Va£V/?y = Sa/?y = sine sin/> e , 
= S/?ya = sin a sinp a , 
= Sya/J = sin b s\np bJ 



■Mijugyi 



GEOMETRIC MULTIPLICATION AND DIVISION. 117 

and, taking the tensor of Y . Va/3V/3y from the last example, 

sine sin a sins = sin a sinp a = sin b sinp b = sine sinp * 
or 

sinp = sine sin b, 

sine sin a . 

sm p h = sin b, 

sin 6 

sinp c = sinasinB. 

12. Show that if abc, a'b'c' be two tri-rectangidar triangles on 
the surface of a sphere, 

COS AA r = COS BB f COS CC f — COS b'c COS BC', 

the triangles being lettered in the same order. 

Let a, /J, y, a! # y' be the vectors to the vertices. These 
being at right angles, in each triangle, we have 

cos aa' = - Saa' = - S . YPyjp V 

or, Equation (122), 

cos a a' = SftS'Syy' - SjS'ySjSy' 

= COS BB' COS CC' — COS B f C COS BC*. 

[The vectors of Equation (122) are arbitrary, but we may 
divide both members by the tensor of the product of the vectors, 
so that 

S(VUaj3VUy8) =SUaSSU/?y - SUaySU/?8, 

for the unit sphere.] 

13. Let abcd be a spherical quadrilateral whose sides are 
ab = a, bc = &, cd = c, da = d, the vectors to the poles of these 
arcs being a, # yj $' respectively. Then 

Va/? = a r sina, 
Vy8 =y'sinc. 



118 QUATERNIONS. 

From Equation (122), 

8 . Ya/3YyS = Sa8S0y - SayS/?8, 
or 

sin a sine Sa'y r = (— cos da) ( — cosbc) — ( — cosdb) ( — cos ac) . 

But Sa f y' = — cosl, l being the angle formed b}' the arcs ab 
and cd where they meet, the arcs being estimated in the 
directions indicated by the order of their terminal letters. 
Hence 

sinAB sin cd cosl = cosac cosbd — cos ad cosbc, 

a formula due to Gauss. 

14. Retaining the above notation, abcd being still a spherical 
quadrilateral, denote the angles at the intersections of the arcs 
ab and cd, ac and db, ad and bc, by l, m and n respectively. 
Then, from Equation (125), 

S[Va/?Vy8 + YayY80 + Ya8Y0y] = 0, 

we have identically 

sinAB sin cd cosl + sinAC suibd cosm + sin ad sinBC cosn = 0. 

Were the points a, b, c, d on the same great circle, the angles 
l, m and n would be zero, and the above reduces to 

sinAB sin cd -\- sinAC suibd + sin ad sinBC = 0, 

and for a line oaJ perpendicular to oa and in the same plane, 
dropping the accent, we have 

cosab sin cd + cos ac sinBD -f- cos ad sinBC = 0, 

which are the results of Example 5 of this article. 



r 



GEOMETRIC MULTIPLICATION AND DIVISION. 119 

68. General Formulae. 

1. We have seen, Equation (86), that SS = 2S and TS = 2V ; 
but (Art. 50, 4) that ST is not equal to T2, nor SU to VX We 
have also seen, Equations (96) and (97), that Tn = IIT and 
Un = IIU; but SII is not equal to IIS, nor VII to nV: for, 
1st, SII is independent of the factors under the II sign, provided 
the product remains the same, while IIS is dependent upon 
them ; and, 2d, because (Art. 55, 5) IIV is not necessarily a 
vector. 

2. Resuming Equation (92), 

Srg = Sgr, 

and, since r is arbitrary, writing rs for r, we have, by the asso- 
ciative law (Art. 52) , 

S(rs)? = Sg(rs), 

.-. Srsg=Ssgr=Sgrs . . . . (126), 

a formula which may evidently be extended. Hence, the tta&xi 
of the product of any number of quaternions is the same, so long 
as the cyclical order is maintained. 

3. Let p, q, r, s be four quaternions, such that 

qr = ps. (rt) 

Operating with Kq x , 

Kq . qr = (Kg . q)r = {qKq)r = Kq . ps, 

since conjugate quaternions are commutative. Hence 

(Tq)*r = Kq.ps, 
or 

Kq. ps 1 

(Tg)* =sB ^ m P 8SS q m P 8 # # * ( '' 

Operating on (a) with xKr, we have 

qr . Kr = />s • Kr, 






1 



120 QUATEBNIONS. 

or 

q(Tr)*=psKr, 

psKr 1 

••• Q = Jfr)2 = psKr = p8- . . . (128). 

Hence, in any equation of the products of two quaternions, 
the first factor of one member may be removed by writing its con- 
jugate as the first factor of the second member, and dividing the 
latter by the square of the tensor, or simply by introducing the 
reciprocal as the first factor in the second member. By substi- 
tuting the word last for first, the above rule will apply to the 
second transformation. 

4. Resuming, for facility of reference, the equations 

q = ^ = '!^(coa<l> + *sm<t>) = Tq.Vq = 8q + Yq, (A) 

1 B T/3 
g- , = - = - = ^(cos^-€sin^), (B) 

K^ = ||(cos<A-csin^) = Sg-yg, (C) 

we observe directly that 

Sg = S(T?.Ug) = Tg.SUg . . . (129), 

Jq = TYq. VYq^Tq.YUq .. . . (130), 

TV? = Tq . TVU? = TYKq .... (131). 

5. It has been already shown (Art. 54, Fig. 40) that 
(Ta) 2 + (T0) 2 = (Ty) 2 , and (Art. 54, Fig. 42) that Ta=Ty . cos<£, 
T/3 = Ty . sin <f> ; and therefore 

(Ty) 2 cos 2 <£ + (Ty) 2 sin 2 <£ = (Ty) 2 , 
or 

sin 2 <£-t-cos 2 <£ = l. 

Hence, from Equations (44), 

(SUg) 2 + (TVU?) 2 = l .... (132). 



GEOMETRIC MULTIPLICATION AND DIVISION. 121 

This important formula might have been written at once by 
assuming the above well-known relation of Plane Trigonometry. 

6. From Equations (129) and (131), we may write Equa- 
tion (132) under the form 

(S?) 2 + (TVg) 2 = (T0) 2 (133), 

or, from Equation (107), 

(Sg) 2 -(Tg) 2 = (Tg) 2 = (^) 2 + (TT^ . (134), 
since c 2 = — 1. 

7. Comparing (A), (B) and (<7), 

SUg = SU- = SUKg . . . (135), 

TYVq = TVU- = TYVKq . . (136), 

and from Equations (129) and (135), 

1 
S? = T?.SU? = Tg.SU-==Tg.SUKg. . (137). 

8. Since Tq = TKg, we have 

'Tq .TKq = (Tq)* (138), 

and Tq being a positive scalar, 

KTq = TKq (139). 

As exercises in the transformation of these and the following 
symbolical equations, some of the results already obtained will 
be deduced anew. Thus, to prove that T(gg') = TgTg', whence 
T • <f = (Tq) 2 , we have 

(Tqq')*= (qq')K(qq') Equation (107) 

= qq'Kq'Kq Equation (99) 

= q(q'Kq')Kq = (Tq'yqKq 

= (iy) 2 (T<z)', 

.'.Tqq' = TqTqi 



122 QUATERNIONS. 

9. Substituting for Hq and TYq their values from Equations 
(79) and (131) 

(SKg) 2 +(TYK( ? ) 2 = (Sg) 2 + (TT< ? ) 2 . . (140). 

10. Resuming from Art. 51, 1, the expressions 

Yrq = SrYq + HqYr + Y . YrYq, (a) 

Yqr = HqYr + SrYq + T . YqYr, (b) 

8qr = SqSr + 8 i YqYr, (c) 

we have, by adding and subtracting, 

Yqr + Yrq=28qYr + 28rYq\ n± . 

Yqr-Yrq = 2Y.YqYr ) ' ' * ' { >' 



And, if q = r, from (a) and (c), 

,} (142), 



Y .<f=28qYq 

S.g 2 = (S^) 2 + (Y^) 8 : 
whence 

<f = (S?) 2 + 2 SqYq + (Yq)>. . . (143). 



Dividing Equations (142) by (Tg)* 

SU. 9 ! ^(SU9) ! +(TUg)») , .. 

YU. 9 s = 2SUg.VUg J • • • • ^**)» 
since, evidently, 

S.g 2 = (Tg) 2 SU.9 2 ) ruM 

Y. 9 2 = (Tg) 2 YU.9 2 i V h 

Again, substituting in the second of Equations (142) the value 
of (Yq) 2 from Equation (134), we have 

S.( ? 2 =2(^) 8 -(Tg) 2 (146), 

and dividing by (Tg) 2 

SU.g 2 =2(SUg) 2 -l (147). 

Substituting (Sg) 2 from the same equation 

S-? 2 =2(Y?) 2 + (T?) 2 (148). 



GEOMETRIC MULTIPLICATION AND DIVISION. 123 

Equations (146) and (148) may be written 

(S + T)g 2 =2(Sg) 2 and (S-T)g* = 2(Y?) 2 . 

Introducing in (a), or (6), the condition that q and r are 
complanar, we have, after substituting versors, 

YVqr = VUgSUr + YUrSUg, 

since, under the condition, Y(YCqYUr) = 0. 
Taking the tensors, since q and r are complanar, 

TYU?r=TYUgSUr + SUgTYUr . . . (149), 

and, interpreting, Art. 51, 6, 

sin (0 + <f>) = sin0 cos <f> + cos0 sin <£. 

Introducing the same condition of complanarity in (c) 

S?r = S?Sr — TTgTVr, 

or, substituting versors as above, 

SUgr = SUgSUr-TVUgTYUr . . . (150), 
or, interpreting, 

cos(0 + <£) = cos0 cos<£ — sin<£ sin0. 

11. Putting Equation (146) under the fbrm 

and writing Vg for #, we have 

SVg = VJ(S^+W) .... (151). 

12. Taking the tensors of the first of Equations (142) , we have 

TT *=-2sr 



whence 
and 



124 QUATEBNIONS. 

and writing \fq for q 

or, by Equations (133) and (151), 

™ r , / (Tg)'-(Sg)» 

TY . V^= Vi(T 9 - Sg) (152), 

13. From the definition of the powers of a quaternion, we have 

q- m q m =l, (q*) n = q m .... (154). 

Hence, since q = Tq . Ug, Til = IIT and UII = IIU, 

Tg -m . Tg w = 1, Vq' m .Vq m =l . . (155) . 

Also, because Vq~ M = UKg m , 

q- m =zTq~ m . Vq- m =Tq- m . UKg w = Tg- 2m Kg m , 

or, since Kpq = KgKp, writing pg for g, and making m = 1, 

(PS)" 1 = T(i>g)- 2 Kpg = T( W )" 2 KgKp 
= T{pqY\Tqf{Tpyq' l p-\ 
Or 

{pqY l = r l p- 1 (156), 

the reciprocal of the product of ttoo quaternions being equal to the 
product of their reciprocals in inverted order. 

This formula may be extended by the Associative principle, by 
a process similar to that emplo}*ed in the deduction of Equation 
(126), so that if IT represent the product of the same factors as 
those of n, in reverse order, 

(Uq)- l = Wq- 1 (157). 



GEOMETRIC MULTIPLICATION AND DIVISION. 125 

The equation lLpq = KgKp may be deduced without reference 
to spherical arcs. For, bj r Art. 44, anj r two quaternions can be 

reduced to the forma g = £- , p = 1, whence 

a p 

and therefore 

Kp . y = Kp . p)8 = (Kp . p)P = (Tp)% 
Now 

(KgKi>)y= Kg(T i >)^ = (Ti ? ) 2 Kg . fi 
= (Tp) 2 Kg . ga = (T/>) 2 (Tg) 2 a = (Tpg) 2 a 
= Kpq .pq.a = Kpg . y 
.'. Kpg = KgKp, 

which, by the Associative law, gives 

KII = n'K ..... . (158). 

14. Show that K(- q) = - Kg. 

15. Show that 

TQ> + g) 2 = Q> + g)(K/> + Kg) 

= (Tp) 2 + (Tg) 2 + 2S.2)Kg 

= (Tp) 2 + (Tg) 2 + 2TpTgSU . pKq 

== (Tp + Tg) 2 - 2Ti>Tg(l-SU . pKg) , 

and therefore that T(|> + g) cannot be greater than the sum or 
less than the difference of Tp and Tg. 

16. Show that gUVg" 1 = TVg - SgUVg. 

59. Applications to Plane Trigonometry. 
1. For formulae involving 20, let 

g = Tg(cos20 + €sin20). 
Then _ 

Vg = q' = VTg(cos0 + csin0). 



126 QUATERNIONS. 

From Equation (142), S . (f = (Sg) 2 + (V?) 2 , we then have 

S? = (S?') 2 + (W, 
or, dividing out T#, 

SUg = (SU?') 2 + (VU?') 2 ; 
and, interpreting, 

cos20 = cos 2 0-sin 2 0. 
Again, from Equation (147), SU . q 2 = 2(Slty) 2 - 1, 

SU^ = 2(SUg , ) 2 -l; 
whence 

cos20 = 2cos 2 — 1. 

Again, from Equation (142), Y . q 2 = 2StfYtf, 

Yq = 2Sq l Yq l , 

or, dividing out Tq and c, 

TVUg = 2Slfy'TYU? f ; 
whence 

sin20 = 2cos0sin0. 

2. Resuming Equations (149) and (150), 

TVUgr = TVlfySUr + SUgTVUr, 
SUgr = SlfySUr - TVltyTVUr, 

which have alreadj r been interpreted as the sine and cosine of 
the sum of two angles, and writing for 

r = Tr(cos<£ + €sin <£) , r" 1 = — (cos<£ — csin<£), 

q and r being complanar, we have 

TVUgr- 1 ==TVUgSUr-SU^TVUr . . (159), 
SU^r-^SU^SUr + TVUgTVUr . . (160), 
or, interpreting, 

sin (0 — <£) = sin cos <f> — sin <£ cos0, 
cos ($ — </>) = cos0 cos<£ + sin0 sin <f>. 



GEOMETRIC MULTIPLICATION AND DIVISION. 127 

3. Adding Equations (149) and (159), 

TYVqr + TYVqr- 1 = 2 SUrTVUg, 

in which, if qr = p, qr' 1 = t, .•. q = -y/pt, r = -y/pt' 1 (Art. 58, 3), 

TVUi> + TVUf==2SU(ypF 1 )TVU(V^) . (161), 

sina;+siny = 2 cos £ (a? — y)sin^(x + y). 

Similarly, by subtracting the same equations, 

TYVqr — TYVqr- 1 = 2 SUgTVUr, 

TVUp - TVU* = 2SU( VjTOTVl^VpF 1 ) . (162), 



or 



or 



sin a; — siny = 2 cos£(a? + y) sin£(a; — y) . 



4. From Equations (150) and (160), by addition and sub- 
traction, we obtain, in a similar manner, 



and 
whence 



SUp + SU* = 2SU(Vz>0SU(ViF 1 ) • • 
SUp - SVt = - 2 TVU(ypOTVU(VpF 1 ) , 

cosa;-f-cos#= 2cos£(a;-|-y) co&i(x — y), 
cosy — cos# = 2 sin £(# -f- y) sin £(# — y) . 



5. Resuming Equation (152), 



TVVg = V4(Tg-S?), 

it may be put under the form 

2(TVUVg) 2 = l-SUg, 
or 

2sin 2 £0 = l — cos0. 

and, in a similar manner, from Equation (151), 



or 



sV? = V£(Sg + Tg), 
2(SUV</) 2 = SUg+l, 

2cos 2 £0 = l+cos0. 



(163), 



128 QUATERNIONS. 

6. From Equation (142) 

(TV:S)g° = _l?ffi_ 9 
' (**/)*+ (V</) 2 

2TYq (S?) 8 



Sq (S<,) 2 -(TVg) 2 
2(TV ; S)q 



l-[(TV:S)g] 2 ' 
or 

tan2fl = 2tan * . 
l-tan 2 

And, in a similar manner, 

cot 2 0-l 



cot20 = - 



2cot0 



7. From Equations (90) and (91), q and r being complanar, 

Sgr = S?Sr + S . YqVr = S?Sr - TVgTVr, 
TVgr=S?TVr + SrTV?, 

we have, by division, 

v '* S?Sr-TV?TVr 

_ (TV; g)r + (TY:S)g 
~"l-(TV:S)g(TV:S)r' 

tan(fl + <fr)= tan * + tan * 
v r/ l-tan</>tan0 

f TV • SW" 1 = (TY;S)g-(TV;S)r 
v * }H l + (TV:S)g(TV: S)r' 

tan(fl-</>) = <"g" <"* . 
l + tan0tan<£ 



or 



Also 



or 



8. From Equation (153) 



(TT:S)Vg = J T ^-^, 

V } * \Tq + Sq' 



GEOMETRIC MULTIPLICATION AND DIVISION. 129 

by substituting in succession p and t for Vg, .•. g — p 8 and 
q = £ 2 , we obtain, after reduction, 

(TY:S)p + (TY:S)t= p ^ ^, 

or, dividing out TpTJ from the second member, and applying 
Equation (149), 

(ty:S)p +( tv:S),= I^, 

or 

, sin(aj + y) 

tana; + tan v = — — . 

cos a; Cosy 

And similarly, by subtraction and applying Equation (159), 

(TY:S)p-(TY:S)t = !^, 
or 

tana — tany = — L_ZJlZ. 
cos x cosy 

9. From Equations (161) and (163) 

«™ /- TYUp + TYU* 



2suypr 



whence 



or 



2suVpr l 
<Tra!8B) ^ =(T r :S) ^_n^±£™ 

cosaj-f- cosy 
And, in a similar manner, from Equations (162) and (163), 

(TT;S)^P = TYU ^- Tm , 

v ' * SUp + SU* ' 

or 

* , • , v sin x — sin y 

tan£(a — y) = . *. 

cosa + cosy 



130 QUATEKNIONS. 

10. Similar formulae may be deduced for functions of other 
ratios of an angle. Thus, from Equation (90), writing rs for 
?•, and making q = r = s all complanar, we have, by Equation 
(142), 

S.g 8 = (S9) 8 -3Sg(TT^) 2 , 
or 

cos 3 $ = cos 8 — 3 cos $ sin 2 0, 

or, under the more familiar form, 

cos 3 = 4 cos 3 — 3 0080. 



CHAPTER III. 

Applications to Loci. 

60. Any vector, as /o, may be resolved into three component 

vectors parallel to any three given vectors, as a, /?, y, no two 

of which are parallel, and which arc not parallel to any one 

plane. Thus 

p = xa + yp + zy (164) 

refers to any point in space. 

If the variable scalars #, y, z are functions of two independ- 
ent variable scalars, as t and w, p is the vector to a surface, 
which, if the functions are linear, will be a plane. We maj r , 

therefore, write 

P=<t>(t, u) (165) 

as the general equation of a surface. 

If #, y and z are functions of one. independent variable scalar, 
as t y p is the vector to a curve, which, if the functions are 
linear, becomes a right line. We may, therefore, write 

P = *(0 (166) 

as the general equation of a curve in space. 

If a, /?, y are complanar, we may replace either two of the 
vectors in Equation (164) by a single vector, in which case 
p = <f>(t) contains but two variable scalars, functions of t, and 
is the equation of a plane curve, or of a straight line if the func- 
tions are linear. 

The essential characteristic of the various * equations of a 
straight line is that they are linear, and involve, explicitly or 
implicitly, one indeterminate scalar. 

181 



132 QUATERNIONS. 

61. Assuming 

p = xa + yp, (a) 

in which x and y are variable scalars, functions of a single vari- 
able and independent scalar, as t, as the general form of the 
equation of a plane curve, bj T substituting in any particular case 
the known functions x = f(t), y=f\t), or x=f"(y), we may 
avail ourselves of the Cartesian forms and apply to the resulting 
function in p the reasoning of the Quaternion method. 

For example, suppose a and ft are unit vectors along the axis 
and directrix of a parabola, the origin being taken at the focus. 
In this case we have the Cartesian relation 

tf=2px+p\ (b) 

or, substituting in (a), 

as the vector equation of the parabola. 

Or, again, a and /? being any given vectors parallel to a diam- 
eter and tangent at its vertex, 

is the vector equation of a parabola, in terms of a single inde- 
pendent scalar t. . 

62. Let /(a) be any scalar function as, for example, 

f(x) = x>. 
Then 

d[f(x)]=2xdx = [f\x)]dx. 

If, however, f(q) be a function of a quaternion q, as, for 
example, in the above case, 

then 

f(q + dq) = (q + dqy = q* + qdq + dq.q + (<lq)\ 
-'-<l[f((l)l=qdq + dq .q, 



APPLICATIONS TO LOCI. 133 

which cannot, however, be written 2qdq, because of the non- 
commutative character of quaternion multiplication. We can- 
not, therefore, write, in general, 

or form, as usual, a differential coefficient. Since vector, as 
well as quaternion, multiplication is non-commutative, the same 
is true of the differentiation of a function of a vector. Thus, if 

f(p) = p\ 
tf[/(p)] = pfy + cfy> . p, 

and in order to write d[/(/o)] = [/'(p)]<fy>, it would be necessary 
to determine a vector <r, such that <rdp = dp • p, or 

<r = dp • pdp~ x , 

or, if c be the versor of dp, since the tensors cancel, 

o- = €/o€ _1 ; 

that is (Art. 56, 18), we must have /o, c and <r complanar, or 
Vco- =s Yp€. Since complanar quaternions are commutative, if q 
and dq are complanar, or if dq or dp is a scalar, this peculiarity 
of quaternion and vector differentiation disappears. In this 
case, dq and dp being scalars, f(q) or f(p) are quaternion or 
vector functions of scalar variables, to which the ordinary rules 
of differentiation are applicable. In fact we have only to assume 
such a function, as 

p = X'a' + x"a" + x"'a'" + = %Xa = </>(*) , 

in which a', a", a'", are constants and the onlj r variables are 

the scalar multipliers, to see that the vectors a', a", a"' are 

to be treated as constants and the usual rules of differentiation 
applied to the scalar coefficients. 

Such equations, then, as those of the parabola, (b) and (c) , 



134 QUATERNIONS. 

Art. 61, in which a and fi are given constant vectors, may be 
differentiated as usual. Thus, from 



we have 






p and p' being any two vectors to the curve, 

is the vector secant ; so that when p and p' become consecutive, 
and the secant a tangent, 

dp = (ta+P)dt 

is a vector along the tangent to the curve at the point corre- 

t 2 
sponding to t. The vector to this point being - a -+- *£, and x 

any variable scalar, we ma}' write the equation of the tangent 
line at that point 

p = ?a + tp + x(ta + /3); 

for anj T given point, x being the only scalar variable. 

63. It has been seen that the usual definition of differential 
coefficients is inapplicable to quaternions in general, for this 
definition involves the commutative property of multiplication, 
which is not, in general, true of quaternions, nor of the vectors 
to which thej' may degrade. It becomes necessary, therefore, to 
give a definition of differentials which shall not involve this prop- 
erty,' yet which shall also be true of quaternions which degrade 
to scalars, and therefore be equally applicable to ordinary scalar 
quantities. 

If p =/(g), such a definition is involved in the formula 

* = n li f i i»[/(? + n- 1 ^) -/(</)] • • (167), 



APPLICATIONS TO LOCI. 135 

for, let/(g, r, s, ) = be an}' relation between a sj'stem of 

quaternions g, r, s, , and let Ag, Ar, As, be finite and 

simultaneous differences, so that g-f-Ag, r + Ar, s + As, 

satisfy the relation /(g, r, s, ) = 0. Then in passing from the 

new system g + Ag, to the old system g, , the simul- 
taneous differences can all be made to approach zero together, 
since they all vanish together. If, while these differences Ag, 
Ar, thus decrease indefinitely together, thej r be all multi- 
plied by the same increasing number, n, the equimultiples nAg,. 

wAr, may tend to finite limits, and these limits are defined 

to be the simultaneous differentials of the related quaternions g, 
r, s, , and are written dq, dr, cfo, Simultaneous differ- 
entials are, therefore, the limits of equimultiples of simultaneous 
decreasing differences. If, then, in Ap=/(g + Ag)— /(g), 
while the finite differences Ap, Ag be indefinitefy decreased, they 
be multiplied by a number, n, ultimatelj' to be made infinity, 
so that 

nAjp = n[/(g + Ag)-/(g)], 

and we pass to the limit, writing dp for nAp, and dq for nAg, 
we have 

*-JS,^/(.+s)-/(,)} 

a formula for the differential of a single explicit function of a 
single variable. 

If Q=F(q,r, ), 

d Q =i = x »[* , (9+*" 1 *i r+n-*dr, )-F(g, r, )] (168). 

In these formulae, dq, dr, are an}' assumed variables, no 

reference having been made to their magnitudes, and n any 
positive whole number conceived so as to tend to infinity. To 
show that these differentials need not be small, as also the ap- 
plication of the formula to the differentiation of ordinar}' scalar 
quantities, let 



136 QTJATEBNIONS. 

then 



(y + Ay) = (x+Ax)*; 



whence, as usual, 



Ay=2xAx + (Ax)*, 

or, n being a positive whole number, 

nAy = 2xn A x + n~ l {n A x) 2 . 

If, now, the differences A y and A x tend together to zero, 
while n increases and tends to infinity in such a manner that 
n A x tends to some finite limit, as a, we have, for the other 
equimultiple nAy, 

nAy = 2xa + n~ l a*. 

But, since a, and therefore a 2 , is finite, n~ l a 2 tends to zero, 
and, at the limit, nAy== 2xa. Hence the limits of the equi- 
multiples n A x and n Ay are respectively a and 2#a, and 
dx = a, dy = 2xa by definition ; from which 

dy= 2xdx. 

For a vector function we should write 

d P = ^ lt K n\J( P + n-*dp)-f(p)l ■ • (169), 
and for a scalar function, p = <f> (t) , 

dp = d[^(0]= M li ™ i ^«^^+^-^(o] • (170), 

in which letter t and dt are independent and arbitrary scalars. 
64. As a further illustration of the definition, let 



APPLICATIONS TO LOCI. 



137 



Fig. 64. 




be the equation of any plane curve in space, and op = p (Fig. 64 ) 
a vector from the origin to a point p 
of the curve ; t being an}' arbitrary sca- 
lar representing time, for example ; so 
that its value, for any other point p' of 
the curve, represents the interval 
elapsed from an} r definite epoch to the 
time when the point generating the 
curve has reached p.' 

If p' be the vector to p; then 

p f — p = pp'= Ap 

is strictly the finite difference between p and p' f and, if the corre- 
sponding change in t be At, 

PP r = ( p + A p) - p == A p= <f>(t + A t) - <£(0 = A <f>(t ) ; 

where op'= <f>(t + A t) , and At is the interval from p to p! 

In £ A 2, p would have reached some point as p", for which 
op"=<£(£ + £A t), on the supposition that pp" is described in 
£ A t. On the basis of this closer approximation to the velocity 
at p, p would have been found at p", had this velocity remained 
unchanged, such that 

pp"= 2pp"= 2(op"- op) = 2[>(* + i A - +(*)]■ 

For a closer approximation to the vector described in A t with 
the velocity at p, suppose at the end of -J A t the point is at p r ", 
for which op'"=<£(£ + £A£). Under this supposition, the vec- 
tor described in A t would have been 



pp'"= 3pp'"= 3(op'"- op) = 3[>(* + iA0- <K0]i 

and, at the limit, representing the multiple of the diminishing 
chord by dp, 

*-™.»K'+*)-*('>} 



138 QUATERNIONS. 

65. Resuming Equation (167), 

dp = df(q) = ^ l K if(q + n^dq) -f(q) ] , (a) , 

the second member may be written /(g, dq) , but not, as ordi- 
narily, f(q) dq. 

In /(g, dq), dq may be composed of parts, as q', g", q"', , 

with reference to which /(g, dg)=/(g, g'+g"+ ) is distrib- 
utive. To prove this, let 

dg = g'+g"; 
we are to prove that 

f(q,q'+q")=f(q,q')+Aq,q"). 

Since before passing to the limit, the second member of (a) 
is a function of n, q and dq, we may express this function by 
the symbol / n (g, dq) , and write 

/(g, dq) = n[/(g + n^dq) -/(g)]=/ n (g, dq), 
or 

X<l + n- l dq)=f(q) + n- l f n (q, dq). 

Replacing dq by q' and g" in succession, we have 

f(q + n-iq')=f(q) + n->Uq,q'), 
f(q + n-»ff» ) =/(<?) + n-y n ( 3 , q") , 

and, following the same law of derivation, 

f(q + rr*q"+ n"V) =/(<? + Wq") + n^f K (q + n"^", q'), 
f(q + n- l q'+ «" V) =/(?) + ""'/-(S. ?'+ <?"), 

from which 

/„(<?, <?'+ ?") =/.(<?, 9") +/.(« + »" Y', <?') , 
the limiting form of which, for n = oo, is 

Xq,q'+q")=f(q,q")+f(q,q') ■ ■ (in), 



APPLICATIONS TO LOCI. 139 

which may, in like manner, be extended to the case of 

d4 = q'+q"+q'"+ 

It follows from the above that, if p =f(q, xdq), 

f(q,xdq) = xf(q,dq) .... (172). 
If Q = F(q y r, ), whence, Equation (168), 

dQ = d[F(q,r, )] 

limit 

'w = oo 



the last member will be a linear and homogeneous function of 

dq, dr, , and distributive with reference to each of them. 

Hence, to differentiate such a function, we do so with reference 
to each factor, and take the sum of the results obtained, as usual ; 
taking care, however, not to make use of the commutative prop- 
erly. Thus d{qr) — dq . r + qdr, but not rdq + qdr. 

66. When q is a function of anj T variable scalar t, represent- 
ing time, for example, then, if t be given a finite increment A f, 
for which the corresponding one of q is A g, we have 

A q = A w + A xi + A yj + A zk ; 

and, if the several parts of the quaternion vary continuousl} r 
with the independent variable 2, at the limit we may form, as 
usual, the differential coefficient 

dq^dw dx ( dy. dz^ 
dt dt dt dt J dt ' 

The successive differential coefficients, as also the partial ones, 

when q = <f>(t, v, ), are derived from the quadrinomial form in 

the same manner. 






= n~l n WV + n- l dq,r+n- 1 dr, ..,..) - F(q, r ) ] , .i 



140 QUATEENIONS. 

67. Examples. 
1. TofinddTg. 

dTg _ d Vw 2 + o? + y 2 + s 8 
d* d* 

TgV <** * ctt d*/ 
1 « dq dq UKgTg 

dq 



or 



or 



dq 

dTg__ d£_ 
~W~ Vq 

2. (Tp) 2 = - P 2 . 

The first member being a scalar, we have 

2TpdTp. 
From the second member 

= limit pd/a + dp . p + 7i _I (dp)* 
= pdp -\-dp • p = 2 Spdp. 
Equating 

T/odT/o = — Spd/o. 

From this we may obtain 

dTp = — S • XJpdp = S |j-» 

dty dp 

3. TofinddUg. We have 

TgUg = g; , 
dTg . U? + dVq . Tg = dq, 



whence 



APPLICATIONS TO LOCI. 141 



dTg . Uq . dUq . Tq = dq 
TqUq TqUq ~" q ' 



or 



dUq dq dTq 
Ug ~~ q Tq' 

and, substituting from Ex. 2, 

dUq _ dq ~ dq . 

Ug ~~ g q ' 

" U</ <? ' 

or 

c?Ug = Y^2.Ug. 
Q 

4. From the above expressions for dTg and dUg, we have 

dq = dTg . Vq + TgdUg 

= / s ^ + Y dq\ v 

\ vq VqJ 

as the form under which the differential of a quaternion may 
alwa}*s be written. 

5. To find dUp. We have, from p = TpUp, 

dp = dTp . Up + TpdUp, 

dp = dTp . riUp 

P T P Up 

= s dp OTg from Ex. 2, 

P V 
or 

<TPp __^P __ G^P — V^= yp(?P Vp<?p __ . 

Up p p p p 2 p 2 

whence, also, 

P (Tp) 3 



142 QUATERNIONS. 

6. From the above expressions for dTp and dUp, 



dp = dTp . Up + p- 



(Tp) 2 



7. That S, V and K are commutative with d is seen from the 
following : 

g = Sg + Vg, 
whence 

dq = d8q + dYq, (a) 

and, since dq is a quaternion, 

dq = Sdg + Ydg, (6) 

hence 

dSg=Sdg and dYg = Ydg. (c) 

Again 

Kg = Sg-Yg, 
whence 

dKg = dSg — dYg, 

and, taking the conjugate of dq in either (b) or (a), we have, 
with or without (c) , 

dKq = Kdg. 

8. (Tg)* = gKg. 

2 TgdTg = n U ™^ n [(g + trtig) (Kg + irHlKg) - gKg] 

= limit [dg(Kg + rr l lLdq) + gKdg] 

= dq . Kg + gKdg 

= K . gKdg + gKdg 

= 2 S . gKdg =28. Kgdg, [Equation (80)] 

or, since Tg = TKg and UKg = V- = — , 

g Ug 

dTg = S . V-dq = S . Ug^dg. 

If g = a vector, as p, then, since Kp = — p, this becomes 

dTp = — S . Updp, 
as in Ex. 2. 



APPLICATIONS TO LOCI. 143 

9. r = g*. 

= limit [qdq + dq . q + n~ 1 (dq) 2 '} 
= qdq + dq .q; 
.'. dr = 2 Sqdq + 2 SgVdg + 2 SdqYq. 

If (/ = a vector, as p, then 8q = 0, Sdg = 0, and 

dO> 2 )=2Sp(fy 
as in Ex. 2. 

10. r = Vg. Then g = r 2 , and, as before, 

dg = ?*dr + dr • r. 

Operating with rx and xKr, in succession, 

rdq = r*dr + rdr • r, 
dg . Kr = rdr . Kr + dr . rKr 
= rdr . Kr + (Tr) 2 dr, 
or, adding, 

rdq + dq . Kr = [V + (Tr) 2 ]c?r + rdr(r + Kr) 
= [? J} + (Tr) 2 + 2Sr.r]dr, 

which gives dr = dVg in terms of dq. 

11. gg" 1 = 1 . We have 

qd(q-*) + dq.q-i = 0. 
Operating with q~ l X 

q- l qd(q- l ) + q- l dq.q- l = O y 

d± = -±dq . i. 
Q Q Q 



144 QUATEBNIONS. 



r 

f : If q = a vector, as p, 

£. . d- = — dp- 

\ P P P 

= — ap- + --cfy> ap 

^ P P P P P 

p p\p pj 

f p 3 p p 

j: P\P P/ P P 

f 12. Differentiate SUg. 

dSVq = SdTCg = S . T y Vq [Exs. 7 and 3.] 

[ -■.**» 

=:S.— VYqTWq 

= -S, dg TVUg. 
gUVg * 

13. Differentiate VUg. 

dYUg = T.dUg=V.VyUg [Exs. 7 and 3.] 
= V.Ug- 1 V(eZg.g- 1 ). 

14. Differentiate TVUg. 

dTVU g = S d( T y [Ex.2.] 

~ S UVg VYq 

do 



APPLICATIONS TO LOCI. 145 



The Bight Line. 

As in Cartesian coordinates, the form of the equations of a 
right line, as of other loci, will depend upon the assumed con- 
stants, and in any given problem one form may be more con- 
veniently used than another. 

68. Bight line through the origin. 

If o be the initial point, or origin, and p = or a variable vec- 
tor in the prolongation of a = oa, then 

p = xa (173) 

is the equation of a right line through the origin in the direction 
of the constant vector a. 
The equations 

%z?\ o«> 

obviously refer to the same right line. 

Since an} r line, represented as a vector by a, is parallel to 
p =a?a, we may say that the above equations are those of a right 
line through the origin parallel to a given line ; or, a being a 
point given by a = oa, they are the equations of a right line 
through the origin and a given point. 

69. Parallel lines. 

If /? = ob be a constant vector to a given point b, then 

p = p + xa (175) 

is the equation of a right line through a given point, and parallel 
to a given line, as p'= xa through the origin. Or, a being a given 
vector, it is the equation of a right line through a given point 
and having a given direction. If a is an undetermined vector, 
it becomes the general equation of an}' one of the infinite num- 
ber of right lines which may be drawn through a given point. If 
and b coincide, /? = 0, and, as before, p = xa. 






146 QUATERNIONS. 

a remaining the same, and /?' = ob' being a vector to any other 
point bJ for the equations of two parallels, we have 

;:#*} < i76 >- 

or, since a and p —ft are parallel, 

70. Bight line through two given points. 

If OA = a (Fig. 65), ob = /? are the vectors to the given 
points, and p the variable vector to any 
point r of the line whose equation is re- 
quired, we have 

ar = #ab = x(/3 — a), 
and 

OR = OA -h AR, 

or, for the required equation, 

p = a + a(0-a) . (178), 

which, if one of the points, as a, coincides with the origin, 
becomes p = x/3, as before. 

We have seen, Art. 55, that if Sa/fy = 0, a, ft and y are com- 
planar. Replacing y by the variable vector p, 

Sa#> = (179) 

is the equation of a plane, since it expresses the condition that p 
is complanar with a and /?. If we have also Sayp = 0, the two 
equations, taken together, represent the line of intersection of 
these two planes. 

These equations may be obtained from the line p = xa by ope- 
rating with S(Va£) x and S(Yay)x ; or, conversely, to find the 
equation of the line in terms of known quantities, having given 

Safip = 0, Sayp = 0, 




APPLICATIONS TO LOCI. 147 

write these latter under the form 

S . pVa/3 = 0, S . pYay = 0, 

whence it appears that p is perpendicular to both Ya/3 and Yay, 
and is consequently parallel to the axis of their product; 
therefore 

p = yj . Va^Vay 

= y (ySa0a - aSa/3y) [Eq. (112) ] 

= - yaSafiy, 

or, putting — ySafiy = a, 

p = xa. 

73- Bight line perpendicular to a given line. 

1. Let 8 = od (Fig. 66) be a vector through the origin. To 
find the equation of DC through its extremity rig. 66. 

and perpendicular to it. Now p — 8 is a d r c 

vector along dr, and therefore by condition 

S8(p-8) = 0. 

Whence S8p = -(T8) 2 , or 

S8p = c, a constant (180). 

In order that p, p — 8 and 8 be complanar, we must have 

S.8/>0>-8) = 0, 
or 

S.(Y8/>)0>-8) = 0. 

2. p — 8, being perpendicular to both 8 and Y8/>, will be 
parallel to the axis of their product, or to V . SYSp. Hence, if 
y = oc be a vector to any point c, in the plane of od and dr, the 
equation of a right line through a given point c, perpendicular to 
a given line od, will be 

p =ry + a?V.8V8y (181). 




148 QUATERNIONS. 

3. If the perpendicular is to pass through the origin, then, 
from Equation (180), 

S8p = (182), 

or, in another form, from Equation (181), y being parallel to 

V . 8V8y, 

p = yV.$y8y (183). 

4. The student will find it useful to translate the Quaternion 
into the Cartesian forms. Thus, from Equation (180) , if rod= 0, 

. S8p = — TSTpcoaO, 

whence, if r and d represent the tensors, 

rdcos$ = cP 1 or ?• = - 



COS0' 

the polar equation of a right line. 

5. Equation (181), of a line through a given point and per- 
pendicular to a given line through the origin, may be otherwise 
obtained, as follows : 

Let y and 8, as before, be vectors to the point and along the 
given line, respectively, and ft a vector along the required per- 
pendicular, whose equation will then be 

P = y + xp. (a) 

To eliminate /3 we have the conditions 

88)8 = 0, 

since 8 and ft are perpendicular to each other, and 

Sy80 = O, 

since y, 8 and /3 are complanar. But Y8y is perpendicular to this 
plane, and therefore V • 8V8y is parallel to /? ; hence, substitut- 
ing in (a), 

p = y + ay.8V8y, 
or simply 

p = y + a?8V8y. 



APPLICATIONS TO LOCI. 149 

If SyfyS ^ 0, y, 8 and ft are not complanar, and the problem is 
indeterminate ; which also appears from (a) , by operating with 
X S . 8, whence, since S/K = 0, 

SpS = Sy8, 

a result which is independent of /?, and an infinite number of 
lines satisf} r the condition. 

* 6. If the line to which the perpendicular is drawn does not 
pass through the origin, let 

p = fi + xa (a) 

be its equation. Then, if p be the vector to the foot of the per- 
pendicular, we have Sa(p — y) = 0, or 

Sa(sa + /5-y) = 0, (6) 

because the line is perpendicular to (a) , or its parallel a. Hence, 
from (6), 

or, for the perpendicular p — y, 

p - y = Xa + P - y = a" 1 Sa (y - /3) - <T l * (y - fi) 
= -a- l Ya(y-/B). 

Its length is evidently 

TY[Ua.(y-j8)] (184). 

7. This perpendicular is the shortest distance from the point 
to the line. The problem may, therefore, be stated thus: to 
find the shortest distance from c to the line p = xa + ft. p being 
the vector to the foot of a line from c to any point of the given 
line, this vector is 

P + Xa — y, 

and, in order that its length be a minimum, 

dT(/3 + sa-y) = 
= T(0 + Xa - y)dT(/3 + xa - y) 
= - S[(/3 + Xa - y)a]dx = 0, 



150 QUATERNIONS, 

or 

that is, the line must be perpendicular to p = xa + p. 

8. If the perpendicular distance from the origin to p — ft + xa 
is required, p, being as before the vector to the foot of the per- 
pendicular, coincides with it ; hence, y being zero, and 8 repre- 
senting this value of p, 

Operating with X S . 8, since Sa8 = 0, 
-(T8) 2 = S08. 



Hence 



or 



n n ' 

T8 = S.0U8 (185). 



72. We are to observe that the foregoing equations of a right 
line are, as remarked in Art. 60, all linear functions involving, 
explicitly or implicitly, a single real and independent variable 
scalar. Such is evidently the case for such equations as 

p = za, [Eq. (173)] 

p = + aa, [Eq. (175)] 

p = a + *08-a). [Eq. (178)] 

So also for the implicit forms, as Yap = [Eq. (174)] ; em- 
ploying the trinomial forms 

a = ai + bj + c&, 
p = xi + yj + zk, 
we have 

ap = (bz — cy)i + (ex — az)j + (ay — bx) k — (ax + by + cz). 

Whence 

Yap = (bz — cy)i + (cx — az)j + (ay — bx)k = ; 
.-. bz = cy, cx = az, ay = bx, 

in which x and y are functions of z. 




APPLICATIONS TO LOCI. 151 

The Plane. 

73* Equation of a plane. 

1. If, in the equation S . 8/J = 0, which denotes that ft is per- 
pendicular to 8, we replace ft by the variable vector p, 

S.8p = (186) 

is the equation of a plane through the origin perpendicular to 8. 

2. Or, let 8 = od (Fig. 66) be the vector- Fig. 66 {bis). 

perpendicular on the plane, and dr an} r line _p r c 

of the plane. 

Then 

S8(p-8)=:0, 

S8p = 8 2 = -(T8) 2 , 
or 

S8/o = c, a constant . . (187) 

is the general equation of a plane perpendicular to 8. Here dr 
is any line of the plane ; and, if \8p = c, 

Sep = an indeterminate quantity . . . (188). 

If the plane pas^ through the origin, we have, as before, 
SS/o = 0. Converse^', if S8/> = c is the equation of a plane, 8 is a 
vector perpendicular to the plane. 

3. The equation of a plane through the origin perpendicular 
to 8 may also be written in terms of any two of its vectors, as 
y and ft ; 

p = x/3 + y y . 

Both of these indeterminate vectors may be eliminated by 
operating with S . 8 x , whence 

S5/> = 

as before ; or one may be eliminated by operating with V . ft X , 
whence 

Yfip = «8, 



152 QUATERNIONS. 

from which we may again derive SSp = b}- operating with 
V . 8 x ; for 

Y .3V^ = Y2;S 2 =0 

= p®/3-pa&p, [Eq. (Ill)] 

whence, since S8/3 = 0, S3p = 0. 

4. The equation of a plane through a point b, for which 
ob = /?, and perpendicular to 8, is 

S8O>-0) = O (189). 

5. Having the equation of a plane, SSp = c, to find its dis- 
tance from the origin, or the length of p when it coincides with 
8, we have p = #8 ; hence 

S8p = c = Sa8 2 = x&, 
or 



which, in p = #8, gives 



•*=§' 



' = 8' 



or 



*p=j5 ( 190 >- 



74. To jtfnd Me equation of a plane through the origin, making 
equal angles ivith three given lines. 

Let a, /?, y be unit vectors along the lines. The equation of 
the plane will be of the form 

S8p = 0. 

By condition, Sa8 = S£8 = SyS = T8 sin <£ = as, </> being the 
common angle made by the lines with the plane. 
Hence 

sin<f> = -rv 
r T8 



APPLICATIONS TO LOCI. 153 

To eliminate 8, we have, from Equation (118), 
8Sa0y = Va/?SyS + V/?ySaS + YyaS08, 
anfl, b} T condition, 

SSa/3y = X ( Va/3 + Y/?y + Vya) . 

The vector represented b}' the parenthesis is, then, the per- 
pendicular on the plane, whose equation, therefore, is 

Sp(Ya£ + Y/3y + Yya) = . . . . (191), 

and the sine of the angle <£ is 

Safty 

T(Va0+Vj8y + Yya)' 

75. Equation of a plane through three given points. 

Let a, /?, y be vectors to the given points ; then are the lines 
joining these points, as (a — /?), (/? — y), lines of the plane. If 
p is the variable vector to any point of the plane, p — a is also a 
line of the plane. Hence 

S(p-«)(«-j8)G8- y ) = 0, 
or 

S ( pa/3 - pay - pf? + pfiy - a 2 /? + a 2 y + a/3 2 - a/?y ) = . 

But 

S(-pj8*) = 0, S(-a 2 0) = O, etc., 
S( — pay) = Spya = S • pYya, 
Spa/?=S . pVafi, etc., 
hence 

S.p(Va/? + V/?y + Yya)-Sa/?y=0 . . (192), 

which, by making the vector-parenthesis = 8, may be written 
under the form 

S/>8-Sa/3y = 0, 



154 QUATERNIONS. 

in which 8 is along the perpendicular from the origin on the 
plane. When p coincides with this perpendicular, p = a:S, and, 
from the above equation, 

zS*=Sa/?y, 

or, for the vector-perpendicular, 

p = *S = S^Sa/Sy = ; ^1 



Ya0 + V0y+Vya 

76. We observe again, from inspection of the equations of a 
plane, that, as remarked in Ait. 60, the} r are linear and func- 
tions of two indeterminate scalars. Thus, for a plane through 
the origin 

S5p = 0, [Eq. (186)] 

employing the trinomial forms §=ai-\-bj+clc and p=xi+yj+zk, 
we obtain 

Sp = (bz — cy)i + (ex — az)j -f (ay — bx)k — (ax -f by -f cz) , 

the last term of which is the scalar part ; hence 

ax -|- by -f cz = 0, 

the equation of a plane through the origin o, perpendicular to a 
line from o to (a, 6, c) , which ma}' be written f(x, y, z) = ; 
or as a function of two indeterminates. In the same way, from 
an inspection of the other forms, 

p = xa + yp, [Art. 73,3] 

p = 8 -f xa +• #0, 
S8/3 — c' = ax -f by -f cz — c' = 0, [Eq. (187)] 

we observe the}' are linear functions of two indeterminate scalars. 

77. Exercises and Problems on the Right Line and 
Plane. 

1. p and y being vectors along two given lines which intersect 
at the point a, to which the vector is oa = a, to write the equation 
of a line perpendicular to each of the tivo given lines at their 
intersection. 



APPLICATIONS TO LOCI. 155 

V/fy is a vector m the direction of the required line, whose 
equation, therefore, is 

p = a + xVPy (193). 

If a ' = oa' be a vector to any other point, as aJ then is 

p = a' + xSpy 

the equation of a line through a given point perpendicular to a 
given plane ; the latter being given by two of its lines. 

2. a and fi being vectors to two given points, a and b, and 
S8/3 = c the equation of a given plane, to find the equation of a 
plane through a and b perpendicular to the given plane. 

8, p — a and a — ($ are lines of the required plane, hence 

S0>-a)(a-/?)8 = 0, 
or 

S p (a-0)S + Sa/38 = (194) 

is the required equation. 

3. oc = y being a vector to a given point c, and p = a -f- xp, 
p = a'-hx'/?' the equations of two given lines, to write the equation 
of a plane through c parallel to the two given lines. 

If lines be drawn through the given point parallel to the given 
lines, the}' will lie in the required plane. As vectors, ft and ft' 
are such lines, and p — y is also a line of the plane. Hence 

W'(p-y) = o (195) 

is the required equation. If y = a, or a', it is the equation of a 
plane through one line parallel to the other. Or, if y is inde- 
terminate, the general equation of a plane parallel to two given 
lines. 

Otherwise : the equation of a plane through the extrcmit}' of 
y parallel to two given lines, whose directions are given by 
a and /?, is evidently p = y -f xa -f yfi. 

4. To find the distance between two points. 
a and p being vectors to the points, 

y = j8-a. 
Squaring 

c 2 = b 2 + a 2 -2a&cosc. 



156 QUATEBNIONS. 

5. A plane being given by two of its lines, /3 and y, to write 
the equation of a right line through a perpendicular to the plane. 

Let oa = a. Draw two lines through a parallel to fi and y. 
Then 

p = a + zV0y (196). 

If the plane is given b}- the equation S8p = c, then 

p = a + xS (197). 

6. Find the length of the perpendicular from a to the plane, 
in the preceding example. 

Operating on Equation (197) with S . 8 x 

or 

aj8 2 = c-S8a; 
.-. xS =8- 1 (c-S8a) (198). 

7. S8(p — /?) = 0, Equation (189), being the equation of a 
plane through b perpendicular to 8, to find the distance from a 
point c to the plane. 

Let y = oc. The perpendicular on the plane from c, being 
parallel to 8, will have for its equation 

P = y + x8. 

To find x, operate with S . 8 x , whence 

S8p = SSy + #8 2 , 

or, from the equation of the plane, 

SSy + a8 2 = S80; 
.-. a:S = -8- 1 SS(y-/3), 
and 

xT8 = T8- 1 S8(y- ) 8) = S[U8 . (y-0)]. 

8. Write the equation of a plane through the parallels 

p = a+xft, 
p = a'+Xp. 



APPLICATIONS TO LOCI. 157 

9. Write the equation of a plane through the line 

perpendicular to the plane 

S8p = 0. 

10. Given the direction of a vector-perpendicular to a plane, 
to find its length so that the plane may meet three given planes in 
a point. 

Let 8 be the given vector-perpendicular, and 

Sap = a, S/?p = 6, Syp = c 

the equations of the given planes. If the equation of the plane 
be written 

SSp = #, 

then x must have such a value that one value of p shall satisfy 
the equations of all four of the planes. From Equation (118) 
we have 

pSajSy = VajSSyp + V^ySap + VyaS/?p 
= cVa0 + aYj3y + bYya. 

Operating with S . 8 X , to introduce #, 

a?Saj8y = cSSa/3 + aSSjgy + 6S8ya. 

11. To find the shortest distance between two given right lines. 
Let the lines be given by the equations 

p = a + z/?, (a) 

p = a'+x% (b) 

The equation of a plane through either line, as (b) , parallel to 
the other (a), is [Equation (195)] 

S/?0'O>-*') = O. (c) 

Y/3/3' is a vector-perpendicular to this plane. Hence, if yVpft 
be the shortest vector distance between the lines, we have, since 
a — a'— yVPf? is a vector complanar with /? and /?,' 

S ^'(a-a'-t/V^') = 0, 



I 



158 QUATERNIONS. 

or 

B(Sfl8' + W) (a - a' - yip?) = 0, 
wtlGDCG 

-y(Y/8/By + S[W(a-a')]=0; 

or, dividing by T(V/3/3'), 

TOT0')=TS[(a-a')U(V,8£')]. . . (199), 

the s}'inbol T denoting that only the positive numerical value of 
the scalar is taken. 

Otherwise : since the distance is to be a minimum, 

dT(p'-p) = 0, 
whence 

S(p'-p)(/3'dx'-/3cfc) = 0, 
or 

S(p'-p)£ = and S(p'-p)0' = O, 

or the shortest distance is their common perpendicular, whose 
length may be found as above. 

12. Given S^p = di and S3 2 p = d 2 , Me equations of two planes, 
to find the equation of their line of intersection. 
This equation will be of the form 

p = mS 1 + n$ 2 + xYS x 8 2 . (a ) 

To find m and w, we have, from (a), 

SS 1 p = m8 1 2 -f wSSjSa, 
S5 2 p = n8 2 2 + rMSS 1 S 2 , 

from which we .obtain 

_ S8, p - ttS^ 8 2 _ S3 2 p - ?*8 2 2 . 
m ~~ 8, 2 ~ SM 2 ' 

SSi&SSip — 8i s SS 2 p 



w = 



But 



(S8 1 8 2 ) 2 -8 1 2 8 2 2 



(S8 1 8 2 ) 2 -8 1 2 8 2 2 = (V8 1 8 2 ) 2 ; 
c? l SS 1 8 2 ~d 2 8 1 2 



APPLICATIONS TO LOCI. 159 

And similarly 

Substituting these values in (a) 

P = (V8,8 2 ) 2 * 2 + (VW 8l + * TVSlS2 ' 

which is the equation of the required line, a less useful form than 
those of the two simple conditions of Art. 70. 

If the two planes pass through the origin, then also does their 
line of intersection ; and since every line in one plane is perpen- 
dicular to 8,, and every line in the other to 8 2 , ¥8,80 is a line 
along the intersection, as in (a), and the equation becomes 

p^xYS^ (200). 

13. The planes being given as in Equation (189), 

S8(p-/8) = 0, («) 

S3'0,-/J') = 0, (6) 

to find the line of intersection. 

The vector p to any point of the line must satisfy both (a) 
and (b). This vector may be decomposed into three vectors 
parallel to 8, 8' and YS8' which are given, and not complanar, 
by Equation (118) ; whence 

P S . SVYW = S P 8V(8' . T88') + Sp8'V(V88' . 8) + S(pV88') VS8', 

or, from (a) and (&), 

- p(TV8S') 2 = S8/?V(8' . Y88') + S8^ , V(V8S' . 8) + m'pYW, 

or, since S88 p is the only indeterminate scalar, putting it equal 
to x, we have 

- p(TV88') 2 = S8/?V(8' . V8S') + S8'/3'V(Y88' . 8) + xYSV. 

If the planes pass through the origin, in which case /? and ft 
are zero, we have, as before, 

p = xm: 



160 QUATERNIONS. 

14. To write the equation of a plane through the origin and 
the line of intersection of 

88(p-0) = O, (a) 

S8'(p-j8') = 0. {b ) 

If p is such that S8p = S8/3, then also SS'p = SS'# and both the 
above equations will be satisfied. Hence, from (a) and (b) 

S8pS37?'-S5/?S5'p = 0, 

which is also a plane through the origin. This equation ma}' 
also be written 

which shows that 

SSS'jS'-S'SSjS 

is a vector-perpendicular to the plane, and therefore to the line 
of intersection of (a) and (&). 

15. To find the equation of condition that four points lie in 
a plane. 

If the vectors to the four points be a } /?, y, 8, then, to meet 
the condition, 

8 — a, 8 — /?, 8 — y 

must be complanar, and therefore 

S(8-a)(8-/?)(8- 7 ) = 0, 
whence 

SS0y + Sa8y + Sa08 = Sa/?y . . . (201), 

which is the equation of condition. 

Or, x and y being indeterminate, we have also 



or 
and 



8 = a + *O8-a) + y(y-0), 

a + (*-l)a + fo-aOj8-yy = 0, 



APPLICATIONS TO LOCI. 161 

Or, in general, 

aa-\ 

a + b- 



aa + bp + cy + €tt = 0\ m2 . 

» + c-fo'=0) • ■ • • V )y 



are the sufficient conditions of complanarity. 

These conditions are analogous to Equation (9). 

16. Given the three planes of a triedral, to find the equations 
of planes through the edges perpendicular to the opposite /aces, 
and to shoio that they intersect in a rig^ht line. 

, Taking the vertex as the initial point, let 

Sap = 0, (a) 

S#> = 0, (6) 

Sy P = (c) 

be the equations of the plane faces. Then Va/? is a vector par- 
allel to the intersection of (a) and (ft), and V • yYa/J is a vector 
perpendicular to the required plane through their common edge. 
Hence the equation of this plane is 

S(pV.yVa0) = O. (a') 

Similarly, or by a c} r clic change of vectors, 

S(pY.aV/? 7 ) = 0, (V) 

S(pV.0Vya) = O (c') 

are the equations of the other two planes. 

If from their common point of intersection normals are drawn 
to the planes, then are V . yVa/?, V . aV/3y and V • /?Vya vector 
lines parallel to them ; but, Equation (123), 

Y (yVa/? + aV/3y + /?Vya) = 0. 

Hence these vectors are complanar, and the planes therefore 
intersect in a right line. 

Otherwise: from Equation (111) 

V(aV/3y) = ySa/3 - /?Say ; 



162 QUATERNIONS. 

hence, from (6'), 

S(pySa0 - p/?Say) = SajGSpy - SaySp0 = 0. 

Similarly, or by cyclic permutation, 

S/fySpa — S/?aSpy = 0, 
SyaSp/? - Sy0Spa = 0. 

But the sum of these three equations is identically zero, either 
two giving the third by subtraction or addition. 

17. To find the locus of a point which divides all right lines 
terminating in two given lines into segments which have a com- 
mon ratio. 

Let da and d'b (Fig. 67) be the two 
given lines, a and /3 unit vectors parallel 
to them, ba an}- line terminating in the 
given lines, and it a point such that 
ra = wibr. Assume dd', a perpendicular 
to both the given lines, o, its middle 
point, as the origin, and od = 8, od' = — 8, or = p. 
Then 

OA = p + RA = 8 -f- Xa. 
OB = p -f RB = — 8 -f y(2. 
Adding 

2p-fRA-fRB = iKa-f yp. (a) 

But 

RA + RB== ??Lzil RA== !!?LzJ:(« p + s + a ; a ) 
m m 

which substituted in (a) gives 

p — 8 — xa = m(y/3-- p — 8), 
whence, since S8/3 = SSa = 0, 

SSp(m + l) = 8 2 (l-m) = c, 
or the locus is a plane perpendicular to dd'. 




APPLICATIONS TO LOCI. 163 

If the given ratio is unity, or br = ra, then m = 1 and 

S8p = 0, 

and the locus is a plane through o perpendicular to dd\ 
If a and f3 are parallel, then (b) becomes 

p — 8 = m (x'a — p — 8) , 
whence 

SSp(m + l) = (l-m)8 2 , 

a right line perpendicular to dd! If at the same time ra = 1 , 

S5p = and p = x"a, 

a right line through the origin parallel to the given lines. 

18. If the sums of the perpendiculars from two given points on 
two given planes are equal, the sum of the perpendiculars from 
any point of the line joining them is the same. 

Let a and b be the given points, oa = a, ob = /?, and SSp=<7, 
SS p = d' be the equations of the planes ; 8 and 8' being unit 
vectors, so that #8 and y& are the vector-perpendiculars from a 
on the planes. Then 

x = Sa8 - d, 
?/ = S a 8'-d; 
and 

a? + y=Sa(8 + 8')-(d + cr). 
Similarly 

x'+tf =*?{* + *')-{* + *')- 

But, by condition, 

Sa(S + S') = S0(8 + S r ), 
or 

S(/?-a)(S + 8')=0. (a) 

The vector from o to any other point of the line ab is 
a -f- z (/? — a) ; whence, for this point, 

a5 »+y' = S[a + *(^-a)](8+8 , )-(cl + cl'), 

for which point, since (a) remains true, the sum therefore is 
unchanged. 



164 QUATERNIONS. 

19. To find the locus of the middle points of the elements of an 
hyperbolic paraboloid. 

Let the equations of the plane director and rectilinear direc- 
trices be 

S5p == 0, 
pz=a + x/3 and p = a' -f x'pl 

Also, let om = fi be the vector to the middle point of an ele- 
ment so chosen that the vectors to the extremities are a-t-x/3 
and a' + x'fi'. Then, since m is the middle point, 

2fi = a + x/3 + a' + *'# (a) 

The vector element is 

-x , /3'-a' + a + xfi, 

and, being parallel to the plane director, 

S8(- a' + a + x/3 - x'p) = 0. 

This is a scalar equation between known quantities from which 
we may find x' in terms of x ; substituting this value in (a) , we 
have an equation of the form 

fl = ttj + xfi x , 

or the locus is a right line. ' 

20. If from any three points on the line of intersection of tivo 
planes, lines be drawn, one in each plane, the triangles formed 
by their intersections are sections of the same pyramid. 

The Circle and Sphere. 

78. Equations of the circle. 

The equation of the circle ma}' be written under various 
forms. If a and ft are vector-radii at right angles to each other, 
and Ta = T/2, we ma}' write 

p = cos0 . a + sin0 .£.... (203) , 

in terms of a single variable scalar 0. 



APPLICATIONS TO LOCI. 
If a and p are unit vectors along the radii, 

or, since x* + y 2 = r 2 , 

The initial point being at the center, 
Tp = Ta 
T^=l 



165 



P 2 = -* 



(204). 



(205) 



are evident!}' all equations of the circle. 

If o (Fig. 68) be any initial point, c the center, to which ilu- 
vector oc = y, p the variable vector to 
any point p, cp = a, then 



whence 



p-y = a, 

(p-y) 2 = -^. .(206), 

the vector equation of the circle whose 
radius is r. 

If Ty = c, it may be put under the form q 




(207). 



p 2_2Spy = c 2 -r 2 .... 

If the initial point is on the circumference, we still have 
(p "" y) 2 = — f 2 ; but y 2 = — j* 2 , hence 

p 2 -2Spy = . 

or, since in this case Spy = Spa, 

p 2 -2Spa = 



(20ft), 



cmii). 



79. Equations of the sphere. 

This surface may be conveniently treated of in co mice I inn 
with the circle ; for, since nothing in the previous article restricts 
the lines to one plane, the equations there deduced for the circle 
are also applicable to the sphere. 



166 



QUATERNIONS. 



Another convenient form of the equation of a sphere is 
(Fig. 68) 

T(p- 7 ) = Ta (210), 

the center being at the extremity of y and Ta the radius. 

80. Tangent line and plane. 

A vector along the tangent being dp, we have, from Equation 
(203), 

dp = — sin0 • a -+- cos0 • /3, 

and for the tangent line tr = p-\- xdp, 

7r = cos0 • a-f sin0 • /? + #[— sin0 . a-f cos0 • /?] (211), 

where ir is an} T vector to the tangent line at the point corre- 
sponding to 0. 

From the above we have directly 

Spdp = 0, 

or the tangent is perpendicular to the radius vector drawn to the 
point of tangency. 
***' m ' By means of this propert\' we ma}* 

7 b write the equation of the tangent as 
follows : let ir be the vector to any point 
of the tangent, as b (Fig. 69), c being 
the initial point and p the vector to p, 
the point of tangency. Then 

S/>(tt — p) = 0, 
S/Hr=-r*l 




S- = l 
P 



. (212), 



are the equations of a tangent. Since nothing restricts the line 
to one plane, they are also the equations of the tangent plane to 
a sphere. 



APPLICATIONS TO LOCI. 167 

The above well-known property may also be obtained by 
differentiating Tp = Ta ; whence, Art. 67, 2, 

Spdp = 0, 

and therefore p is perpendicular to the tangent line or plane. 

8L CJiords of contact. 

In Fig. 69 let cb = ft be the vector to a given point. The 
equation of the tangent bp must be satisfied for this point; 
hence, from Equation 212, 

Sft> = -r*, 

or 

S/fc^-r 2 (213), 

which is equally true of the other point of tangency p,' and being 
the equation of a right line, it is that of the chord of contact pp.' 
And for the reason previously given, it is also the equation of 
the plane of the circle of contact of the tangent cone to the 
sphere, the vertex of the cone being at b. 

82. Exercises and Problems on the Circle and the 
Sphere. 

In the following problems the various equations of the pla\ae, 
line, circle and sphere are empk^ed to familiarize the student 
with their use. Other equations than those selected in any 
special problem might have been used, leading sometimes more 
directly to the desired result. It will be found a useful exercise 
to assume forms other than those chosen, as also to transform 
the equations themselves and interpret the results. Thus, for 
example, the equation of the circle (209), 

p 2 -2Spa = 
may be transformed into 

S,>0)-2a) = 0, 



168 



QUATERNIONS. 




which gives immediately (Fig. 70) the property of the circle, 
that the angle inscribed in a semi-circle is 
a right angle. Obviously, this includes the 
case of chords drawn from any point in a 
sphere to the extremities of a diameter, and 
the above equation is a statement of the prop- 
osition that, p being a variable vector, the 
locus of the vertex of a right angle, whose 
sides pass through the extremities of a and 
p ' — a, is a sphere. 

Again, with the origin at the center, we have (Fig. 71), 

(p + a) + (a-p)=2a, 

and, operating with x S . (p — a) , 

S(p + a)(p-a) = 0; 

.-. p is a right angle. This also follows from 
Tp=Ta, whence p 2 =a 2 and S(p+a)(p— a)=0. 
Again, from Tp = Ta, 

T(p + a)(p-a)=2TVap. 

The first member is the rectangle of the chords pd, pd' (Fig. 71), 
and the second member is 

2od • op sinDOP. 
i 
Hence the rectangle on the chords drawn from any point of a 
circle to the extremities of a diameter is four times the area of 
a triangle whose sides are p and a. 
Also, from Tp = Ta, 

and for any other point 

••• Hp'+ P )(p'-p) = 0. 

But p 1 — p is a vector along the secant, and p'+p is a vector 
along the angle-bisector ; now when the secant becomes a tan- 




APPLICATIONS TO LOCI. 



169 



gent, the angle-bisector becomes the radius ; therefore the radius 
to the point of contact is perpendicular to the tangent. 

1. The angle at the center of a circle is double that at the cir- 
cumference standing on the same arc. 



T P = Ta, 



We have 
and therefore, Art. 56, 18, 

p = 0> + a)~ 1 a(p + a), 

whence the proposition. 

2. In any circle, the square of the tangent equals the product 
of the secant and its external segment. 

Fig. 69 (bis). 

In Fig. 69 we have p 

cb = cp -t- PB, 
.«. cb 2 =cp 2 +pb 2 , 
or 

PB 2 =CB 2 — CP 2 

= cb 2 — cd 2 , as lines, 

= BD • BD'. 

3. The right line joining the points of intersection of two circles 
is perpendicular to the line joining their centers. 

Let (Fig. 72) cc r = a, cp = p, cp' = p', and r, r' be the radii 
of the circles. Then 

2 — _ r ' 2 • 




also 

Hence 
or 



o>-*) 2 = 



Spa = Sp'a, 
Sa(p-p') = 0; 

hence pp' and cc' are at right angles. 




170 QUATERNIONS. 

4. A chord is draivn parallel to the diameter of a circle; the 
radii to the extremities of the chord make equal angles with the 
diameter. 

If p and p be the vector-radii, 2 a the vector-diameter, then 
xa = the vector-chord, and 

(p'-aa)* = -f*, 

whence the proposition. 

5. If abc is a triangle inscribed in a circle, show that the vector 
of the product of the three sides in order is parallel to the tangent 
at the initial point. [Compare Art. 55.] * 

If ab = /3, ca = y, and o is the center of the circle, then 

-V(ab .bc . ca) = V. /?(/? + y)y 
= V(/3 2 y + /V) 

Or, c and b being points of the circumference satisfying 
p 2 — 2Spa = [Eq. (209)], substituting and operating with 

S.oX 

S . aV(AB . BC • CA) = 2 Sa/3Say - 2 Sa/3Say = 0. 

Hence V(ab . bc • ca) is perpendicular to a, or parallel to the 
tangent at a. 

6. The sum of the squares of the lines from any point on a 
diameter of a circle to the extremities of a parallel chord is equal 
to the sum of the squares of the segments of the diameter. 

Let pp' (Fig. 73) be the chord parallel to the diameter dd! 
Fig. 73. o the given point, and c the center of the 

circle. Let cp = p, cp' = p[ oc = a, op = ft 
and op' = # Then 

0P 2 = -^=-(a 2 + 2Sap+p 2 ), 

OP' 2 = -0' 2 =-(a 2 +2Sap'+p' 2 ) ; 

.• . OP 2 -f- op' 2 = 2 OC 2 + 2 DC 2 - 2(Sap-f Sap'). 




APPLICATIONS TO LOCI. 171 

But 

S(p - p') (p + p') = S(p + p')sa = 0. 

Therefore 

Sap + Sap' = 0, 
and 

op 2 + op' 2 = do 2 + od' 2 . 

7. To find the intersection of a plane and a sphere. 

Let p 2 = — r 2 be the equation of the sphere, 8 a vector-perpen- 
dicular from its center on the plane and T8 = d. Then, if /? be 
a vector of the plane, 

Substituting in the equation of the sphere, since S/?8 = 0, we 

have 

/^-(r 2 -^ 2 ), 

the equation of a circle whose radius is Vr 2 — d 2 , and which is 
real so long as d < r. 

8. To find the intersection of two spheres. 

Let the equations of the given spheres be (Eq. 207) 

p 2 -2Spy = c 2 -r 2 , 
p 2 -2Spy'=c' 2 -r' 2 . 

Subtracting, we have 

2 Sp(y — y f ) = a constant. 

The intersection is therefore a circle whose plane is perpen- 
dicular to y— y| the vector-line joining the centers of the spheres. 
Assuming (Eq. 210) 

T(p-y) = Ta and T(p-y') = Ta,' 

show that 2 Sp(y — y f ) = a constant, as above. 



172 QUATERNIONS. 

9. The planes of intersection of three spheres intersect in a 
right line. 

Let y\ y", y r " be the vector-lines to the centers of the spheres, 
and their equations 

p 2 -2Spy' = c', 
p 2 - 2Spy" = c", 
p 2 -2Spy'"=C r ". 

The equations of the planes of intersection are, from the pre- 
ceding problem, 

2ty(yW)=c"-c\ (a) 

2S,o(y'-y'")=c'"-c', (6) 

2Sp(y"-y'")=c'"-e". (c) 

Now, if p be so taken as to satisfy (a) and (6), we shall 
obtain their line of intersection. But if p satisfies (a) and (&), 
it will also satisfy their difference, which is (c) ; the planes there- 
fore intersect in a right line. 

10. To find the locus of the intersections of perpendiculars from 
a fixed point upon lines through another fixed point. 

Let p and p' be the points, pp' = a, and 8 a vector-perpen- 
dicular on any line through pj as p = a -f- xfi. Then 

and operating with S . 8 x 

S 2 = S3a, 

which is the equation of a circle (Eq. 209) whose diameter is pp'. 

11. From a fixed point p, lines are drawn to points, as 

p', p", of a given right line. Required the locus of a point o 

on these lines, such that pp' • po = m 2 . 

Let the variable vector po=p ; then pp'=#p. By the condition 

T(pp'.po) = m 2 , 
or 

T(xp • p) = m 2 ; 
.•. #p 2 = — m 2 . 



APPLICATIONS TO LOCI. 173 

If 8 be the vector-perpendicular from p on the given line, 

and TS = d, 

SS(z P -S)=0, 
or 

xSSp = - d 2 ; 

hence the locus is a circle through p. 

12. If through any point chords be drawn to a circle, to Jftirf 
the locus of the intersection of the pairs of tangents drawn at the 
points of section of the chords and circle. 

Let the point a be given by the vector oa = a, o heiug the 
initial point taken at the center of the circle. Let p = ur be 
the vector to one point of intersection r. The locus of n is 
required. The equation of the chord of contact is (Eq, 213) 

which, since the chord passes through a, may be written 

Sp'a^-r 2 , 

where a is a constant vector. The locus is therefore a straight 
line perpendicular to oa (Eq. 180). 

13. To find the locus of the feet ofperpendiadars drawn throwfh 
a given point to planes passing through another given point. 

Let the initial point be taken at the origin of perpendiculars,, 
a the vector to the point through which the planes are passed, 
and 8 a perpendicular. Then 

S8(8-a)=0, 
or 

8 2 -Sa8=0 

is true for any perpendicular. Hence the locus is a sphere whose 
diameter is the line joining the given points. 



174 QUATERNIONS. 

Otherwise : if the origin be taken at the point common to the 
planes, and the equation of one of the planes is S8p = 0, then 
the vector-perpendicular is (Eq. 198) 

S^SSa, 
and, if p be the vector to its foot, 

p = a-8- 1 S8a, 
or 

p— a = — 8 -1 S8a, 
whence 

(p-a) 2 - > 2 (S8a) 2 , 

and 

Sap-a 2 = -8- 2 (S8a) 2 . 

Adding the last two equations 

p 2 — Sap = 0, 
or 

T(p-£a) = T£a, 

which is the equation of a sphere whose radius is T* and center 

is at the extremity of -, or whose diameter is the line joining 

the points. 

14. To find the locus of a point p which divides any line os 
drawn from a given point to a given plane, so that 

op . os = m, a constant. 

Let op = p and os = o- ; also let SSo- = c be the equation of the 
plane. We have, by condition, 

TpTo- = m, 
and 

Up = Ucr ; 

.-. Tcr= ™, 
Tp 
and 

mUp 

__ mp 

— 7"* 



APPLICATIONS TO LOCI. 175 

Substituting in the equation of the plane 

mS8p + cp 2 = 0, 

which is the equation of a sphere passing through o and having 
— 

OD 



AY) 

— for a diameter 



15. To find the locus of a point the ratio of whose distances 
from two given points is constant. 

Let o and a be the two given points, oa = a, or = p, r being 
a point of the locus. Then, by condition, if m be the given 
ratio, 

T(p — a) = mTp, 



whence 



or 



/) 2 -2Spa-fa 2 = mV, 

(l-m 2 )p 2 =2Sap-a 2 

G 1 — m 2 2 

1 — m 

2 2 Sap a 2 = m 2 a 2 

9 l-m 2 (1-m 2 ) 2 (1-m 2 ) 2 ' 



*('-n3.)-i5 



ma 
V 



m 

which is the equation of a sphere whose radius is T -a, and 

.. 1— wr 

whose center c is on the line oa, so that oc = ; a. (Eq. 210) . 

1— m 1 

16. Given two points a and b, to find the locus of? when 

ap 2 -f bp 2 = OP 2 . 

o being the origin, let OA = a, ob = /3, OP=p. Then, by 
condition, 

^ = (p-a)* + (p-0)*, 
whence 

P ! -28/>(a + /8) = -(a s + i 8 2 ), 

[p-(a + /J)] 2 = 2Sa/^_ 

T[ P -(a + J 8)] = V-2SayS, 



176 QUATERNIONS. 

which is the equation of a sphere whose center is at the extremity 
of (a-h/?), if Sa/3 is negative, or the angle aob acute. If this 
angle is obtuse, there is no point satisfying the condition. If 
aob = 90°, the locus is a point. 

83. Exercises in the transformation and interpretation of 
elementary symbolic forms. 

1 . From the equation 

(p + a)* = (p-a)' 

derive in succession the equations 

T(p + a) = T(p-a) and T^±^= 1, 
. p — a 

and state what locus they represent. 

2. From the equation 

a a 
derive s}'mbolicall3' the equations 

ap + pa = 0, s£ = 0, Su£=0, fu£Y=-l, and TVu£=l, 

and interpret them as the equations of the same locus. 

3. Transform 

p — a 
S- = 

to the forms 

and interpret. a a P 

4. Transform S^^ = to S-=S-> and interpret. 

a a a 

5. Transform (p - 0) 2 = (p - a) 2 to T (p -/3) = T(p- a), 
and interpret. 

p p 

6. What locus is represented by K - = ? 



J-=l and SU-=T* 



APPLICATIONS TO LOCI. 177 

7. WhatbyQ 2 =-l? By^Y=-a 2 ? 

8. Whatbyu£=U^? Up = U0? u|=l? 

P P, 

9. U- = — U-? 

a a 

io. (u0 = u£? 

11. V— - = 0? v- = v-? 

a a a 

12. V-=0? 

a 

13. ?■*?■ = a>? 

a a 

H. su- = su-? su- = -su-? (su-]=(su-j? 

a a a a \ a/ \ ay 

15. Tp = l? 

16. Transform (p — a) 2 = a 2 to T(p — a) = Ta, and interpret. 

17. Under what other form ma}* we write (p — a) 2 =(ft— a) 2 ? 
Of what locus is it the equation ? 

18. p 2 + a 2 = ? p 2 + 1 = ? Translate the latter into Car- 
tesian coordinates, by means of the trinomial form, and so deter- 
mine the locus anew. 

19. T(p-/3) = T(0-a)? 

20. Compare SU - = T- and S- = 1 with the forms of Ex. 3. 

pap 

21 . What locus is represented by S/?p + p 2 = when Tft = 1 ? 

22 - ( 8 -0 l -( T i=i? 

23. (t^J=-1? 

24. Show that Y . Va/JYap = is the equation of a plane. 
What plane? [Eq. (112)]. 



178 QUATERNIONS. 

The Conic Sections. 
Cartesian Forms. 
84. Tlie Parabola. 
Resuming the general form of the equation of a plane curve 

P = xa + y/3, 
from the relation y 2 = 2px, we obtain 

p = |^+m3 ; (2i4) 

for the vector equation of the parabola when the vertex is the 
initial point. If the latter is taken anywhere on the curve, from 
the relation y* = 2p'#, we obtain 

P = jp« + VP ..... (215); 

and if the initial point is at the focus, then y 2 = 2px -f/) 2 gives 

P~tf-lt)* + yP .... (216); 
ip 

or again, in terms of a single scalar £, 

p^a + tfi (217). 

In Equations (21 4), (215) and (216), a and ft are unit vectors 
parallel to a diameter and tangent at its vertex, being at right 
angles to each other in Equations (214) arid (216) ; in Equation 
(217) a and /? are any given vectors parallel to a diameter and 
tangent at its vertex, the initial point being on the curve. 

85. Tangent to the parabola. 

From Equation (216) we have for the vector along the tangent 
(Art. 62) 



Fi',» .v'V^i.KQ 



APPLICATIONS TO LOCI. 179 

and, therefore, the equation of the tangent is 

* = ^tf-?)* + yP + Y(Za + l3) . . (218). 
From Equation (217) the vector along the tangent is 

ta + fr 
and the equation of the tangent is 

,r = -a + tp + x(ta + fi) .... (219). 

If p be the vector to a point on the diameter of a parabola, the 
point being given by the equation 

p = ma + n/J, (a) 

and a tangent to the curve be drawn through this point, then 
(a) must satisf}' the equation of the tangent-line and 

ma + nfi = *- a + tp + x(ta + p) , 

whence , 2 

m = - + xt and n = t + x, 

t = n± V/t 2 — 2 m ; (ft) 

hence, in general, two tangents can be drawn to the curve through 
the given point. When « 2 = 2m, they coincide ; in this case, 
from (a), ^ 

the point being on the curve. If 2m > n 2 , t is imaginary, and 
no tangent can be drawn ; in this case (a) becomes 

p=r±+a\a + np, 

the point being within the curve. 



180 



QUATERNIONS. 



rig. 74. 



86. Examples on the parabola. 

1 . The intercept of the tangent on the diameter is equal to the 
abscissa of the point of contact. 

Since the tangent is parallel to 
the vector fa + /?, or to any multiple 
of it, it is parallel to t 2 a + t/3 or to 

L a + */? + '- a, that is, to (Fig. 74) 




But 



OP + ox. 

TP = TO -f op ; 
TO = OX. 



2. If from any point on a di- 
ameter produced, tangents be drawn, 
the chord of contact is parallel to the 
tangent at the vertex of the diameter. 

If t' and t" correspond to the 
points of tangenc}-, we have for the vector-chord of contact 



which is parallel to 



412 fH2 

t'+-t" 



or, from Equation (&), Art. 85, to 

/3 + na, 
which is independent of m. 

3. To find the locus of the extremity of the diagonal of a rect- 
angle whose sides are two chords drawn from the vertex. 

Let op and op' be the chords. Then 



V 2 

or'= P >=fa-y% 
2p 






APPLICATIONS TO LOCI. 181 

The vector-diagonal w' is p -t- p\ or 

which may be put under the form of the equation of the parabola 
by adding and subtracting -p^ a, giving 

But, by condition, Spp' = 0. Hence, from (a) and (6), Sa/J 
being zero, 

yy'-^- = 0, .: yy' = (2 P )\ (d) 

which in (c) gives 

Z'^-^p^a + iy-y^ + ipa. 

Changing the origin to the extremity of 4 pa, g 

- {y-y'Y 



2p 



-« + (y-y')0- 



Hence the locus is a similar parabola whose vertex is at a 
distance of twice the parameter of the given parabola from its 
vertex. 

Moreover, from (rf), xx' = (2p) 2 . Hence the parameter is a 
mean proportional betiveen the ordinates and the abscissas of the 
extremities of chords at right angles. 

4. If tangents be drawn at the vertices of an inscribed triangle, 
the sides of the triangle produced will intersect the tangents in three 
points of a right line. 

Let opp' (Fig. 74) be the inscribed triangle, and one of the 
vertices, as o, the initial point. Then, for the points p and p' 
respectively, we have 

P =| 2 a + ^, 
t' 2 



182 QUATERNIONS. 

Let 7r 2 , 7r 2 , tt 3 be the vectors to the points of intersection ; then 

t 2 

/2 T i f t2 

.-. £ + ** = ^-, * + * = *'*,' 
* 2 



Also 



2tt'-*' 2 
Hence 



'■bMt ,+, )-E5(J- + '} 



In a similar manner 
But 



Also 



7I* 3 = OP + yPP' = OP + y (/)'— />) 
/2 /'2_*2 

'• 1 + ^^ = °' t+y(t'-t) = z, 



Hence 



Now 



* + «' 



ft' « 



2*-«' 2*'-* fi-t* 



-*i jr- 7r 2- 



/*-{r+*)-«(sH*)^-w=o. 



Also 

2*-*' 2t'-t t 2 -t' 2 



= 0. 



Hence ir 1? 7r 2 and 7r 3 terminate in a straight line. 



APPLICATIONS TO LOCI. 



183 



5. TJie principal tangent is tangent to all circles described on 
the radii vectores as diameters. 

Fig. 75. 

Let ap = p (Fig. 75) , a and /? 
being unit vectors along the axis d 
and principal tangent. Then, if 
the circle cut the tangent in t, 
and tc be drawn to the center, 

T(tc) = T(fc) = T(£fp) ; 
.•. TC 2 = i(p — ma) 2 . 

Also 

TC = TA -f AF -J- FC 

= — zfi + ma -I- i(p — ma) , 
TC 2 = [— z/3+ ma-|-i(p— Wla)] 2 . 

Equating these values of tc 2 , 
we have, since S/ito = 0, 

z*p* - z$p P + mSap = 0, 
... *t_*y + ?£ = o, 




which gives but one value for z. 



6. To find the length of the curve. 

It has been seen (Art. 62) that, if p = <f>(t) be the equation 
of a plane curve, the differential coefficient is the tangent to the 
curve. Hence, if this be denoted by p' = <£'(£), Tp'dt is an 
element of the curve whose length will be found by integrating 
Tp' with reference to the scalar variable involved between proper 
limits ; or 



For the parabola 



,- So =jTV 



■S-+* 



184 QUATERNIONS, 

we have 



7. To find the area of tlie curve. 

With the notation of the previous example, twice the area 
swept over by the radius vector will be measured by (Art. 41,7) 
TYpp'dt. The area will then be found by integrating TYpp' with 
reference to the involved scalar between proper limits and taking 
one-half the result ; or 

A-Ao = ifT\pp'. 

For the parabola 



A 

or, since a/3 = 90' 



*X*p up UJ *' 



From the origin, where y = 0, to any point whose ordinate is 
y, the area of the sector swept over by p is -- — t/ 3 = %xy ; adding 

the area %xy of the triangle, which, with the sector, makes up 
the total area of the half curve, we have £ xy, or two-thirds that 
of the circumscribing rectangle. The origin may be changed to 
any point in the plane of the curve, to which the vector is y, by 
substituting the value p = y + p x in the equation of the curve, 
p x being the new radius vector ; we may thus find any sector area 
limited by two positions of p x , the vertex of the sector being at 
the new origin. Thus, transferring to an origin on the principal 
tangent, distant b from the vertex, p = b(2 + p x ; which, in the 
equation of the parabola, gives 



APPLICATIONS TO LOCI. 185 

integrating, as before, between the limits y = b and y = 0, 

87. Relations betioeen three inter&ectiiig tangents to tlie Parabo- 
la. ["Am. Journal of 
Math.," vol. i. p. 379. 
M. L. Holman and 
E. A. Engler.] 

Let p x , p 2 , p 3 be the 
vectors to the three 
points of tangency, p x , 
p 2 , p 3 [Fig. 76], and 
tti, 7r 2 , ir z the vectors to 
Su Sa, Sg, the points of 
intersection of the tan 
gents. Resuming Equa- 
tion (216), where the 
focus is the initial 
point, and a and /? are 
unit vectors along the 
axis and the directrix, 

P = ^(/-^)* + ^ (a). 

Since p 2 = — (Tp) 2 , and 8a@ = 0, we have for the three points 
Pi? P25 P3 



• 




The vector along the taugeut is 



(b). 



186 QUATERNIONS. 

and therefore 

whence, equating the coefficients of a and /?, 

whence, substituting, and by the cyclic permutation of the sub- 
scripts, 



»« = T"(»iy« -p*) a + t(y* + yJP 

2p 



From (6) 



(c). 



Tp 1 Tp 2 = i ^(y 1 2 + i > 2 )(y 2 2 + p 2 ) 
Tp2Tp 8 = i p(y2 2 -hl> 2 )(y3 2 +P 2 ) 
T * T * = A^3 2 +2^i 2 +i> 2 ) 



and from (c) 



W, 



(T,r0 * = 4? (y/+ ^ )(yil!+J5S) 



(«), 



and from (d) and (e) 



(T7r 2 )» = Tp,T ftl 



(/)• 



APPLICATIONS TO LOCI. 



187 



From (c) , it appears that tJie distance of the point of intersec- 
tion of two tangents from tlie axis is the arithmetical mean of the 
ordinates to their points of contact. From (/), that the distance 
from the focus to the point of intersection of two tangents is a 
mean proportional to the radii vectores to the points of contact. 

1st. If fH becomes a multiple of /?, 

P * = 2p( p * ""^ a + V *P = *P 5 
.-. z = y 2 = ±p. 

Or, the parameter is the double ordinate tnrough the focus, or 
twice the distance from the focus to the directrix. 



Fig. 77, 




2d. If pi is the multiple of p 2 (Fig. 77), then pj — pi is a focal 

chord, and 

*P2 = Pi? 
or, from (a) , 



whence 



yi -tf yi 



188 QUATERNIONS, 

or 

yiivxy* +p*) = y*{yiy* +i>*)i. 

and 

yiy2+P*=0. (g) 

From (a) and (c) 

= s -i?3(»i i +j')(yiyt+i')= 5 w 

or, a line from the focus to the intersection of the tangents at the 
extremities of a focal chord is perpendicular to Hie focal chord. 
The vectors along the tangents are 

Pi — ttz and ps — ir$, 
and, from (7i), 

S(pi — w 3 ) (pa ~ t s ) = Sp,Ps + *s 2 = 0, 

or, tAe tangents at the extremities of the focal chord are perpen- 
dicular to each other. 
Since, from (g) , 

&& =— 1*, 

we have 

**=j-(yiy2-p i )<>-+i(yi+y2)P 
= -.p* + i(yi+y*)j8, 

or, £Ae tangents at the extremities of a focal chord intersect on 
the directrix. 

3d. If P2 becomes a multiple of a (Fig. 78) , y 2 = 0, and from 

(«) 

or, f/ie subtangent is bisected at the vertex. 



APPLICATIONS TO LOCI. 



189 



Also 



--»=-f+M^*H 



2p 2 P 

Operating with S . v t x 

St S (t S -P,) = ^-£ = 0, 
4 4 

or, a perpendicular from the focus on the tangent intersects it on 
the tangent at the vertex. 

Fig. 78. 




Again, since w 3 is parallel to the normal at p 2 , the latter maj- 
be written, from (i), 



whence 



or 



«^ = *(-|a+-| I /s) = » + yii8; 



P Vi 



a? = 2, 



» = — 2>S 



190 QUATERNIONS. 

hence, the subnormal is constant; and tlie normal is twice the 
perpendicular on the tangent from the focus. 
The normal at Pi may be written 

Xir s = — Za + pi, 

or 

whence, from (5), 

= 2, and 2=^(yi 2 +i> 2 ) = T /0l ; 

or, the distance from the foot of the normal to the focus equals 
the radius vector to the point of contact, or the distance from tJue 
point of contact to the directrix, or the distance from the focus to 
the foot of the tangent. 

The portion of the tangent from its foot to the point of con- 
tact maj T be written za + p^ in which z has just been found. 
Hence 

or 

.Vi 2 
Za+p^^a + yrf, (J) 

the portion of the tangent from the foot of the focal perpendicu- 
lar to the point of contact is 

or 

or, comparing (j) and (k) , the tangent is bisected by the focal 
perpendicular, and hence the angles between the tangent and the 
axis and the tangent and the radius vector are equal, and the 
tangent bisects the angle between the diameter and radius vector 
to the point of contact. 



APPLICATIONS TO LOCI. 191 

(k) is also the perpendicular from the focus on the normal, 
and shows that the locus of the foot of the perpendicular from the 
focus on the normal is a parabola, ichose vertex is at the focus of 
the given parabola and whose parameter is one-fourth that of the 
given parabola, 

88. The Ellipse. 

1. Substituting in the general equation p = xa + yp the value 
of y from the equation of the ellipse referred to center and axes 

we. have 

p=sa + mKa 2 --s 2 )i0 .... (220), 

in which m = — and a and /? are unit vectors along the axes. 

For unit vectors along conjugate diameters, the equation of the 
ellipse becomes 

p = j»a-fm'4(a' 2 -i» 2 )4^ .... (221). 

Again, if <f> be the eccentric angle, the equation of the ellipse 
may be written in terms of a single scalar variable, 

p = cos<f> . a + sin<£ . P .... (222). 

2. From Eq. (220) we have, for the vector along the tangent, 

a-ml(a*-x*)-hxp = a ^ X /3 = a-^/? 

Vm Va 2 — a* y 

= X(ya — mxfi) ; 
hence, for the equation of the tangent line, 

irz=xa + yP+X(ya — mxl$) . . . (223); 
or, more simply, from Eq. (222) , the vector-tangent is 
— sin ^ . a + cos ^ . 0, 



192 QUATERNIONS, 

and the equation of the tangent is 

7T = cos^ • a + sin^ • /? + x( — sin^ . a -f cos^ • /?), (224). 

Since — sin^ . a 4- cos^ . /? is along the tangent, cos<£ • a + 
sin <f> • /? and — sin <f> . a + cos <£ • /8 are vectors along conjugate 
diameters. 



89. Examples on the Ellipse. 

1. The area of the parallelogram formed by tangents drawn 
through the vertices of any pair of conjugate diameters is constant. 

We have directly 

TY[2(cos<£ • a + sin<f> . /?)2-( — sin<£ . a + cos<£ . /?)] 
= 4 TVa/J = a constant; 

namely, the rectangle on the axes. 

2. 2%e sum of the squares of conjugate diameters is constant, 
and equal to the sum of the squares on the axes. 

For, since Soft = 0, 

(cos^ • a + sin <f> • /?)* -f ( — sin tf> . a -f- cos ^ • /?)* = a 2 -f /?. 

3. 7%e eccentric angles of the vertices of conjugate diameters 
differ by 90? 

The vector tangent at the extremity of 

p = cos <f> • a + sin ^ . p (a) 

is 

— sin ^ . a -h cos<£ . £. 

This is also a vector along the diameter conjugate to p, and is 
seen to be the value of p when in (a) we write <f> + 90° for <f>. 

4. The eccentric angle of the extremity of equal conjugate diam- 
eters is 45? and the diameters fall upon the diagonals of the 
rectangle on the axes. 



APPLICATIONS TO LOCI. 193 

5. The line joining the points of contact of tangents is 
parallel to the line joining the extremities of parallel diam- 
eters. 

6. Tangents at rigltit angles to each other intersect in the cir- 
cumference of a circle. 

7. If an ordinate pd to the major axis be produced to meet the 
circumscribed circle in q, then 

qd : pd : : a : b. 

8. If an ordinate pd to the minor axis meets the inscribed circle 

in q, then 

qd : pd : : b : a. 

9. Any semi-diameter is a mean proportional between the dis- 
tances from the center to the points where it meets the ordinate of 
any point and the tangent at that point. 

For the point p (Fig. 82) we have 

p = cos <f> • a -f- sin <j> • /?. 
Also 

OT = SOP = OQ + QT 

= x(cos<£. a -f- sin <j> • ft) 

= cos<£' . a + sin <£'. j8 + t(— sin^' • a-hcos</>' • /?). 



Eliminating I, 



— x 



cos(<£— <f>') 



or 



But 



OT = BOP = —OP. 

COS(<£ — <£') 



ON = flj'OP = OQ + QN 

= a?'(cos ^ • a + sin <j> • /?) 

= cos<£' . a + sin<£' . P + J'(— sin<£ . a + cos<£ • /?). 

Eliminating # 

aj' = cos(<£ — <£'), 

or 

on = cos(<£ — <£')op ; 

.'. ON • OT = OP 2 . 



194 QUATERNIONS. 

10. To find the length of the curve. 

With the notation of Ex. 6, Art. 86, we obtain, from Eq. 
(222), 

p'= — sin <f> • a + cos <f> . /?, 

Tp'= V(a 2 -6 2 )sin 2 <£ + 6 2 , 
8-s = fV(a 2 -& 2 ) sin 2 <£ + 6 2 , 



•0 

which involves elliptic functions. If a = &, we have, for the 
circle, 



b, s — s = |r = r(<£ — <fo). 
From Eq. (220), we obtain 

p'= a — ml (a 2 ■— x 2 ) ""*#/?, 

> a 2 -* 2 Va 2 -* 2 ^ a 2 



which ma} r be expanded and integrated ; giving for the entire 
curve 

2 7ra [ 1 etc. ), 

^ 2.2 2.2.4.4 2.2.4.4.6.6 J 

a converging series. If e = 0, we have, for the circle, 2irr. 

11. To find the area of the ellipse. 
With the notation of Ex. 7, Art. 86, 

TYpp'= TY(cos<£ . a + sin<£ . /3) (— sin<£ . a + cos <£ • /?) 
= TY(cos 2 <£ . a/3 - sin 2 <£ . £a) = TVa/3 ; 

A 

or, since a/3 = 90, ^ 

if y TY PP '=j7ra6. 

The whole area is therefore nab. 



APPLICATIONS TO LOCI. 195 

90. The Hyperbola. 

1. Let a and ft be unit vectors parallel to the asymptotes. 
Then, from the equation, 

we have, for the equation of the hyperbola, 

fft 

p = xa + -p ' (225); 

x 

or, if a and /8 are given vectors parallel to the asymptotes, 

P = 'a+! (226); 

or, again, in terms of the eccentric angle, 

p = sec<£ • a + tan<£ . 0. . . . (227). 

2. The equation of the tangent, obtained as usual, is from 
Eq. (226), 

p = ta+£ + x(ta-£\ . . . (228), 
where ta — f- is a vector along the tangent. 

9SL Examples on the Hyperbola. 

1. If when the hyperbola is referred to its asymptotes, one 
diagonal of a parallelogram whose sides are the coordinates is 
the radius vector, the other diagonal is parallel to the tangent. 

If (Fig. 79) ex = ta, xp = £, then 

> , P P 

CP = ta+Z, Q X =fa— ^; 



196 



QUATERNIONS. 



but ta — E. is parallel to the tangent at p (Art. 90). ta -f £ and 

R ^ 

fa — " are evidently conjugate semi-diameters, 
t 

Fig. 70. 




2. -4 diameter bisects all chords parallel to the tangent at its 
vertex. 

Let (Fig. 79) cp be the diameter, t corresponding to the 

point p. The tangent at p is parallel to fa — £- and cp = ta -f- I-. 
p'p" being the parallel chord, 

CP r = CO + OP = xfta -f-£- j +y(ta — £ \ 

Also, if t' correspond to pj 

cp'=<'a+£; 

... (x + y)t = t', ^ = l„ 
or a?-»» = l. 



APPLICATIONS TO LOCI. 197 

Hence, for every point, as o, determined bj- x, there are two 
points p' and p", determined by the two corresponding values 
of y, which are equal with opposite signs. 

3. The tangent at p 2 to the conjugate hyperbola is parallel to 
cp (Fig. 79). 

4. The portion of the tangent limited by the asymptotes is 
bisected at the point of contact. 

5. If, from the point d (Fig. 79) , where the tangent at p meets 
the asymptote, dn be draton parallel to the other asymptote, then 
the portion of pn produced, which is limited by the asymptotes, is 
trisected at p and n. 

We have ~ ^ 

CP = *a + ^; 
. • . pn = CN — CP = ta — £-, 

and the equation of ss' is 

p . fa+ |+,( fa _0 

whence, for the points s, s) 

x = — l, x=2. 

6. The intercepts of the secant betioeen the hyperbola and its 
asymptotes are equal. 

The vector along the tangent parallel to the secant is ta — -. 
Hence (Fig. 79) ' 



CR 

cr' 






y = -2/ f ; 



but op" = op' (Ex. 2) , and therefore p"r" = p'r! 



198 QUATERNIONS. 

7. If through any point p" (Fig. 79) a line r"p'r' be drawn 
in any direction, meeting the asymptotes in r" and r', then 

p' r" . p'r' = pd' 2 . 

8. If through p,' p" (Fig. 79) lines be draivn parallel to the 
asymptotes, forming a parallelogram of which p'p" is one diagonal, 
the other diagonal will pass through the center. 

The vector from c to the farther extremity of the required 
diagonal is 

But t"a + & is the vector from c to the other extremity of the 
required diagonal. 

9. If the tangent at any point p meet the transverse axis in t, 
and pn be the ordinate of the point p ; then 

ct . cn = a 2 , 

c being the center and a the semi-transverse axis. 
From Eq. (-227), substituting in ct = cp -f pt, 

3sec<£ . a = sec^ . a + tan<£ . fi -f #(tan<£ sec<£ . a + sec 2 ^ . 0) ; 

• x- X 
sec 2 ^ 
and 

ct . CN = (asec<£ . a)(sec<£ . a) = a 8 , 
or 

ct . cn = a 2 . 

10. If the tangent at any point p meet the conjugate axis at t, 
and pn' be the ordinate to the conjugate axis, then 

ct' . CN f = b 2 , 
c being the center and b the semi-conjugate axis. 



APPLICATIONS TO LOCI. 199 

92. The preceding examples on the conic sections involve 
directly the Cartesian forms. A method will now be briefly 
indicated peculiar to Quaternion analysis and independent of 
these forms. 

1 . The general form of an equation of the first degree, or as 
it may be called from analog} 7 , a linear equation in quaternions, is 

aqb + a'qb'+ a"qb"+ = c, 

or 

%aqb = c, (a) 

in which q is an unknown quaternion, entering once, as a factor 

only, in each term, and a, b, a\ b\ , c are given quaternions. 

It ma} r evidently be written 

SSagfc + SVagft = Sc + Vc, 
whence 

SSagfc = Sc, (b) 

2Yaqb=Yb. (c) 

But 

Sag& = Sqba = StfS&a + S . YqYba, 
and 

Yaqb = V(Sa + Va) (Sg + Yq) (S6 + Yb) 

= V . Sg(Sa + Va) (S6 + Yb) 

+ V(SaV?S6 + SaVgVfc + VaVgSfc 4- YaYqYb) 
= SqYab + V(SaS6 - SaYb + $bYa)Yq 

+ V . YaYqYb + V . YaYbYq - Y . YaYbYq 
[Eq. (116)] = SgVafc + V(SaS6 - SaVfc + S&Va - YaYb) Yq 

+ 2VaS.VgV6 
= SgVafc + V . a (K6) Vg + 2 VaS . VgVfc. 

We have therefore, from (b) and (c) , 

Sc = SgSS&a + S . Yq^Yba, 

Vc = SqVfab + 2V . a(Kb) Yq + 2 SVaS . VgVfc, 

or, writing 

SSa& = a% 2Va& = 8, 2V&a = $; Sg = w, Vg = p, 



200 QUATERNIONS. 

we obtain 

Sc = wd + SpS,' 

Vc = iv$ + 2V . a(lib) P + 2 SVaS . pYb. 

We may now eliminate to between these equations, obtaining 

Vc . cl - Sc . 8 = d2Va(K6)p - 8SpS'+ d2SVaS . pV& 

which involves only the vector of the unknown quaternion q, and 
which, since V and 2 are commutative, may be written under 
the general form 

in which y, a, aj , /?, # are known vectors, r a known 

quaternion, but p an unknown vector. This equation is the 
general form of a linear vector equation. The second member, 
being a linear function of p, may be written 

Vrp + 2/3Sap = <f>p = y .... (229), 

where <f>p designates any linear function of p. If we define the 
inverse function <£ _1 by the equation 

the determination of p is made to depend upon that of <£~ 1 . 

2. Without entering upon the solution of linear equations, it 
is evident on inspection that the function <j> is distributive as 
regards addition, so that 

*(/> + />'+ ) = <fr> + <fr>'+ . . • (230). 

Also that, a being any scalar, 

} = atf>p (231), 



and 

d<f>p = <f>dp (232). 

3. Furthermore, if we operate upon the form 

<f>p = 20Sap + Vrp 



APPLICATIONS TO LOCI. 



201 



with S « <r X , <r being any vector whatever, 

S<r<fr> = 2S(<r/?Sap) + So-(Vrp) . 
S(<70Sap) = Scr/SSap = SpaS/fc- = S(paS/&r) , 



But 

and 



S(oTrp) = S[oT(Sr + Vr)p] = SrScrp + S<r(Vr)p 
= SrSpo- - Sp ( Vr) o- = S[pV (Kr) <r] . 



Hence, if we designate by <£'o-, 

<£'<r = SaS0<r + V(Kr)<r, 

a new linear function differing from <f> bj- the interchange of the 
letters a and /?, and Kr for r, we shall have, whatever the vectors 
p and <r, 

S(<r*p)=S(p<£'cr). 

Functions, which, like <f> and <f>', enjoy this property, are called 
conjugate functions. The function <£ is said to be self-conjugate, 
that is, equal to its conjugate <£, when for any vectors p, <r, 

8<T<f>p = Sp<f>(T. 

93. In accordance with Boscovich's definition, a conic sec-, 
tion is the locus of a point so moving that the ratio of its dis- 
tances from a fixed point and a fixed right line is constant. 



1. Let f (Fig. 80) be the fixed point or 
focus, do the fixed line or directrix, and p 

FP D 

any point such that — = e, the constant ratio 

DP 

or eccentricity. Draw fo perpendicular to 
the directrix, and let FO=a, OD=yy, pd = x\x 
and fp = p. By definition, 



Fig. 80. 




or 



Also 



T(pd) 
p 2 = eVa*. 

p + xa = a 4- yy. 



(a) 



202 QUATERNIONS. 

Operating with S . a X , we have, since Say = 0, 
Sap + 2*** = a 2 , 
aV = (a 2 -Sap) 2 . 



or 



Substituting in (a) 



ay = e 2 (a*-Sap) 2 .... (233), 



in which e may be less, greater than, or equal to unity, corre- 
sponding to the ellipse, hyperbola and parabola. 



Fig. 81. 




2. For the ellipse, Fig. 81, putting p = xa for the points 
a and a[ we have 



x = - 



1+e 



and x = ■ 



1-e 



or, since p = xa = jcfo, 



fa = 



FO, 



; — ; ryj i 



FA'=- 



1-e 



FO, 



whence 



and therefore 



AA f = 2a = 



2e 



FO = - 



1-e 2 
1-e 2 



FO, 



^ 



APPLICATIONS TO LOCI. 203 

which furnish the well-known properties of the ellipse, 

FA = a(l— e), 
FA f =a(l + e), 

cf = ae, 

1 — e 

ao = a, 

e 

a 
co = -. 

e 

3. Changing the origin to the center of the curve, let cf = a' ; 

then cp= p' and p = p'— a\ a 1 = f ] a ; whence 

1 _ g2 \1 — e 3 1 + ej 

a = — — - a. Substituting these values of p and a in 

a y = e 2 (a 2 --Saf>) 2 > 

remembering that a' 2 = — a 2 e* we obtain 

aV 2 + (Say) 2 = -a 4 (l-e 2 ), 

or, dropping the accents, c being the initial point, 

aV + (Sa f >) 2 = -a 4 (l-e 2 ) . . . (234), 

the equation of the ellipse in terms of the major axis with the 
origin at the center. If p coincides with the axes, Tp = a or 6, 
as it should. 

4. Equation (234) may be deduced directly from Newton's 
definition, thus: let cf =a (Fig. 81) as before, f and f' being 
the foci, and cp = p. Then 

fp = p — a, f'p = p + a, 
and, by definition, 

fp + f'p = 2 a 

as lines ; or 

T(p-a) + T0> + a) = 2a, 



204 QUATERNIONS. 

a being the semi-major axis. Whence 



V-(p-a) 2 = 2 a - V-(p + a) 2 . 

Squaring 



- p 2 + 2 Spa - a 2 = 4a 2 - 4a V- (p + a) 2 - p 2 - 2 Spa - a 2 , 

Spa — a 2 = — a V— (p-fa) 2 . 
Squaring again 

(Spa) 2 - 2 a 2 Spa + a 4 = - ar(p 2 + 2 Spa + a 2 ) , 

a y + (Spa) 2 = -a 4 -a 2 a 2 , 

or, as before, 

a 2 p 2 -f-(Spa) 2 = -a 4 (l-e 2 ). 

94. 1. The equation of the ellipse 

a 2 p 2 + (Sap) 2 =-a 4 (l-e 2 ) 
may be put under the form 

o S « 2 p+aSap1 , 

or, in the notation of Art. 92, writing 

<X 2 p -f" aSap , 

a\l-e 2 ) ~^' 

the equation of the ellipse becomes 

Sp<£p=l (235). 

2. By inspection of the value of <t>p it is seen that, when p 
coincides with either axis, p and <f>p coincide. 
Operating on <£p with S . o- x , we have 

« i tt 2 Stp -fr~ StraSap . 



APPLICATIONS TO LOCI. 205 

a • , a. u -4- aSctcr . , , ^ 

operating on <£<r = -f- — — with S . p x , we have 

cr(l — e A ) 

Spfrr = - a2 fy°" + S P aSa(r ■ 
a 4 (l — e 2 ) ' 
hence 

Sp<£o- = Sor<£p (23G), 

and </> is self-conjugate. 

3. Differentiating Equation (235), we have 

Sdp<£p + Spd<f>p = 0, 

Mp<l>p + Sp^dp = 0, [Eq. ( 232 ) ] 

&p<l>dp + Sp<£dp = 2 Sp<£dp = 0. [Eq. { 236 ) ] 

If ?r be a vector to any point of the tangent line, 

7T = p -|- xdp, 
whence 

Sp<£(tt — p) = S(tt — p)<f> p = 0, (a) 

or 

S?r<£p = Sp<£p = Sp<^7r = 1 .... (237) 

is the equation of the tangent line. 

From (a) we see that <£p is a vector parallel to the normal at 
the point of contact, being parallel to p only when p coincides 
with the axes, as already remarked. 

4. To transform the preceding equations into the usual Car- 
tesian forms, let i be a unit vector along ca (Fig. 81), and j*.u 
unit vector perpendicular to it. If the coordinates of p fire v 
and y, then, since a = cf, 

p = xi + yj, 
and 

__ ft 2 p + aSap a 2 (xi + #/) + aei'S * «es(ag -f- &/ ) 

* p ~ a 4 (l-e*) " a 4 (l-e 2 ) '" 

__ a 2 iw ( 1 — e 2 ) + a 2 yj 
"" a 4 (l-e 2 ) ' 



206 QTJATEBNIONS. 



1 2 & 2 

or, since 1 — «r = — 
or 






... Sp<fr> = l=-S. (si+^)g + |j, 
and 

Again, if x' and #' are the coordinates of a point in the tangent, 
ir = x'i + y'j; 

and 

a 2 ##'+ b 2 xx' = o 2 6 2 . 

The above applies to the hyperbola when e > 1, that is, when 

b 2 

1 — e 2 = ;, giving the corresponding equations 

a* 

a*tf-Va? = -a 2 6 2 , 
a 2 yy'— b 2 xx' = — a 2 ft 2 . 

95. Examples. 

1. To find the locus of the middle points of parallel choi*ds. 

Let /? bea vector along one of the chords, as rq (Fig. 82) , 
the length of the chord being 2#, and let y be the vector to its 
middle point ; then 

P = y + VP and P = y-yp 
are vectors to points of the ellipse, and 

S(y+y/3)«Ky+M9) = l, 

whence, expanding, subtracting, and applying Equation (236), 

Sy<fc8=0, 



APPLICATIONS TO LOCL 207 

the equation of a straight line through the origin. Since <f>@ 
is parallel to the normal at the extremity of a diameter parallel 
to /?, the locus is the diameter parallel to the tangent at that 
point. 

Fig. 82. 




2. Equation of condition for conjugate diameters. 

Denote the diameter op (Fig. 82) of the preceding problem, 
bisecting all chords parallel to /3, by a. Then 

Sa<£0 = 0, 
or 

S/fya = 0. 

In the latter, /? is perpendicular to the normal <f>a at the ex- 
tremity of a, and is therefore parallel to the tangent at that 
point; hence this is the equation of the diameter bisecting all 
chords parallel to a. Therefore, diameters which satisf}' the 
equation Sa<£/? = are conjugate diameters. 

3. Supplementary chords. 

Let pp' (Fig. 82) and dd' be conjugate diameters, and the 
chords pd, pd' be drawn. Then, with the above notation, 

dp = a — j8, 
D'p=a + /?, 

and 

S(a+jB)*(a-jB)==S(a+P)(*a-flB) 

= S(a<£a - *<f>$ + P<t>a - j3<£/3) . 



208 
But 



QUATEBNIONS. 

Sa0a=l, S0<£0=1, Sa4£ = Sj94a; 
••• S(a + j8)*(a-j8) = 0. 



Hence, if dp is parallel to a diameter, pd' is parallel to its 
conjugate. 

4. If two tangents be drawn to the ellipse, the diameter parallel 
t ) the chord of contact and the diameter through the intersection 
of the tangents are conjugate. 

Fig 82. 




Let tq (Fig. 82) and tr be the tangents at the extremities of 
the chord parallel to /?, and ot = ir. Then 

OQ = Xa + y/3, OR = Xa -f y'fi. 

From the equation of the tangent Sn-^/3 = 1 , we have 

H7r<t>(xa + yP)=l, 
87r<t>(xa+y'P)=l. 

Expanding and subtracting 

Stt<£/? = 0. 

Hence, Ex. 2, ir and j8, or op and od, are conjugate. The 
locus of t for "parallel chords is the diameter conjugate to the 
chord through the center. , 



APPLICATIONS TO LOCI. 209 

5. If qoq' (Fig. 82) be a diameter and qr a chord of contact, 
then is q'r parallel to ot. 

rq being parallel to /?, and oq' = — oq, we have 
rq = 2#£, rq' = y/3-x<L — xa-y/3; 

whence, directly rq' = — 2xa ; as also Srq<£rq' = 0, rq and rq' 
being supplementary chords. 

6. The points in which any tivo parallel tangents as q'tJ qt 
(Fig. 82) are intersected by a third tangent, as ttJ lie on conju- 
gate diameters. 

The equation of rt' is Sn-^p = 1, and that of q't' is Stt'<^/)'=1. 
For the point tJ it = ir' ; whence, by subtraction, 

S7T<£(p-/>') = 0. 

7. Chord of contact. 

The equation of the tangent, 

Sp<^7T = 1 , 

is linear, and satisfied for both q and r. Hence, writing <r for p 
as the variable vector, ir being constant, 

St^tt = 1 

is the equation of the chord of contact. 

8. To find the locus of t for all chords through a fixed point 
(Fig. 82). 

Let s be a fixed point of the chord rq, so that os = <r = a 
constant. Then 

So"07T = S7T<£o" = 1 , 

a right line perpendicular to <£cr, or parallel to the tangent at the 
extremity of os, and the locus of t^ for all chords through s. 



210 



QUATERNIONS. 



9. Any semi-diameter is a mean proportional between the dis- 
tances from the center to the points where it meets tJie ordinate of 
any point and the tangent at tliat point. 




od (Fig. 82) and op being still represented by /J and a, let 
ot = x'a and oq = p = xa + yfi. Then from the equation of the 
tangent, Stt<£p= 1, we obtain 

whence, since Sa<£/3 = 0, 

xx'Sa<f>a = 1, 
or 

xx' = 1 ; 

.*. Xa • x'a = a 2 , 
ON . OT = OP 2 . 



or 



10. T^dd' (Fig. 82) and pp' are conjugate diameters, then are 
pd and pd' propoftional to the diameters parallel to them. 



With the same notation 
whence 



DP = a-0, D r P = a-f0, 

OE = m(a — £), OP = ?i(a +/?). 



and 



Similarly, from (b) , 



APPLICATIONS TO LOCI. 211 

From the equation of the ellipse 

Sm(a - 0)0m(a - /3) = 1, (a) 

Sn(a + £)<£» (a + 0)=1. (b) 

Now, from (a) , since S/fy/2 = Sa<£a = 1 and S/?<£a = Sa<££ = 0, 
2m 2 =l. * 

2?i 2 =l; 

.•. m — n. 
Also 

DP:D'p::T(a— /?) : T(a + /3) 

::Tm(a-/3) : Tn(a + 0) 
: : OE : OF. 

1 1 . The diameters along the diagonals of the parallelogram on 
the axes are conjugate; and the same is true of diameters along 
the diagonals of any parallelogram tohose sides are the tangents at 
the extremities of conjugate diameters. 

12. Diameters parallel to the sides of an inscribed parallelo- 
gram are conjugate. 

Fig. 83. 

R 




Let the sides of the parallelogram (Fig. 83) be 

PP' = a, PQ = /?, 



and let 
Then 



OP = p, OQ = p. 
OP' = p+a, OQ' = p r + a, p' - p = 0. 



212 QUATERNIONS. 

From the equation of the ellipse, Sp<£p= 1 , we have for Q f and p' 

S(p'+a)<Kp'+a)=l, 
S(p+a)</>(p+a)=l; 

whence, since Sp<£p = Sp'<£p' = 1, 



Subtracting 



or 



2&typ'+Sa</>a = 0, 
2 Sa</>p -|- Sa</>a = 0. 

Sp' <f>a — $pcf>a = 0, 

S(p' — p)<f>a = S£</>a = 0. 



13. The rectangle of the x>erpendiculars from the foci on the 
tangent is constant, and equal to the square of the semi-conjugate 
axis. 



Fig. 83. 




Let the tangent be drawn at r (Fig. 83) and or = p. Then 
<t>p is parallel to the normal at r, that is, to the perpendiculars 
fd, f'd! Hence, of being a, 

ov>'=x'<f>p — a, 
OD = a + X<f>p, 

which, since d and d' are on the tangent, in S?r<£p = 1 give 

&(x'<l>p — a) <f>p = 1 , 
S(a -f #<£p)<£p = 1, 
or 

*'(4>p) 2 =l+S a </>p, 
X (<£p) 2 = 1 — Sa</>p; 



whence 



and 



APPLICATIONS TO LOCI. 213 



Tx<f>p = FD =T T . " » 



But 



, , l-(Sa<f>p) 2 
FD X fV = T /, 7? J . 



. xf / a*p + a8apV _ a 2 (a 2 P 2 ) + 2a 2 (Sap) 2 + a 2 (Sap) g 
( W = V~ a 4 (l -«*),/" a 8 (l-e 2 ) 2 

or, substituting a 2 p 2 from Equation (234) and a 2 = — aV, 

^ (Sap) 2 -a 4 
a«(l-e 2 )' 
A.lso 

.•■,x^-."-y)' ff ) r^ -^-o-K 



14. 27*e /oof o/ the perpendicular from the focus on the tangent 
is in the circumference of the circle described on the major axis. 

To prove this we have to show that the line od (Fig. 83) is 
equal to a. Now 

OD = a + %<f>p 

<fto(l-S a<frp) 

= *+— (55*— 

from the preceding example. Hence 

(OD) =a + fep + <*)' 

_ l-(Sa^)' a«-(Sap)»a«(l-e') 

~ (#? - -a<r + a 4 (Sap)*-* 

=a _ a * e »-o*(l-e J ) = -a , ; 
1 .*. oi> = a. 



214 QUATERNIONS. 



The Parabola. 

96. 1. Resuming Equation (233) and making e = l, the 
equation of the parabola is 

a 2 p 2 = (a 2 -Sap) 2 (238), 

which maj r be written 

p 2 + 2Sap-a- 2 (Sap) 2 __ , . 

~2 — 1 » 



a 2 



or 



Sp r P + 2a-a-'aSap1 1 , 
in which, if we put 

<f>p - 2 — 1 

we have for the equation of the parabola 

Sp(</>p + 2a- 1 )=l (239), 

and, as in the case of the ellipse, 

S<r^p = Sp</wr (240). 

Operating on <£p by S . a x , we obtain 

Sa</>p = (241); 

hence, <f>p is a perpendicular to the axis. 
Operating on <f>p by S • p x 



^- ^- tf 7* >0 ' .rf(fr)«. • • (242). 



2. Differentiating Equation (239), we have 
2Sp<£dp + 2Sdpa- 1 = 0. 

For any point of the tangent line to which the vector is tt, 
ir=z p -\- axZp, 



APPLICATIONS TO LOCI. 215 

from which, substituting dp in the above, 

Hp<t>(ir-p) -f S(7r-p)a- 1 =0, 

8(p<f>ir — p<l>p -\-7ra' 1 — pa" 1 ) =0; (a) 

or, since Sp<£p = 1 -2 Spa' 1 [Eq. (239)], 

87r<f>p -1+2 Spa" 1 + STra- 1 - Spa" 1 = 0, 

whence 

Stt(^p + a- 1 ) + Spa" 1 = 1 . . . . (243) , 

the equation of the tangent line. 

3. From (a) we obtain 

S(7r-p)(^>p-fa- 1 )=0; 

or, since it — p is a vector along the tangent, 

+P + *' 1 
is in the direction of the normal. 

4. If cr be a vector to any point of the normal, the equation 
of the normal will be 

< r = p4-aj(^p-f-a- 1 ) (244). 

5. The Cartesian form of Equation (239) is obtained by 
making 

p = xi + yj, a = fo (Fig. 80) = — pi ; 

xpi 

xi -f yj 

-rw p yj 

•• <fo> = —* = --*' 

whence, Equation (239) becomes 

P 2 P 
.-. y 2 =2px+p 2 , 

the equation of the parabola referred to the focus. 



216 QUATERNIONS. 

97. Examples. 

1. The sub tangent is bisected at the vertex. 

Fig. 84. 




gives 



(a) 



We have (Fig. 84) ft = aa, which in the equation of the tangent 

S7r(</>p + a- 1 )H-S/)a- 1 =l 

SxaOfro + a-^-f Sa" 1 p=l. 

But Sa<£p = ; hence 

multiplying b}' a 

Xa -+- aSa _1 p = a, 

(x — £)a = a — \a — aSa -1 p 
= \a — aSa -1 p, 
AT = — AF — aSa -1 p. 

But the value of <f>p gives 

a 2 <f>p = p — a" 1 Sap ; 



APPLICATIONS TO LOCI. 217 

and, since <f>p is a vector along mp and a" 1 Sap a vector along fm, 
from p = fm 4- mp we have 

FM = a" 1 Sap = aSa^pi (P) 

MP = a 2 <f>p; (c) 

.*. AT = — AF — FM = — AM, 

or, as lines, 

AT = AM. 

2. The distances from the focus to the point of contact and the 
intersection of the tangent with the axis are equal. 



or (Fig. 84), 



Xa = a — aSa *p, 

(FT) 2 = (a-aSa- 1 p) 2 
= (a-a" 1 Sap) 2 
__ (a 2 - Sap) 2 
a 2 
[Eq.(238)] =p 2 ; 

.'. FP= FT. 

3. The subnormal is constant and equal to half the parameter. 

The vector-normal being <£p -f- a" 1 (Art. 96, 3), we have 
(Fig. .84) 

PN = z(<fo> + cr 1 ); 

but 

PN = PM + MN 

= — a 2 </>p + Xa ; [Ex. 1, (c)] 

.-. S^p + a-^-afyp-ftfa, 
z = — a 2 = #a 2 , 



or 



0? = — 1 , xa = — a ; 



or, the distances mn and fo are equal, and the subnormal = />, 
a constant. 

4. The perpendicular from the focus on the tangent intersects 
it on the tangent at the vertex^ and aq = £mp (Fig. 84). 



218 



QUATBBNIONS. 



But 



Since (Ex. 2) fp = ft == pd, fd is perpendicular to pt or par- 
allel to pn. Otherwise : 

NP = -*(</>« + a" 1 ) = a 2 (</>p + a- 1 ) (Ex. 3) 

= a 2 <fr> + « = MP + fo, [Ex. 1, (c) ] 

a 2 <£p -f a = FO + OD = FD. 

|FD = FQ = i<L*<f>p + £a 
= |a 2 <fy> + FA; 
.-. £a 2 <fr> = AQ = |mp. 

5. 2b ^nd the locus of the intersection of the perpendicular 
from the vertex on the tangent and the diameter produced 
through the point of contact. 



Fig. 84. 




Let fs = o- (Fig. 84) be a vector to a point of the locus. 
Then 

FS = FA + AS = FP -f- PS, 
<r = i a + Z (<fr> -f a" 1 ) = p + »<*• 



') 



APPLICATIONS TO LOCI. 219 

Operating with X S . <f>p, then, since Sa<£p = [Eq. (241)], 

z(<f> P y = S^p = a 2 (<fr>) 2 ; [Eq. (242)] 

.". Z = a 2 , 

and 

o- = b" + a 2 (</>/> + a" 1 ) = fa + a 2 <£p, 
or 

<r — | a = a 2 <£p. 

Operating with x S . a 

S(cr — |-a)a=0, 
S<m = -f(Ta) 2 , 

or [Eq. (180)], the locus is a right line perpendicular to the 
axis and f p distant from the focus. 

6. To find the locus of the intersection of the tangent and the 
perpendicular from the vertex. 

If the origin be taken at the vertex, then since <f>p -f- a -1 is a 
vector along the normal, the equation of the locus will be 

fl- = .T(<^p + a- 1 ). (a) 

To eliminate a?, operate with S . a x which gives 

x 

x = Scwr, whence Sa" ir = -. 

a 2 

To eliminate /o, the equation of the tangent, fhr(fa> + a~ v ) + 
Spa" 1 = 1 , for the new origin becomes 

s ( 7r4 "i) ( ^ +a " 1)+Spa " 1=1 ' 

or 

2 Stt^p + 2 ScrV + 2 Sa" V = 1 . 

Operating on (a) with x S ..<£/», whence 8Tr<f>p = x(<f>p) 2 , the 
preceding equation becomes 

2*(ftO f -?f + 2Sa-V=l. (6) 



220 QUATERNIONS. 

Also [Eq. (242)] Sp<^> = a 2 (<£/o) 2 , which, in the equation of 

the parabola 8p(<f>p 4- 2a" 1 ) = 1, gives 

i 

a 2 (<fo>) 2 + 2Sa-V=l. (c) 

Whence, from (b) and (c) , by subtraction, 

But, from (a), 

V9P; z 2 s 2 a 2 

Equating these values of (<fr>) 2 , and substituting the value of .t, 

2 ^SttTT - a 2 7r 2 + (SaTr) 2 = 0, 

which is the equation of the locus required. To transform to 
Cartesian coordinates, make 

7T = xi + yj, and a = ai, 
whence 

7T 2 = — (a 2 -f- y 2 ) , Scwr = — OX, a 2 = — a 2 , 

and 

. s 8 

a 

a? 

2 

the equation of the cissoid to the circle whose diameter is the 
distance from the vertex to the directrix. 

7. If pp' (Fig. 75) be a focal chord, and pa, pa' produced 
meet the directrix in d', d, then will pd and p'd' be parallel to af. 

ad'= — xxp = AO + OD, f 
or 



•(H-i 



+ yy- 



Operating with S . a x 

8(a'-2Sap) = a s . (a) 



APPLICATIONS TO LOCI. 



221 



Fig. 75 (bis). 



Now fp = p and fp' = — x p are vectors to points on the 
curve, and hence satisfy its equa- 
tion. Whence [Eq. (238)] 

a 2 p 2 = (a 2 --Sap> 2 , . 
aVp 2 = (a 2 + a'Sap) 2 ; 

... ^(a 2 -Sap) 2 = (a 2 +a;'Sap) 2 ; 

or 

0?'(a 2 -Sap) = a 2 -f-a; , Sa /0 , 

.-. x'(a 2 -28ap) = a 2 . 

Hence, comparing with (a), 

x = x\ 

or, the sides produced of the 
triangle apf are cut propor- 
tionately, and therefore d'p' is 
parallel to af. 

8. If with a diameter equal to three times the focal distance, 
a circle be described with its center at the vertex, the common 
chord bisects the line joining the focus and vertex. 

. The equation of the curve being 

a 2 p 2 = (a 2 -Sap) 2 , («) 

that of the circle whose center is a (Fig. 75), referred to f, is 
of the form [Eq. (210)] 

T(p-y) = T0, 
or, b} r condition, 




H)' 



.•■(p-D-ft* 



which, in (a), gives 
which is the proposition. 



p 2 = Sap -f- ft a -> 



Sap=-, 



222 QUATEBNIONS. 

96L The Cycloid. 

1. Let a and /? be vectors along the base and axis of the 
cycloid and T/? = Ta = r, the radius of the generating circle. 
Then, for any point p of the curve, 

x = rO — r sin = r(0 — sin 0) , 
y=zr — rcos0 = r(l — cos#), 

and the equation of the cycloid is 

p = (0 - sin0)a + (1 - cos0)0. 

2. The vector along the tangent is 

(1 — cos 0)a 4- sin0 . /?, 
and the equation of the tangent is 
a- = (0 - sin0)a -f (1 - cos0)/? + *[(1- cos0)a + sintf . /?]. 

3. The vector from p to the lower extremity of the vertical 
diameter of the generating circle through p is 

pc = — (1 — cos0)/? + sin • a, 

and, from the above expression, for the vector-tangent pt, 

S(pc . pt) = ; 

hence pc is perpendicular to the tangent, or the normal passes 
through the foot of the vertical diameter of the generating cir- 
cle for the point to which the normal is drawn, and the tangent 
passes through the other extremity. 

4. If, through p, a line be drawn parallel to the base, 
intersecting the central generating circle in q, show that 
pq = ?•(*■ — 6) = arcQA, a being the upper extremity of the 
axis. 



APPLICATIONS TO LOCI. 223 

5. With the notation of Ex. 6, Art. 86, 

p' =(1 — cos0)a+sin0 . 0, 
^ = -[(l-cos^) 2 4-sin 2 ^]r 2 , 
Tp' = r Vl— 2cos0 + cos 2 + sin 2 = r V2 — 2cos0 
= 2rsin|0; 
s-8 =z f2rsin£0 = [4r cositf]^ = 8r, 

•/27T 

the length of the entire curve. 

6. With the notation of Ex. 7, Art. 86, 

TYpp'= TV[(0 - sin0)sin0 . a/? + (1 - co80) 2 0a] 
= TV[(0 sin0 - sin 2 - (1- cos0) 2 ]a/3 

= ^(08^0+20080-2). 

.4 --4 = ** f (* sin0 + 2 cos0 - 2) 

•/27T 

= | — (sin0 - cos0 + 2 sin0 - 20) T 

L 2 J27T 

= ["-(3 sin0 - cos0 - 20)T= 3 ttt 2 , 
the whole area of the curve. 

99. Elementary Applications to Mechanics. 

1. If b be the magnitude of any force acting in a known di- 
rection, the force, as having magnitude and direction, may be 
represented by the vector symbol /?, which is independent of 
the point of application of the force. In order, completely, to 
define the force with reference to any origin o, the vector OA=a, 
to its point of application a, must also be given. For concur- 
ring forces, whose magnitudes are b[ 6" , we have, for the 

resultant, /J = 2/JJ which is true, whether the forces are compla- 
nar or not, and is the theorem of the 'polygon of forces extended. 
For two forces, yS = ^ , + yS rf ; whence 2 = p 2 + /3" 2 + 2 S/?'/3", or 



224 QUATERNIONS. 

6* = b n + b" 2 + 2b'b" cos0, which is the theorem of the parallelo- 
gram of forces. For any number of concurring forces, the con- 
dition of equilibrium will be 2/3'= 0. For a particle constrained 
to move on a plane curve whose equation is p= <f>(t), dp being 
in the direction of the tangent, since the resultant of the ex- 
traneous forces must be normal to the curve for equilibrium, we 
have 

S<fy>20'= 8d P p = 0. (a) 

2. If oa'= a( and ft is a force acting at aJ then TVa'£'= a'&' sin 
is the numerical value of the moment of the couple ft at a' and 
— ft at o. Representing, as usual, the couple by its axis, its 
vector symbol will be Yd ft. If — ft act at some point other 
than the origin, as c', and oc'= y 1 , the couple will be denoted by 
V(a'— y')ft. From this vector representation of couples, it fol- 
lows that their composition is a process of vector addition ; hence 
the resultant couple is 2V(a'— y')# and, for equilibrium, 
2V(a f — y')/?'=0. If the couples are in the same or parallel 
planes, their axes are parallel and T2 = 2T. Since a'— y' is 
independent of the origin, the moment of tlie couple is the same 
for all points. Since V(a'— y')ft= Va'ft-Vy'ft, the moment of 
a couple is the algebraic sum of the moments of its component 
forces. If the forces are concurring, and a' is the vector to 
their common point of application, 2Va'/?'= V2a'/?'= Va':S/?'= 
Va'/?, or the moment of the resultant about any point is the sum 
of the moments of the component forces. When the origin is on 
the resultant, a 1 coincides with ft in direction, and Va'/? = ; or 
the algebraic sum of the moments about any point of the resultant 
is zero. If a single force ft acts at a[ we may, as usual, intro- 
duce two equal and opposite forces at the origin, or at any other 
point c', and thus replace ft K , \>y ft and Ya'ft, or by ft > and 
V(a'— y')/?! If £ be a unit vector along anj axis oz through the 
origin, then the moment of ft acting at aJ with reference to the 
axis oz, will be - S/?V£, or — S . £J/3'a'. If ft and £ are in the 
same plane, in which case they either intersect or are parallel ; 
or, if the axis passes through a' there will be no moment : in 
these cases, a\ ft and £ are complanar, and — S/3'a'£ = 0. 



APPLICATIONS TO LOCI. 225 

3. If the forces are parallel, their resultant p = 2£'= 2&'U/?' 
= U/326' ; and, therefore, for equilibrium, 2T/?'= 26'= 0. The 
moment of a force with reference to any axis oz through the 
origin being — S/3'a'£ , and the moment of the resultant being 
equal to the sum of the moments of the components, we have 
S0a£=2S/?'a'£, which, for parallel forces, becomes 8(26' . U/? . a£) 
= S(U/?26'a' • £) , which, being true for any axis, is satisfied 
for2&'.a=2&V; 

26V 



26' 



(») 



which is independent of U/?, and hence is the vector to the cen- 
ter of parallel forces. When 26'= 0, the above equations give 
/3 = and a = oo, the system reducing to a couple. For a sj-s- 

tem of particles whose weights are w' y w',' , we have the vec- 

Sw'a' 
tor to the center of gravity a = f . From this equation, 

2w'(a — a') = ; whence, if the particles are equal, the mm of 

the vectors from the center of gravity to each particle is zero; and, 

if unequal, and the length of each vector is increased propor- 

tionatel}- to the weight of each particle, their sum is zero. For 

w'1a f 
equal particles, a= , or the center of gravity of a system of 

2wr 

equal particles is the mean point (Art. 18) of the polyedron of 
which the particles are the vertices. For a continuous bbd}- 
whose weight is w, volume v, and density d at the extremity of 

a, a = — — — , in which 2 ma} r be replaced b} r the integral sign 
if the density is a known function of the volume. For a homo- 
geneous bod}% a=r a , which is applicable to lines, surfaces 
2,dv 

or solids, v representing a line, area or volume. Thus, for a 
plane curve p = <£(£) = a,' dv = ds = Tdp = T<f>\t)dt and 



(V(0T<//(0<fc 
J T<f>'(t)dt 



226 QUATERNIONS. 

4. General conditions of equilibrium of a solid body. Let 

the forces $ #' , act at the points a J a," of a solid body, 

and OA'=a, f OA" = a," Replacing each force by an equal 

one at the origin and a couple, the given system will be equiva- 
lent to a system of concurring forces at the origin and a system 
of couples. Hence, for equilibrium, 

2£'=0, (d) 

2Va'/3'=0. (e) 

Let £ be the vector to any point x. Then, from (d), 
V . £2/3'= 0, and therefore, from (e) , V . £20'= 2Va'0' ; whence 

2V0V- 2V0'£ = 2V0'(a'- £ ) = 0. (/) 

Conversely, £ being a vector to any point, the resultant couple, 
for equilibrium, is 2V(a'— £)/?'= ; .-. 2Va'0'= and 20'= 0. 
Therefore (/) is the necessary and sufficient condition of equi- 
librium. 

This condition may- be otherwise expressed by the principle 

of virtual moments. Let 8 f , 8" be the displacements. Then 

the virtual moment of 0' is — S0'8' ; and, for equilibrium, 
2S0'8'=O. This equation involves (d) and (e). Thus, if the 
displacement corresponds to a simple translation, 8'= 8"= 8'" 
= etc. = a constant, and we may write 2S0'8' = S320' = ; 
whence, since 8 is real, 20'= 0. Again, if the displacement 
corresponds to a rotation about an axis £, £ being a unit vector 
along the axis, 

a'= Z-Ha'=£-\SW + Vfa') = - £S£a'- £V£a,' 

the last term being a vector perpendicular to the axis. For a 
rotation about this axis through an angle 0, this term becomes 
- ^"fT£a f = - £ cos0 Y£a'+ sin0 V£aJ and a' becomes 

a'i= - £S£a'- £ cos0 V£a'+ sin0 V£aJ 

which, for an infinitely small displacement, 

= -£S£a'-£V£a'+0V£a.' 



APPLICATIONS TO LOCI. 227 

Placing the scalar factor under the vector sign and writing. f 
simply for 0f, to denote the indefinitely short vector along oz, 

a'+S'=a'+V£a'; 

or, 8'= VW Hence 2S0'S'= 2S0'V£a'= S£2Va'0' ; or, since £ is 
not zero, 2Va'0'= 0. 

5. Illustrations. 

(1) Three concurrent forces, represented in magnitude and 
direction by the medials of any triangle, are in equilibrium. 
(See Ex. 2, Art. 18.) 

(2) If three concurring forces are in equilibrium, they are 
complanar. By condition, /3'+/?"+/3'"= 0. Operating with 
S . 07?" x , we have S/?7?"0'"= 0. 

(3) In the preceding case, operating with V. /3'x, we have 
YP l P"+Y/3'/3 ,,t = ; whence, since the forces are complanar, 
TV/?'/3"= TY/3'Pl" or b'b" sin(# /3") = bV" sin(fl /?'") . A sim- 
ilar relation may be found for any two of the forces ; whence 

b' : b" : V" : : sin(#' £'") : sin(# 0'") : sin(# /3") . 

(4) If two forces are represented in magnitude and position 
by two chords of a semicircle drawn from a point on the circum- 
ference, the diameter through the point represents the resultant; 

(5) A weight, w\ rests on the arc of a vertical plane curve, 
and is connected, by a cord passing over a pulley, with another 
weight, wl 1 Find the relation between the weights for equili- 
brium. 

(a) Let the curve be a parabola, and the pulley at the focus. 
Then, from Eq. (a) of this article, the equation of the curve be- 
ing p = — (y 2 — p 2 ) a +T//S, we have 



228 QUATERNIONS, 

in which r = radius vector. Hence 

p ' pr ' r ' 

or, since r = x + p, w'=w" Hence, if the weights are equal, 
equilibrium will exist at all points of the curve. 

(b) Let the curve be a circle and the pulley at a distance m 
from the curve on the vertical diameter produced. With the 
origin at the highest point of the circle, p = xa + V2r# — x 2 ^. 
Hence, r being the distance of the pulley from w\ 

S (^p/? + a) („»• - yg + (» + *>V A = ; 

(c) Let w be placed on the concave arc of a vertical circle, 
and acted upon by a repulsive force varying inversely as the 
square of the distance from the lowest point of the circle. To 
find the position of equilibrium. The origin being at the lowest 
point of the circle, and r the distance required, let p be the 

intensity of the force at a unit's distance ; then ^ will be its 
intensity for any distance r, and 

_/ , K — x\fxa + yBp \ - 

\ a+ -^-^){—r E ?- wa ) =0i 

whence 

r = 

(d) Let w 1 rest on a right line inclined at an angle to the 
horizontal, and connected with w" by a cord passing over a pul- 
ley at the upper end of the line. Find the relation between the 
weights. With the origin at the lower end of the line, its equa- 
tion is p = xa. If p is in the direction of w\ then &a(w , fi+w n a) 
= ; .*. w"=w' sin0. 

(6) To find the center of gravity of three equal particles at 
the vertices of a triangle, a, b, c being the vertices, the vector 



APPLICATIONS TO LOCI. 22& 

from A to the center of gravity of the weights at A and b is 
£ab = ad. The vector to the center of gravity of the three 
weights is £(ab + ac) = % ab + xdc = £ab -|- x( — £ ab + ac) ; 
.*. #=£, and the required point is the center of gravity of the 
triangle. 

(7) Find the center of gravity of the perimeter of a triangle. 

(8) Find the center of gravity of four equal particles at the 
vertices of a tetraedron. 

(9) Show that the center of gravity of four equal particles 
at the angular points of any quadrilateral is at the middle point 
of the line joining the middle points of a pair of opposite sides. 

(10) The center of gravity of the triangle formed by joining 
the extremities of perpendiculars, erected outwards, at the mid- 
dle points of an} T triangle, and proportional to the corresponding 
sides, coincides with that of the original triangle. Let abc be 
the triangle, bc = 2 a, ca = 2/8 and € a vector perpendicular to 
the plane of the triangle. Then, if m is the given ratio, b the 
initial point, and r x , r 2 , r 3 the extremities of the perpendiculars 
to bc, ca, ab, respectively, 

BR! = a + mca, BRg = 2 a + ft + mtfi, BR 3 = a + /? — me(a +fi) ; 
.-. i(BR 1 + BR 2 + BR d ) = i(4a+2 i g) = i[2a-h2(aH-^)]. 

(11) To find the center of gravit} r of a circular arc. The 
equation of the circle p = r(cos0 . a-|-sin0 . /?), gives dp = 
r(— sin0 • a + cos0 • p)dfl\ 

C<t>(0)T<t>'(d)dO fr(cos0 . a + sin0 . p)d$ 

.-. a x = ^ =^ 

CT<t>'(6)dO CdO 

For an arc of 90? integrating between the limits - and 0, 

2 T 2v/~ 

aj = — (a + ($) , the distance from the center being — V2 ; which 

IT IT 



230 QUATERNIONS. 

may be obtained directly also by integrating between the limits 
j and — -. For a semicircumference or arc of 60° we have, in 

ri 4 2r . 3r 

like manner, — and — . 

IT IT 

(12) If a, (3, y are the vector edges of any tetraedron, the 
origin being at the vertex, then p — a, /? — y, a — /J are lines of 
the base, p being any vector to its plane. Hence this plane is 
represented by S (p - a) (£ - y) (a - £) = ; .-. Sp (Va/? + 
Vya 4- V/fy) — Sa/fy = 0. If 8 be the vector perpendicular on 
the base, 

and, taking the tensors, 

T(Va0 + V0y + Vya) = 6 X 1 V ° L = 2 area base, 
alt. 

But Va0 + V0y + Vya + Y0a + Yy0 + Vay = O, in which the 
last terms are twice the vector areas of the plane faces. The 
sum of the vector areas of all the faces is therefore zero. Since 
any potyedron may be divided into tetraedra bj T plane sections, 
whose vector areas will have the same numerical coefficient, but 
have opposite signs two and two, the sum of the vector areas of 
any polyedron is zero. These vector areas represent the pres- 
sures on the faces of a polyedron immersed in a perfect fluid 
subjected to no external forces. For rotation, since the points 
of application of these pressures are the centers of gravity of 
the faces, to which the vectors are 

we have the couples 

-iY\(a+P+y){YaP+Jfi r +Yya) + (a + P)YPa + (fi + y) 
Vy/3 + (y + a )Vay} 
= -iV(aVj3y + /3Vya + yYa/2), 

since aYa/? + aV/?a = 0, etc. But, Equation (123), this sum is 
zero. Hence there is no rotation. • 



MISCELLANEOUS EXAMPLES. 231 

100. Miscellaneous Examples. 

1. In Fig. 58, f, a and k are collinear. 

2. In Fig. 58, ad^ae^ab^ac 2 . 

3. In Fig. 13, if the lines from the vertices of the parallelo- 

gram through o and p are angle-bisectors, omitp is a 
rectangle. 

4. If the corresponding sides of two triangles are in the same 

ratio, the triangles are similar. 

5. /3, a, y being the vector sides of a plane triangle, if /?=a-|-y, 

show that b 2 =c 2 — ca cos B+ab cose. 

6. The sides bc, ca, ab of a triangle are produced to d, e, f, 

so that cd = mBC, ae = ncA, bf =^ab. Find the inter- 
1 sections q 2 , q 2 , q 3 of eb, fc ; fc, da ) da, eb. 

7. In any right-angled, triangle, four times the sum of the 

squares of the medials to the sides about the right angle 
is equal to five times the square of the hypothenuse. 

8. If abc be any triangle,. m it$ mean point, and o any point 

in space, then : : 

AB 2+ bc 2 + ca 2 = 3(oa 2 +ob 3 + oc 2 ) -(3qm) z . 

9. If abcd be any quadrilateral, m its mean point, and o any 

point in space, then 

ab 2 -|- bc 2 + cb 2 + DA 2 

= 4(OA 2 + OB 2 -|- OC 2 + OD 2 ) - (40M) 2 - AC 2 - BD 2 . 

10. If abc be any triangle, and c', b', a' the middle points of 

ab, ac, cb, then, o being any point in space, 

ab 2 +bc 2 +ca 2 =4(oa 2 +ob 2 +oc 2 )-4(ob' 2 +oc' 2 +oa' 2 ). 

11. If abc be any triangle and m its mean point, then 

AB 2 -|- BC 2 -|- CA 2 = 3 (AM 2 + BM 2 -|- CM 2 ) . 

12. Points p, q, r, s are taken in the sides ab, bc, cd, da of a 

parallelogram, so that ap= wab, bq = wise, etc. Show 
that pqrs is a parallelogram whose mean point coincides 
with that of abcd. 



232 QUATERNIONS. 

13. The sides of any quadrilateral are divided equably at p, q, 

r, s, and the points of division joined in succession. If 
pqrs is a parallelogram, the original quadrilateral is a 
parallelogram, 

14. The middle points of the three diagonals of a complete 

quadrilateral are collinear. 

15. If any quadrilateral be divided into two quadrilaterals by 

any cutting line, the centers of the three are collinear. 

16. If a circle be described about the mean point of a paral- 

lelogram as a center, the sum of the squares of the lines 
drawn from any point in its circumference to the four 
angular points of the parallelogram is constant. 

17. A quadrilateral possesses the following property : any point. 

being* taken, and four triangles formed by joining this 
point with the angular points of the figure, the centers 
of gravity of these triangles lie in the circumference of a 
circle. Prove that the diagonals of this quadrilateral are 
at right angles to each other. 

18. The sum of the vector perpendiculars from a, b, c, .... on 

an}; line through their mean point is zero. 

19. a, b, c are the three adjacent edges of a rectangular paral- 

lelopiped. Show that the area of the triangle formed by 
joining their extremities is %^/b 2 c?+ a-<?-\- a 2 b 2 . 

20. Given the co-ordinates of a, b, c, d referred to rectangular 

axes. Find the volume of the pyramid o— abcd, o being 
the origin. 

21. Any plane through the middle points of two opposite edges 

of a tetraedron bisects the latter. 

22. The chord of contact of two tangents to a circle drawn 

from the same point is perpendicular to the line joining 
that point with the center. 

23. If two circles cut each other and from one point of section 

a diameter be drawn to each circle, the line joining their 
extremities is parallel to the line joining their centers, 
and passes through the other point of section. 



MISCELLANEOUS EXAMPLES. 233 

24. The square of the sum of the diameters of two circles, tan- 

gent at a common point, is equal to the sum of the 
squares of any two common chords through the point of 
tangency, at right angles to each other. 

25. t is any point without a circle whose centre is c ; from t 

draw two tangents tp, tq, also any line cutting the circle 
in v, and pq in r ; draw cs perpendicular to tv. Then 
sr . st = sv 2 . 

26. If a series of circles, tangent at a common point, are cut 

by a fixed circle, the lines of section meet in a point. 

27. In Ex. 26, the intersections of the pairs of tangents to the 

fixed circle, at the points of section, lie in a straight 
line. 

28. If three given circles are cut by any circle, the lines of 

section form a triangle, the loci of whose angular points 
are right lines perpendicular to the lines joining the 
centers of the given circles. 

29. The three loci of Ex. 28 meet in a point. 

30. Given the base of an isosceles triangle, to find the locus of 

the vertex. 

31. Find the locus of the center of a circle which passes through 

two given points. 

32. Find the locus of the center of a sphere of given radius, 

tangent to a given sphere. 

33. The locus of the point from which two circles subtend 

equal angles is a circle, or a right line. 

34. Given the base of a triangle, and m times the square of 

one side plus n times the square of the other, to find the 
locus of the vertex. 

35. Given the base and the sum of the squares of the sides of 

a triangle, to find the locus of the vertex. 

36. In Ex. 35, given the difference of the squares, to find the 

locus. 



234 QUATEBNIONS. 

37. ob and oa are any two lines, and mp is a line parallel to 

ob. Find the locus of the intersection of oq and bq 
drawn parallel to ap and op, respectively. 

38. From a fixed point p, on the surface of a sphere, chords 

pp', pp", .... are drawn. Find the locus of a point o on 
these chords, such that pp'. po = m 2 . 

39. A line of constant length moves with its extremities cm two 

straight lines at right angles to each other. Find the 
locus of its middle point. 

40. Find the locus of a point such that if straight lines be 

drawn to it from the four corners of a square, the sum 
of their squares is constant. 

41. Find the locus of a point the square of whose distance 

from a given point is proportional to its distance from a 
given line. 

42. Find the locus of the^ feet of perpendiculars from the origin 

on planes cutting off pyramids of equal volume from 
three rectangular co-ordinate axes. 

43. Given the base of a triangle and the ratio of the sides, to 

find the locus of. the vertex. 

44. Show that YapYpp = (Vafi) 2 is the equation of a* hyperbola 

whose asymptotes are parallel to a and'/?* '•.■'« 

45. Find the point on an ellipse the tangent to which cuts off 

equal distances on the axes. 

46. a and b are two similar, similarly situated, and concentric 

ellipses ; c is a third ellipse similar to a and b, its center 
being on the circumference of b, and its axes parallel to 
those of a and b : show that the chord of intersection of 
a and b is parallel to the tangent to b at the center of c. 



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