CHAPTER 2 THE DOT CROSS PRODUCTS AND  00 ;then sin 00 0 If  A  B  AB   900 ;then sin 900 and if  A  B  AB  AB if and are given in terms of position vectors, then A  A1iˆ  A2  A3kˆ and A B ˆj    B B1iˆ B2 B3kˆ ˆj Then  B1ˆ      A B B2 ˆ B3k       A1B1ˆ A1B2 A1B3 k ˆ ˆ ˆ ˆ       A2 B1 iˆ A2 B2 A2 B3 kˆ ˆj ˆj ˆj ˆj       A3 B1k A3 B2k A3 B3k k ˆ ˆ Now, since , we have . Also in 0 0 0 sin 900 1             iˆ iˆ iˆ iˆ ˆj ˆj kˆ kˆ iˆ ˆj direction OZ, i.e.   iˆ ˆj kˆ iˆ ; kˆ  iˆ  kˆ   ˆj ˆj ˆj iˆ   kˆ ˆj ;   iˆ kˆ ˆj   ; kˆ iˆ  iˆ  ˆj ˆj kˆ b1c2 iˆ a2c1  a1b2 kˆ Thus    A B b2c1 a1c2 ˆj a2b1 b1c2  a1c2  a1b2 k or   b2c1 a2c1 a2b1 ˆ ˆ Which we may recognize as the expansion of the determinant iˆ kˆ ˆj A B B1 B2 B3  A1 A2 A3 Example 2.11 If and then 3iˆ 2 4kˆ 2iˆ 3 2kˆ A   B  ˆj ˆj iˆ kˆ ˆj 3 2 4 16iˆ 14 5kˆ     A B ˆj 2 3 2
CHAPTER 2 THE DOT AND CROSS PRODUCTS If   00 ;then A  B  AB sin 00  0 and if   900 ;then A  B  AB sin 900  AB if A and B are given in terms of position vectors, then A  A1iˆ  A2 ˆj  A3kˆ and B  B1iˆ  B2 ˆj  B3kˆ Then A  B   B1ˆ  B2 ˆ  B3k   A1B1ˆ  ˆ  A1B2 ˆ  ˆ  A1B3 ˆ  k  A2 B1 ˆj  iˆ  A2 B2 ˆj  ˆj  A2 B3 ˆj  kˆ  A3 B1k  ˆ  A3 B2k  ˆ  A3 B3k  k Now, since iˆ  iˆ  0  0 , we have iˆ  iˆ  ˆj  ˆj  kˆ  kˆ  0 . Also iˆ  ˆj  sin 900  1 in direction OZ, i.e. iˆ  ˆj  kˆ iˆ ˆj kˆ iˆ  ˆj  kˆ ˆj  kˆ  iˆ kˆ  iˆ  ˆj ; ˆj  iˆ  ; kˆ  ˆj  ; iˆ  kˆ  ˆj Thus or A B  b1c2 b2c1 iˆ  a2c1 a1c2  ˆj  a1b2 a2b1 kˆ  b1c2 b2c1  ˆ a1c2 a2c1  ˆ  a1b2 a2b1 k Which we may recognize as the expansion of the determinant iˆ ˆj kˆ A B  A1 B1 A2 B2 A3 B3 Example 2.11 If A  3iˆ 2 ˆj  4kˆ and B  2iˆ 3 ˆj 2kˆ then iˆ ˆj kˆ A  B  3 2 4  16iˆ  14 ˆj 5kˆ 2 3 2