CHAPTER 1 VECTOR SCALARS AND   t After hours the displacement of P from its starting point is t , hence     4 ˆ 8  4      P t t t ˆ Similarly 4 iˆ 3  2t 5t     q ˆj Hence     5  4t PQ q p ˆj 2 2 Hence the distance between the boats is given by PQ  2 40t 25  km To find the least distance, we consider 20t 40t 25 20(t 1)2 5     PQ 2 2 2 5 positive. Hence least distance PQ  . Example 1.15 Given that A,B,C are the points (1,1,1), (5,0,0) and (3,2,1) respectively, find the (x,y,z) P equation which must be satisfied by the coordinates of any point in the ABC. plane P ABC,   As lies in the plane we may write AP m AB n AC . Then, since ; 1 AB  AC  and AP  y 0 z   1 y 0 z Thus  1  4m  x  
CHAPTER 1 VECTOR AND SCALARS After t hours the displacement of P from its starting point is t   , hence P     t   4  4 t ˆ  8  t  ˆ Similarly q  4  2t iˆ  3  5t  ˆj Hence PQ  q p   5  4t  ˆj 2 2 Hence the distance between the boats is given by PQ  2 40t  25 km To find the least distance, we consider PQ 2  20t 2 40t  25  20(t 1)2  5 2 positive. Hence least distance PQ  5 . Example 1.15 Given that A,B,C are the points (1,1,1), (5,0,0) and (3,2,1) respectively, find the equation which must be satisfied by the coordinates (x,y,z) of any point P in the plane ABC. As P lies in the plane ABC, we may write AP  m AB  n AC . Then, since AB  ; AC  1 and AP  y 0 z y   1 z 0 Thus x  1  4m   