问题
解答题
设平面内的向量
|
答案
由题意,可设
=(x,y),∵点P在直线OM上,OP
∴
与OP
共线,而OM
=(2,1),OM
∴x-2y=0,即x=2y,故
=(2y,y),OP
又
=PA
-OA
=(1-2y,7-y),OP
=PB
-OB
=(5-2y,1-y),OP
所以
•PA
=(1-2y)(5-2y)+(7-y)(1-y)=5y2-20y+12,PB
当y=-
=2时,-20 2×5
•PA
=5y2-20y+12取最小值-8,PB
此时
=(4,2),OP
=(-3,5),PA
=(1,-1),PB
∴cos∠APB=
=
•PA PB |
||PA
|PB
=--8 34 2 4 17 17