Question

已知向量→op=(2,1),→oa=(1,7),→ob=(5,1),设x是直线OP上的一点（O为坐标原点）,（1）求使→

Answer 1

(1)

OP=(2,1),OA=(1,7),OB=(5,1),

let M be(x,y)

M is on OP

=> x/y = 2/1

x=2y

MA.MB

=(OA-OM).(OB-OM)

=(1-x,7-y).(5-x,1-y)

=(1-x)(5-x)+(7-y)(1-y)

=(1-2y)(5-2y)+(7-y)(1-y)

=12-20y+5y^2

=5(y-2)^2-8

min MA.MB = -8

(2)

y= 2,x = 4

OM = (4,2),MA=(-3,5),MB=(1,-1)

MA.MB=|MA||MB|cos∠AMB

-8=√34√2 cos∠AMB

cos∠AMB = -4√17/17