问题 解答题
设平面内的向量
OA
=(1,7)
OB
=(5,1)
OM
=(2,1)
,点P是直线OM上的一个动点,求当
PA
PB
取最小值时,
OP
的坐标及∠APB的余弦值.
答案

由题意,可设

OP
=(x,y),∵点P在直线OM上,

OP
OM
共线,而
OM
=(2,1)

∴x-2y=0,即x=2y,故

OP
=(2y,y),

PA
=
OA
-
OP
=(1-2y,7-y),
PB
=
OB
-
OP
=(5-2y,1-y),

所以

PA
PB
=(1-2y)(5-2y)+(7-y)(1-y)=5y2-20y+12,

当y=-

-20
2×5
=2时,
PA
PB
=5y2-20y+12取最小值-8,

此时

OP
=(4,2),
PA
=(-3,5),
PB
=(1,-1),

∴cos∠APB=

PA
PB
|
PA
||
PB
|
=
-8
34
2
=-
4
17
17

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