问题 解答题
设平面内的向量
OA
=(-1,-3)
OB
=(5,3)
OM
=(2,2)
,点P在直线OM上,且
PA
PB
=16

(Ⅰ)求
OP
的坐标;
(Ⅱ)求∠APB的余弦值;
(Ⅲ)设t∈R,求|
OA
+t
OP
|
的最小值.
答案

(Ⅰ)设

OP
=(x,y).

由点P在直线OM上,可知

OP
OM
共线.

OM
=(2,2),

所以2x-2y=0,即x=y,有

OP
=(x,x).

PA
=
OA
-
OP
=(-1-x,-3-x),
PB
=
OB
-
OP
=(5-x,3-x)

所以

PA
PB
=(-1-x)(5-x)+(-3-x)(3-x),

PA
PB
=2x2-4x-14.

PA
PB
=16,所以2x2-4x-14=16.

可得x=5或-3.

所以

OP
=(5,5)或(-3,-3).…(4分)

OP
=(5,5)时,

PA
=(-6,-8),
PB
=(0,-2)满足
PA
PB
=16

OP
=(3,3)时,

PA
=(-4,-6),
PB
=(2,0)不满足
PA
PB
=16

所以

OP
=(5,5)

(Ⅱ)由

PA
=(-6,-8),
PB
=(0,-2),

可得|

PA
|=10,|
PB
|=2.

PA
PB
=16.

所以cos∠APB=

PA
PB
|
PA
|•|
PB
|
=
16
10×2
=
4
5
.…(8分)

(Ⅲ)

OA
+t
OP
=(-1+5t,-3+5t),|
OA
+t
OP
|=
50t2-40t+10

t=

2
5
时,|
OA
+t
OP
|
的最小值是
2
.         …(12分)

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