设平面内的向量
(Ⅰ)求
(Ⅱ)求∠APB的余弦值; (Ⅲ)设t∈R,求|
|
(Ⅰ)设
=(x,y).OP
由点P在直线OM上,可知
与OP
共线.OM
而
=(2,2),OM
所以2x-2y=0,即x=y,有
=(x,x).OP
由
=PA
-OA
=(-1-x,-3-x),OP
=PB
-OB
=(5-x,3-x),OP
所以
•PA
=(-1-x)(5-x)+(-3-x)(3-x),PB
即
•PA
=2x2-4x-14.PB
又
•PA
=16,所以2x2-4x-14=16.PB
可得x=5或-3.
所以
=(5,5)或(-3,-3).…(4分)OP
当
=(5,5)时,OP
=(-6,-8),PA
=(0,-2)满足PB
•PA
=16,PB
当
=(3,3)时,OP
=(-4,-6),PA
=(2,0)不满足PB
•PA
=16,PB
所以
=(5,5)OP
(Ⅱ)由
=(-6,-8),PA
=(0,-2),PB
可得|
|=10,|PA
|=2.PB
又
•PA
=16.PB
所以cos∠APB=
=
•PA PB |
|•|PA
|PB
=16 10×2
.…(8分)4 5
(Ⅲ)
+tOA
=(-1+5t,-3+5t),|OP
+tOA
|=OP
.50t2-40t+10
当t=
时,|2 5
+tOA
|的最小值是OP
. …(12分)2