设O是直角坐标系的原点,向量OA=（2,3）,向量OB=（4,-1）,在x轴上求一点P,使向量AP与向量BP的数量积最
问题描述：
设O是直角坐标系的原点,向量OA=（2,3）,向量OB=（4,-1）,在x轴上求一点P,使向量AP与向量BP的数量积最
小,求此时的∠APB
答
Let P be (x.0) ,as P is on x-axis
AP .BP
= (-OA + OP) .( -OB + OP)
=(x-2,-3) .(x-4,1)
=(x-2)(x-4) - 3
Let S = AP .BP
S' = (x-2) + (x-4) = 0
x = 3
S'' = 2 > 0 (min)
P(3,0 ) #
for x=3
AP = (1,-3),BP=(-1,1)
AP .BP = (x-2)(x-4) - 3
= -4
also
AP .BP = |AP||BP|cos∠APB
-4 = 2 √5 cos∠APB
cos∠APB = -2/√5
∠APB = arc cos ( -2√5/5)