Question

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

Answer 1

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)