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
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![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](/uploads/image/z/5200473-57-3.jpg?t=%E8%AE%BEO%E6%98%AF%E7%9B%B4%E8%A7%92%E5%9D%90%E6%A0%87%E7%B3%BB%E7%9A%84%E5%8E%9F%E7%82%B9%2C%E5%90%91%E9%87%8FOA%3D%EF%BC%882%2C3%EF%BC%89%2C%E5%90%91%E9%87%8FOB%3D%EF%BC%884%2C-1%EF%BC%89%2C%E5%9C%A8x%E8%BD%B4%E4%B8%8A%E6%B1%82%E4%B8%80%E7%82%B9P%2C%E4%BD%BF%E5%90%91%E9%87%8FAP%E4%B8%8E%E5%90%91%E9%87%8FBP%E7%9A%84%E6%95%B0%E9%87%8F%E7%A7%AF%E6%9C%80%E5%B0%8F%2C%E6%B1%82%E6%AD%A4%E6%97%B6%E7%9A%84%E2%88%A0APB)
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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
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
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
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)