Question

已知O为坐标原点,向量OA=(sinα,1),向量OB=(cosα,9),OC=(-sinα,2),点P满足向量AB=向量BP

1.记函数f(α）=向量PB*向量CA,α∈（-π/8,π/2)讨论函数的单调性并求其值域

2，若OPC三点共线 ，求|向量OA+向量OB|的值

Answer 1

AB
=OB - OA
=( cosα-sinα,8)
f(α）
=PB.CA
= -AB.CA
= -( cosα-sinα,8) . ( -2sinα,1)
=2sinαcosα-2(sinα)^2 + 8
= sin2α +(1-2(sinα)^2) + 7
= sin2α + cos2α + 7
= √2(sin(2α+π/4) + 7
值域 = [7-√2,7+√2]
增加 (-π/8,π/8]
减小[π/8, π/2)
(2)
let P be (x,y)
OPC三点共线
-2/sinα = y/x
-2x = ysinα
x = -ysinα/2

AB= BP
(cosα-sinα, 8) = (-ysinα/2-cosα, y-9)
=> y-9 = 8 and cosα-sinα= -ysinα/2-cosα
= y = 17 and cosα-sinα= -17sinα/2-cosα

cosα-sinα= -17sinα/2-cosα
2cosα = -15sinα/2
cosα = -15sinα/2
tanα = -2/15

|OA+OB|^2
=|(sinα+cosα,10)|^2
= (sinα+cosα)^2 +100
= 2sinαcosα + 101
= -2(2/√229)(15/√229) +101
= 23069/229
|OA+OB| = √(23069/229)