已知向量a=(sinθ,1)向量b=(1,cosθ),-2/π<θ<2/π 若a模长 =b模长=a-b模长。求a与a+b夹角的大小

Question

如题所述

Answer 1

let a,(a+b)夹角 = x
|a|= |b|
=> (sinθ)^2 + 1 = (cosθ)^2 +1
=> (tanθ)^2 =1
=> tanθ = 1 or -1

|a| = |a-b|
=> (sinθ)^2 + 1 = (sinθ-1)^2 + (1-cosθ)^2
= 3 - 2(sinθ+cosθ)
3 - 2(sinθ+cosθ) = 1/2 +1
(sinθ+cosθ) = 3/4

|a|^2 = (sinθ)^2 + 1
= 1/2 +1
|a| = √6/2

|a+b|^2
= (sinθ+1)^2 + (cosθ+1)^2
= 3 + 2(sinθ+cosθ)
= 3+2(3/4) = 9/2
|a+b| = 3√2/2

a.(a+b) = |a||a+b|cosx
|a|^2 + a.b = |a||a+b|cosx
3/2 + sinθ + cosθ = (√6/2)(3√2/2) cos x
3/2 + 3/4 = (6√3/4) cosx
cosx = √3/2
x = π/6
a,(a+b)夹角 = π/6