(a+b)·(a-b)=|a+b|·|a-b|·cosθ
|a+b|=|a-b| = 根号2
(a+b)·(a-b)= (cos+sin,2,sin+cos)·(cos-sin,0,sin-cos)
=cos 2a - cos 2a =0
cos θ=0
θ=π/2
(a+b)·(a-b)=|a+b|·|a-b|·cosθ
|a+b|=|a-b| = 根号2
(a+b)·(a-b)= (cos+sin,2,sin+cos)·(cos-sin,0,sin-cos)
=cos 2a - cos 2a =0
cos θ=0
θ=π/2