Question

高一数学啊！求详细过程！

Answer 1

设夹角为θ，则
cosθ=(OA*OB)/(|OA|*|OB|)
=(λ*(cosa*cos(π/2+b)+sina*sin(π/2+b))/(|λ|)
=λcos(π/2+b-a)/λ
=cos(π/2+b-a) 因为 b=a-π/6 所以π/2+b-a=π/3
=cosπ/3
=1/2
所以 夹角为θ=π/3

向量AB=向量OB-OA向量=(cos(π/2+b)-λcosa，sin(π/2+b)-λsina)
|AB|^2
=1+λ^2-2λ(cos(π/2+b)cosa+sin(π/2+b)sina)
=1+λ^2-2λcos(π/2+b-a)
4|OB|^2=4

所以 λ^2-2λcos(π/2+b-a)>=3
因为 λ^2-2λcos(π/2+b-a)>= λ^2-2λ
所以 λ^2-2λ>=3
λ^2-2λ-3>=0
λ<=-1或λ>=3