Question

已知向量a=(cosα,sinα),向量b=(cosβ,sinβ),其中0＜α＜β＜π 若ka+b与a-kb的长度相等,

Answer 1

|ka+b|^2

=k^2|a|^2+|b|^2+2ka.b (1)

|a-kb|^2

=|a|^2+k^2|b|^2-2ka.b (2)

(1)=(2)

(k^2-1)|a|^2+(1-k^2)|b|^2 + 4ka.b=0

(k^2-1)+(1-k^2)+4k(cosαcosβ +sinαsinβ)=0

4kcos(β-α) =0

β-α = π/2