A = (COS α, sin α), B = (COS β, sin β), 0 < β < α < π if | A-B | = √ 2 prove a ⊥ B

A = (COS α, sin α), B = (COS β, sin β), 0 < β < α < π if | A-B | = √ 2 prove a ⊥ B

prove:
|a-b|=√2
That is (a-b) ² = 2
That is a & # 178; + B & # 178; - 2A. B = 2
∵ a²=cos²α+sin²α=1
b²=cos²β+sin²β=1
∴ 1+1-2a.b=2
B = 0
∴ a⊥b