a⊥b =>
a • b = (sinθ) * 1 + 1 * sinθ =0
=>
sinθ = cosθ
又 -π/2
a+b= ( sin(π/4) +1 ,1+cos(π/4))
|a+b|= √{[sin(π/4)+1]^2 + [1+cos(π/4)]^2}
= 1+√2