已知向量a=(sinθ,1),b=(1,cosθ),-π/2<θ<π/2 (1)若a⊥b,求θ （2）求| a+b|的最大值
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a⊥b =>
a • b = (sinθ) * 1 + 1 * sinθ =0
=>
sinθ = cosθ
又 -π/2<θ<π/2
=>
θ=π/4
=>
a+b= ( sin(π/4) +1 , 1+cos(π/4))
|a+b|= √{[sin(π/4)+1]^2 + [1+cos(π/4)]^2}
= 1+√2
a • b = (sinθ) * 1 + 1 * sinθ =0
=>
sinθ = cosθ
又 -π/2<θ<π/2
=>
θ=π/4
=>
a+b= ( sin(π/4) +1 , 1+cos(π/4))
|a+b|= √{[sin(π/4)+1]^2 + [1+cos(π/4)]^2}
= 1+√2
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