题目内容
已知向量a=(cosλθ,cos(10-λ)θ),b=(sin(10-λ)θ,sinλθ),λ、θ∈R.
(1) 求|a|2+|b|2的值;
(2) 若a⊥b,求θ;
(3) 若θ=
,求证:a∥b.
(1) 解:∵ |a|=![]()
|b|=
,
∴ |a|2+|b|2=2.
(2) 解:∵ a⊥b,
∴ cosλθ·sin(10-λ)θ+cos(10-λ)θ·sinλθ=0,
∴ sin[(10-λ)θ+λθ]=0,∴ sin10θ=0,
∴ 10θ=kπ,k∈Z,∴ θ=
,k∈Z.
(3) 证明:∵ θ=
,
cosλθ·sinλθ-cos(10-λ)θ·sin[(10-λ)θ]
=cos
·sin
-cos
·sin![]()
=cos
·sin
-sin
·cos
=0,∴ a∥b.
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