已知向量a=(cosα,sinα)b=(cosβ,sinβ)求 a.(a+2b) 若α-β求|a+2b|
α-β=π/3
人气:152 ℃ 时间:2019-11-07 19:43:20
解答
∵a=(cosα,sinα),b=(cosβ,sinβ)
==>a+2b=(cosα+2cosβ,sinα+2sinβ)
∴a*(a+2b)=cosα(cosα+2cosβ)+sinα(sinα+2sinβ)
=cos²α+2cosαcosβ+sin²α+2sinαsinβ
=(cos²α+sin²α)+2(cosαcosβ+sinαsinβ)
=1+2cos(α-β)
∵α-β=π/3
∴|a+2b|=√[(cosα+2cosβ)²+(sinα+2sinβ)²]
=√[(cos²α+4cosαcosβ+4cos²β)+(sin²α+4sinαsinβ+4sin²β)]
=√[(cos²α+sin²α)+4(cos²β+sin²β)+4(cosαcosβ+sinαsinβ)]
=√[1+4+4cos(α-β)]
=√[5+4cos(π/3)]
=√[5+4(1/2)]
=√7
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