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幼苗
共回答了18个问题采纳率：88.9% 举报
|a|=1,|b|=2,a,b夹角=π/3
let a+b与a-b夹角 =x
|a+b|^2 = |a|^2+|b|^2 +2a.b = 1+4+2= 7
|a+b|=√7
|a-b|^2 = |a|^2+|b|^2 -2a.b = 1+4-2= 3
|a-b|=√3
(a+b).(a-b) = |a+b||a-b|cosx
|a|^2-|b|^2 = |a+b||a-b|cosx
1-4 = √7.√3 cosx
cosx = -√21/7
1年前
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