Question

已知向量a,b的夹角为β,|a+b|=2根号3,|a-b|=2,则β的取值范围是

Answer 1

|a+b|^2 + |a-b|^2 = 2*(|a|^2 + |b|^2)

|a|^2 + |b|^2=8>=2|a||b|

|a||b|=1/2

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