Question

已知非零向量a,b满足|a|=1,且(a-b)×（a+b）=1/2，当ab=1/2,求向量a与b的夹角的值

Answer 1

(a-b)×（a+b）=1/2
a^2-b^2=1/2
|a|^2-|b|^2=1/2
1-|b|^2=1/2
|b|^2=1/2
|b|=√2/2

cos(a,b)=ab/|a||b|
=1/2/(1*√2/2)
=√2/2
所以夹角是45°