Question

已知|a|=3,|b|=4,a与b的夹角为60°,求（1）|a+b|的值：（2）a+b与a-b的夹角的余弦值

Answer 1

a·b=|a||b|cos60=3*4*1/2=6

|a+b|^2=a^2+2a·b+b^2=9+2*6+16=37

所以|a+b|=根号37

(2)

|a-b|^2=a^2-2a·b+b^2=9-12+16=13

(a+b)·(a-b)=a^2-b^2=9-16=-7

设夹角是m,则有cos(m)=(a+b)·(a-b)/[|a+b||a-b|]=-7/(根号37*根号13）=-7/根号481