Question

已知单位向量a,b,c满足a+b+c=0,求a与b的夹角?

Answer 1

单位向量a,b,c

则|a|=|b|=|c|=1

设向量a与b的夹角是θ

a+b=-c

(a+b)^2=c^2

a^2+b^2+2ab=c^2

|a|^2+|b|^2+2ab=|c|^2

1+1+2cosθ=1

cosθ=-1/2

θ=2π/3=120°