Question

已知,IaI=√3,IbI=3,IcI=2√3,且a+b+c=0,则a*b+b*c+c*a=?

Answer 1

(a+b+c)^2=a^2+b^2+c^2+2(a*b+b*c+c*a)=24+2(a*b+b*c+c*a)=0

故a*b+b*c+c*a=-12