|a|=|b|=1,a,(b-a)的夹角为120度(a-c).(b-c)=0To find :|c|a.(b-a) = a.b-|a|^2=|a||b-a|cos120度=>a.b-1=(-1/2)|b-a|(a.b)^2-2a.b+1 = (1/4)(|b|^2+|a|^2-2a.b)2(a.b)^2-3a.b+1=0(2(a.b)-1)(a.b-1)=0a.b =1/2 or 1 ...