题目就是求让(a-b)与(a-tb)夹角为π/4的t值,因为向量a⊥向量b所有ab=0,设它们夹角为f.cosf=2分之根号2=(a-b)*(a-tb)/根号【(a-b)^2】*根号【(a-tb)^2】,a^2=5,b^2=1整理得
(5-t)/根号[6*(5+t^2)],===>t^2+5t-5=0,t=(-5+-3根号5)/2
题目就是求让(a-b)与(a-tb)夹角为π/4的t值,因为向量a⊥向量b所有ab=0,设它们夹角为f.cosf=2分之根号2=(a-b)*(a-tb)/根号【(a-b)^2】*根号【(a-tb)^2】,a^2=5,b^2=1整理得
(5-t)/根号[6*(5+t^2)],===>t^2+5t-5=0,t=(-5+-3根号5)/2