设两相两x y 满足|x|=2|y|=1 x y的夹角为60度若向量2tx+7y与向量x+ty的夹角为钝角 求t的取值范围
參考答案:sqr表示根号
(2tx+7y)(x+ty)=2t|x|^2+(2t^2+7)|x||y|cos60+7t|y|^2
=2t+(2t^2+7)/4+7t/4=t^2/2+15t/4+7/4
|2tx+7y|^2=4t^2|x|^2+28t|x||y|cos60+49|y|^2
=4t^2+7t+49/4
|x+ty|^2=|x|^2+2t|x||y|cos60+t^2|y|^2=1+t/2+t^2/4
(2tx+7y)(x+ty)/|2tx+7y||x+ty|
=(t^2/2+15t/4+7/4)/sqr(4t^2+7t+49/4)sqr(1+t/2+t^2/4)<0
2t^2+15t+7<0
(t+7)(2t+1)<0
-7<t<-1/2
