e1+e2与e3共线
=> e1+e2=k1e3 (1)
e2+e3与e1共线
=>e2+e3=k2e1 (2)
(1)+(2)
e1-e3= k1e3-k2e1
(1+k2)e1= (1+k1)e3
=> 1+k2=0 and 1+k1=0
k2=-1 and k1=-1
from (1)
e1+e2=-e3
e1+e2+e3
=-e3+e3
=0其实我在你发答案之前就自己解出来了 不过还是很谢谢你 谢谢！ 以后有问题在咨询你