已知向量a≠e

选C
|a-te|≥|a-e|,
a^2- 2t*a*e+t^2*1≥a^2-2ae+1
t^2-1-2ae(t-1)=(t-1)(t+1-2ae)≥0
根据题意,对任意t属于R,恒有|a-te|≥|a-e|,就要求(t-1)(t+1-2ae)≥0,对任意t属于R恒成立
即t-1=t+1-2ae,ae=1
则 e(a-e)= ea-e^2=1-1=0,即 e垂直(a-e)