Question

在棱长为A的正方体ABCD-A1B1C1D1正方体中，EF分别是棱AB，BC上的动点，且AE=BF，求证A1F⊥C1E

用立体几何中的向量方法来做

Answer 1

设D(0,0,0),A(0,a,0),B(a.a,0),B1(a,a,a)
则A1(0,a,a), F(a,a-x,0) A1F(a,-x,-a)
C1(a,0,a), E(x,a,0) C1E(x-a,a,-a)
a1a2+b1b2+c1c2=a(x-a)-ax+a^2=0
A1F⊥C1E