Question

已知正方体ABCD-A1B1C1D1中,求证：AC1⊥面A1BD

Answer 1

建立直角坐标系,设变长为1,

A1(0,0,1)D(0,1,0)B(1,0,0,)A(0,0,0)C1(1,1,1),

AC1(1,1,1)A1D(0,1,-1)DB(-1,1,0)

AC1*A1D=1+（-1）=0

AC*DB=(-1)+1=0

AC1交DB=D

所以AC1⊥A1DB