证明:(1)∵A1A=A1C,且O为AC的中点,
∴A1O⊥AC.
又侧面AA1C1C⊥底面ABC,其交线为AC,且A1O∈平面AA1C1C,
所以A1O⊥底面ABC.…..(2分)
以O为坐标原点,OB,OC,OA1所在直线分别为x,y,z轴建立空间直角坐标系.
由已知可得:O(0,0,0),A(0,-1,0),A1(0,0,
),C(0,1,0),C1(0,2,
3
),B(1,0,0),E(
3
,1,1 2
).则有:
3
2
=(0,1,?
A1C
),
3
=(0,1,AA1
),
3