(1)证明:∵AE⊥A1B1,A1B1∥AB,∴AE⊥AB,
又∵AA1⊥AB,AA1⊥∩AE=A,∴AB⊥面A1ACC1,
又∵AC⊂面A1ACC1,∴AB⊥AC,
以A为原点建立如图所示的空间直角坐标系A﹣xyz,
则有A(0,0,
—We'd better eat up the rest as soon as possible.