已知正方体ABCD-A’B’C’D,求证：AC’⊥B’ CAC’⊥平面CB’D’_数学解答

已知正方体ABCD-A’B’C’D,求证：
AC’⊥B’ C
AC’⊥平面CB’D’

人气：227 ℃　时间：2019-11-20 04:57:19

解答

AB⊥面BCC'B'==>AB⊥B'C;
正方形对角线垂直,BC'⊥B'C;
故面ABC'⊥B'C;
从而 AC'⊥B'C.
AA'⊥面A'B'C'D'==>AA'⊥B'D';
正方形对角线垂直,A'C'⊥B'D';
故面AA'C'⊥B'D';
从而 AC'⊥B'D'.
于是 AC'⊥面CB'D'.