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见解析
【证明】如图,以C
1 点为原点,C
1 A
1 ,C
1 B
1 ,C
1 C所在直线分别为x轴、y轴、z轴建立空间直角坐标系.
设AC=BC=BB
1 =2,
则A(2,0,2),B(0,2,2),C(0,0,2),A
1 (2,0,0),B
1 (0,2,0),
C
1 (0,0,0),D(1,1,2).
(1)由于
![](https://img.yulucn.com/upload/b/44/b44d491670ee20c5f0f6ea977986ad58_thumb.jpg)
=(0,-2,-2),
![](https://img.yulucn.com/upload/0/75/0751f3f18f9edebd80b54839e1d44224_thumb.jpg)
=(-2,2,-2),
所以
![](https://img.yulucn.com/upload/b/44/b44d491670ee20c5f0f6ea977986ad58_thumb.jpg)
·
![](https://img.yulucn.com/upload/0/75/0751f3f18f9edebd80b54839e1d44224_thumb.jpg)
=0-4+4=0,
因此
![](https://img.yulucn.com/upload/b/44/b44d491670ee20c5f0f6ea977986ad58_thumb.jpg)
⊥
![](https://img.yulucn.com/upload/0/75/0751f3f18f9edebd80b54839e1d44224_thumb.jpg)
,
故BC
1 ⊥AB
1 .
(2)取A
1 C的中点E,连接DE,由于E(1,0,1),
所以
![](https://img.yulucn.com/upload/7/40/740ab8c28bdb680471b0c142ce3a05d5_thumb.jpg)
=(0,1,1).
又
![](https://img.yulucn.com/upload/b/44/b44d491670ee20c5f0f6ea977986ad58_thumb.jpg)
=(0,-2,-2),
所以
![](https://img.yulucn.com/upload/7/40/740ab8c28bdb680471b0c142ce3a05d5_thumb.jpg)
=-
![](https://img.yulucn.com/upload/8/34/834e56be8e99f82cef8b1ae0d852c81f_thumb.jpg)
.
又ED和BC
1 不共线,所以ED∥BC
1 .
又DE⊂平面CA
1 D,BC
1 ⊄平面CA
1 D,
故BC
1 ∥平面CA
1 D.
1年前
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