pp82229630
幼苗
共回答了18个问题采纳率:94.4% 举报
(1)详见解析;(2)30°;(3)
![](https://img.yulucn.com/upload/1/e0/1e0203e8ff3c7cd6820eab6015460b59_thumb.jpg)
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试题分析:方法一:向量法以A为原点,AB,AD,AO分别x轴,y轴,z轴建立空间直角坐标系,A-xyz (1)利用向量的数量积的坐标运算与垂直的关系,∵
![](https://img.yulucn.com/upload/2/80/280187372546864451d9acb43b07d76b_thumb.jpg)
=(-1,1,0),
![](https://img.yulucn.com/upload/9/0a/90ae6bd96ec637c9be2fbb5a50ae9e01_thumb.jpg)
=(0,0,2),
![](https://img.yulucn.com/upload/c/2d/c2d7c01dfdc7234f10201e00d1964a6e_thumb.jpg)
=(1,1,0)∴
![](https://img.yulucn.com/upload/e/9a/e9ae7cc8b74317e5a8eb02af3b057ff7_thumb.jpg)
=0,
![](https://img.yulucn.com/upload/f/f5/ff5d12c2b1586a5dd630bf880aa57f6f_thumb.jpg)
=-1+1=0∴BD⊥AD,BD⊥AC,又AO∩AC=A故BD⊥平面OAC ;
(2)取平面OAC的法向量
![](https://img.yulucn.com/upload/f/0d/f0d6b9aa4cdc2200ab3a03058fca1c43_thumb.jpg)
=(-1,1,0),又
![](https://img.yulucn.com/upload/a/fb/afb671cffc9d5d474b7c3d9301443565_thumb.jpg)
=(0,1,-1)[ K则:
∴
![](https://img.yulucn.com/upload/a/19/a1977d845bb284bcf5261afbd5b84b81_thumb.jpg)
=60°故:MD与平面OAC所成角为30°;
(3)设平面OBD的法向量为
![](https://img.yulucn.com/upload/8/d3/8d3355e8521fdcd05c7aee2ac9751de4_thumb.jpg)
=(x,y,z),则
取
![](https://img.yulucn.com/upload/8/d3/8d3355e8521fdcd05c7aee2ac9751de4_thumb.jpg)
=(2,2,1)则点A到平面OBD的距离为d=
![](https://img.yulucn.com/upload/a/f0/af02107c0234e4120eaf64e4701bc774_thumb.jpg)
;
方法二:几何法(1)由线面垂直的的判断定理证明,由OA⊥底面ABCD,OA⊥BD,∵底面ABCD是边长为1的正方形∴BD⊥AC ∴BD⊥平面OAC ;(2)先构造线面所成的角,设AC与BD交于点E,连结EM,则∠DME是直线MD与平面OAC折成的角,又由于∵MD=
![](https://img.yulucn.com/upload/4/ee/4ee97d7fc4eb9f8315df404fa4412631_thumb.jpg)
,DE=
![](https://img.yulucn.com/upload/f/5f/f5f4c526f2e5b3671a9b2cd8986cd6db_thumb.jpg)
∴直线MD与平面OAC折成的角为30°;(3)构造点到面的距离,作AH⊥OE于点H,∵BD⊥平面OAC∴BO⊥AH
线段AH的长就是点A到平面OBD的距离,有AH=
![](https://img.yulucn.com/upload/8/d1/8d1042e2c5db194101be3bcaf1cd20a5_thumb.jpg)
可知点A到平面OBD的距离为
![](https://img.yulucn.com/upload/1/e0/1e0203e8ff3c7cd6820eab6015460b59_thumb.jpg)
.
试题解析:方法一:以A为原点,AB,AD,AO分别x轴,y轴,z轴建立空间直角坐标系,A-xyz。
(1)∵
![](https://img.yulucn.com/upload/2/80/280187372546864451d9acb43b07d76b_thumb.jpg)
=(-1,1,0),
![](https://img.yulucn.com/upload/9/0a/90ae6bd96ec637c9be2fbb5a50ae9e01_thumb.jpg)
=(0,0,2),
![](https://img.yulucn.com/upload/c/2d/c2d7c01dfdc7234f10201e00d1964a6e_thumb.jpg)
=(1,1,0)
∴
![](https://img.yulucn.com/upload/e/9a/e9ae7cc8b74317e5a8eb02af3b057ff7_thumb.jpg)
=0,
![](https://img.yulucn.com/upload/f/f5/ff5d12c2b1586a5dd630bf880aa57f6f_thumb.jpg)
=-1+1=0
∴BD⊥AD,BD⊥AC,又AO∩AC=A
故BD⊥平面OAC 4分
(2)取平面OAC的法向量
![](https://img.yulucn.com/upload/f/0d/f0d6b9aa4cdc2200ab3a03058fca1c43_thumb.jpg)
=(-1,1,0),又
![](https://img.yulucn.com/upload/a/fb/afb671cffc9d5d474b7c3d9301443565_thumb.jpg)
=(0,1,-1)
则:
∴
![](https://img.yulucn.com/upload/a/19/a1977d845bb284bcf5261afbd5b84b81_thumb.jpg)
=60°
故:MD与平面OAC所成角为30° 8分
(3)设平面OBD的法向量为
![](https://img.yulucn.com/upload/8/d3/8d3355e8521fdcd05c7aee2ac9751de4_thumb.jpg)
=(x,y,z),则
取
![](https://img.yulucn.com/upload/8/d3/8d3355e8521fdcd05c7aee2ac9751de4_thumb.jpg)
=(2,2,1)
则点A到平面OBD的距离为d=
![](https://img.yulucn.com/upload/a/f0/af02107c0234e4120eaf64e4701bc774_thumb.jpg)
12分
方法二:(1)由OA⊥底面ABCD,OA⊥BD。
∵底面ABCD是边长为1的正方形
∴BD⊥AC ∴BD⊥平面OAC 4分
(2)设AC与BD交于点E,连结EM,则∠DME是直线MD与平面OAC折成的角
∵MD=
![](https://img.yulucn.com/upload/4/ee/4ee97d7fc4eb9f8315df404fa4412631_thumb.jpg)
,DE=
∴直线MD与平面OAC折成的角为30° 8分
(3)作AH⊥OE于点H。
∵BD⊥平面OAC
∴BO⊥AH
线段AH的长就是点A到平面OBD的距离。
∴AH=
∴点A到平面OBD的距离为
![](https://img.yulucn.com/upload/1/e0/1e0203e8ff3c7cd6820eab6015460b59_thumb.jpg)
12分
1年前
7