题目
题型:不详难度:来源:
![](http://img.shitiku.com.cn/uploads/allimg/20191021/20191021030916-22470.png)
![](http://img.shitiku.com.cn/uploads/allimg/20191021/20191021030916-57405.png)
(2) 求点E到平面O1BC的距离.
答案
![](http://img.shitiku.com.cn/uploads/allimg/20191021/20191021030916-32276.png)
解析
![](http://img.shitiku.com.cn/uploads/allimg/20191021/20191021030917-85506.png)
(1) 过O作OF⊥BC于F,连接O1F,
∵OO1⊥面AC,∴BC⊥O1F,
∴∠O1FO是二面角O1-BC-D的平面角,········ 3分
∵OB = 2,∠OBF = 60°,∴OF =
![](http://img.shitiku.com.cn/uploads/allimg/20191021/20191021030917-49989.png)
在Rt△O1OF中,tan∠O1FO =
![](http://img.shitiku.com.cn/uploads/allimg/20191021/20191021030917-54071.png)
∴∠O1FO="60°" 即二面角O1—BC—D的大小为60°············· 6分
(2) 在△O1AC中,OE是△O1AC的中位线,∴OE∥O1C
∴OE∥O1BC,∵BC⊥面O1OF,∴面O1BC⊥面O1OF,交线O1F.
过O作OH⊥O1F于H,则OH是点O到面O1BC的距离,··········· 10分
∴OH =
![](http://img.shitiku.com.cn/uploads/allimg/20191021/20191021030916-32276.png)
![](http://img.shitiku.com.cn/uploads/allimg/20191021/20191021030916-32276.png)
解法二:
(1) ∵OO1⊥平面AC,
∴OO1⊥OA,OO1⊥OB,又OA⊥OB,········· 2分
建立如图所示的空间直角坐标系(如图)
∵底面ABCD是边长为4,∠DAB = 60°的菱形,
∴OA = 2
![](http://img.shitiku.com.cn/uploads/allimg/20191021/20191021030917-49989.png)
则A(2
![](http://img.shitiku.com.cn/uploads/allimg/20191021/20191021030917-20392.png)
![](http://img.shitiku.com.cn/uploads/allimg/20191021/20191021030917-49989.png)
设平面O1BC的法向量为
![](http://img.shitiku.com.cn/uploads/allimg/20191021/20191021030917-49303.png)
![](http://img.shitiku.com.cn/uploads/allimg/20191021/20191021030917-49303.png)
![](http://img.shitiku.com.cn/uploads/allimg/20191021/20191021030918-86734.png)
![](http://img.shitiku.com.cn/uploads/allimg/20191021/20191021030917-49303.png)
![](http://img.shitiku.com.cn/uploads/allimg/20191021/20191021030918-93218.png)
∴
![](http://img.shitiku.com.cn/uploads/allimg/20191021/20191021030918-39485.png)
![](http://img.shitiku.com.cn/uploads/allimg/20191021/20191021030917-49989.png)
∴
![](http://img.shitiku.com.cn/uploads/allimg/20191021/20191021030917-49303.png)
![](http://img.shitiku.com.cn/uploads/allimg/20191021/20191021030917-49989.png)
![](http://img.shitiku.com.cn/uploads/allimg/20191021/20191021030918-24834.png)
∴ cos<
![](http://img.shitiku.com.cn/uploads/allimg/20191021/20191021030917-49303.png)
![](http://img.shitiku.com.cn/uploads/allimg/20191021/20191021030918-24834.png)
![](http://img.shitiku.com.cn/uploads/allimg/20191021/20191021030919-30394.png)
设O1-BC-D的平面角为α, ∴cosα=
![](http://img.shitiku.com.cn/uploads/allimg/20191021/20191021030919-17195.png)
故二面角O1-BC-D为60°.······················ 6分
(2) 设点E到平面O1BC的距离为d,
∵E是O1A的中点,∴
![](http://img.shitiku.com.cn/uploads/allimg/20191021/20191021030919-93248.png)
![](http://img.shitiku.com.cn/uploads/allimg/20191021/20191021030917-49989.png)
![](http://img.shitiku.com.cn/uploads/allimg/20191021/20191021030919-11832.png)
则d=
![](http://img.shitiku.com.cn/uploads/allimg/20191021/20191021030920-57673.png)
∴点E到面O1BC的距离等于
![](http://img.shitiku.com.cn/uploads/allimg/20191021/20191021030919-11832.png)
核心考点
试题【 如图,直四棱柱ABCD—A1B1C1D1的高为3,底面是边长为4且∠DAB = 60°的菱形,ACBD = O,A1C1B1D1 = O1,E是O1A的中点.】;主要考察你对空间几何体的结构特征等知识点的理解。[详细]
举一反三
A.3 | B.4 | C.6 | D.8 |
![](http://img.shitiku.com.cn/uploads/allimg/20191021/20191021030905-25815.gif)
![](http://img.shitiku.com.cn/uploads/allimg/20191021/20191021030906-86719.gif)
![](http://img.shitiku.com.cn/uploads/allimg/20191021/20191021030906-35823.gif)
![](http://img.shitiku.com.cn/uploads/allimg/20191021/20191021030906-97460.gif)
则
![](http://img.shitiku.com.cn/uploads/allimg/20191021/20191021030907-34878.gif)
![](http://img.shitiku.com.cn/uploads/allimg/20191021/20191021030907-79100.gif)
如图,在几何体ABCDE中,DA⊥平面EAB,CB∥DA,EA⊥AB,M是EC的中点,EA=DA=AB=2CB.
(1)求证:DM⊥EB; (2)求异面直线AB与CE所成角的余弦值.
![]() |
19. (本小题满分13分)
如右图所示,已知正方形
![](http://img.shitiku.com.cn/uploads/allimg/20191021/20191021030850-78627.gif)
![](http://img.shitiku.com.cn/uploads/allimg/20191021/20191021030850-52337.gif)
![](http://img.shitiku.com.cn/uploads/allimg/20191021/20191021030850-70484.gif)
![](http://img.shitiku.com.cn/uploads/allimg/20191021/20191021030851-40289.gif)
(1)求证:
![](http://img.shitiku.com.cn/uploads/allimg/20191021/20191021030851-27766.gif)
![](http://img.shitiku.com.cn/uploads/allimg/20191021/20191021030851-27129.gif)
(2)求证:
![](http://img.shitiku.com.cn/uploads/allimg/20191021/20191021030851-72568.gif)
![](http://img.shitiku.com.cn/uploads/allimg/20191021/20191021030851-37758.gif)
(3)求二面角
![](http://img.shitiku.com.cn/uploads/allimg/20191021/20191021030852-82953.gif)
![]() |
四棱锥P-ABCD中,底面ABCD是正方形,
边长为
![](http://img.shitiku.com.cn/uploads/allimg/20191021/20191021030844-28304.gif)
![](http://img.shitiku.com.cn/uploads/allimg/20191021/20191021030844-28304.gif)
(1)求证: AC⊥PB ;
(2)求二面角A-PB-D的大小;
(3)求四棱锥外接球的半径.
(4)在这个四棱锥中放入一个球,求球的最大半径;
![](http://img.shitiku.com.cn/uploads/allimg/20191021/20191021030844-64798.gif)
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