题目
题型:不详难度:来源:
![](http://img.shitiku.com.cn/uploads/allimg/20191022/20191022050721-58405.png)
![](http://img.shitiku.com.cn/uploads/allimg/20191022/20191022050721-92226.png)
![](http://img.shitiku.com.cn/uploads/allimg/20191022/20191022050721-63966.png)
![](http://img.shitiku.com.cn/uploads/allimg/20191022/20191022050722-79534.png)
![](http://img.shitiku.com.cn/uploads/allimg/20191022/20191022050722-64184.png)
(1)求证:
![](http://img.shitiku.com.cn/uploads/allimg/20191022/20191022050722-56937.png)
(2)求证:
![](http://img.shitiku.com.cn/uploads/allimg/20191022/20191022050722-99408.png)
(3)求二面角
![](http://img.shitiku.com.cn/uploads/allimg/20191022/20191022050723-61269.png)
答案
![](http://img.shitiku.com.cn/uploads/allimg/20191022/20191022050723-15233.png)
根据FA=FC,得到
![](http://img.shitiku.com.cn/uploads/allimg/20191022/20191022050723-58028.png)
![](http://img.shitiku.com.cn/uploads/allimg/20191022/20191022050722-56937.png)
(Ⅱ)由四边形
![](http://img.shitiku.com.cn/uploads/allimg/20191022/20191022050723-41655.png)
![](http://img.shitiku.com.cn/uploads/allimg/20191022/20191022050724-87498.png)
得到
![](http://img.shitiku.com.cn/uploads/allimg/20191022/20191022050724-59831.png)
![](http://img.shitiku.com.cn/uploads/allimg/20191022/20191022050724-83063.png)
平面
![](http://img.shitiku.com.cn/uploads/allimg/20191022/20191022050724-69536.png)
![](http://img.shitiku.com.cn/uploads/allimg/20191022/20191022050724-12769.png)
(Ⅲ)二面角A-FC-B的余弦值为
![](http://img.shitiku.com.cn/uploads/allimg/20191022/20191022050725-28900.png)
解析
试题分析:(Ⅰ)证明:设AC与BD相交于点O,连结FO.
因为四边形ABCD为菱形,所以
![](http://img.shitiku.com.cn/uploads/allimg/20191022/20191022050723-15233.png)
又FA=FC,所以
![](http://img.shitiku.com.cn/uploads/allimg/20191022/20191022050723-58028.png)
因为
![](http://img.shitiku.com.cn/uploads/allimg/20191022/20191022050725-44613.png)
所以
![](http://img.shitiku.com.cn/uploads/allimg/20191022/20191022050722-56937.png)
![](http://img.shitiku.com.cn/uploads/allimg/20191022/20191022050725-29452.png)
(Ⅱ)证明:因为四边形
![](http://img.shitiku.com.cn/uploads/allimg/20191022/20191022050723-41655.png)
![](http://img.shitiku.com.cn/uploads/allimg/20191022/20191022050724-87498.png)
所以
![](http://img.shitiku.com.cn/uploads/allimg/20191022/20191022050724-59831.png)
因为
![](http://img.shitiku.com.cn/uploads/allimg/20191022/20191022050725-96142.png)
所以
![](http://img.shitiku.com.cn/uploads/allimg/20191022/20191022050724-83063.png)
又
![](http://img.shitiku.com.cn/uploads/allimg/20191022/20191022050726-56235.png)
所以平面
![](http://img.shitiku.com.cn/uploads/allimg/20191022/20191022050724-69536.png)
又
![](http://img.shitiku.com.cn/uploads/allimg/20191022/20191022050726-22490.png)
所以
![](http://img.shitiku.com.cn/uploads/allimg/20191022/20191022050724-12769.png)
(Ⅲ)解:因为四边形BDEF为菱形,且
![](http://img.shitiku.com.cn/uploads/allimg/20191022/20191022050726-83913.png)
![](http://img.shitiku.com.cn/uploads/allimg/20191022/20191022050726-37698.png)
因为
![](http://img.shitiku.com.cn/uploads/allimg/20191022/20191022050727-14882.png)
![](http://img.shitiku.com.cn/uploads/allimg/20191022/20191022050727-65466.png)
![](http://img.shitiku.com.cn/uploads/allimg/20191022/20191022050727-13255.png)
![](http://img.shitiku.com.cn/uploads/allimg/20191022/20191022050727-31269.png)
![](http://img.shitiku.com.cn/uploads/allimg/20191022/20191022050727-11193.png)
由
![](http://img.shitiku.com.cn/uploads/allimg/20191022/20191022050728-85217.png)
![](http://img.shitiku.com.cn/uploads/allimg/20191022/20191022050728-21865.png)
设AB=2.因为四边形ABCD为菱形,
![](http://img.shitiku.com.cn/uploads/allimg/20191022/20191022050728-59599.png)
![](http://img.shitiku.com.cn/uploads/allimg/20191022/20191022050728-77444.png)
所以
![](http://img.shitiku.com.cn/uploads/allimg/20191022/20191022050729-75831.png)
所以
![](http://img.shitiku.com.cn/uploads/allimg/20191022/20191022050729-39133.png)
设平面BFC的法向量为
![](http://img.shitiku.com.cn/uploads/allimg/20191022/20191022050729-41629.png)
![](http://img.shitiku.com.cn/uploads/allimg/20191022/20191022050729-32768.png)
![](http://img.shitiku.com.cn/uploads/allimg/20191022/20191022050729-38541.png)
取
![](http://img.shitiku.com.cn/uploads/allimg/20191022/20191022050730-92898.png)
![](http://img.shitiku.com.cn/uploads/allimg/20191022/20191022050730-38002.png)
易知平面
![](http://img.shitiku.com.cn/uploads/allimg/20191022/20191022050730-86570.png)
![](http://img.shitiku.com.cn/uploads/allimg/20191022/20191022050730-34554.png)
由二面角A-FC-B是锐角,得
![](http://img.shitiku.com.cn/uploads/allimg/20191022/20191022050731-25500.png)
所以二面角A-FC-B的余弦值为
![](http://img.shitiku.com.cn/uploads/allimg/20191022/20191022050725-28900.png)
![](http://img.shitiku.com.cn/uploads/allimg/20191022/20191022050731-46402.png)
点评:典型题,立体几何题,是高考必考内容,往往涉及垂直关系、平行关系、角、距离的计算。证明过程中,往往需要将立体几何问题转化成平面几何问题加以解答。本题解答,通过建立适当的空间直角坐标系,利用向量的坐标运算,简化了繁琐的证明过程,实现了“以算代证”,对计算能力要求较高。
核心考点
举一反三
![](http://img.shitiku.com.cn/uploads/allimg/20191022/20191022050652-91625.png)
![](http://img.shitiku.com.cn/uploads/allimg/20191022/20191022050652-62387.png)
![](http://img.shitiku.com.cn/uploads/allimg/20191022/20191022050653-79488.png)
![](http://img.shitiku.com.cn/uploads/allimg/20191022/20191022050653-79983.png)
现以
![](http://img.shitiku.com.cn/uploads/allimg/20191022/20191022050653-52672.png)
![](http://img.shitiku.com.cn/uploads/allimg/20191022/20191022050653-74427.png)
![](http://img.shitiku.com.cn/uploads/allimg/20191022/20191022050653-52672.png)
![](http://img.shitiku.com.cn/uploads/allimg/20191022/20191022050653-74427.png)
![](http://img.shitiku.com.cn/uploads/allimg/20191022/20191022050653-74427.png)
![](http://img.shitiku.com.cn/uploads/allimg/20191022/20191022050652-91625.png)
![](http://img.shitiku.com.cn/uploads/allimg/20191022/20191022050653-55988.png)
![](http://img.shitiku.com.cn/uploads/allimg/20191022/20191022050654-50821.png)
(1)求证:
![](http://img.shitiku.com.cn/uploads/allimg/20191022/20191022050654-99846.png)
![](http://img.shitiku.com.cn/uploads/allimg/20191022/20191022050654-77081.png)
(2)求证:
![](http://img.shitiku.com.cn/uploads/allimg/20191022/20191022050654-64074.png)
![](http://img.shitiku.com.cn/uploads/allimg/20191022/20191022050655-38619.png)
(3)求点
![](http://img.shitiku.com.cn/uploads/allimg/20191022/20191022050655-64148.png)
![](http://img.shitiku.com.cn/uploads/allimg/20191022/20191022050654-77081.png)
![](http://img.shitiku.com.cn/uploads/allimg/20191022/20191022050655-43307.png)
![](http://img.shitiku.com.cn/uploads/allimg/20191022/20191022050655-52892.png)
图
![](http://img.shitiku.com.cn/uploads/allimg/20191022/20191022050655-62820.png)
![](http://img.shitiku.com.cn/uploads/allimg/20191022/20191022050656-41049.png)
![](http://img.shitiku.com.cn/uploads/allimg/20191022/20191022050646-38969.png)
![](http://img.shitiku.com.cn/uploads/allimg/20191022/20191022050646-35095.png)
(1)试建立适当的坐标系,并写出点P、B、D的坐标;
(2)问当实数a在什么范围时,BC边上能存在点Q,使得PQ⊥QD?
(3)当BC边上有且仅有一个点Q使得PQ⊥QD时,求二面角Q-PD-A的大小.
(1)求证:平面ACB1⊥平面BB1C1C;
(2)在A1B1上是否存在一点P,使得DP与平面ACB1平行?证明你的结论.
![](http://img.shitiku.com.cn/uploads/allimg/20191022/20191022050643-85512.png)
![](http://img.shitiku.com.cn/uploads/allimg/20191022/20191022050628-87751.png)
![](http://img.shitiku.com.cn/uploads/allimg/20191022/20191022050628-20443.png)
![](http://img.shitiku.com.cn/uploads/allimg/20191022/20191022050628-62767.png)
![](http://img.shitiku.com.cn/uploads/allimg/20191022/20191022050628-20885.png)
![](http://img.shitiku.com.cn/uploads/allimg/20191022/20191022050628-20443.png)
![](http://img.shitiku.com.cn/uploads/allimg/20191022/20191022050628-75808.png)
![](http://img.shitiku.com.cn/uploads/allimg/20191022/20191022050629-75658.png)
![](http://img.shitiku.com.cn/uploads/allimg/20191022/20191022050629-39089.png)
(Ⅰ)求证:
![](http://img.shitiku.com.cn/uploads/allimg/20191022/20191022050629-15646.png)
![](http://img.shitiku.com.cn/uploads/allimg/20191022/20191022050629-94587.png)
(Ⅱ)若
![](http://img.shitiku.com.cn/uploads/allimg/20191022/20191022050629-57681.png)
![](http://img.shitiku.com.cn/uploads/allimg/20191022/20191022050629-89039.png)
![](http://img.shitiku.com.cn/uploads/allimg/20191022/20191022050629-59799.png)
![](http://img.shitiku.com.cn/uploads/allimg/20191022/20191022050630-83083.png)
![](http://img.shitiku.com.cn/uploads/allimg/20191022/20191022050618-34782.png)
![](http://img.shitiku.com.cn/uploads/allimg/20191022/20191022050618-63355.png)
![](http://img.shitiku.com.cn/uploads/allimg/20191022/20191022050618-46053.png)
![](http://img.shitiku.com.cn/uploads/allimg/20191022/20191022050618-69275.png)
![](http://img.shitiku.com.cn/uploads/allimg/20191022/20191022050618-63355.png)
![](http://img.shitiku.com.cn/uploads/allimg/20191022/20191022050618-63282.png)
![](http://img.shitiku.com.cn/uploads/allimg/20191022/20191022050618-63355.png)
![](http://img.shitiku.com.cn/uploads/allimg/20191022/20191022050619-63508.png)
![](http://img.shitiku.com.cn/uploads/allimg/20191022/20191022050619-83483.png)
![](http://img.shitiku.com.cn/uploads/allimg/20191022/20191022050619-33587.png)
(Ⅰ)求证:
![](http://img.shitiku.com.cn/uploads/allimg/20191022/20191022050619-62096.png)
![](http://img.shitiku.com.cn/uploads/allimg/20191022/20191022050619-94863.png)
(Ⅱ)求三棱锥
![](http://img.shitiku.com.cn/uploads/allimg/20191022/20191022050619-23756.png)
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