题目
题型:不详难度:来源:
![](http://img.shitiku.com.cn/uploads/allimg/20191105/20191105110643-99822.png)
(Ⅰ)求四面体ABCD的体积;
(Ⅱ)求二面角C﹣AB﹣D的平面角的正切值.
答案
![](http://img.shitiku.com.cn/uploads/allimg/20191105/20191105110643-94909.png)
![](http://img.shitiku.com.cn/uploads/allimg/20191105/20191105110643-81427.png)
解析
试题分析:法一:几何法,
(Ⅰ)过D作DF⊥AC,垂足为F,由平面ABC⊥平面ACD,由面面垂直的性质,可得DF是四面体ABCD的面ABC上的高;设G为边CD的中点,可得AG⊥CD,计算可得AG与DF的长,进而可得S△ABC,由棱锥体积公式,计算可得答案;
(Ⅱ)过F作FE⊥AB,垂足为E,连接DE,分析可得∠DEF为二面角C﹣AB﹣D的平面角,计算可得EF的长,由(Ⅰ)中DF的值,结合正切的定义,可得答案.
法二:向量法,
(Ⅰ)首先建立坐标系,根据题意,设O是AC的中点,过O作OH⊥AC,交AB与H,过O作OM⊥AC,交AD与M;易知OH⊥OM,因此可以以O为原点,以射线OH、OC、OM为x轴、y轴、z轴,建立空间坐标系O﹣XYZ,进而可得B、D的坐标;从而可得△ACD边AC的高即棱住的高与底面的面积,计算可得答案;
(Ⅱ)设非零向量
![](http://img.shitiku.com.cn/uploads/allimg/20191105/20191105110644-91052.png)
![](http://img.shitiku.com.cn/uploads/allimg/20191105/20191105110644-91052.png)
![](http://img.shitiku.com.cn/uploads/allimg/20191105/20191105110644-96245.png)
![](http://img.shitiku.com.cn/uploads/allimg/20191105/20191105110644-91052.png)
![](http://img.shitiku.com.cn/uploads/allimg/20191105/20191105110644-96245.png)
![](http://img.shitiku.com.cn/uploads/allimg/20191105/20191105110644-91052.png)
![](http://img.shitiku.com.cn/uploads/allimg/20191105/20191105110644-49679.png)
解:法一
(Ⅰ)如图:过D作DF⊥AC,垂足为F,由平面ABC⊥平面ACD,
可得DF⊥平面ABC,即DF是四面体ABCD的面ABC上的高;
设G为边CD的中点,由AC=AD,可得AG⊥CD,
则AG=
![](http://img.shitiku.com.cn/uploads/allimg/20191105/20191105110644-34004.png)
![](http://img.shitiku.com.cn/uploads/allimg/20191105/20191105110644-49638.png)
![](http://img.shitiku.com.cn/uploads/allimg/20191105/20191105110644-57556.png)
由S△ADC=
![](http://img.shitiku.com.cn/uploads/allimg/20191105/20191105110645-21701.png)
![](http://img.shitiku.com.cn/uploads/allimg/20191105/20191105110645-21701.png)
![](http://img.shitiku.com.cn/uploads/allimg/20191105/20191105110645-92508.png)
![](http://img.shitiku.com.cn/uploads/allimg/20191105/20191105110645-96748.png)
在Rt△ABC中,AB=
![](http://img.shitiku.com.cn/uploads/allimg/20191105/20191105110645-98088.png)
![](http://img.shitiku.com.cn/uploads/allimg/20191105/20191105110646-54682.png)
S△ABC=
![](http://img.shitiku.com.cn/uploads/allimg/20191105/20191105110646-67346.png)
![](http://img.shitiku.com.cn/uploads/allimg/20191105/20191105110646-11333.png)
故四面体的体积V=
![](http://img.shitiku.com.cn/uploads/allimg/20191105/20191105110646-53947.png)
![](http://img.shitiku.com.cn/uploads/allimg/20191105/20191105110643-94909.png)
(Ⅱ)如图,过F作FE⊥AB,垂足为E,连接DE,
由(Ⅰ)知DF⊥平面ABC,由三垂线定理可得DE⊥AB,故∠DEF为二面角C﹣AB﹣D的平面角,
在Rt△AFD中,AF=
![](http://img.shitiku.com.cn/uploads/allimg/20191105/20191105110647-41601.png)
![](http://img.shitiku.com.cn/uploads/allimg/20191105/20191105110647-58623.png)
![](http://img.shitiku.com.cn/uploads/allimg/20191105/20191105110647-12757.png)
在Rt△ABC中,EF∥BC,从而
![](http://img.shitiku.com.cn/uploads/allimg/20191105/20191105110647-61950.png)
![](http://img.shitiku.com.cn/uploads/allimg/20191105/20191105110647-92032.png)
在Rt△DEF中,tan∠DEF=
![](http://img.shitiku.com.cn/uploads/allimg/20191105/20191105110648-25348.png)
![](http://img.shitiku.com.cn/uploads/allimg/20191105/20191105110643-81427.png)
则二面角C﹣AB﹣D的平面角的正切值为
![](http://img.shitiku.com.cn/uploads/allimg/20191105/20191105110643-81427.png)
解法二:(Ⅰ)如图(2)
设O是AC的中点,过O作OH⊥AB,交AB与H,过O作OM⊥AC,交AD与M;
由平面ABC⊥平面ACD,知OH⊥OM,
因此以O为原点,以射线OH、OC、OM为x轴、y轴、z轴,建立空间坐标系O﹣XYZ,
已知AC=2,故A、C的坐标分别为A(0,﹣1,0),C(0,1,0);
设点B的坐标为(x1,y1,0),由
![](http://img.shitiku.com.cn/uploads/allimg/20191105/20191105110648-88097.png)
![](http://img.shitiku.com.cn/uploads/allimg/20191105/20191105110648-71587.png)
![](http://img.shitiku.com.cn/uploads/allimg/20191105/20191105110648-71587.png)
有
![](http://img.shitiku.com.cn/uploads/allimg/20191105/20191105110648-80554.png)
![](http://img.shitiku.com.cn/uploads/allimg/20191105/20191105110649-14418.png)
解可得
![](http://img.shitiku.com.cn/uploads/allimg/20191105/20191105110649-41460.png)
![](http://img.shitiku.com.cn/uploads/allimg/20191105/20191105110649-79397.png)
即B的坐标为(
![](http://img.shitiku.com.cn/uploads/allimg/20191105/20191105110649-93053.png)
![](http://img.shitiku.com.cn/uploads/allimg/20191105/20191105110650-82915.png)
又舍D的坐标为(0,y2,z2),
由|
![](http://img.shitiku.com.cn/uploads/allimg/20191105/20191105110650-79219.png)
![](http://img.shitiku.com.cn/uploads/allimg/20191105/20191105110650-87962.png)
解可得
![](http://img.shitiku.com.cn/uploads/allimg/20191105/20191105110650-64342.png)
![](http://img.shitiku.com.cn/uploads/allimg/20191105/20191105110651-70272.png)
则D的坐标为(0,
![](http://img.shitiku.com.cn/uploads/allimg/20191105/20191105110651-84208.png)
![](http://img.shitiku.com.cn/uploads/allimg/20191105/20191105110651-14022.png)
从而可得△ACD边AC的高为h=|z2|=
![](http://img.shitiku.com.cn/uploads/allimg/20191105/20191105110651-94277.png)
又|
![](http://img.shitiku.com.cn/uploads/allimg/20191105/20191105110652-33846.png)
![](http://img.shitiku.com.cn/uploads/allimg/20191105/20191105110652-61578.png)
![](http://img.shitiku.com.cn/uploads/allimg/20191105/20191105110652-15978.png)
故四面体的体积V=
![](http://img.shitiku.com.cn/uploads/allimg/20191105/20191105110652-26354.png)
![](http://img.shitiku.com.cn/uploads/allimg/20191105/20191105110653-58327.png)
![](http://img.shitiku.com.cn/uploads/allimg/20191105/20191105110652-33846.png)
![](http://img.shitiku.com.cn/uploads/allimg/20191105/20191105110652-15978.png)
![](http://img.shitiku.com.cn/uploads/allimg/20191105/20191105110653-14066.png)
(Ⅱ)由(Ⅰ)知
![](http://img.shitiku.com.cn/uploads/allimg/20191105/20191105110653-71037.png)
![](http://img.shitiku.com.cn/uploads/allimg/20191105/20191105110653-25673.png)
![](http://img.shitiku.com.cn/uploads/allimg/20191105/20191105110654-15400.png)
![](http://img.shitiku.com.cn/uploads/allimg/20191105/20191105110654-24984.png)
![](http://img.shitiku.com.cn/uploads/allimg/20191105/20191105110654-57274.png)
![](http://img.shitiku.com.cn/uploads/allimg/20191105/20191105110654-64023.png)
设非零向量
![](http://img.shitiku.com.cn/uploads/allimg/20191105/20191105110655-27741.png)
![](http://img.shitiku.com.cn/uploads/allimg/20191105/20191105110655-19225.png)
![](http://img.shitiku.com.cn/uploads/allimg/20191105/20191105110655-56056.png)
![](http://img.shitiku.com.cn/uploads/allimg/20191105/20191105110655-43090.png)
![](http://img.shitiku.com.cn/uploads/allimg/20191105/20191105110656-97322.png)
由
![](http://img.shitiku.com.cn/uploads/allimg/20191105/20191105110655-19225.png)
![](http://img.shitiku.com.cn/uploads/allimg/20191105/20191105110656-27643.png)
![](http://img.shitiku.com.cn/uploads/allimg/20191105/20191105110656-23645.png)
![](http://img.shitiku.com.cn/uploads/allimg/20191105/20191105110656-11741.png)
取m=﹣1,由(1)(2)可得,l=
![](http://img.shitiku.com.cn/uploads/allimg/20191105/20191105110657-59103.png)
![](http://img.shitiku.com.cn/uploads/allimg/20191105/20191105110657-93042.png)
![](http://img.shitiku.com.cn/uploads/allimg/20191105/20191105110657-94794.png)
![](http://img.shitiku.com.cn/uploads/allimg/20191105/20191105110657-59103.png)
![](http://img.shitiku.com.cn/uploads/allimg/20191105/20191105110657-93042.png)
显然
![](http://img.shitiku.com.cn/uploads/allimg/20191105/20191105110657-42019.png)
从而cos<
![](http://img.shitiku.com.cn/uploads/allimg/20191105/20191105110657-94794.png)
![](http://img.shitiku.com.cn/uploads/allimg/20191105/20191105110657-42019.png)
![](http://img.shitiku.com.cn/uploads/allimg/20191105/20191105110657-72262.png)
故tan<
![](http://img.shitiku.com.cn/uploads/allimg/20191105/20191105110658-82757.png)
![](http://img.shitiku.com.cn/uploads/allimg/20191105/20191105110658-96362.png)
![](http://img.shitiku.com.cn/uploads/allimg/20191105/20191105110658-42534.png)
则二面角C﹣AB﹣D的平面角的正切值为
![](http://img.shitiku.com.cn/uploads/allimg/20191105/20191105110658-42534.png)
![](http://img.shitiku.com.cn/uploads/allimg/20191105/20191105110659-66965.png)
点评:本题是立体几何综合题目,此类题目一般有两种思路即几何法与向量法,注意把握两种思路的特点,进行选择性的运用.
核心考点
试题【(12分)(2011•重庆)如图,在四面体ABCD中,平面ABC⊥平面ACD,AB⊥BC,AC=AD=2,BC=CD=1(Ⅰ)求四面体ABCD的体积;(Ⅱ)求二】;主要考察你对空间向量的基本概念等知识点的理解。[详细]
举一反三
(1)求证:DE∥平面FGH;
(2)若点P在直线GF上,
![](http://img.shitiku.com.cn/uploads/allimg/20191105/20191105110630-18419.png)
![](http://img.shitiku.com.cn/uploads/allimg/20191105/20191105110630-24168.png)
![](http://img.shitiku.com.cn/uploads/allimg/20191105/20191105110631-76276.png)
![](http://img.shitiku.com.cn/uploads/allimg/20191105/20191105110631-55539.png)
![](http://img.shitiku.com.cn/uploads/allimg/20191105/20191105110618-90338.png)
![](http://img.shitiku.com.cn/uploads/allimg/20191105/20191105110618-72114.png)
![](http://img.shitiku.com.cn/uploads/allimg/20191105/20191105110618-26973.png)
![](http://img.shitiku.com.cn/uploads/allimg/20191105/20191105110618-22830.png)
![](http://img.shitiku.com.cn/uploads/allimg/20191105/20191105110618-47431.png)
![](http://img.shitiku.com.cn/uploads/allimg/20191105/20191105110618-99354.png)
![](http://img.shitiku.com.cn/uploads/allimg/20191105/20191105110618-16880.png)
![](http://img.shitiku.com.cn/uploads/allimg/20191105/20191105110619-46990.png)
![](http://img.shitiku.com.cn/uploads/allimg/20191105/20191105110619-66888.png)
![](http://img.shitiku.com.cn/uploads/allimg/20191105/20191105110619-87150.png)
(1)证明:
![](http://img.shitiku.com.cn/uploads/allimg/20191105/20191105110619-65646.png)
![](http://img.shitiku.com.cn/uploads/allimg/20191105/20191105110619-69201.png)
(2)若
![](http://img.shitiku.com.cn/uploads/allimg/20191105/20191105110619-23875.png)
![](http://img.shitiku.com.cn/uploads/allimg/20191105/20191105110619-51217.png)
![](http://img.shitiku.com.cn/uploads/allimg/20191105/20191105110619-32365.png)
![](http://img.shitiku.com.cn/uploads/allimg/20191105/20191105110619-76339.jpg)
![](http://img.shitiku.com.cn/uploads/allimg/20191105/20191105110602-64326.png)
![](http://img.shitiku.com.cn/uploads/allimg/20191105/20191105110602-64326.png)
(1)求证:AB//平面DEG;
(2)求证:BD
![](http://img.shitiku.com.cn/uploads/allimg/20191105/20191105110602-64326.png)
(3)求二面角C—DF—E的正弦值.
![](http://img.shitiku.com.cn/uploads/allimg/20191105/20191105110603-28156.jpg)
![](http://img.shitiku.com.cn/uploads/allimg/20191105/20191105110547-10229.png)
![](http://img.shitiku.com.cn/uploads/allimg/20191105/20191105110547-96274.png)
![](http://img.shitiku.com.cn/uploads/allimg/20191105/20191105110547-93610.png)
![](http://img.shitiku.com.cn/uploads/allimg/20191105/20191105110547-52989.png)
![](http://img.shitiku.com.cn/uploads/allimg/20191105/20191105110547-77907.png)
![](http://img.shitiku.com.cn/uploads/allimg/20191105/20191105110547-17524.png)
![](http://img.shitiku.com.cn/uploads/allimg/20191105/20191105110547-93744.png)
![](http://img.shitiku.com.cn/uploads/allimg/20191105/20191105110547-42778.png)
![](http://img.shitiku.com.cn/uploads/allimg/20191105/20191105110547-54295.png)
![](http://img.shitiku.com.cn/uploads/allimg/20191105/20191105110548-91244.png)
![](http://img.shitiku.com.cn/uploads/allimg/20191105/20191105110548-95328.png)
![](http://img.shitiku.com.cn/uploads/allimg/20191105/20191105110548-36017.png)
![](http://img.shitiku.com.cn/uploads/allimg/20191105/20191105110548-10944.png)
![](http://img.shitiku.com.cn/uploads/allimg/20191105/20191105110547-42778.png)
![](http://img.shitiku.com.cn/uploads/allimg/20191105/20191105110547-54295.png)
![](http://img.shitiku.com.cn/uploads/allimg/20191105/20191105110548-91244.png)
![](http://img.shitiku.com.cn/uploads/allimg/20191105/20191105110548-66003.png)
![](http://img.shitiku.com.cn/uploads/allimg/20191105/20191105110548-10777.png)
![](http://img.shitiku.com.cn/uploads/allimg/20191105/20191105110547-93744.png)
![](http://img.shitiku.com.cn/uploads/allimg/20191105/20191105110548-10944.png)
![](http://img.shitiku.com.cn/uploads/allimg/20191105/20191105110548-97175.png)
![](http://img.shitiku.com.cn/uploads/allimg/20191105/20191105110547-42778.png)
![](http://img.shitiku.com.cn/uploads/allimg/20191105/20191105110548-62049.png)
(1)求证:
![](http://img.shitiku.com.cn/uploads/allimg/20191105/20191105110549-57751.png)
![](http://img.shitiku.com.cn/uploads/allimg/20191105/20191105110549-23477.png)
(2)若点E为四边形BCQP内一动点,且二面角E-AP-Q的余弦值为
![](http://img.shitiku.com.cn/uploads/allimg/20191105/20191105110549-72595.png)
![](http://img.shitiku.com.cn/uploads/allimg/20191105/20191105110549-80788.png)
![](http://img.shitiku.com.cn/uploads/allimg/20191105/20191105110526-59018.png)
![](http://img.shitiku.com.cn/uploads/allimg/20191105/20191105110526-51497.png)
![](http://img.shitiku.com.cn/uploads/allimg/20191105/20191105110526-41364.png)
(1)求二面角
![](http://img.shitiku.com.cn/uploads/allimg/20191105/20191105110526-66714.png)
(2)证明:在AB上存在一个点Q,使得平面
![](http://img.shitiku.com.cn/uploads/allimg/20191105/20191105110527-13342.png)
![](http://img.shitiku.com.cn/uploads/allimg/20191105/20191105110527-91703.png)
并求出
![](http://img.shitiku.com.cn/uploads/allimg/20191105/20191105110527-21078.png)
![](http://img.shitiku.com.cn/uploads/allimg/20191105/20191105110527-44456.png)
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