题目
题型:不详难度:来源:
![](http://img.shitiku.com.cn/uploads/allimg/20191106/20191106004715-48331.jpg)
答案
解析
又CF⊂平面AC1,∴B1D⊥CF,
故若CF⊥平面B1DF,则必有CF⊥DF.
设AF=x(0<x<3a),则CF2=x2+4a2,
DF2=a2+(3a-x)2,又CD2=a2+9a2=10a2,
∴10a2=x2+4a2+a2+(3a-x)2,
解得x=a或2a.
法二:分别以BA、BC、BB1所在直线为x轴、y轴、z轴建立空间直角坐标系B-xyz,
则B(0,0,0),B1(0,0,3a),设F(
![](http://img.shitiku.com.cn/uploads/allimg/20191106/20191106004715-87059.png)
![](http://img.shitiku.com.cn/uploads/allimg/20191106/20191106004716-90484.png)
![](http://img.shitiku.com.cn/uploads/allimg/20191106/20191106004715-87059.png)
![](http://img.shitiku.com.cn/uploads/allimg/20191106/20191106004716-46514.png)
![](http://img.shitiku.com.cn/uploads/allimg/20191106/20191106004715-87059.png)
![](http://img.shitiku.com.cn/uploads/allimg/20191106/20191106004715-87059.png)
![](http://img.shitiku.com.cn/uploads/allimg/20191106/20191106004716-23402.png)
![](http://img.shitiku.com.cn/uploads/allimg/20191106/20191106004716-41780.png)
![](http://img.shitiku.com.cn/uploads/allimg/20191106/20191106004717-86102.png)
![](http://img.shitiku.com.cn/uploads/allimg/20191106/20191106004715-87059.png)
∵CF⊥面B1DF,∴CF⊥B1F,
![](http://img.shitiku.com.cn/uploads/allimg/20191106/20191106004716-46514.png)
![](http://img.shitiku.com.cn/uploads/allimg/20191106/20191106004716-23402.png)
![](http://img.shitiku.com.cn/uploads/allimg/20191106/20191106004716-46514.png)
![](http://img.shitiku.com.cn/uploads/allimg/20191106/20191106004716-23402.png)
![](http://img.shitiku.com.cn/uploads/allimg/20191106/20191106004716-46514.png)
![](http://img.shitiku.com.cn/uploads/allimg/20191106/20191106004717-86102.png)
可得2a2+m(m-3a)=0,解得m=a或2a.
核心考点
试题【如图所示,在直三棱柱ABC-A1B1C1中,底面是∠ABC为直角的等腰直角三角形,AC=2a,BB1=3a,D是A1C1的中点,点F在线段AA1上,当AF=__】;主要考察你对平面的法向量等知识点的理解。[详细]
举一反三
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A.![]() | B.![]() | C.![]() | D.![]() |
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A.![]() | B.![]() | C.![]() | D.![]() |
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