题目内容
如图,在矩形ABCD中,AB=2,BC=a,又PA⊥平面ABCD,PA=4.
(Ⅰ)若在边BC上存在一点Q,使PQ⊥QD,求a的取值范围;
(Ⅱ)当边BC上存在唯一点Q,使PQ⊥QD时,求二面角A-PD-Q的余弦值.
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/201408231639221761496.gif)
(Ⅰ)若在边BC上存在一点Q,使PQ⊥QD,求a的取值范围;
(Ⅱ)当边BC上存在唯一点Q,使PQ⊥QD时,求二面角A-PD-Q的余弦值.
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/201408231639221761496.gif)
(1)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823163922191420.gif)
(2)二面角A-PD-Q的余弦值为
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823163922223268.gif)
解法1:(Ⅰ)如图,连
,由于PA⊥平面ABCD,则由PQ⊥QD,必有
.
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/201408231639222852375.gif)
设
,则
,
在
中,有
.
在
中,有
. ……4分
在
中,有
.
即
,即
.
∴
.
故
的取值范围为
.……6分
(Ⅱ)由(Ⅰ)知,当
,
时,边BC上存在唯一点Q(Q为BC边的中点),
使PQ⊥QD.
过Q作QM∥CD交AD于M,则QM⊥AD.
∵PA⊥平面ABCD,∴PA⊥QM.∴QM⊥平面PAD.
过M作MN⊥PD于N,连结NQ,则QN⊥PD.
∴∠MNQ是二面角A-PD-Q的平面角. ……10分
在等腰直角三角形
中,可求得
,又
,进而
.
∴
.
故二面角A-PD-Q的余弦值为
. ……12分
解法2:(Ⅰ)以
为x.y.z轴建立如图的空间直角坐标系,则
B(0,2,0),C(a,2,0),D(a,0,0),
P(0,0,4), ……2分
设Q(t,2,0)(
),则
=(t,2,-4),
=(t-a,2,0). ……4分
∵PQ⊥QD,∴
=0.
即
.
∴
.
故
的取值范围为
. ……6分
(Ⅱ)由(Ⅰ)知,当
,
时,边BC上存在唯一点Q,使PQ⊥QD.
此时Q(2,2,0),D(4,0,0).
设
是平面
的法向量,
由
,得
.
取
,则
是平面
的一个法向量.
而
是平面
的一个法向量, ……10分
由
.
∴二面角A-PD-Q的余弦值为
. ……12分
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823163922238252.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823163922254353.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/201408231639222852375.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/201408231639223012263.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823163922316298.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823163922347423.gif)
在
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823163922363480.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823163922379499.gif)
在
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823163922410491.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823163922425627.gif)
在
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823163922457479.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823163922472489.gif)
即
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823163922488608.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823163922503444.gif)
∴
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823163922519479.gif)
故
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823163922535192.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823163922191420.gif)
(Ⅱ)由(Ⅰ)知,当
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823163922581229.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823163922597236.gif)
使PQ⊥QD.
过Q作QM∥CD交AD于M,则QM⊥AD.
∵PA⊥平面ABCD,∴PA⊥QM.∴QM⊥平面PAD.
过M作MN⊥PD于N,连结NQ,则QN⊥PD.
∴∠MNQ是二面角A-PD-Q的平面角. ……10分
在等腰直角三角形
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823163922613264.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823163922628456.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823163922644426.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823163922659448.gif)
∴
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/201408231639226751037.gif)
故二面角A-PD-Q的余弦值为
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823163922223268.gif)
解法2:(Ⅰ)以
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823163922784408.gif)
B(0,2,0),C(a,2,0),D(a,0,0),
P(0,0,4), ……2分
设Q(t,2,0)(
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823163922800242.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823163922815261.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823163922847255.gif)
∵PQ⊥QD,∴
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823163922862653.gif)
即
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823163922503444.gif)
∴
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823163922519479.gif)
故
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823163922535192.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823163922191420.gif)
(Ⅱ)由(Ⅰ)知,当
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823163922581229.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823163922597236.gif)
此时Q(2,2,0),D(4,0,0).
设
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823163923003474.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823163923018290.gif)
由
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823163923049754.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823163923065683.gif)
取
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823163923081215.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823163923096458.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823163923018290.gif)
而
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823163923127543.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823163922613264.gif)
由
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/201408231639231591034.gif)
∴二面角A-PD-Q的余弦值为
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823163922223268.gif)
![](http://thumb2018.1010pic.com/images/loading.gif)
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