【答案】
分析:解法一(几何法)(1)连AC,设AC∩BD=O,连接OC,OC
1,可得∠COC
1即为二面角C
1-DB-C的平面角,解Rt△COC
1,即可得到二面角C
1-DB-C的正切值.
(2)设AP与面BDD
1B
1交于点G,连OG,可得∠AGO即为AP与面BDD
1B
1所成的角,解Rt△AOG,即可得到一个关于m的方程,解方程即可得到满足条件的m的值.
解法二(向量法)(1)建立如图所示的空间直角坐标系,分别求出平面C
1DB和平面DBC的法向量,代入向量夹角公式,即可得到二面角C
1-DB-C的正切值;
(2)分别求出直线AP的方向向量与平面BDD
1B
1的法向量,根据根据直线AP与平面BDD
1B
1所成角的正切值为3
![](http://thumb.1010pic.com/pic6/res/gzsx/web/STSource/20131103172740535273058/SYS201311031727405352730017_DA/0.png)
,构造一个关于m的方程,解方程即可得到满足条件的m的值.
解答:![](http://thumb.1010pic.com/pic6/res/gzsx/web/STSource/20131103172740535273058/SYS201311031727405352730017_DA/images1.png)
解法一(几何法):
解:(1)如图,连AC,设AC∩BD=O,连接OC,OC
1,
则AC⊥BD,CC
1⊥BD,
∴BD⊥平面CC
1O,
∴BD⊥CC
1,
故∠COC
1即为二面角C
1-DB-C的平面角
在Rt△COC
1中,CC
1=1,CO=
![](http://thumb.1010pic.com/pic6/res/gzsx/web/STSource/20131103172740535273058/SYS201311031727405352730017_DA/1.png)
则tan∠COC
1=
![](http://thumb.1010pic.com/pic6/res/gzsx/web/STSource/20131103172740535273058/SYS201311031727405352730017_DA/2.png)
=
![](http://thumb.1010pic.com/pic6/res/gzsx/web/STSource/20131103172740535273058/SYS201311031727405352730017_DA/3.png)
故二面角C
1-DB-C的正切值为
![](http://thumb.1010pic.com/pic6/res/gzsx/web/STSource/20131103172740535273058/SYS201311031727405352730017_DA/4.png)
(2)设AP与面BDD
1B
1交于点G,连OG,
因为PC∥面BDD
1B
1,而BDD
1B
1∩面APC=OG,
故OG∥PC,
所以OG=
![](http://thumb.1010pic.com/pic6/res/gzsx/web/STSource/20131103172740535273058/SYS201311031727405352730017_DA/5.png)
PC=
![](http://thumb.1010pic.com/pic6/res/gzsx/web/STSource/20131103172740535273058/SYS201311031727405352730017_DA/6.png)
.
又AO⊥DB,AO⊥BB
1,
所以AO⊥面BDD
1B
1,
故∠AGO即为AP与面BDD
1B
1所成的角
在Rt△AOG中,tan∠AGO=
![](http://thumb.1010pic.com/pic6/res/gzsx/web/STSource/20131103172740535273058/SYS201311031727405352730017_DA/7.png)
=3
![](http://thumb.1010pic.com/pic6/res/gzsx/web/STSource/20131103172740535273058/SYS201311031727405352730017_DA/8.png)
即m=
![](http://thumb.1010pic.com/pic6/res/gzsx/web/STSource/20131103172740535273058/SYS201311031727405352730017_DA/9.png)
.?
故当m=
![](http://thumb.1010pic.com/pic6/res/gzsx/web/STSource/20131103172740535273058/SYS201311031727405352730017_DA/10.png)
时,直线AP与平面BDD
1B
1所成角的正切值为3
![](http://thumb.1010pic.com/pic6/res/gzsx/web/STSource/20131103172740535273058/SYS201311031727405352730017_DA/11.png)
.?
![](http://thumb.1010pic.com/pic6/res/gzsx/web/STSource/20131103172740535273058/SYS201311031727405352730017_DA/images13.png)
解法二(向量法)
解:(1)建立如图所示的空间直角坐标系,则A(1,0,0),B(1,1,0),P(0,1,m),C(0,1,0),D(0,0,0),B
1(1,1,1),D
1(0,0,1)?
则
![](http://thumb.1010pic.com/pic6/res/gzsx/web/STSource/20131103172740535273058/SYS201311031727405352730017_DA/12.png)
=(0,0,1)为平面DBC一个法向量,
设
![](http://thumb.1010pic.com/pic6/res/gzsx/web/STSource/20131103172740535273058/SYS201311031727405352730017_DA/13.png)
=(x,y,z)为平面C
1DB的一个法向量,则
![](http://thumb.1010pic.com/pic6/res/gzsx/web/STSource/20131103172740535273058/SYS201311031727405352730017_DA/14.png)
即
![](http://thumb.1010pic.com/pic6/res/gzsx/web/STSource/20131103172740535273058/SYS201311031727405352730017_DA/15.png)
则
![](http://thumb.1010pic.com/pic6/res/gzsx/web/STSource/20131103172740535273058/SYS201311031727405352730017_DA/16.png)
=(1,-1,1)
设二面角C
1-DB-C的平面角为θ
则cosθ=
![](http://thumb.1010pic.com/pic6/res/gzsx/web/STSource/20131103172740535273058/SYS201311031727405352730017_DA/17.png)
=
![](http://thumb.1010pic.com/pic6/res/gzsx/web/STSource/20131103172740535273058/SYS201311031727405352730017_DA/18.png)
则sinθ=
![](http://thumb.1010pic.com/pic6/res/gzsx/web/STSource/20131103172740535273058/SYS201311031727405352730017_DA/19.png)
,tanθ=
![](http://thumb.1010pic.com/pic6/res/gzsx/web/STSource/20131103172740535273058/SYS201311031727405352730017_DA/20.png)
即二面角C
1-DB-C的正切值为
![](http://thumb.1010pic.com/pic6/res/gzsx/web/STSource/20131103172740535273058/SYS201311031727405352730017_DA/21.png)
(2)∵
![](http://thumb.1010pic.com/pic6/res/gzsx/web/STSource/20131103172740535273058/SYS201311031727405352730017_DA/22.png)
=(-1,1,0),
![](http://thumb.1010pic.com/pic6/res/gzsx/web/STSource/20131103172740535273058/SYS201311031727405352730017_DA/23.png)
=(0,0,1),?
![](http://thumb.1010pic.com/pic6/res/gzsx/web/STSource/20131103172740535273058/SYS201311031727405352730017_DA/24.png)
=(-1,1,m),
![](http://thumb.1010pic.com/pic6/res/gzsx/web/STSource/20131103172740535273058/SYS201311031727405352730017_DA/25.png)
=-1,1,0),?
又由
![](http://thumb.1010pic.com/pic6/res/gzsx/web/STSource/20131103172740535273058/SYS201311031727405352730017_DA/26.png)
•
![](http://thumb.1010pic.com/pic6/res/gzsx/web/STSource/20131103172740535273058/SYS201311031727405352730017_DA/27.png)
=0,
![](http://thumb.1010pic.com/pic6/res/gzsx/web/STSource/20131103172740535273058/SYS201311031727405352730017_DA/28.png)
•
![](http://thumb.1010pic.com/pic6/res/gzsx/web/STSource/20131103172740535273058/SYS201311031727405352730017_DA/29.png)
=0知,
![](http://thumb.1010pic.com/pic6/res/gzsx/web/STSource/20131103172740535273058/SYS201311031727405352730017_DA/30.png)
为平面BB
1D
1D的一个法向量.?
设AP与平面BB
1D
1D所成的角为θ,?
则sinθ=cos(
![](http://thumb.1010pic.com/pic6/res/gzsx/web/STSource/20131103172740535273058/SYS201311031727405352730017_DA/31.png)
-θ)=
![](http://thumb.1010pic.com/pic6/res/gzsx/web/STSource/20131103172740535273058/SYS201311031727405352730017_DA/32.png)
=
![](http://thumb.1010pic.com/pic6/res/gzsx/web/STSource/20131103172740535273058/SYS201311031727405352730017_DA/33.png)
依题意有
![](http://thumb.1010pic.com/pic6/res/gzsx/web/STSource/20131103172740535273058/SYS201311031727405352730017_DA/34.png)
=
![](http://thumb.1010pic.com/pic6/res/gzsx/web/STSource/20131103172740535273058/SYS201311031727405352730017_DA/35.png)
,?
解得m=
![](http://thumb.1010pic.com/pic6/res/gzsx/web/STSource/20131103172740535273058/SYS201311031727405352730017_DA/36.png)
,??
故当m=
![](http://thumb.1010pic.com/pic6/res/gzsx/web/STSource/20131103172740535273058/SYS201311031727405352730017_DA/37.png)
时,直线AP与平面BDD
1B
1所成角的正切值为3
![](http://thumb.1010pic.com/pic6/res/gzsx/web/STSource/20131103172740535273058/SYS201311031727405352730017_DA/38.png)
.
点评:本题考查的知识点是二面角的平面角及求法,直线与平面所成的解,其中解法一的关键是找到线面夹角和二面角的平面角,将空间线面夹角问题和二面角问题转化为解三角形问题;而解法二的关键是建立空间坐标系,求出直线的方向向量和平面的法向量,将空间线面夹角问题和二面角问题转化为向量夹角问题.