试题分析:解:(1)取AB中点M,EF//AD//MG EFGM共面, 由EM//PB,PB 面EFG,EM 面EFG,得PB//平面EFG ………………4分 (2)如图建立直角坐标系,E(0,0,1),F(1,0,1),G(2,1,0) ="(1,0,0)," =(1,1,-1),
![](http://img.shitiku.com.cn/uploads/allimg/20191021/20191021132954-94480.png) 设面EFG的法向量为 =(x,y,z)由![](http://img.shitiku.com.cn/uploads/allimg/20191021/20191021132955-47797.png) ![](http://img.shitiku.com.cn/uploads/allimg/20191021/20191021132955-92075.png) 得出x="0," 由![](http://img.shitiku.com.cn/uploads/allimg/20191021/20191021132955-47797.png) ![](http://img.shitiku.com.cn/uploads/allimg/20191021/20191021132955-92075.png) 得出x+y-z=0 从而 =(0,1,1),又 =(0,0,1),得cos = = ( 为 与 的夹角)![](http://img.shitiku.com.cn/uploads/allimg/20191021/20191021132953-56261.png) =45o ……………8分 (3)设Q(2,b,0),面EFQ的法向量为 =(x,y,z), =(2,b,-1) 由![](http://img.shitiku.com.cn/uploads/allimg/20191021/20191021132956-15736.png) ![](http://img.shitiku.com.cn/uploads/allimg/20191021/20191021132955-92075.png) 得出x="0," 由![](http://img.shitiku.com.cn/uploads/allimg/20191021/20191021132956-15736.png) ![](http://img.shitiku.com.cn/uploads/allimg/20191021/20191021132955-92075.png) 得出2x+by-z=0,从而 =(0,1,b) 面EFD的法向量为 =(0,1,0),所以 ,解得,b=![](http://img.shitiku.com.cn/uploads/allimg/20191021/20191021132957-42595.png) CQ= ……………12分 点评:解决该试题的关键是利用向量法合理的建立直角坐标系,然后借助于平面的法向量,以及直线的方向向量来求解二面角的问题。同时能熟练的运用线面的垂直的判定呢性质定理解题,属于中档题。 |