(1)∵AB=1,PA=2,∠PAB=60°,∴在△PAB中,由余弦定理得 PB2=PA2+AB2-2AB·PAcos600=4+1-2×1×2× =3 ∴PA2=PB2+AB2,即AB⊥PB ∵DA⊥面ABP,CB∥DA ∴CB⊥面ABP CB⊥AB ,∴AB⊥面PBC 又DC∥AB,∴DC∥面PBC ∵DC 面PDC,∴平面PBC⊥面PDC (2)如图建立空间直角坐标系
![](http://img.shitiku.com.cn/uploads/allimg/20191021/20191021120925-68214.png) 则A(0,1,0),P( ,0,0),C(0,0,1) 设E(x,y,z), = ![](http://img.shitiku.com.cn/uploads/allimg/20191021/20191021120926-11828.png) (0< <1) 则(- ,0,1)= (x- ,y,z) x= (1- ),y=0,z=![](http://img.shitiku.com.cn/uploads/allimg/20191021/20191021120926-62286.png) 设面ABE的法向量为n=(a,b,c),则![](http://img.shitiku.com.cn/uploads/allimg/20191021/20191021120926-67849.png) ![](http://img.shitiku.com.cn/uploads/allimg/20191021/20191021120925-82151.png) ![](http://img.shitiku.com.cn/uploads/allimg/20191021/20191021120927-24269.png) 令c=![](http://img.shitiku.com.cn/uploads/allimg/20191021/20191021120925-94784.png) ![](http://img.shitiku.com.cn/uploads/allimg/20191021/20191021120925-82151.png) ![](http://img.shitiku.com.cn/uploads/allimg/20191021/20191021120927-20219.png) n=( ,0, ) 同理可求平面PAE的法向量为m=(1, , ) ∵cos<n,m>= = = =![](http://img.shitiku.com.cn/uploads/allimg/20191021/20191021120924-10430.png) ∴ = 或 =1(舍去) ∴E( ,0, )为PC的中点,其竖坐标即为点E到底面PAB的距离 ∴VE-PAB= × ×1× × =![](http://img.shitiku.com.cn/uploads/allimg/20191021/20191021120924-69165.png) |