方法一 如图所示,以D为原点,DA、DC、DD1所在直线分别为x轴、y轴、z轴建立空间直角坐标系,
![](http://img.shitiku.com.cn/uploads/allimg/20191106/20191106044512-13444.gif) 设正方体的棱长为1,则可求得 M(0,1, ),N( ,1,1), D(0,0,0),A1(1,0,1),B(1,1,0), 于是 =( ,0, ),
=(1,0,1), =(1,1,0). 设平面A1BD的法向量是 n=(x,y,z). 则n· =0,且n· =0, 得![](http://img.shitiku.com.cn/uploads/allimg/20191106/20191106044513-77704.gif) 取x=1,得y=-1,z=-1. ∴n=(1,-1,-1). 又 ·n=( ,0, )·(1,-1,-1)=0, ∴ ⊥n, 又∵![](http://img.shitiku.com.cn/uploads/allimg/20191106/20191106044513-97806.gif) 平面A1BD,∴MN∥平面A1BD. 方法二 ∵ = - =![](http://img.shitiku.com.cn/uploads/allimg/20191106/20191106044513-47163.gif) -![](http://img.shitiku.com.cn/uploads/allimg/20191106/20191106044513-47163.gif) ![](http://img.shitiku.com.cn/uploads/allimg/20191106/20191106044514-70357.gif) = ( - )=![](http://img.shitiku.com.cn/uploads/allimg/20191106/20191106044513-47163.gif) , ∴ ∥ ,又∵MN 平面A1BD. ∴MN∥平面A1BD. |