证明:建立如图所示的空间直角坐标系.
设A(a,0,0),S(0,0,b),则B(a,a,0),C(0,a,0),E,F.\s\up6(→(→)=.
取SD的中点G,连接AG,则\s\up6(→(→)=.
因为\s\up6(→(→)=\s\up6(→(→),所以EF∥AG,
又AG平面SAD,EF⃘平面SAD,
所以EF∥平面SAD.
如图,正方体ABCDA1B1C1D1中,E,F,G分别是B1B,AB,BC的中点.证明:D1F⊥平面AEG.
证明:设\s\up6(→(→)=a,\s\up6(→(→)=b,\s\up6(→(→)=c,则\s\up6(→(→)=\s\up6(→(→)+\s\up6(→(→)+\s\up6(→(→)=-\s\up6(→(→)-\s\up6(→(→)+\s\up6(→(→)=a-b-c,
\s\up6(→(→)=\s\up6(→(→)+\s\up6(→(→)=\s\up6(→(→)+\s\up6(→(→)=a+b,
\s\up6(→(→)=\s\up6(→(→)+\s\up6(→(→)=\s\up6(→(→)+\s\up6(→(→)=a+c,
∴\s\up6(→(→)·\s\up6(→(→)=(a-b-c)·(a+b)=0,
∴\s\up6(→(→)⊥\s\up6(→(→),即D1F⊥AG.
\s\up6(→(→)·\s\up6(→(→)=(a-b-c)·(a+c)=0,