已知正方体ABCD-A1B1C1D1的棱长是1,求直线DA1与AC间的距离.
建立如如图所示坐标系,则A(0,0,0),C(1,1,0),D(0,1,0),
A1(0,0,1),
=(1,1,0),AC
=(0,-1,1),DA1
设MN为直线DA1与AC的公垂线段,且
=(x,y,z),MN
且
⊥MN
,AC
⊥MN
,x+y=0,-y+z=0,令y=t,则DA1
=(-t,t,t),MN
而另可设M(m,m,0),N(0,a,b),则
=(-m,a-m,b),MN
∴
∴N(0,2t,t).-m=-1 a-m=t b=t.
又2t+t=1,∴t=
.1 3
∴
=(-MN
,1 3
,1 3
,|1 3
|=MN
=
+1 9
+1 9 1 9
.3 3
即直线DA1与AC间的距离为
.3 3