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),
AC
=(1,1,0),
DA1
=(0,-1,1),
设MN为直线DA1与AC的公垂线段,且
MN
=(x,y,z),
且
MN
⊥
AC
,
MN
⊥
DA1
,x+y=0,-y+z=0,令y=t,则
MN
=(-t,t,t),
而另可设M(m,m,0),N(0,a,b),则
MN
=(-m,a-m,b),
∴
-m=-1a-m=tb=t.
∴N(0,2t,t).
又2t+t=1,∴t=
13
.
∴
MN
=(-
13
,
13
,
13
,|
MN
|=
19+19+19
=
33
.
即直线DA1与AC间的距离为
33
.