Because a 1c 1∑AC

While A 1C 1 is not in the ACD 1 plane, and AC is in the ACD 1 plane.

Therefore, a1c1Σ plane ACD 1.

So the distance between AD 1 and A 1C 1 is the distance from C 1 to ACD 1 plane.

Let the height of the triangular pyramid C 1-ACD 1 be h,

H equals the distance from C 1 to the plane ACD 1.

Because the volume of triangular pyramid C 1-ACD 1 = the volume of triangular pyramid A-CC 1D 1.

Therefore, the area of h×△ACD 1 /3 = the area of 3 =AD×△CC 1D 1+0/3.

Therefore, h = (area of ad ×△ cc1d1) ÷ area of acd65438 +0.

Because AD= 1, the area of △CC 1D 1 = 0.5 ×1= 0.5.

AC=CD 1=AD 1= root number 2, area of △ACD 1 = square of root number 3 × (root number 2) =0.5× root number 3,

Therefore, h = 0.5(0.5× root number 3)

That is, the root number 3 of h =.

So the distance between AD 1 and A 1C 1 is the root sign of 3/3.