Example 1. As shown in Figure 4. Let the edge length of the cube be $2, M, N$ be the centers of faces $A_{1} C_{1}, C_{1} B$, respectively. Find the distance between $A N$ and $D M$, the angle they form, and the position of their common perpendicular.
It is known that $|A N|=|D M|=\sqrt{6}$, hence
$$
\left(\begin{array}{l}
x^{2}+y^{2}+d^{2}-2 x y \cos \theta=|M N|^{2}=2, \\
(\sqrt{6}-x)^{2}+(\sqrt{6}-y)^{2}+d^{2}-2 \\
\cdot(\sqrt{6}-x)(\sqrt{6}-y) \cos \theta=|A D|^{2} \\
=4, \\
x^{2}+(\sqrt{6}-y)^{2}+d^{2}+2 x(\sqrt{6}-y) \\
\cdot \cos \theta=|A M|^{2}=6, \\
(\sqrt{6}-x)^{2}+y^{2}+d^{2}+2(\sqrt{6}-x) y \\
\cdot \cos \theta=|D N|^{2}=6 .
\end{array}\right.
$$

Solving, we get
$$
\left\{\begin{array}{l}
x=\frac{\sqrt{6}}{3} \\
y=\frac{\sqrt{6}}{3} \\
\cos \theta=\frac{1}{2} \\
d=\frac{2 \sqrt{3}}{3}
\end{array}\right.
$$