Example 20 Given in the rectangular prism $A B C D-A_{1} B_{1} C_{1} D_{1}$, $A B=4, A D=3, A A_{1}=2$, if $M, N$ are the midpoints of $D C, B B_{1}$ respectively, find the distance between the skew lines $M N$ and $A_{1} B$.
To establish a spatial rectangular coordinate system $A-xyz$ as shown in Figure 8-10, we have $A(0,0,0), A_{1}(0,0,2), B(0,4,0), B_{1}(0,4,2), C(3,4,0), D(3,0,0), M(3,2,0), N(0,4,1)$, so
$$
\begin{array}{l}
\overrightarrow{M N}=(-3,2,1), \overrightarrow{A_{1} B}=(0,4,-2), \\
\overrightarrow{M A_{1}}=(-3,-2,2) .
\end{array}
$$

Let $\vec{n}=(x, y, z)$, and $\vec{n} \perp \overrightarrow{M N}, \vec{n} \perp A_{1} B$,
$$
\text { then }\left\{\begin{array}{l}
\vec{n} \cdot \overrightarrow{M N}=-3 x+2 y+z=0, \\
\vec{n} \cdot \overrightarrow{A_{1} B}=4 y-2 z=0,
\end{array}\right.
$$

Solving, we get $\left\{\begin{array}{l}x=\frac{4}{3} y, \\ z=2 y .\end{array}\right.$
Taking $y=3$, then $\vec{n}=(4,3,6)$.
Let the projection length of vector $\overrightarrow{M A_{1}}$ on vector $\vec{n}$ be $d$, which is the distance between the skew lines $M N$ and $A_{1} B$, then
$$
d=\left|\overrightarrow{M A_{1}}\right| \cdot\left|\cos \left\langle\overrightarrow{M A_{1}} \cdot \vec{n}\right\rangle\right|=\frac{\left|\overrightarrow{M A_{1}} \cdot \vec{n}\right|}{|\vec{n}|}=\frac{|-6|}{\sqrt{61}}=\frac{6 \sqrt{61}}{61} .
$$