Example 3: In a cube $A B C D-A_{1} B_{1} C_{1} D_{1}$ with edge length 1, $A C_{1}$ is a diagonal. $M, N$ are the midpoints of $B B_{1}$ and $B_{1} C_{1}$, respectively, and $P$ is the midpoint of line segment $M N$. Find the distance between the skew lines $D P$ and $A C_{1}$.
Solve As shown in Figure 15-3, in the plane $A B_{1} C_{1} D$, draw a line through $D$ parallel to $A C_{1}$, intersecting the extension of $B_{1} C_{1}$ at $E$. Then $A C_{1} / /$ plane $D P E$.

Therefore, the distance between $D P$ and $A C_{1}$ is the distance from $A C_{1}$ to the plane $D P E$, which is also the height of the tetrahedron $C_{1}-D E P$, denoted as $d$.
From $V_{C, \text { UEP }}=V_{D-C, E P}$,

we get $\quad \frac{1}{3} \cdot d \cdot S_{\triangle \text { EEP }}=\frac{1}{3} \cdot D C \cdot S_{\triangle C E T}$
and $S_{\therefore, c \in:}=\frac{1}{2} \cdot C_{1} E \cdot P H=\frac{1}{8}$
Given $D E=\sqrt{3}, P E=\sqrt{H E^{2}+P H^{2}}=\frac{5}{4} \sqrt{2}$
$$
D P=\sqrt{P F^{2}+D F^{2}}=\frac{\sqrt{34}}{4}
$$

Using Heron's formula, we get $\quad S_{\therefore D E P}=\frac{\sqrt{86}}{8}$
Combining (1)(2)(3) we get $\quad d=\frac{D C \cdot S_{\triangle C E P}}{S_{\text {IDEP }}}=\frac{\sqrt{86}}{86}$
Therefore, the distance between the skew lines $D P$ and $A C_{1}$ is $\frac{\sqrt{86}}{86}$