Example 2 Find the distance between the face diagonals $A_{1} B$ and $B_{1} D_{1}$ in a cube $A B C D-A_{1} B_{1} C_{1} D_{1}$ with edge length 1.
Solution 1: As shown in Figure 15-3, connect $A C_{1}$. By the Three Perpendiculars Theorem, it is easy to see that $A C_{1} \perp B_{1} D_{1}, A C_{1} \perp A_{1} B$, and $A C_{1}$ is perpendicular to both $B_{1} D_{1}$ and $A_{1} B$ in a skew manner. We need to find a line parallel to $A C_{1}$ that intersects both $B_{1} D_{1}$ and $A_{1} B$. For this, we use the theorem of proportional segments cut by parallel lines in a triangle.

Take the midpoint $M$ of $A_{1} B_{1}$, and connect $C_{1} M$ intersecting $B_{1} D_{1}$ at $E$. Clearly, $E$ is the intersection of the medians of $\triangle A_{1} B_{1} C_{1}$ on sides $A_{1} C_{1}$ and $A_{1} B_{1}$, so
$$
C_{1} E: E M=2: 1.
$$

Connect $A M$ intersecting $A_{1} B$ at $F$. Similarly, $A F: F M=2: 1$. In $\triangle M A C_{1}$, $E F \parallel C_{1} A$, so $E F$ is the distance between the skew lines $B_{1} D_{1}$ and $A_{1} B$.
Since $A C_{1}=\sqrt{3}$, then $E F=\frac{\sqrt{3}}{3}$.

Solution 2: The distance between two skew lines is the minimum distance between any two points on the respective lines. Therefore, we can use the method of finding the minimum value of a function to determine the distance between $B_{1} D_{1}$ and $A_{1} B$.

Take a point $E$ on $D_{1} B$, and draw $E G \perp D_{1} B$ intersecting $A_{1} B_{1}$ at $G$. Draw $G F \perp A_{1} B$ in the plane $A_{1} B$ intersecting $A_{1} B$ at $F$, then $F G \perp G E$.

Connect $E F$. Let $B_{1} G=x$, then $A_{1} G=1-x$, $G E^{2}=\frac{x^{2}}{2}$, $G F=1-x$. Thus, $E F^{2}=G E^{2}+G F^{2}=\frac{x^{2}}{2}+(1-x)^{2}=\frac{3}{2}\left(x-\frac{2}{3}\right)^{2}+\frac{1}{3}$. When $x=\frac{2}{3}$, $E F$ has a minimum value of $\frac{\sqrt{3}}{3}$.

Solution 3: Connect $A_{1} D, B D$. Since $D_{1} B_{1} \parallel D B$, it follows that $D_{1} B_{1} \parallel$ plane $A_{1} D B$. Thus, the problem of finding the distance between the skew lines $D_{1} B_{1}$ and $A_{1} B$ is converted to finding the distance between $D_{1} B_{1}$ and the plane $A_{1} D B$, which is the distance from $B_{1}$ to the plane $A_{1} D B$, denoted as $h$.

Since $V_{B_{1}}-A_{1} D B=V_{D}-A_{1} B_{1} B$, we have $\frac{1}{3} h \frac{\sqrt{3}}{4}(\sqrt{2})^{2}=\frac{1}{3} \times 1 \times \frac{1}{2}$, which gives $h=\frac{\sqrt{3}}{3}$.

Solution 4: Clearly, plane $D_{1} B_{1} C \parallel$ plane $A_{1} B D$. The required distance can be converted to finding the distance between the planes $D_{1} B_{1} C$ and $A_{1} B D$. Connect $A C_{1}$. It is easy to see that $A C_{1}$ is perpendicular to both planes. Using the volume method to find the distance from $A$ to the plane $A_{1} D B$, the result can be obtained (readers can solve it themselves).