Example 5 (2001 National High School Competition Question) The edge length of the cube $A B C D-A_{1} B_{1} C_{1} D_{1}$ is 1, then the distance $d$ between the line $A_{1} C_{1}$ and $B D_{1}$ is $\qquad$ .
Solve: Fill in $\frac{\sqrt{6}}{6}$. Reason: As shown in Figure 8-3, establish a spatial rectangular coordinate system $B-x y z$. It is easy to know that $A_{1}(1,0,1), C_{1}(0,1,1), D_{1}(1,1,1)$, so $\overrightarrow{A_{1} C_{1}}(-1,1,0), \overrightarrow{A_{1} D_{1}}=(0,1,0), \overrightarrow{B D_{1}}=(1,1,1)$. Let $\vec{n}=(x, y, z)$, and $\vec{n} \perp \overrightarrow{A_{1} C_{1}}, \vec{n} \perp \overrightarrow{B D_{1}}$. From $\left\{\begin{array}{l}\vec{n} \cdot \overrightarrow{A_{1} C_{1}}=0, \\ \vec{n} \cdot \overrightarrow{B D_{1}}=0,\end{array}\right.$ we get $\quad\left\{\begin{array}{l}-\dot{x}+y=0, \\ x+y+z=0 .\end{array}\right.$ Let $x=1$, then $\vec{n}=(1,1,-2)$. According to Conclusion 6 of this chapter, $\quad d=\frac{\left|\overrightarrow{A_{1} D_{1}} \cdot \vec{n}\right|}{|\vec{n}|}=\frac{\sqrt{6}}{6}$.