9. For the cube $A B C D-A_{1} B_{1} C_{1} D_{1}$ with edge length 1, the distance between the lines $A_{1} C_{1}$ and $B D_{1}$ is $\qquad$
$$
\frac{\sqrt{6}}{6}
$$
9. 【Analysis and Solution】This is a problem of finding the distance between two skew lines.

To ensure that the line segment representing the distance is perpendicular to both $A_{1} C_{1}$ and $B D_{1}$, we can first place one of the lines in a plane perpendicular to the other. For this purpose, construct the diagonal plane $B D D_{1} B_{1}$ of the cube, then $A_{1} C_{1} \perp$ plane $B D D_{1} B_{1}$, and $B D_{1}$ lies in plane $B D D_{1} B_{1}$. Let $A_{1} C_{1} \cap B_{1} D_{1}=O$, and in plane $B D D_{1} B_{1}$, construct $O H \perp B D_{1}$, with the foot of the perpendicular at $H$, then the length of line segment $O H$ is the distance between the skew lines $A_{1} C_{1}$ and $B D_{1}$. In the right triangle $\triangle B B_{1} D_{1}$, $O H$ is half the altitude on the hypotenuse $B D_{1}$, i.e., $O H=\frac{\sqrt{6}}{6}$.