Example 1. The edge length of the cube $\mathrm{ABCD}-\mathrm{A}_{1} \mathrm{~B}_{1} \mathrm{C}_{1} \mathrm{D}_{1}$ is $\mathrm{a}$, find the distance between $\mathrm{A}, \mathrm{C}$ and $\mathrm{BD}$.
Solve as shown in Figure 1, on line segment $\mathrm{A}_{1} \mathrm{C}$, take any point $\mathrm{M}$, draw $M N \perp$ the base $\mathrm{AC}$, the foot of the perpendicular falls on the diagonal $A C$ of the base square $\mathrm{ABCD}$, the intersection of $A \subset$ and $\mathrm{BD}$ is denoted as $\mathrm{P}$, connect $M P$, and let $M N=x$.
From $\frac{C N}{\mathbf{A C}}=\frac{M N}{\mathbf{A}_{4} \mathbf{A}}$, we get $C N=\sqrt{2} x$.

Then $P N=\frac{\sqrt{2}}{2} a-\sqrt{2} x$.
In $\mathrm{R} \triangle \mathrm{MNP}$, $\mathrm{MP}^{2}=\mathrm{MN}^{2}+\mathrm{PN}^{2}$
$$
\begin{aligned}
= & \mathrm{x}^{2}+(\sqrt{2} \mathrm{a}-\sqrt{2} \mathrm{x})^{2}=3\left(\mathrm{x}-\frac{\mathrm{a}}{3}\right)^{2} \\
& +\frac{1}{6}{ }^{-\mathrm{a}^{2}} .
\end{aligned}
$$

Since $\mathrm{MP}>0, \therefore \mathrm{MP}$'s minimum value $\vee_{-}^{\prime} \frac{6}{6} \mathrm{a}$ is the distance between $\mathbf{A}_{1} \mathrm{C}$ and $\mathrm{BD}$.