1. In the rectangular prism $A B C D-A_{1} B_{1} C_{1} D_{1}$, $A B=5, B C=4, B B_{1}=3$, find the distance between $A B$ and $D B_{1}$.
1. Connect $B D$, take a point $P$ on $B D$, draw a line through $P$ parallel to $B B_{1}$, intersecting $B D$ at $M$, then $P M \perp$ plane $A B C D$. Draw $M N \perp A B$ intersecting $A B$ at $N$, connect $P N$, let $P M=x$, then from $\frac{P M}{B B_{1}}$ $=\frac{D M}{D B}$, we get $D M=\frac{\sqrt{41}}{3} x$. Therefore, $B M=\sqrt{41}\left(1-\frac{x}{3}\right)$.
Also, $\frac{M N}{A D}=\frac{B M}{B D}$, so $M N=4\left(1-\frac{x}{3}\right), N P^{2}=\frac{25}{9}\left(x-\frac{48}{25}\right)^{2}+\frac{144}{25}$.
Thus, we derive $N P_{\min }=\frac{12}{5}$, which means the distance between $A B$ and $D B$ is $\frac{12}{5}$.
Note: To find the distance between $A B$ and $D B_{1}$, it can be converted to finding the distance between $A B$ and the plane $D B_{1} C$. Using the volume $V_{B-C B_{1} D}=V_{B_{1}-B V D}$ can yield the result.