Example 5. The base of the prism $\mathrm{ABCD}-\mathrm{A}^{\prime} \mathrm{B}^{\prime} \mathrm{C}^{\prime} \mathrm{D}^{\prime}$ is an isosceles trapezoid, $\mathrm{AB} / / \mathrm{CD}, \angle \mathrm{DAB}=60^{\circ}$, $\mathrm{CD}=5$, the leg length $\mathrm{BC}=3$, and the two non-parallel sides are perpendicular to the base. Find the distance between $\mathrm{CO}^{\prime}$ and $\mathrm{BD}^{\prime}$.
Solve as shown in Figure 6, let the intersection of line $B^{\prime} \mathrm{C}^{\prime}$ and line $\mathrm{A}^{\prime} \mathrm{D}^{\prime}=\mathrm{E}^{\prime}$, the intersection of line $\mathrm{BC}$ and line $\mathrm{AD}=\mathrm{E}$, connect $\mathrm{E}^{\prime} \mathrm{E}$.
Plane $\mathrm{B}^{\prime} \mathrm{C}^{\prime} \mathrm{CB} \perp$ plane $\mathrm{ABCD}$ $\left.\begin{array}{l}\text { Plane } \mathrm{B}^{\prime} \mathrm{C}^{\prime} \mathrm{CB} \perp \text { plane } \mathrm{ABCD} \\ \text { Plane } \mathrm{A}^{\prime} \mathrm{D}^{\prime} \mathrm{DA} \perp \text { plane } \mathrm{ABCD}\end{array}\right\} \Longrightarrow$
$\mathrm{E}^{\prime} \mathrm{E} \perp$ plane $\mathrm{ABCD}$.
It is clear that this prism is a right prism, so $\mathrm{C}^{\prime} \perp \mathrm{BC}$, $\mathrm{CC}^{\prime} \perp \mathrm{DC}$.
$\because \angle \mathrm{BCD}=180^{\circ}-60^{\circ}=120^{\circ}$,
$\therefore$ the dihedral angle $\mathrm{B}-\mathrm{CC}^{\prime}-\mathrm{D}^{\prime}$ is $120^{\circ}$.
The distances from $\mathrm{B}$ and $\mathrm{D}^{\prime}$ to $\mathrm{CC}^{\prime}$ are $3$ and $5$ respectively. Using the formula (1) for the distance between skew lines, the distance between $\mathrm{C}^{\prime} \mathrm{C}$ and $\mathrm{BD}^{\prime}$ is
$$
\mathrm{h}=\frac{15}{14} \sqrt{3} \text {. }
$$