Example 1 (2003 National High School Competition Question) In tetrahedron $A B C D$, let $A B=1, C D=\sqrt{3}$, the distance between lines $A B$ and $C D$ is 2, and the angle between them is $\frac{\pi}{3}$. Then the volume of tetrahedron $A B C D$ is equal to ( ).
A. $\frac{\sqrt{3}}{2}$
B. $\frac{1}{2}$
C. $\frac{1}{3}$
D. $\frac{\sqrt{3}}{3}$
Solution: Choose B. Reason: As shown in Figure 7-2, draw $C E \Perp A B$, and construct a prism $A B F-E C D$ with $\triangle C D E$ as the base and $B C$ as the side edge. Then the volume $V_{1}$ of the required tetrahedron is equal to $\frac{1}{3}$ of the volume $V_{2}$ of the above prism. And $S_{\triangle C D E}=\frac{1}{2} C E \cdot C D \cdot \sin \angle E C D, A B$ and $C D$'s common perpendicular $M N$ is the height of the prism $A B F-E C D$, so
$$
\begin{aligned}
V_{2} & =\frac{1}{2} M N \cdot C E \cdot C D \cdot \sin \angle E C D=\frac{3}{2} \text {, therefore } V_{1}= \\
\frac{1}{3} V_{2} & =\frac{1}{2} .
\end{aligned}
$$