Example 15 (2003 National High School Competition Question) In tetrahedron $ABCD$, let $AB=1, CD=\sqrt{3}$, the distance between line $AB$ and line $CD$ is 2, and the angle between them is $\frac{\pi}{3}$. Then the volume of tetrahedron $ABCD$ is ( ).
A. $\frac{\sqrt{3}}{2}$
B. $\frac{1}{2}$
C. $\frac{1}{3}$
D. $\frac{\sqrt{3}}{3}$
Solution: Choose B. Reason: Perform shape completion and reduction. As shown in Figure 2-10, draw $CE$ parallel to $AB$, and construct a prism $ABF-ECD$ with $\triangle CDE$ as the base and $BC$ 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 $S_{\triangle ODE}=\frac{1}{2} CE \cdot CD \sin \angle ECD, AB$ and $CD$'s common perpendicular $MN$ is the height of the prism $ABF-ECD$, so
$$
V_{2}=\frac{1}{2} MN \cdot CE \cdot CD \sin \angle ECD=\frac{3}{2} \text {. }
$$

Therefore, $V_{1}=\frac{1}{3} V_{2}=\frac{1}{2}$.