1. In tetrahedron $ABCD$, let $AB=1$, $CD=\sqrt{3}$, the distance between lines $AB$ and $CD$ is 2, and the angle between them is $\frac{\pi}{3}$. Then the volume of tetrahedron $ABCD$ is ( ).
(2003 National High School League Question)
A. $\frac{\sqrt{3}}{2}$
B. $\frac{1}{2}$
C. $\frac{1}{3}$
D. $\frac{\sqrt{3}}{3}$
1. Choose B. Reason: The area of the parallelogram formed by $\overrightarrow{A B}$ and $\overrightarrow{C D}$ is $|\overrightarrow{A B}| \cdot|\overrightarrow{C D}| \cdot \sin 60^{\circ}=\frac{3}{2}$. If a line $C E \mathbb{E}$ is drawn through point $C$ parallel to $A B$, then $S_{\triangle C D E}=\frac{1}{2} \cdot \frac{3}{2}=\frac{3}{4}$, and the volume of the prism $A B F-E C D$ with $\triangle C D E$ as the base and $B C$ as the side edge is 3 times that of $V_{A B C D}$, and $V_{A B F-E C D}=S_{\triangle C D E} \cdot 2=\frac{3}{2}$, hence $V_{A B C D}=\frac{1}{2}$.