6. 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
A. $\frac{\sqrt{3}}{2}$
B. $\frac{1}{2}$
C. $\frac{1}{3}$
D. $\frac{\sqrt{3}}{3}$
Answer
As shown in the figure, draw $A D_{1} \stackrel{\|}{=} C D$, and connect $B D_{1}$.
Then $S_{\triangle A B D_{1}}=\frac{1}{2} A B \cdot A D_{1} \sin \frac{\pi}{3}=\frac{3}{4}$.
Thus, $V_{A B C D}=V_{B-A C D}=V_{B-A D_{1} D}=V_{D-A B D_{1}}$ $=\frac{1}{3} \cdot \frac{3}{4} \cdot 2=\frac{1}{2}$. Therefore, the answer is $B$.