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) $\frac{\sqrt{3}}{3}$
(B)
6.【Analysis and Solution】Solution 1 As shown in the figure, draw $B E \Perp C D$, then quadrilateral $B C D E$ is a parallelogram, the angle formed by skew lines $A B$ and $C D$ is $\angle A B E=60^{\circ}$, thus $S_{\triangle A B C}=\frac{1}{2} A B \cdot B E \sin$ $60^{\circ}=\frac{3}{4}$.
$\because C D / / B E$,
$\therefore C D / /$ plane $A B E$.
$\therefore$ The distance $h$ from point $D$ to plane $A B E$ is equal to the distance between the skew lines $A B$ and $C D$, i.e., $h=2$.
Therefore, $V_{A}-{ }_{B C D}=V_{A}-{ }_{B D E}=V_{D}-{ }_{A B E}=\frac{1}{3} \times \frac{3}{4} \times 2=\frac{1}{2}$.
Solution 2 As shown in the figure, take the midpoints $P 、 Q 、 R 、 S$ of $B C 、 C A 、 A D 、 D B$ respectively, then quadrilateral $P Q R S$ is a parallelogram with $\angle Q P S=60^{\circ}$, and $P Q=\frac{1}{2} A B=\frac{1}{2}, P S=\frac{1}{2} C D=\frac{\sqrt{3}}{2}, A B$ $/ /$ plane $P Q R S / / C D$.
Let the distances from $A B 、 C D$ to plane $P Q R S$ be $h_{1} 、 h_{2}$ respectively, then $h_{1}+h_{2}=2$. By the volume formula of a frustum, we get
$$
\begin{aligned}
V_{A B C D} & =V_{A B}-{ }_{F Q R S}+V_{C D}-P Q R S \\
& =\frac{1}{3}\left(h_{1}+h_{2}\right) \cdot S_{\triangle P Q R S} \\
& =\frac{1}{3} \times 2 \times \frac{3}{4}=\frac{1}{2} .
\end{aligned}
$$