3. In tetrahedron $ABCD$, $AB=CD=2, AD=AC=BC=BD=3$, and $M, N$ are moving points on $AB, CD$ respectively. Then the minimum value of $MN$ is
A. $\sqrt{6}$
B. $\sqrt{7}$
C. $2 \sqrt{2}$
D. 3
Solve:
As shown in the figure, by extending the tetrahedron $ABCD$ to form a rectangular prism, let its length, width, and height be $x, y, z$. Then we have
$$
\left\{\begin{array}{l}
x^{2} + y^{2} = 4, \\
y^{2} + z^{2} = 9, \\
x^{2} + z^{2} = 9
\end{array} \Rightarrow \left\{\begin{array}{l}
x=\sqrt{2}, \\
y=\sqrt{2} \\
z=\sqrt{7}
\end{array}\right.\right.
$$

Thus, the minimum value of $MN$ is $\sqrt{7}$, so the answer is $B$.