3. In a planar convex quadrilateral $ABCD$, $M$ and $N$ are the midpoints of sides $AD$ and $BC$, respectively. If $|\overrightarrow{AB}|=2, |\overrightarrow{MN}|=\frac{3}{2}$, $\overrightarrow{MN} \cdot (\overrightarrow{AD}-\overrightarrow{BC})=\frac{3}{2}$, then $\overrightarrow{AB} \cdot \overrightarrow{CD}=$ . $\qquad$
3. -2 .
$$
\begin{array}{l}
\text { Given } \overrightarrow{M N}=\frac{1}{2}(\overrightarrow{M B}+\overrightarrow{M C}) \\
=\frac{1}{2}(\overrightarrow{M A}+\overrightarrow{A B}+\overrightarrow{M D}+\overrightarrow{D C}) \\
=\frac{1}{2}(\overrightarrow{A B}-\overrightarrow{C D}), \\
\overrightarrow{A D}-\overrightarrow{B C}=\overrightarrow{A B}+\overrightarrow{B C}+\overrightarrow{C D}-\overrightarrow{B C} \\
=\overrightarrow{A B}+\overrightarrow{C D},
\end{array}
$$

we know
$$
\begin{array}{l}
\frac{3}{2}=\overrightarrow{M N} \cdot(\overrightarrow{A D}-\overrightarrow{B C}) \\
=\frac{1}{2}(\overrightarrow{A B}-\overrightarrow{C D}) \cdot(\overrightarrow{A B}+\overrightarrow{C D}) \\
=\frac{1}{2}\left(|\overrightarrow{A B}|^{2}-|\overrightarrow{C D}|^{2}\right) .
\end{array}
$$

Given $|\overrightarrow{A B}|=2$, we know $|\overrightarrow{C D}|=1$.
$$
\begin{array}{l}
\text { Also, } \frac{9}{4}=|\overrightarrow{M N}|^{2}=\frac{1}{4}(\overrightarrow{A B}-\overrightarrow{C D})^{2} \\
=\frac{1}{4}\left(|\overrightarrow{A B}|^{2}+|\overrightarrow{C D}|^{2}-2 \overrightarrow{A B} \cdot \overrightarrow{C D}\right),
\end{array}
$$

solving for $\overrightarrow{A B} \cdot \overrightarrow{C D}=-2$.