(7) As shown in the figure, the side length of square $A B C D$ is $3, E$ is the midpoint of $D C, A E$ intersects $B D$ at $F$, then the value of $\overrightarrow{F D} \cdot \overrightarrow{D E}$ is $\qquad$ .
(7) $-\frac{3}{2}$ Hint: $\frac{D F}{F B}=\frac{D E}{A B}=\frac{1}{2}$, so
$$
\begin{aligned}
\overrightarrow{F D} \cdot \overrightarrow{D E} & =-D F \cdot D E \cdot \cos \angle F D E \\
& =-\frac{1}{3} \times 3 \sqrt{2} \times \frac{3}{2} \times \frac{\sqrt{2}}{2}=-\frac{3}{2} .
\end{aligned}
$$