Example 7 As shown in the figure, in the right triangle $\triangle ABC$, it is known that $BC=a$. If the line segment $PQ$ of length $2a$ has point $A$ as its midpoint, for what value of the angle between $\overrightarrow{P Q}$ and $\overrightarrow{B C}$ is the value of $\overrightarrow{B P} \cdot \overrightarrow{C Q}$ maximized? Also, find this maximum value.
Analyze: Combining the diagram, according to the rules of vector operations, transform the problem into solving using function knowledge.

Solution: Since $\overrightarrow{A B} \perp \overrightarrow{A C}$,
it follows that $\overrightarrow{A B} \cdot \overrightarrow{A C}=0$.
Since $\overrightarrow{A P}=-\overrightarrow{A Q}, \overrightarrow{B P}=\overrightarrow{A P}-\overrightarrow{A B}, \overrightarrow{C Q}=\overrightarrow{A Q}-\overrightarrow{A C}$,
it follows that $\overrightarrow{B P} \cdot \overrightarrow{C Q}=(\overrightarrow{A P}-\overrightarrow{A B}) \cdot(\overrightarrow{A Q}-\overrightarrow{A C})$
$$
\begin{array}{l}
=\overrightarrow{A P} \cdot \overrightarrow{A Q}-\overrightarrow{A P} \cdot \overrightarrow{A C}-\overrightarrow{A B} \cdot \overrightarrow{A Q}+\overrightarrow{A B} \cdot \overrightarrow{A C} \\
=-a^{2}-\overrightarrow{A P} \cdot \overrightarrow{A C}+\overrightarrow{A B} \cdot \overrightarrow{A P} \\
=-a^{2}+\overrightarrow{A P}(\overrightarrow{A B}-\overrightarrow{A C}) \\
=-a^{2}+\frac{1}{2} \overrightarrow{P Q} \cdot \overrightarrow{B C} \\
=-a^{2}+a^{2} \cdot \cos \theta
\end{array}
$$

Therefore, when $\cos \theta=1$, i.e., $\theta=0$ (which means $\overrightarrow{P Q}$ is in the same direction as $\overrightarrow{B C}$), $\overrightarrow{B P} \cdot \overrightarrow{C Q}$ is maximized, and its maximum value is 0.