Question 82, Given vectors $\overrightarrow{\mathrm{OA}}$ and $\overrightarrow{\mathrm{OB}}$ have an angle of $120^{\circ}$ between them, $|\overrightarrow{\mathrm{OA}}|=2|\overrightarrow{\mathrm{OB}}|=6$. A line segment $\mathrm{PQ}$ of length $2|\overrightarrow{\mathrm{AB}}|$ is drawn through 0, and $\mathrm{P}, \mathrm{Q}$ are symmetric about 0. When the angle between vector $\overrightarrow{\mathrm{PQ}}$ and $\overrightarrow{\mathrm{AB}}$ is what, will $\overrightarrow{\mathrm{AP}} \cdot \overrightarrow{\mathrm{BQ}}$ be maximized, and find this maximum value.
Question 82, Answer: According to the cosine theorem, we know: $|\overrightarrow{\mathrm{AB}}|=\sqrt{|\overrightarrow{\mathrm{OA}}|^{2}+|\overrightarrow{\mathrm{OB}}|^{2}-2|\overrightarrow{\mathrm{OA}}| \cdot|\overrightarrow{\mathrm{OB}}| \cos 120^{\circ}}=3 \sqrt{7}$, so $|\overrightarrow{\mathrm{OP}}|=|\overrightarrow{\mathrm{OQ}}|=3 \sqrt{7}$.

Since $\overrightarrow{\mathrm{AP}}=\overrightarrow{\mathrm{OP}}-\overrightarrow{\mathrm{OA}}, \overrightarrow{\mathrm{BQ}}=\overrightarrow{\mathrm{OQ}}-\overrightarrow{\mathrm{OB}}$, thus $\overrightarrow{\mathrm{AP}} \cdot \overrightarrow{\mathrm{BQ}}=(\overrightarrow{\mathrm{OP}}-\overrightarrow{\mathrm{OA}}) \cdot(\overrightarrow{\mathrm{OQ}}-\overrightarrow{\mathrm{OB}})=\overrightarrow{\mathrm{OP}} \cdot \overrightarrow{\mathrm{OQ}}-$ $\overrightarrow{\mathrm{OA}} \cdot \overrightarrow{\mathrm{OQ}}-\overrightarrow{\mathrm{OB}} \cdot \overrightarrow{\mathrm{OP}}+\overrightarrow{\mathrm{OA}} \cdot \overrightarrow{\mathrm{OB}}=-(3 \sqrt{7})^{2}+(\overrightarrow{\mathrm{OB}}-\overrightarrow{\mathrm{OA}}) \cdot \overrightarrow{\mathrm{OQ}}+6 \times 3 \times \cos 120^{\circ}=\overrightarrow{\mathrm{AB}} \cdot \overrightarrow{\mathrm{OQ}}-72 \leq$ $|\overrightarrow{\mathrm{AB}}| \cdot|\overrightarrow{\mathrm{OQ}}|-72=(3 \sqrt{7})^{2}-72=-9$. Therefore, the maximum value of $\overrightarrow{\mathrm{AP}} \cdot \overrightarrow{\mathrm{BQ}}$ is -9, and this maximum value is achieved when $\overrightarrow{\mathrm{PQ}}$ is parallel to $\overrightarrow{\mathrm{AB}}$.