Question 88, In rectangle $A B C D$, $A B=2, A D=1$, the moving point $P$ on side $D C$ (including $D, C$) and the moving point $Q$ on the extension of $C B$ (including point $\mathrm{B}$) satisfy $|\overrightarrow{\mathrm{DP}}|=|\overrightarrow{\mathrm{BQ}}|$, then the dot product $\overrightarrow{\mathrm{PA}} \cdot \overrightarrow{\mathrm{PQ}}$ of vector $\overrightarrow{\mathrm{PA}}$ and vector $\overrightarrow{\mathrm{PQ}}$ has the minimum value of $\qquad$ -
Question 88, Solution: As shown in the figure, let $D P=B Q=x$, then
$$
\overrightarrow{\mathrm{PA}} \cdot \overrightarrow{\mathrm{PQ}}=\frac{\mathrm{AP}^{2}+\mathrm{PQ}^{2}-A Q^{2}}{2}=\frac{\left(1+\mathrm{x}^{2}\right)+\left[(2-\mathrm{x})^{2}+(1+x)^{2}\right]-\left(2^{2}+x^{2}\right)}{2}=x^{2}-x+1=\left(x-\frac{1}{2}\right)^{2}+\frac{3}{4} \geq \frac{3}{4}
$$

When $\mathrm{x}=\frac{1}{2}$, we have $\overrightarrow{\mathrm{PA}} \cdot \frac{2}{\mathrm{PQ}}=\frac{3}{4}$.