1. Let $P$ be any point on the graph of the function $y=x+\frac{2}{x}(x>0)$, and draw perpendiculars from $P$ to the line $y=x$ and the $y$-axis, with the feet of the perpendiculars being $A$ and $B$, respectively. Then the value of $\overrightarrow{P A} \cdot \overrightarrow{P B}$ is $\qquad$ .
Let $P\left(x_{0}, x_{0}+\frac{2}{x_{0}}\right)$, then $|P A|=\frac{\sqrt{2}}{x_{0}},|P B|=x_{0}, \angle A P B=\frac{3 \pi}{4}$. So $\overrightarrow{P A} \cdot \overrightarrow{P B}=\frac{\sqrt{2}}{x_{0}} \cdot x_{0} \cdot\left(-\frac{\sqrt{2}}{2}\right)=-1$.