6. Let $|\overrightarrow{A B}|=10$, if a point $P$ on the plane satisfies, for any $t \in \mathbf{R}$, $|\overrightarrow{A P}-t \overrightarrow{A B}| \geq 3$, then the minimum value of $\overrightarrow{P A} \cdot \overrightarrow{P B}$ is
$\qquad$ , and at this time $|\overrightarrow{P A}+\overrightarrow{P B}|=$ $\qquad$
Rong囊 $-16,6$.
Analysis From $|\overrightarrow{A P}-t \overrightarrow{A B}| \geq 3$ we know that the distance from point $P$ to line $A B$ is 3. Let the midpoint of $A B$ be $O$. By the polarization identity, we get $\overrightarrow{P A} \cdot \overrightarrow{P B}=\frac{1}{4}\left\{(\overrightarrow{P A}+\overrightarrow{P B})^{2}-(\overrightarrow{P A}-\overrightarrow{P B})^{2}\right\}=\frac{1}{4}\left\{(2 P O)^{2}-10^{2}\right\} \geq \frac{1}{4}\{36-100\}=-16$ at this time $|\overrightarrow{P A}+\overrightarrow{P B}|=6$.