6. A moving point $P$ on a plane is at a distance of 3 from the line containing the segment $A B$ of length 10. When $P A \cdot P B$ reaches its minimum value, then $P A+P B=$ $\qquad$
$6.4 \sqrt{10}$.

From $\frac{1}{2} P A \cdot P B \geqslant S_{\triangle P A B}=\frac{1}{2} \times 10 \times 3=15$, we know $P A \cdot P B \geqslant 30$, and the equality holds when $\angle A P B=90^{\circ}$. At this time,
$$
\begin{array}{l}
P A \cdot P B=30, \text{ and } P A^{2}+P B^{2}=10^{2} . \\
\text{Therefore, } P A+P B=\sqrt{P A^{2}+P B^{2}+2 P A \cdot P B} \\
=\sqrt{160}=4 \sqrt{10} .
\end{array}
$$