13. Given vector $\vec{a}=\{2,-3\}$, vector $\overrightarrow{A B}$ is perpendicular to $\vec{a}$, and $|\overrightarrow{A B}|=3 \sqrt{13}$, point $A$ has coordinates $(-3,2)$, find the coordinates of the position vector $\overrightarrow{O B}$.
13. Let $\overrightarrow{O B}=\{x, y\}, \overrightarrow{A B}=\{x+3, y-2\}$. Since $\overrightarrow{A B}$ is perpendicular to $\vec{a}$, and $|\overrightarrow{A B}|=3 \sqrt{13}$, we have $\left\{\begin{array}{l}2(x+3)-3(y-2)=0 \\ (x+3)^{2}+(y-2)^{2}=(3 \sqrt{13})^{2},\end{array}\right.$ which gives $\left\{\begin{array}{l}x=6 \\ y=8\end{array}\right.$ or $\left\{\begin{array}{l}x=-12 \\ y=-4 .\end{array}\right.$