Let $$\vec p$$ and $$\vec q$$ be the position vectors of $$P$$ and $$Q$$ respectively, with respect to $$O$$ and $$\left| {\vec p} \right| = p,\,\left| {\vec q} \right| = q.$$    The points $$R$$ and $$S$$ divide $$PQ$$  internally and externally in the ratio 2 : 3 respectively. If $$OR$$  and $$OS$$  are perpendicular then :

Solution :
We have $$\overrightarrow {OR} = \frac{{3\vec p + 2\vec q}}{{3 + 2}} = \frac{1}{2}\left( {3\vec p + 2\vec q} \right)\,\,\,\,\left[ {{\text{Internal division}}} \right]$$
and $$\overrightarrow {OS} = \frac{{3\vec p - 2\vec q}}{{3 - 2}} = 3\vec p - 2\vec q\,\,\,\,\left[ {{\text{External division}}} \right]$$
$$\eqalign{ & {\text{Given }}\overrightarrow {OR} \, \bot \,\overrightarrow {OS} \Rightarrow \overrightarrow {OR} \,.\,\overrightarrow {OS} = 0 \cr & \Rightarrow \frac{1}{5}\left[ {3\vec p + 2\vec q} \right].\left( {3\vec p - 2\vec q} \right) = 0 \cr & \Rightarrow 9{\left| {\vec p} \right|^2} = 4{\left| {\vec q} \right|^2} \cr & \Rightarrow 9{p^2} = 4{q^2} \cr} $$