Question

If P is orthocentre, Q is circumcenter and G is centroid of a triangle ABC, then prove that $\overline{QP}=3\overline{QG}$.

Answer 1

Let $\overline{p}$ and $\overline{g}$ be the poistion vectors of $P$ and $G$ w.r.t. the circumcentre $Q.$

i.e. $\overline{QR}=\overline{p}$ and $\overline{QG}=\overline{g}$

We know that $Q,G,P$ are collinear and $G$ diviced segment $QP$ internally in the ratio $1:2.$

$\therefore$ by section formula for internal division,

$\overline{g}=\frac{1.\overline{p}+2\overline{q}}{1+2}=\frac{\overline{p}}{3}$ ......$[\because \overline{q}=0]$

$\therefore \overline{p}=3\overline{g}$

$\overline{QP}=3\overline{QG}.$