Question

If O and O' are circumcentre and orthocentre of ∆ ABC, then $\overrightarrow{OA}+\overrightarrow{OB}+\overrightarrow{OC}$ equals
(a) 2 $\overrightarrow{OO}$'
(b) $O\overrightarrow{O\text{'}}$
(c) $\overrightarrow{O\text{'}O}$
(d) $2\overrightarrow{O\text{'}O}$

Answer 1

Option (b).
Given: $O$ be the circumcentre and $O\text{'}$ be the orthocentre of $△ABC$. Let $G$ be the centroid of the triangle.
We know that $O,G$ and $H$ are collinear and by geometry $\overrightarrow{O\text{'}G}=2\overrightarrow{OG}.$ This yields,
$\overrightarrow{O\text{'}O}=\overrightarrow{O\text{'}G}+\overrightarrow{GO}=2\overrightarrow{GO}+\overrightarrow{GO}=3\overrightarrow{GO}.$
In other words $\overrightarrow{OO\text{'}}=3\overrightarrow{OG}.$
Since, $\overrightarrow{OG}=\frac{\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}}{3}$.

∴ $\overrightarrow{OO\text{'}}=3\times \frac{\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}}{3}=\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}=\overrightarrow{OA}+\overrightarrow{OB}+\overrightarrow{OC}.$