Q. If $S$ is the circumcentre, $O$ is the orthocentre of$\mathrm{\Delta}ABC$, then $\overrightarrow{\mathbf{SA}}\mathbf{+}\overrightarrow{\mathbf{SB}}\mathbf{+}\overrightarrow{\mathbf{SC}}$ is equal to
A) $\overrightarrow{\mathbf{SO}}$
B) 2 $\overrightarrow{\mathbf{SO}}$
C) $\overrightarrow{\mathbf{OS}}$
D) 2 $\overrightarrow{\mathbf{OS}}$
Explanation
(a)
From geometry
$\overrightarrow{\mathbf{AO}} = 2\overrightarrow{\mathbf{SD}}$
Where $D$ is the mind point of $BC$
[image:10_image14.emf]c
$\therefore\overrightarrow{\mathbf{SA}}\mathbf{+}\overrightarrow{\mathbf{SB}}\mathbf{+}\overrightarrow{\mathbf{SC}}$
$= \overrightarrow{\mathbf{SA}}\mathbf{+}2\overrightarrow{\mathbf{SD}}\ \ \ (\because\overrightarrow{\mathbf{SB}}\mathbf{+}\overrightarrow{\mathbf{SC}} = 2\overrightarrow{\mathbf{SD}})$
$= \overrightarrow{\mathbf{SA}}\mathbf{+}\overrightarrow{\mathbf{AO}}$
$= \overrightarrow{\mathbf{SO}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.