Q. If $A,B$ and $C$ are the vertices of a triangle whose position vectors are $\overrightarrow{\mathbf{a}},\ \overrightarrow{\mathbf{b}}\ and\ \overrightarrow{\mathbf{c}}\ \$respectively $G$ is the centroid of the $\mathrm{\Delta}ABC,$ then $\overrightarrow{\mathbf{GA}} + \overrightarrow{\mathbf{GB}} + \overrightarrow{\mathbf{GC}}\ is\$
Explanation
(a)
Position vectors of vertices $A,B\ and\ C$ of the triangle $ABC$ are $\overrightarrow{\mathbf{a}},\overrightarrow{\mathbf{b}}\ and\ \overrightarrow{\mathbf{c}}$
$\therefore$Centroid of triangle
$(G) = \frac{\overrightarrow{\mathbf{a}} + \overrightarrow{\mathbf{b}} + \overrightarrow{\mathbf{c}}}{3}$
Now, $\overrightarrow{\mathbf{GA}} + \overrightarrow{\mathbf{GB}} + \overrightarrow{\mathbf{GC}}$
$= \left( \overrightarrow{\mathbf{a}} - \frac{\overrightarrow{\mathbf{a}} + \overrightarrow{\mathbf{b}} + \overrightarrow{\mathbf{c}}}{3} \right) + \left( \overrightarrow{\mathbf{b}} - \frac{\overrightarrow{\mathbf{a}} + \overrightarrow{\mathbf{b}} + \overrightarrow{\mathbf{c}}}{3} \right) + \left( \overrightarrow{\mathbf{c}} - \frac{\overrightarrow{\mathbf{a}} + \overrightarrow{\mathbf{b}} + \overrightarrow{\mathbf{c}}}{3} \right)$
$= \overrightarrow{\mathbf{0}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.