If G is the centroid of a triangle ABC, then $\vec {GA} +\vec{GB}+\vec{GC}$ equals

$\vec 0$ $3\vec {GA}$ $3\vec {GB}$ $3\vec {GC}$

$\vec 0$

We have, $\vec {GB}+\vec {GC}=(1+1)\vec {GD}$ $⇒\vec {GB}+\vec {GC}=2\vec {GD}$, where D is the mid-point of BC. $⇒\vec {GA}+\vec {GB}+\vec {GC}=\vec {GA}+2\vec {GD}$ ∵ G divides AD in the ratio 2 : 1 $∴ 2 \vec {GD} =-\vec {GA}$ $∴\vec {GA} +\vec{GB}+\vec{GC}=\vec {GA}-\vec {GA}$