If G is the centroid of the triangle ABC,then | vedclass

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(A) Let the position vectors of the vertices $A, B,$ and $C$ be $\vec{a}, \vec{b},$ and $\vec{c}$ respectively.Since $G$ is the centroid of $	riangle

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(A) Let the position vectors of the vertices $A, B,$ and $C$ be $\vec{a}, \vec{b},$ and $\vec{c}$ respectively.Since $G$ is the centroid of $	riangle ABC$,its position vector $\vec{g}$ is given by $\vec{g} = \frac{\vec{a} + \vec{b} + \vec{c}}{3}$.This implies $\vec{a} + \vec{b} + \vec{c} = 3\vec{g}$.Now,the sum of the vectors $\vec{GA} + \vec{GB} + \vec{GC}$ can be written in terms of position vectors as:$\vec{GA} + \vec{GB} + \vec{GC} = (\vec{a} - \vec{g}) + (\vec{b} - \vec{g}) + (\vec{c} - \vec{g})$$= (\vec{a} + \vec{b} + \vec{c}) - 3\vec{g}$Substituting $\vec{a} + \vec{b} + \vec{c} = 3\vec{g}$,we get:$= 3\vec{g} - 3\vec{g} = 0$.

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

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