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If A,B and C are the vertices of a trian...

If A,B and C are the vertices of a triangle with position vectors `vec(a) , vec(b) ` and `vec(c )` respectively and G is the centroid of `Delta ABC`, then `vec( GA) + vec( GB ) + vec( GC )` is equal to

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Let `veca,vecb` and `vecc` be the position vectors of the vertices A,B and C of a triangle ABC, respectively. Then the position vector of the centroid G is `(veca+vecb+vecc)/(3)`
Now `GvecA+GvecB+GvecC`
`=veca-((veca+vecb+vecc)/(3))+vecb((veca+vecb+vecc)/(3))+vecc((veca+vecb+vecc)/(3))`
`=(veca+vecb+vecc)-(veca+vecb+vecc)=vec0`
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