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Question: If G and G' be the centroids of the triangles ABC and \(A'B'C'\) respectively, then \(\overset{\righ...

If G and G' be the centroids of the triangles ABC and ABCA'B'C' respectively, then AA+BB+CC=\overset{\rightarrow}{AA}' + \overset{\rightarrow}{BB'} + \overset{\rightarrow}{CC}' =

A

23GG\frac{2}{3}\overset{\rightarrow}{GG}'

B

GG\overset{\rightarrow}{GG}'

C

2GG2\overset{\rightarrow}{GG}'

D

3GG3\overset{\rightarrow}{GG}'

Answer

3GG3\overset{\rightarrow}{GG}'

Explanation

Solution

GA+GB+GC=0\overset{\rightarrow}{GA} + \overset{\rightarrow}{GB} + \overset{\rightarrow}{GC} = \mathbf{0} and GA+GB+GC=0\overset{\rightarrow}{G^{'}A^{'}} + \overset{\rightarrow}{G^{'}B^{'}} + \overset{\rightarrow}{G^{'}C^{'}} = \mathbf{0}

(GAGA)+(GBGB)+(GCGC)=0\Rightarrow (\overset{\rightarrow}{GA} - \overset{\rightarrow}{G^{'}A^{'}}) + (\overset{\rightarrow}{GB} - \overset{\rightarrow}{G^{'}B^{'}}) + (\overset{\rightarrow}{GC} - \overset{\rightarrow}{G^{'}C^{'}}) = \mathbf{0}

(GA+GGGA)+(GB+GGGB)+(GC+GGGC)=3GG\Rightarrow (\overset{\rightarrow}{GA} + \overset{\rightarrow}{G^{'}G} - \overset{\rightarrow}{G^{'}A^{'}}) + (\overset{\rightarrow}{GB} + \overset{\rightarrow}{G^{'}G} - \overset{\rightarrow}{G^{'}B^{'}}) + (\overset{\rightarrow}{GC} + \overset{\rightarrow}{G^{'}G} - \overset{\rightarrow}{G^{'}C^{'}}) = 3\overset{\rightarrow}{G^{'}G}

(GAGA)+(GBGB)+(GCGC)=3GG\Rightarrow (\overset{\rightarrow}{GA} - \overset{\rightarrow}{GA^{'}}) + (\overset{\rightarrow}{GB} - \overset{\rightarrow}{GB^{'}}) + (\overset{\rightarrow}{GC} - \overset{\rightarrow}{GC^{'}}) = 3\overset{\rightarrow}{G^{'}G}

AA+BB+CC=3GGAA+BB+CC=3GG\Rightarrow \overset{\rightarrow}{A^{'}A} + \overset{\rightarrow}{B^{'}B} + \overset{\rightarrow}{C^{'}C} = 3\overset{\rightarrow}{G^{'}G} \Rightarrow \overset{\rightarrow}{AA^{'}} + \overset{\rightarrow}{BB^{'}} + \overset{\rightarrow}{CC^{'}} = 3\overset{\rightarrow}{GG^{'}}.