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Question: If \(A,B,C\) are the vertices of a triangle whose position vectors are **a**, **b**, **c** and G is ...

If A,B,CA,B,C are the vertices of a triangle whose position vectors are a, b, c and G is the centroid of the ΔABC,\Delta ABC, then GA→+GB→+GC→\overset{\rightarrow}{GA} + \overset{\rightarrow}{GB} + \overset{\rightarrow}{GC} is

A

0

B

A→+B→+C→\overset{\rightarrow}{A} + \overset{\rightarrow}{B} + \overset{\rightarrow}{C}

C

a+b+c3\frac{\mathbf{a} + \mathbf{b} + \mathbf{c}}{3}

D

a+b−c3\frac{\mathbf{a} + \mathbf{b} - \mathbf{c}}{3}

Answer

0

Explanation

Solution

Position vectors of vertices A, B and C of the triangle ABC = a, b and c. We know that position vector of centroid of the triangle (G) =a+b+c3\frac{\mathbf{a} + \mathbf{b} + \mathbf{c}}{3}.

Therefore ,GA→+GB→+GC→\overset{\rightarrow}{GA} + \overset{\rightarrow}{GB} + \overset{\rightarrow}{GC}

=(a−a+b+c3)+(b−a+b+c3)+(c−a+b+c3)= \left( \mathbf{a} - \frac{\mathbf{a} + \mathbf{b} + \mathbf{c}}{3} \right) + \left( \mathbf{b} - \frac{\mathbf{a} + \mathbf{b} + \mathbf{c}}{3} \right) + \left( \mathbf{c} - \frac{\mathbf{a} + \mathbf{b} + \mathbf{c}}{3} \right)

=13[2a−b−c+2b−a−c+2c−a−b]=0= \frac{1}{3}\lbrack 2\mathbf{a} - \mathbf{b} - \mathbf{c} + 2\mathbf{b} - \mathbf{a} - \mathbf{c} + 2\mathbf{c} - \mathbf{a} - \mathbf{b}\rbrack = \mathbf{0}.