Using vector method, if $Q$ is the point of concurrence of the medians of the triangle $A B C$ ,then prove that $\to Q A + \to Q B + \to Q C = \to 0$

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Solution

Let $\to a, \to b, \to c$ be the position vector of vertices
A,B.C respectively. then position vector of
centroid Q is $\frac{\to a + \to b + \to c}{3}$
LHS $= - - \to Q A + - - \to Q B + - - \to Q C$
$= (- - \to Q A - - - \to O Q) + (- - \to O B - - - \to O Q) + (- - \to O C - - - \to O Q)$
$= - - \to O A + - - \to O B + - - \to O C - 3 - - \to O Q$
$= \to a + \to b + \to c - 3 (\frac{\to a + \to b + \to c}{3})$
$= \to a + \to b + \to c - \to a - \to b - \to c = \to 0$

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