Let $D, E, F$ be the middle points of the sides $B C, C A, A B$ respectively of a triangle $A B C$ . Then $\to A D + \to B E + \to C F$ equals

\to 0

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2 \to A B

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3 \to A B

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None of these

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Solution

The correct option is A $\to 0$
Let the position vector of $A, B, C$ be $\to a, \to b, \to c$ respectively. THen, the position vectors of $D, E, F$ are $\frac{\to b + \to c}{2}, \frac{\to c + \to a}{2}, \frac{\to a + \to b}{2}$ respectively. Therefore.
$\to A D + \to B E + \to C F$
$= (\frac{\to b + \to c}{2} - \to a) + (\frac{\to c + \to a}{2} - \to b) + (\frac{\to a + \to b}{2} - \to c)$
$= \frac{1}{2} [\to b + \to c - 2 \to a + \to c + \to a - 2 \to b + \to a + \to b - 2 \to c] = \to 0$

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