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If $D, E, F$ are midpoints of the sides $B C, C A, A B$ respectively of a triangle $A B C$ and $O$ is any point then $\vec{A D} + \frac{2}{3} \vec{B E} + \frac{1}{3} \vec{C F} =$

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$2 \vec{A C}$

$\vec{A C}$

$\frac{1}{2} \vec{A C}$

$\vec{0}$

answer is C.

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Detailed Solution

$\bar{A D} = \frac{\bar{A B} + \bar{A C}}{2}, \bar{B E} = \frac{\bar{B A} + \bar{B C}}{2}, \bar{C F} = \frac{\bar{C A} + \bar{C B}}{2} .$

$Now \bar{A D} + \frac{2}{3} \bar{B E} + \frac{1}{3} \bar{C F}$

$(\frac{\bar{A B} + \bar{A C}}{2}) + \frac{1}{3} (\bar{B A} + \bar{B C}) + \frac{1}{6} (\bar{C A} + \bar{C B})$

$= \frac{1}{6} [\bar{A B} + \bar{B C} + 2 \bar{A C}] = \frac{3 \bar{A C}}{6} = \frac{1}{2} \bar{A C}$

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