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(B) Let the position vectors of $A, B, C, D, E, F$ be $\overline{a}, \overline{b}, \overline{c}, \overline{d}, \overline{e}, \overline{f}$ respectivel

Vedclass · Accepted Answer

$\frac{1}{2} \overline{AC}$

Vedclass · Answer

(B) Let the position vectors of $A, B, C, D, E, F$ be $\overline{a}, \overline{b}, \overline{c}, \overline{d}, \overline{e}, \overline{f}$ respectively.$	herefore \overline{d} = \frac{\overline{b} + \overline{c}}{2}, \overline{e} = \frac{\overline{c} + \overline{a}}{2}, \overline{f} = \frac{\overline{a} + \overline{b}}{2}$.Now,$\overline{AD} + \frac{2}{3} \overline{BE} + \frac{1}{3} \overline{CF} = (\overline{d} - \overline{a}) + \frac{2}{3}(\overline{e} - \overline{b}) + \frac{1}{3}(\overline{f} - \overline{c})$.Substituting the values:$= \frac{\overline{b} + \overline{c}}{2} - \overline{a} + \frac{2}{3}\left(\frac{\overline{c} + \overline{a}}{2} - \overline{b}ight) + \frac{1}{3}\left(\frac{\overline{a} + \overline{b}}{2} - \overline{c}ight)$.$= \frac{\overline{b} + \overline{c} - 2\overline{a}}{2} + \frac{\overline{c} + \overline{a} - 2\overline{b}}{3} + \frac{\overline{a} + \overline{b} - 2\overline{c}}{6}$.$= \frac{3(\overline{b} + \overline{c} - 2\overline{a}) + 2(\overline{c} + \overline{a} - 2\overline{b}) + (\overline{a} + \overline{b} - 2\overline{c})}{6}$.$= \frac{3\overline{b} + 3\overline{c} - 6\overline{a} + 2\overline{c} + 2\overline{a} - 4\overline{b} + \overline{a} + \overline{b} - 2\overline{c}}{6}$.$= \frac{-3\overline{a} + 3\overline{c}}{6} = \frac{3(\overline{c} - \overline{a})}{6} = \frac{1}{2}(\overline{c} - \overline{a}) = \frac{1}{2} \overline{AC}$.

If $D, E$ and $F$ are the mid-points of the sides $BC, CA$ and $AB$ of triangle $ABC$ respectively,then $\overline{AD} + \frac{2}{3} \overline{BE} + \frac{1}{3} \overline{CF} =$

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