Question

Let $A\left(\overline{a}\right),B\left(\overline{b}\right),C\left(\overline{c}\right)$ be the vertices of the triangle $ABC$ and let $D,E,F$ be the mid points of the sides $BC,CA,AB$ respectively. If $P$ divides the median $AD$ in the ratio $2:1$ then the position vector of $P$ is

• A. $\overline{0}$
• B. $\overline{a}+\overline{b}+\overline{c}$
• C. $\frac{\overline{b}+\overline{c}+\overline{d}}{3}$
• D. $\frac{2\overline{a}+\overline{b}+\overline{c}}{3}$

Answer 1

The correct option is C $\frac{\overline{b}+\overline{c}+\overline{d}}{3}$

From section formula,

$\overrightarrow{P}=\frac{2\left(\frac{\overrightarrow{b}+\overrightarrow{c}}{2}\right)+\overrightarrow{a}}{2H}=\frac{\overrightarrow{b}+\overrightarrow{c}+\overrightarrow{a}}{3}$

$\Rightarrow \left(C\right)$