Question

Let $\Delta OAB$ be an equilateral triangle with side length unity ( $O$ being the origin). Also, $M$ and $N$ are the points of trisection of $AB$ , $M$ being closer to $A$ and $N$ being closer to $B$ . Position vectors of $A,B,M$ and $N$ are $\overrightarrow{a},\overrightarrow{b},\overrightarrow{m}$ and $\overrightarrow{n}$ respectively, then the value of $\overrightarrow{m}\cdot \overrightarrow{n}$ is equal to

• A. $\frac{1}{2}$
• B. $\frac{2}{3}$
• C. $\frac{13}{18}$
• D. $\frac{4}{9}$

Answer 1

Answer: (C)

$\overrightarrow{m}=\frac{2\overrightarrow{a}+\overrightarrow{b}}{3},\overrightarrow{n}=\frac{\overrightarrow{a}+2\overrightarrow{b}}{3}$
$\overrightarrow{m}\cdot \overrightarrow{n}=\frac{1}{9}\left(2\overrightarrow{a}+\overrightarrow{b}\right)\cdot \left(\overrightarrow{a}+2\overrightarrow{b}\right)$
$=\frac{1}{9}\left[2{\left|\overrightarrow{a}\right|}^2+2{\left|\overrightarrow{b}\right|}^2+5\overrightarrow{a}\cdot \overrightarrow{b}\right]$
$=\frac{1}{9}\left[2+2+5\times 1\times 1\times \frac{1}{2}\right]=\frac{13}{18}$