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 $\overset{ \rightarrow }{a}, \, \overset{ \rightarrow }{b}, \, \overset{ \rightarrow }{m}$ and $\overset{ \rightarrow }{n}$ respectively, then the value of $\overset{ \rightarrow }{m}\cdot \overset{ \rightarrow }{n}$ is equal to

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

Answer 1

Answer: (C)

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