Question

In $△OAB,$ if $\overrightarrow{OA}=\overrightarrow{a},\overrightarrow{OB}=\overrightarrow{b},L$ is mid point of $\overrightarrow{OA}$ and $M$ is a point on $\overrightarrow{OB}$ such that $\overrightarrow{OM}:\overrightarrow{MB}=2:1.$ If $P$ is the mid point of $\overrightarrow{LM},$ then $\overrightarrow{AP}=$

• A. $\frac{\overrightarrow{a}}{4}+\frac{4\overrightarrow{b}}{3}$
• B. $-\frac{3\overrightarrow{a}}{4}-\frac{4\overrightarrow{b}}{3}$
• C. $-\frac{3\overrightarrow{a}}{4}+\frac{\overrightarrow{b}}{3}$
• D. $\frac{\overrightarrow{a}}{4}+\frac{\overrightarrow{b}}{3}$

Answer 1

The correct option is C $-\frac{3\overrightarrow{a}}{4}+\frac{\overrightarrow{b}}{3}$

Given : $L$ is mid point of $\overrightarrow{OA}$
$\Rightarrow \overrightarrow{OL}=\frac{\overrightarrow{a}}{2}$

$\overrightarrow{OM}:\overrightarrow{MB}=2:1.$
P.V. of $M=\frac{2\overrightarrow{b}}{3}$
Since $P$ is mid point of $LM$
P.V. of $P=\frac{\frac{\overrightarrow{a}}{2}+\frac{2\overrightarrow{b}}{3}}{2}$
$\overrightarrow{AP}=\frac{\overrightarrow{a}}{4}+\frac{\overrightarrow{b}}{3}-\overrightarrow{a}$
$\therefore \overrightarrow{AP}=-\frac{3\overrightarrow{a}}{4}+\frac{\overrightarrow{b}}{3}$