Question

In $△OAB,$ $E$ is the mid point of $\overrightarrow{AB}$ and $F$ is point on $\overrightarrow{OA}$ such that $\overrightarrow{OF}=2\overrightarrow{FA}.$ If $C$ is the point of intersection of $\overrightarrow{OE}$ and $\overrightarrow{BF},$ then which of the following is/are correct ?

• A. $\overrightarrow{OC}:\overrightarrow{CE}=4:1$
• B. $\overrightarrow{OC}:\overrightarrow{CE}=2:1$
• C. $\overrightarrow{BC}:\overrightarrow{CF}=3:2$
• D. $\overrightarrow{BC}:\overrightarrow{CF}=2:3$

Answer 1

Answer: (A), (C)

$\overrightarrow{BC}:\overrightarrow{CF}=3:2$

From the given figure,
let $\overrightarrow{OA}=\overrightarrow{a},\overrightarrow{OB}=\overrightarrow{b}$
$\Rightarrow \overrightarrow{OF}=\frac{2\overrightarrow{a}}{3},\overrightarrow{OE}=\frac{\overrightarrow{a}+\overrightarrow{b}}{2}$
suppose, $\overrightarrow{BC}:\overrightarrow{CF}=\lambda :1,\overrightarrow{OC}:\overrightarrow{CE}=k:1$
$\Rightarrow \overrightarrow{OC}=\frac{\lambda \overrightarrow{OF}+\overrightarrow{OB}}{\lambda +1}=\frac{k}{k+1}\cdot \overrightarrow{OE}\Rightarrow \overrightarrow{OC}=\frac{\frac{2\lambda \overrightarrow{a}}{3}+\overrightarrow{b}}{\lambda +1}=\frac{\frac{k\overrightarrow{a}+k\overrightarrow{b}}{2}}{k+1}$
now, equating the coefficients of like vectors,
$\Rightarrow \frac{2\lambda }{3(\lambda +1)}=\frac{k}{2(k+1)}\cdots \left(i\right)$ and $\frac{1}{\lambda +1}=\frac{k}{2(k+1)}\cdots \left(ii\right)$
solving $\left(i\right)$ and $\left(ii\right),$
we get, $\lambda =3:2,k=4:1$