Q69 / 160 · TS EAMCET 2022 (20 Jul Shift 2) · Mathematics
Let $\mathrm{ABC}$ be a triangle and $\bar{a}, \bar{b}, \bar{c}$ be the position vectors of $\mathrm{A}, \mathrm{B}, \mathrm{C}$ respectively, Let $\mathrm{D}$ divide $\mathrm{BC}$ in the ratio $3: 1$ internally and $\mathrm{E}$ divide $\mathrm{AD}$ in the ratio $4: 1$ internally. Let $\mathrm{BE}$ meet $\mathrm{AC}$ in $\mathrm{F}$. If $\mathrm{E}$ divides $\mathrm{BF}$ in the ratio $3: 2$ internally then the position vector of $\mathrm{F}$ is
- A$\frac{\bar{a}+\bar{b}+\bar{c}}{3}$
- B$\frac{\bar{a}-2 \bar{b}+3 \bar{c}}{2}$
- C$\frac{\bar{a}+2 \bar{b}+3 \bar{c}}{2}$
- Correct answerD$\frac{\bar{a}-\bar{b}+3 \bar{c}}{3}$
Solution
Here we are given that $\overline{\mathrm{OA}}=\overrightarrow{\mathrm{a}}, \overline{\mathrm{OB}}=\overrightarrow{\mathrm{b}}, \overline{\mathrm{OC}}=\overline{\mathrm{c}}$
Now P.V of D i.e $\overline{\mathrm{OD}}=\frac{1 \times \overline{\mathrm{OB}}+3 \times \overline{\mathrm{OC}}}{1+3}$
$\begin{aligned}<br />& \overline{\mathrm{OD}}=\frac{\overrightarrow{\mathrm{b}}+3 \overrightarrow{\mathrm{c}}}{4} \\<br />& \overline{\mathrm{OE}}=\frac{4 \overline{\mathrm{OD}}+\overline{\mathrm{OA}}}{4+1}=\frac{\frac{4(\overrightarrow{\mathrm{b}}+3 \overrightarrow{\mathrm{c}})}{4}+\overrightarrow{\mathrm{a}}}{5} \Rightarrow \overline{\mathrm{OE}}=\frac{\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{c}}}{5}<br />\end{aligned}$
Now, $\overline{\mathrm{OE}}=\frac{2 \overline{\mathrm{OB}}+3 \overline{\mathrm{OF}}}{2+3}$
$\Rightarrow \overline{\mathrm{OF}}=\frac{5 \overline{\mathrm{OE}}-2 \overline{\mathrm{OB}}}{3}$
$\Rightarrow \overline{\mathrm{OF}}=\frac{\frac{5(\overline{\mathrm{a}}+\overrightarrow{\mathrm{b}}+3 \overrightarrow{\mathrm{c}}}{5}-\overline{2 \mathrm{~b}}}{3}$
$\Rightarrow \overline{\mathrm{OF}}=\frac{\overrightarrow{\mathrm{a}}-\overrightarrow{\mathrm{b}}+3 \overrightarrow{\mathrm{c}}}{3}$
$\Rightarrow$ P.V. of F is $\frac{\vec{a}-\vec{b}+3 \vec{c}}{3}$
Now P.V of D i.e $\overline{\mathrm{OD}}=\frac{1 \times \overline{\mathrm{OB}}+3 \times \overline{\mathrm{OC}}}{1+3}$
$\begin{aligned}<br />& \overline{\mathrm{OD}}=\frac{\overrightarrow{\mathrm{b}}+3 \overrightarrow{\mathrm{c}}}{4} \\<br />& \overline{\mathrm{OE}}=\frac{4 \overline{\mathrm{OD}}+\overline{\mathrm{OA}}}{4+1}=\frac{\frac{4(\overrightarrow{\mathrm{b}}+3 \overrightarrow{\mathrm{c}})}{4}+\overrightarrow{\mathrm{a}}}{5} \Rightarrow \overline{\mathrm{OE}}=\frac{\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{c}}}{5}<br />\end{aligned}$
Now, $\overline{\mathrm{OE}}=\frac{2 \overline{\mathrm{OB}}+3 \overline{\mathrm{OF}}}{2+3}$
$\Rightarrow \overline{\mathrm{OF}}=\frac{5 \overline{\mathrm{OE}}-2 \overline{\mathrm{OB}}}{3}$
$\Rightarrow \overline{\mathrm{OF}}=\frac{\frac{5(\overline{\mathrm{a}}+\overrightarrow{\mathrm{b}}+3 \overrightarrow{\mathrm{c}}}{5}-\overline{2 \mathrm{~b}}}{3}$
$\Rightarrow \overline{\mathrm{OF}}=\frac{\overrightarrow{\mathrm{a}}-\overrightarrow{\mathrm{b}}+3 \overrightarrow{\mathrm{c}}}{3}$
$\Rightarrow$ P.V. of F is $\frac{\vec{a}-\vec{b}+3 \vec{c}}{3}$
Hence, the correct answer is option: 'D'.