Question

The points $D, E, F$ divide $BC, CA, AB$ of triangle $ABC$ in the ratio $1 : 4, 3 : 2$ and $3 : 7$ respectively and the point K divides AB in the ratio $1 : 3$. Let $\overrightarrow{R}_1$ be the resultant of the vectors $\overrightarrow{AD},\overrightarrow{BE},\overrightarrow{CF}$ and let the vector $CK$ be denoted by $\overrightarrow{R}_2$. Then

• A. $\overrightarrow{R}_1=\overrightarrow{R}_2$
• B. $5\,\overrightarrow{R}_1=2\,\overrightarrow{R}_2$
• C. $2\,\overrightarrow{R}_1=5\,\overrightarrow{R}_2$
• D. none of these.

Answer 1

Answer: (B)

Let $\vec{a}, \vec{b}, \vec{c}$ be the position vectors of vertices $A, B, C$ of the $\Delta\,ABC$. Then, the position vectors of $D, E, F$ are respectively
$D\left(\frac{4\,\vec{b}+\vec{c}}{5}\right), E\left(\frac{3\,\vec{a}+2\,\vec{c}}{5}\right), F\left(\frac{7\,\vec{a}+3\,\vec{b}}{10}\right)$
$\therefore \overrightarrow{AD}=\frac{4\,\vec{b}+\vec{c}}{5}-\vec{a}$
$\overrightarrow{BE}=\frac{3\,\vec{a}+2\,\vec{c}}{5}-\vec{b}, \overrightarrow{CF}=\frac{7\,\vec{a}+3\,\vec{b}}{10}-\vec{c}$
Now $\overrightarrow{AD}+\overrightarrow{BE}+\overrightarrow{CF}=\frac{1}{10}(8\,\vec{b}+2\,\vec{c})$
$-10\,\vec{a}+6\,\vec{a}+4\,\vec{c}-10\,\vec{b}+7\,\vec{a}+3\,\vec{b}-10\,\vec{c}$
$=\frac{1}{10}\left(3\,\vec{a}+\vec{b}-4\,\vec{c}\right)$
$\therefore \vec{R}_{1}=\frac{1}{10}\left(3\,\vec{a}+\vec{b}-4\,\vec{c}\right)$
$\Rightarrow 10\,\vec{R}_{1}=3\,\vec{a}+\vec{b}-4\,\vec{c}$
The position vector of the point $K$ is $\frac{\vec{b}+3\,\vec{a}}{4}$
$\therefore \overrightarrow{CK}=\frac{\vec{b}+3\,\vec{a}}{4}-\vec{c}=\frac{1}{4}\left(\vec{b}+3\,\vec{a}-4\,\vec{c}\right)$
$\therefore \vec{R}_{2}=\frac{1}{4}\left(\vec{b}+3\,\vec{a}-4\,\vec{c}\right)$
$\therefore 4\,\vec{R}_{2}=\vec{b}+3\,\vec{a}-4\,\vec{c}$
$\therefore 10\,\vec{R}_{1}=4\,\vec{R}_{2}$
$\Rightarrow 5\,\vec{R}_{1}=2\,\vec{R}_{2}$