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 $\frac{\vec{b}+3\vec{a}}{4}$
$\therefore \overrightarrow{CK}=\frac{\vec{b}+3\vec{a}}{4}-\vec{c}=\frac{1}{4}\left(\vec{b}+3\vec{a}-4\vec{c}\right)$
$\therefore {\vec{R}}_2=\frac{1}{4}\left(\vec{b}+3\vec{a}-4\vec{c}\right)$
$\therefore 4{\vec{R}}_2=\vec{b}+3\vec{a}-4\vec{c}$
$\therefore 10{\vec{R}}_1=4{\vec{R}}_2$
$\Rightarrow 5{\vec{R}}_1=2{\vec{R}}_2$