Question

A point $C=\frac{5\overrightarrow{a}+4\overrightarrow{b}-5\overrightarrow{c}}{3}$ divides the line joining $A=\overrightarrow{a}-2\overrightarrow{b}+3\overrightarrow{c}$ and $B$ in the ratio $2:1,$ then $\overrightarrow{AB}$ is

• A. $\overrightarrow{a}+3\overrightarrow{b}-4\overrightarrow{c}$
• B. $2\overrightarrow{a}-3\overrightarrow{b}+4\overrightarrow{c}$
• C. $\overrightarrow{a}+5\overrightarrow{b}-7\overrightarrow{c}$
• D. $2\overrightarrow{a}+3\overrightarrow{b}-4\overrightarrow{c}$

Answer 1

The correct option is C $\overrightarrow{a}+5\overrightarrow{b}-7\overrightarrow{c}$

Let position vector of point $B$ be $(x,y,z)$
Then position vector of point $C$ is
$\Rightarrow \frac{2x+1}{3}=\frac{5}{3}\Rightarrow x=2$
$\Rightarrow \frac{2y-2}{3}=\frac{4}{3}\Rightarrow y=3$
$\Rightarrow \frac{2z+3}{3}=\frac{-5}{3}\Rightarrow z=-4$
$\Rightarrow \overrightarrow{AB}=\overrightarrow{a}+5\overrightarrow{b}-7\overrightarrow{c}$