Question

If $\overline{a},\overline{b},\overline{c}$ are the position vectors of the points $A,B,C$ respectively and $2\overline{a}+3\overline{b}-5\overline{c}=\overline{0}$, then find the ratio in which the point $C$ divides line segment $AB$.

Answer 1

Let the ratio be $\lambda :1$
Position vector of point $\left(C\right)$
$\overrightarrow{c}=\frac{\lambda \overrightarrow{a}+\overrightarrow{b}}{\lambda +1}$
Put the value of $\overrightarrow{c}$ in $2\overrightarrow{a}+3\overrightarrow{b}-5\overrightarrow{c}=0$
$2\overrightarrow{a}+3\overrightarrow{b}-5\times \left(\frac{\lambda \overrightarrow{a}+\overrightarrow{b}}{\lambda +1}\right)=0$
$\Rightarrow \frac{(\lambda +1)(2\overrightarrow{a}+3\overrightarrow{b})-5\lambda \overrightarrow{a}-5\overrightarrow{b}}{(\lambda +1)}=0$
$\Rightarrow 2\overrightarrow{a}\lambda +3\overrightarrow{b}\lambda +2\overrightarrow{a}+3\overrightarrow{b}-5\lambda \overrightarrow{a}-5\overrightarrow{b}=0$
$\Rightarrow 3\overrightarrow{b}\lambda -3\overrightarrow{a}\lambda +2\overrightarrow{a}-2\overrightarrow{b}=0$
$\Rightarrow 3\lambda (\overrightarrow{b}-\overrightarrow{a})=2\overrightarrow{b}-2\overrightarrow{a}$
$\Rightarrow 3\lambda =\frac{2(\overrightarrow{b}-\overrightarrow{a})}{(\overrightarrow{b}-\overrightarrow{a})}\Rightarrow \frac{\lambda }{1}=\frac{3}{2}$
hence the point $C$ divides line segment $AB$ in the ratio $2:3$.