Question

If $2a+3b-5c=0$ then prove that the points $A\left(\overline{a}\right),B\left(\overline{b}\right)$ and $C\left(\overline{c}\right)$ are collinear. Hence, find the ratio in which the point $C$ divides the line segment $AB$.

Answer 1

Given,

$\overrightarrow{2a}+\overrightarrow{3b}-\overrightarrow{5c}=0$

$\Rightarrow \overrightarrow{2a}+\overrightarrow{3b}=\overrightarrow{5c}$

$\Rightarrow \frac{2}{5}\overrightarrow{a}+\frac{3}{5}b=c$

$\Rightarrow \frac{2\overrightarrow{a}+3\overrightarrow{b}}{2+3}=\overrightarrow{c}$

$\therefore$ point $C\left(\overrightarrow{c}\right)$ divides $A\left(\overrightarrow{a}\right)$ and $B\left(\overrightarrow{b}\right)$ internally in the ratio $2:3$ and hence they are collinear as well.