If $\vec{a}, \vec{b}, \vec{c}$ are position vectors of points $\mathrm{A}, \mathrm{B}, \mathrm{C}$…
If $\vec{a}, \vec{b}, \vec{c}$ are position vectors of points $\mathrm{A}, \mathrm{B}, \mathrm{C}$ respectively, with $2 \vec{a}+3 \vec{b}-5 \vec{c}=\overrightarrow{0}$, then the ratio in which point $\mathrm{C}$ divides segment $\mathrm{AB}$ is
2:3 internally
2:3 externally
3:2 internally
3:2 externally
Solution
$2 \vec{a}+3 \vec{b}=5 \vec{c} \Rightarrow \vec{c}=\frac{2 \vec{a}+3 \vec{b}}{5}=\frac{2 \vec{a}+3 \vec{b}}{2+3}$
i.e., $\vec{c}$ divides $\vec{a}$ and $\vec{b}$ in the ratio 3:2 internally