Question

Let position vector of the points A,B and C are $\overrightarrow{a},\overrightarrow{b}$ and $\overrightarrow{c}$ respectively. Point D divides line segment BC internally in the ratio $2:1$. Find vector AD

Answer 1

P.V of $\overrightarrow{D}=\frac{2(\overrightarrow{\left(c\right)}+1(\overrightarrow{b})}{2+1}=\frac{2\overrightarrow{c}+\overrightarrow{b}}{3}$

we have to find $\overrightarrow{AD}$

we know by triangular law of addition

$\overrightarrow{BA}+\overrightarrow{AD}=\overrightarrow{BD}$

$\overrightarrow{AD}=\overrightarrow{BD}-\overrightarrow{BA}$

P.V of $\overrightarrow{BD}=$ P.V of $D-$ P.V of $B$

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

P.V of $\overrightarrow{BA}=(\overrightarrow{a}-\overrightarrow{b})$

$\therefore \overrightarrow{AD}=\frac{2\overrightarrow{c}-2\overrightarrow{b}}{3}-\overrightarrow{a}+\overrightarrow{b}$

$=\frac{2}{3}\overrightarrow{c}-\overrightarrow{a}+\frac{1}{3}\overrightarrow{b}$