# Find the position vector of a point R which divides the line joining two points P and Q whose position vectors are

Question:

Find the position vector of a point $R$ which divides the line joining two points $P$ and $Q$ whose position vectors are $(2 \vec{a}+\vec{b})$ and $(\vec{a}-3 \vec{b})$ externally in the ratio $1: 2$. Also, show that $P$ is the mid point of the line segment $R Q$.

Solution:

It is given that $\overrightarrow{\mathrm{OP}}=2 \vec{a}+\vec{b}, \overrightarrow{\mathrm{OQ}}=\vec{a}-3 \vec{b}$.

It is given that point R divides a line segment joining two points P and Q externally in the ratio 1: 2. Then, on using the section formula, we get:

$\overrightarrow{\mathrm{OR}}=\frac{2(2 \vec{a}+\vec{b})-(\vec{a}-3 \vec{b})}{2-1}=\frac{4 \vec{a}+2 \vec{b}-\vec{a}+3 \vec{b}}{1}=3 \vec{a}+5 \vec{b}$

Therefore, the position vector of point $\mathrm{R}$ is $3 \vec{a}+5 \vec{b}$.

Position vector of the mid-point of $R Q=\frac{\overrightarrow{O Q}+\overrightarrow{O R}}{2}$

$=\frac{(\vec{a}-3 \vec{b})+(3 \vec{a}+5 \vec{b})}{2}$

$=2 \vec{a}+\vec{b}$

$=\overrightarrow{\mathrm{OP}}$

Hence, P is the mid-point of the line segment RQ.