The position vectors of two points $A$ and $B$ are $(2 \vec{a}+\vec{b})$ and $(\vec{a}-3 \vec{b})$ respectively. Find the position vector of a point $\mathrm{C}$ which divides $\mathrm{AB}$ externally in the ratio $1: 2$. Also, show that $\mathrm{A}$ is the mid-point of the line segment $\mathrm{CB}$.

Home » The position vectors of two points and are and respectively. Find the position vector of a point which divides externally in the ratio . Also, show that is the mid-point of the line segment .

The position vectors of two points $A$ and $B$ are $(2 \vec{a}+\vec{b})$ and $(\vec{a}-3 \vec{b})$ respectively. Find the position vector of a point $\mathrm{C}$ which divides $\mathrm{AB}$ externally in the ratio $1: 2$ . Also, show that $\mathrm{A}$ is the mid-point of the line segment $\mathrm{CB}$ .

CBSE | Class 12 | Maths | RS Aggarwal | Vectors and Their Properties

The position vectors of two points $A$ and $B$ are $(2 \vec{a}+\vec{b})$ and $(\vec{a}-3 \vec{b})$ respectively. Find the position vector of a point $\mathrm{C}$ which divides $\mathrm{AB}$ externally in the ratio $1: 2$ . Also, show that $\mathrm{A}$ is the mid-point of the line segment $\mathrm{CB}$ .

Solution:

Given: $\overrightarrow{O A}=(2 \vec{a}+\vec{b})$
$\overrightarrow{O B}=(\vec{a}-3 \vec{b})$
Position vector of $\mathrm{C}$ which divides $\mathrm{AB}$ in the ratio $1: 2$ externally is given by
$\begin{array}{l} \overrightarrow{O C}=\frac{(\vec{a}-3 \vec{b})-2(2 \vec{a}+\vec{b})}{1-2}=3 \vec{a}+5 \vec{b} \\ \overrightarrow{O A}=2 \vec{a}+\vec{b} \end{array}$
$\begin{array}{l} \overrightarrow{O B}=\vec{a}-3 \vec{b} \\ \overrightarrow{O C}=3 \vec{a}+5 \vec{b} \end{array}$
If A is mid-point, then
OA can also be given by $\frac{(3 \vec{a}+5 \vec{b})+(\vec{a}-3 \vec{b})}{2}$ $\overrightarrow{O A}=2 \vec{a}+\vec{b}$ Which is true
Hence A is mid-point of CB