Find the position vector of a point $R$ which divides the line joining $A(-2,1,3)$ and $B(3,5,-2)$ in the ratio $2: 1$ (i) internally (ii) externally.

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Find the position vector of a point $R$ which divides the line joining $A(-2,1,3)$ and $B(3,5,-2)$ in the ratio $2: 1$ (i) internally (ii) externally.

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

Find the position vector of a point $R$ which divides the line joining $A(-2,1,3)$ and $B(3,5,-2)$ in the ratio $2: 1$ (i) internally (ii) externally.

Solution:

$\begin{array}{l} \overrightarrow{O A}=-2 \hat{\imath}+\hat{\jmath}+3 \hat{k} \\ \overrightarrow{O B}=3 \hat{\imath}+5 \hat{\jmath}-2 \hat{k} \end{array}$
(i) $\mathrm{R}$ divides $\overrightarrow{A B}$ internally in the ratio $2: 1$
So $R$ can be given by $\frac{2(3 \hat{\imath}+5 \hat{\jmath}-2 \hat{k})+1(-2 \hat{\imath}+\hat{\jmath}+3 \hat{k})}{3}$ $\overrightarrow{O R}=\frac{(4 \hat{\imath}+11 \hat{\jmath}-\hat{k})}{3}$
(ii) $\mathrm{R}$ divides $\overrightarrow{A B}$ externally in the ratio $2: 1$
So $R$ can be given by $\frac{2(3 \hat{\imath}+5 \hat{\jmath}-2 \hat{k})-1(-2 \hat{\imath}+\hat{\jmath}+3 \hat{k})}{1}$
$\overrightarrow{O R}=8 \hat{\imath}+9 \hat{\jmath}-7 \hat{k}$