Solutions of Question 11 of Exercise 3.2 of Unit 03: Vectors. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.
Find the position vectors of the point of division of the line segments joining point $C$ with position vector $5\hat{j}$ and point $D$ with position vector $4\hat{i}+\hat{j}$ in the ratio $2:5$ internally.
Position vector of $C$ is $\overrightarrow{OC}=5\hat{j}$
Position vector of $D$ is $\overrightarrow{OD}=4\hat{i}+\hat{j}$
Let $H$ be the point divides the line segment $\overline{CD}$ in the ratio $2:5$internally,
then by ratio theorem, we have position vector $H$ is: \begin{align}\overrightarrow{OH}&=\dfrac{5\overrightarrow{OC}+2\overrightarrow{OD}}{5+2}\\ &=\dfrac{5(5\hat{j})+2(4\hat{i}+\hat{j})}{7}\\ &=\dfrac{1}{7}(8\hat{i}+27\hat{j})\\ \implies \overrightarrow{OH}&=\dfrac{8}{7}\hat{i}+\dfrac{27}{7}\hat{j}\end{align}
Find the position vectors of the point of division of the line segments joining point $E$ with position vector $2\hat{i}-3\hat{j}$ and point $F$ with position vector $3\hat{i}+2\hat{j}$ in the ratio $4:3$ externally.
Position vector of $E$ is $\overrightarrow{OE}=2\hat{i}-3\hat{j}$
Position vector of $F$ is $\overrightarrow{OF}=3\hat{i}+2\hat{j}$
Let $K$ be the point with position vector $\overrightarrow{OK}$ that divides the line segment $\overline{EF}$ externally in the ratio $4:3$, then by ratio theorem, \begin{align}\overrightarrow{OK}&=\dfrac{3\overrightarrow{OE}-4\overrightarrow{OF}}{3-4}\\ &=-[3(2\hat{i}-3\hat{j})-4(3\hat{i}+2\hat{j})]\\ &=-(6-12)\hat{i}-(-9-8)\hat{j}\\ \implies \overrightarrow{OK}&=6\hat{i}+17\hat{j}\end{align}