7.0 Section Formula

Let $A$ and $B$ be the two points having position vectors $\overrightarrow a\ and\ \overrightarrow b $ respectively. Let point $C$ be a point dividing the line joining the points $A$ and $B$ in the ratio $m:n$ internally. So the position vector of point $C$ is given as $$\overrightarrow {OC} = \frac{{m\overrightarrow b + n\overrightarrow a }}{{m + n}}$$

Proof: Let $O$ be the origin. Since we know that

$$\begin{equation} \begin{aligned} \overrightarrow {OA} = \overrightarrow a \\ \overrightarrow {OB} = \overrightarrow b \\\end{aligned} \end{equation} $$

Let $\overrightarrow {OC} = \overrightarrow c $ be the position vector of point $C$ which divides the point $C$ in the ratio of $m:n$ internally.

$$\begin{equation} \begin{aligned} \frac{{AC}}{{CB}} = \frac{m}{n} \\ n\overrightarrow {AC} = m\overrightarrow {CB} \\\end{aligned} \end{equation} $$$$n({\text{position vector of }}\vec C - {\text{position vector of }}\vec A) = m(\vec B - \vec C)$$

$$n(\overrightarrow c - \overrightarrow a ) =m(\overrightarrow b -\overrightarrow c )$$$$n\overrightarrow c - n\overrightarrow a = m\overrightarrow b - m\overrightarrow c $$$$\overrightarrow c (m+n)=m(\overrightarrow b+a )$$$$\overrightarrow c = \frac{{m\overrightarrow b + n\overrightarrow a }}{{m + n}}$$$$\overrightarrow {OC} = \frac{{m\overrightarrow b + n\overrightarrow a }}{{m + n}}$$

Let $A$ and $B$ be the two points having position vectors $\overrightarrow a$ and $\overrightarrow b $ respectively. Let point $C$ be a point dividing the line joining the points $A$ and $B$ in the ratio $m:n$ externally. So the position vector of point $C$ is given as $$\overrightarrow {OC} = \frac{{m\overrightarrow b - n\overrightarrow a }}{{m - n}}$$

Proof : Let $O$ be the origin. Since we know that

$$\begin{equation} \begin{aligned} \overrightarrow {OA} = \overrightarrow a \\ \overrightarrow {OB} = \overrightarrow b \\\end{aligned} \end{equation} $$

Take $\overrightarrow {OC} = \overrightarrow c $ be the position vector of point $C$ which divides the point $C$ in the ratio of $m:n$ externally

$$\begin{equation} \begin{aligned} \frac{{AC}}{{BC}} = \frac{m}{n} \\ n\overrightarrow {AC} = m\overrightarrow {BC} \\\end{aligned} \end{equation} $$$$n({\text{position vector of }}\vec C - {\text{position vector of }}\vec A) = m(\vec C - \vec B)$$$$n(\overrightarrow c - \overrightarrow a ) =m(\overrightarrow c -\overrightarrow b )$$$$n\overrightarrow c - n\overrightarrow a = m\overrightarrow c - m\overrightarrow b $$$$\overrightarrow c (m-n)=m(\overrightarrow b-a )$$$$\overrightarrow c = \frac{{m\overrightarrow b - n\overrightarrow a }}{{m - n}}$$$$\overrightarrow {OC} = \frac{{m\overrightarrow b - n\overrightarrow a }}{{m - n}}$$

i) If $C$ is the midpoint of line $AB$, it means point $C$ divide the line $AB$ in the ratio $1:1$ then position vector of $C$ is given by $$\begin{equation} \begin{aligned} \overrightarrow c = \frac{{1.\overrightarrow b + 1.\overrightarrow a }}{{1 + 1}} \\ \overrightarrow c = \frac{{\overrightarrow b + \overrightarrow a }}{2} \\\end{aligned} \end{equation} $$

ii) We have $\overrightarrow c = \frac{{m\overrightarrow b + n\overrightarrow a }}{{m + n}}$ or we can write as $$\begin{equation} \begin{aligned} \overrightarrow c = \frac{m}{{m + n}}\overrightarrow b + \frac{n}{{m + n}}\overrightarrow a \\ \overrightarrow c = \lambda \overrightarrow b + \mu \overrightarrow a \\ where,\lambda = \frac{m}{{m + n}};\mu = \frac{n}{{m + n}} \\\end{aligned} \end{equation} $$

So position vector of $C$ can be taken as $$\overrightarrow c = \lambda \overrightarrow b + \mu \overrightarrow a $$ where $\lambda + \mu = 1$

Question 14. Let $\overrightarrow p ,\overrightarrow q ,\overrightarrow r $ be three vectors such that $\overrightarrow p + \overrightarrow q $ ,$\overrightarrow q + \overrightarrow r $ and are collinear with $\overrightarrow r ,\overrightarrow p $ respectively. Find the value of $\overrightarrow p + \overrightarrow q + \overrightarrow r $.

Solution: According to given condition

$$\begin{equation} \begin{aligned} \overrightarrow p + \overrightarrow q = t\overrightarrow r ...(i) \\ \overrightarrow q + \overrightarrow r = s\overrightarrow p ...(ii) \\\end{aligned} \end{equation} $$

Subtracting $(ii)$ from $(i)$, we get

$$\begin{equation} \begin{aligned} \overrightarrow p - \overrightarrow r = t\overrightarrow r - s\overrightarrow p \\ \overrightarrow p (1 + s) = \overrightarrow r (1 + t) \\\end{aligned} \end{equation} $$

but, $\overrightarrow r ,\overrightarrow p $ are non-collinear

$$(1 + s)=0$$ $$(1 + t)=0$$$$\overrightarrow p + \overrightarrow q = - \overrightarrow r $$

Hence, $$\overrightarrow p + \overrightarrow q + \overrightarrow r = \overrightarrow 0 $$

Question 15. Consider two points $P$ and $Q$ with position vectors $\overrightarrow {OP} = 3\overrightarrow b + 2\overrightarrow a $ and $\overrightarrow {OQ} = 4\overrightarrow b - 6\overrightarrow a $. Find the position vector of $R$ which divides line $PQ$ in the ratio of $2:1$ internally as well as externally.

Solution: The position vector of the point $R$ dividing the join of $P$ and $Q$ internally in the ratio $2:1$ is $$\begin{equation} \begin{aligned} \overrightarrow {OR} = \frac{{2(4\overrightarrow b - 6\overrightarrow a ) + (3\overrightarrow b + 2\overrightarrow a )}}{{2 + 1}} \\ \overrightarrow {OR} = \frac{{8\overrightarrow b - 12\overrightarrow a + 3\overrightarrow b + 2\overrightarrow a }}{3} \\ \overrightarrow {OR} = \frac{{11\overrightarrow b - 10\overrightarrow a }}{3} \\\end{aligned} \end{equation} $$

The position vector of the point $R$ dividing the join of $P$ and $Q$ externally in the ratio $2:1$ is$$\begin{equation} \begin{aligned} \overrightarrow {OR} = \frac{{2(4\overrightarrow b - 6\overrightarrow a ) - (3\overrightarrow b + 2\overrightarrow a )}}{{2 - 1}} \\ \overrightarrow {OR} = \frac{{8\overrightarrow b - 12\overrightarrow a - 3\overrightarrow b - 2\overrightarrow a }}{3} \\ \overrightarrow {OR} = \frac{{5\overrightarrow b - 14\overrightarrow a }}{3} \\\end{aligned} \end{equation} $$