Section Formula - Practice Questions & MCQ

Let A and B be two points represented by the position vectors $\overrightarrow{O A}$ and $\overrightarrow{O B}$, respectively, with respect to the origin O. 

Let R be a point that divides the line segment joining the points A and B in the ratio m: n. 

Internal Division

If R divides AB internally in the ratio m: n, then the position vector of R is given by  $\overrightarrow{\mathrm{OR}}=\frac{m \vec{b}+n \vec{a}}{m+n}$

Proof : 

Let $O$ be the origin. Then $\overrightarrow{\mathrm{OA}}=\tilde{\mathbf{a}}$ and $\overrightarrow{\mathrm{OB}}=\tilde{\mathbf{b}}$. Let $\tilde{\mathbf{r}}$ be the position vector of R which divides AB internally in the ratio $\mathrm{m}: \mathrm{n}$. Then
or
$
\begin{aligned}
& \frac{\mathrm{AR}}{\mathrm{RB}}=\frac{\mathrm{m}}{\mathrm{n}} \\
& \mathrm{n}(\overrightarrow{A R})=\mathrm{m}(\overrightarrow{R B})
\end{aligned}
$

Now from triangles ORB and OAR, we have
$
\overrightarrow{\mathrm{RB}}=\overrightarrow{\mathrm{OB}}-\overrightarrow{\mathrm{OR}}=\tilde{\mathrm{b}}-\tilde{\mathrm{r}}
$
and,
$
\overrightarrow{\mathrm{AR}}=\overrightarrow{\mathrm{OR}}-\overrightarrow{\mathrm{OA}}=\tilde{\mathrm{r}}-\tilde{\mathrm{a}}
$

Therefore, we have
or
$
\begin{gathered}
\mathrm{m}(\tilde{\mathrm{~b}}-\tilde{\mathrm{r}})=\mathrm{n}(\tilde{\mathrm{r}}-\tilde{\mathrm{a}}) \\
\vec{r}=\frac{m \vec{b}+n \vec{a}}{m+n}
\end{gathered}
$

Hence, the position vector of the point R which divides A and B internally in the ratio of m: n is given by

$\overrightarrow{\mathrm{OR}}=\frac{m \vec{b}+n \vec{a}}{m+n}$

External Division

If R divides AB externally in the ratio m: n, then the position vector of R is given by $\overrightarrow{\mathrm{OR}}=\frac{m \vec{b}-n \vec{a}}{m-n}$

NOTE:

If R is the midpoint of AB, then m = n. And therefore, the midpoint R of $\overrightarrow{A B}$ , will have its position vector as$\overrightarrow{\mathrm{OR}}=\frac{\vec{a}+\vec{b}}{2}$