 # Section Formula

The physical quantities that have the magnitude, as well as the direction attached to them, are called as vectors. Position vectors denote the location or the position of a point in three-dimensional Cartesian system with respect to an origin. Let us take a look in the upcoming discussion on how you can apply the section formula in vectors. The concept of section formula is implemented for finding the coordinates of a point dividing a line segment either internally or externally in a particular ratio. For locating the position of a point in space, you need a coordinate system.

Consider the figure given below:

Here, P and Q are the points that are represented through the position vectors $\overrightarrow{OP}$ and $\overrightarrow{OQ}$ respectively, with respect to the origin 0. You can divide the line segment that divides the two points P and Q with the help of a third point R in two different points: internally and externally. Let us take a look at booth these cases individually.

Section Formula In Coordinate Geometry

If you wish to find the position vector $\overrightarrow{OQ}$ consider the following cases one by one.

Case 1: When R divides the segment PQ internally

Take a look at this figure again.

From the figure, the point R divides $\overrightarrow{PQ}$ in a way that

m $\overrightarrow{RQ}$ - n $\overrightarrow{PR}$ ... (1)

Here, m and n are called as positive scalars and you can say that the point R divides

$\overrightarrow{PQ}$ internally in m:n ratio.

Now, take a look at the triangles ORQ and OPR. You have

$\overrightarrow{RQ}$ = $\overrightarrow{OQ}$ - $\overrightarrow{OR}$ = $\overrightarrow{b}$  - $\overrightarrow{r}$ and

$\overrightarrow{PR}$ = $\overrightarrow{OR}$ -  $\overrightarrow{OP}$ = $\overrightarrow{r}$ - $\overrightarrow{a}$

When you replace the values of $\overrightarrow{RQ}$ and $\overrightarrow{PR}$ in the equation 1, you get,

m( $\overrightarrow{b}$ - $\overrightarrow{r}$ ) = n($\overrightarrow{r}$ - $\overrightarrow{a})$

or

$\overrightarrow{r}$  = $\frac{\overrightarrow{mb} + \overrightarrow{na}}{m + n}$ ...(2)

Therefore, the position vector formula of point R that divides PQ internally in m:n ratio is given by

$\overrightarrow{OR}$ = $\frac{\overrightarrow{mb} + \overrightarrow{na}}{m + n}$

Case 2: When R divides the segment externally

Consider the figure given below:

In the given figure, the point R divides PQ externally in the m:n ratio. Hence, you can say that the point Q divides PR internally in the (m - n): n ratio.

Hence,

$\frac{PQ}{QR}$ = $\frac{m - n }{n}$

When you use the equation 2, you get,

$\overrightarrow{b}$ = $\frac{(m - n)\overrightarrow{r} + \overrightarrow{na}}{(m - n) + n}$

$\overrightarrow{b}$ = $\frac{(m - n)\overrightarrow{r} + \overrightarrow{na}}{m}$

$\overrightarrow{mb}$ = (m - n)$\overrightarrow{r}$ + $\overrightarrow{na}$

You can also write this as

$\overrightarrow{mb}$ - $\overrightarrow{na}$ = (m - n) $\overrightarrow{r}$

Therefore,

$\overrightarrow{r}$ = $\frac{\overrightarrow{mb} - \overrightarrow{na}}{m - n}$ ...(3)

When R is the midpoint of PQ, m = n.

Hence, from equation 2 you have

$\overrightarrow{OR}$ = $\frac{\overrightarrow{mb} + \overrightarrow{na}}{m + n}$

or,

$\overrightarrow{OR}$ = $\frac{m(\overrightarrow{b} + \overrightarrow{a})}{2m}$

Therefore,

$\overrightarrow{r}$ = $\frac{\overrightarrow{b} + \overrightarrow{a}}{2}$

Therefore, the position vector formula of the midpoint R of PQ is given by,

$\overrightarrow{OR}$ =  $\frac{\overrightarrow{b} + \overrightarrow{a}}{2}$

Section Formula Examples

Example 1:

Let two points P and Q have position vectors $\overrightarrow{OP}$ = $\overrightarrow{3a}$ - $\overrightarrow{2b}$ and $\overrightarrow{OQ}$ = $\overrightarrow{a}$ + $\overrightarrow{b}$

Find the position vector formula of the point R that divides the line joining P and Q in the 2:1 ratio internally and externally.

Solution:

Since the point R divides PQ in the 2:1 ratio, you have m = 2, n = 1.

When R divides PQ internally.

From equation 2, you have

$\overrightarrow{r}$ = $\frac{\overrightarrow{mb} + \overrightarrow{na}}{m + n}$

Hence,

$\overrightarrow{r}$ = $\frac{2(\overrightarrow{a} + \overrightarrow{b}) + (\overrightarrow{3a} - \overrightarrow{2b})}{2 + 1}$  = $\frac{\overrightarrow{5a}}{3}$

When R divides PQ externally,

Considering equation 3,

$\overrightarrow{r}$ = $\frac{\overrightarrow{mb} - \overrightarrow{na}}{m - n}$

Hence,

$\overrightarrow{r}$ = $\frac{2(\overrightarrow{a} + \overrightarrow{b}) - (\overrightarrow{3a} - \overrightarrow{2b})}{2-1}$ = $\overrightarrow{4b}$ - $\overrightarrow{a}$

Example 2

Consider A=(−3, 1), B = (3, -6). Determine the coordinates of the point P (x, y) that divides the line segment AB internally in the ratio 1:2.

Solution:

The given point P is $\frac{1}{1 + 2}$ x AB far from the point A.

If you measure it parallel to the x-axis, you get,

x = -3 + $\frac{1}{3}$ x (3 - (-3)) = -1

If you measure it parallel to the y-axis, you get,

y = 1 + $\frac{1}{3}$ x (-6 -1) = $\frac{-4}{3}$

Hence, the coordinates of the point P are ( -1, $\frac{-4}{3}$ )