Section Formula in 3 Dimension Made Easy

The section formula is used in two-dimensional geometry to get the coordinates of a point by dividing a line segment internally in a certain ratio. A coordinate system is required to locate the position of a point in space. After selecting a three-dimensional fixed coordinate system, the coordinates of any point P in that system can be determined. It is supplied by an ordered 3-tuple in the case of a rectangular coordinate system (x, y, z). We can also simply locate the point P in space if the coordinates (x, y, z) are already known. The section formula can be applied to three-dimensional geometry as well as determining the coordinates of a point that divides a line in a specific ratio.

In this article we will understand the section formula in 3d and how to prove three points are collinear by section formula with the help of an example.

Section Formula in 3d

Consider the following two points: $A(x_1, y_1, z_1)$ and $B(x_2, y_2, z_2)$. Consider a point $P(x, y, z)$ divided by AB in the $m : n$ ratio, as indicated in the diagram below.

The stages to determining the coordinates of point P are as follows:

Draw AL, PN, and BM perpendicular to the XY plane, as indicated above: $AL || PN || BM$.

The points L, M, and N are on a straight line produced by the intersection of an AL, PN, and BM plane with the XY-plane.

A line segment ST is drawn from point P that is parallel to LM.

Externally, ST intersects AL at S, and internally, it intersects BM at T.

The quadrilaterals LNPS and NMTP qualify as parallelograms since ST is parallel to LM and $AL || PN || BM$.

Also, $\Delta ASP \sim \Delta BTP$ therefore,

$\dfrac{m}{n} = \dfrac{AP}{BP} = \dfrac{AS}{BT} = \dfrac{SL - AL}{BM - TM} = \dfrac{NP - AL}{BM - PN} = \dfrac{z - z_1}{z_2 - z}$

Rearranging the above equation we get,

$z = \dfrac{mz_2 + nz_1}{m + n}$

The x and y coordinates of the point P that internally divides the line segment AB in the ratio m:n can be obtained by drawing perpendiculars to the XZ and YZ planes.

$x = \dfrac{mx_2 + nx_1}{m + n}$

$y = \dfrac{my_2 + ny_1}{m + n}$

Types of Section Formula 3 Dimension:

• Internal Section Formula

• External Section Formula

Sectional Formula (Internally)

When the coordinates of the point P(x, y, z) divides the line segment joining the points $A(x_1, y_1, z_1)$ and $B(x_2, y_2, z_2)$ in the ratio m:n internally are given by the formula:

$(\dfrac{mx_2 + nx_1}{m + n}, \dfrac{my_2 + ny_1}{m + n}, \dfrac{mz_2 + nz_1}{m + n})$

Sectional Formula (Externally)

If the given point P divides the line segment joining the points $A(x_1, y_1, z_1)$ and $B(x_2, y_2, z_2)$ externally in the ratio m:n, then the coordinates of P are given by replacing n with –n as:

$(\dfrac{mx_2 - nx_1}{m - n}, \dfrac{my_2 - ny_1}{m - n}, \dfrac{mz_2 - nz_1}{m - n})$

This represents the section formula 3 dimension.

If point P internally divides the line segment between points A and B in the ratio $k : 1$, then point P's coordinates are

$(\dfrac{kx_2 + x_1}{k + 1}, \dfrac{ky_2 + y_1}{k + 1}, \dfrac{kz_2 + z_1}{k + 1})$

What if the point P dividing the line segment joining points $A(x_1, y_1, z_1)$ and $B(x_2, y_2, z_2)$ is the midpoint of line segment AB?

If P is the midpoint in this case, P splits the line segment AB in a $1:1$ ratio, i.e. $m=n=1$.

Coordinates of point P will be given as:

$(\dfrac{1 \times x_2 + 1 \times x_1}{1 + 1}, \dfrac{1 \times y_2 + 1 \times y_1}{1+1}, \dfrac{1 \times z_2 + 1 \times z_1}{1 + 1})$

Therefore, the coordinates of the midpoint of line segment joining points $A(x_1, y_1, z_1)$ and $B(x_2, y_2, z_2)$ are given by,

Solved Examples:

How to prove three points are collinear by section formula:

1. Prove that the three given positions (– 4, 6, 10), (2, 4, 6) and (14, 0, –2) are collinear using the section formula.

Solution: Assume A(– 4, 6, 10), B(2, 4, 6), and C(14, 0, –2) are the three points. When the point P divides the line segment AB in the ratio $k : 1$, the coordinates of p are as follows:

$\Rightarrow \left[\dfrac{2k-4}{k+1}, \dfrac{4k+6}{k+1}, \dfrac{6k+10}{k+1}\right]$

Let's see if, for some value of k, the point P and the point C are the same.

Now, by equating $\dfrac{2k-4}{k+1} = 14$,

We get the value of k as $-\dfrac{3}{2}$.

If $k=-\dfrac{3}{2}$, then

$\Rightarrow \dfrac{4k+6}{k+1}$

$\Rightarrow \dfrac{4\left(-\dfrac{3}{2}\right)+6}{\left(-\dfrac{3}{2}\right)+1} = 0$

Similarly,

$\Rightarrow \dfrac{6k+10}{k+1}$

$\Rightarrow \dfrac{6\left(-\dfrac{3}{2}\right)+10}{\left(-\dfrac{3}{2}\right)+1} = -2$

As a result, the point C(14, 0, -2) divides the line segment externally in a 3:2 ratio, and is the same as the point P.

Hence, the points A, B, and C are all collinear.

2. Determine the coordinate points that divide the line segment connecting the points (1, -2, 3) and (3, 4, -5) in a 2:3 internal and external ratio.

Solution: Let’s assume that the point P(x, y, z) divides the line segment that joins A(1, -2, 3) and B(3, 4, -5) internally in the ratio of 2:3.

Here, m=2, and n=3

Now, substitute the values in the section formula,

$\Rightarrow \left[\dfrac{mx_2+nx_1}{m+n}, \dfrac{my_2+ny_1}{m+n}, \dfrac{mz_2+nz_1}{m+n}\right]$

$\Rightarrow \left[\dfrac{2(3)+3(1)}{2+3}, \dfrac{2(4)+3(-2)}{2+3}, \dfrac{2(-5)+3(3)}{2+3} \right]

$\Rightarrow \left[\dfrac{9}{5}, \dfrac{2}{5}, -\dfrac{1}{-5}\right]$

Hence, the required point is $\left[\dfrac{9}{5}, \dfrac{2}{5}, -\dfrac{1}{-5}\right]$

Suppose the point P(x, y, z) divides the line segment that joins the points A(1, -2, 3) and B(3, 4, -5) externally in the ratio of 2:3, then

$\Rightarrow \left[\dfrac{mx_2-nx_1}{m-n}, \dfrac{my_2-ny_1}{m-n}, \dfrac{mz_2-nz_1}{m-n}\right]$

$\Rightarrow \left[\dfrac{2(3)-3(1)}{2-3}, \dfrac{2(4)-3(-2)}{2-3}, \dfrac{2(-5)-3(3)}{2-3}\right]$

$\Rightarrow (-3, -14, 19)$

Hence, the required point is (-3, -14, 19).

Conclusion:

Section Formula in 3D, Points are the most fundamental characteristics of geometry, and they have no dimensions. The section formula for internal division is well explained, however the external division formula is obtained by applying the internal division formula. The formula's exceptional instances, such as points of trisection and the midpoint formula, are also addressed with clarity.