JEE Main - Section formula in three dimensions E-Book Study Material

Section Formula in Three Dimensions

In coordinate geometry, the section formula is a fundamental concept used to determine the coordinates of a point that divides a line segment in a given ratio. When we extend this to three dimensions, we are dealing with points in space, and the formula helps us find the coordinates of a point that divides the line segment joining two points in 3D space internally or externally.

Internal Division

Consider two points $A(x_1, y_1, z_1)$ and $B(x_2, y_2, z_2)$ in three-dimensional space. Let $P(x, y, z)$ be a point that divides the line segment $AB$ internally in the ratio $m:n$. This means that the distance $AP$ to the distance $PB$ is $m/n$.

To derive the coordinates of point $P$, we can think of this in terms of vectors or by extending the 2D section formula.

Derivation using Vectors:

Let $\vec{a}$ be the position vector of point $A$, and $\vec{b}$ be the position vector of point $B$. So, $\vec{a} = \hat{i}x_1 + \hat{j}y_1 + \hat{k}z_1$ and $\vec{b} = \hat{i}x_2 + \hat{j}y_2 + \hat{k}z_2$.

Let $\vec{p}$ be the position vector of point $P$. Since $P$ divides $AB$ internally in the ratio $m:n$, we have $\vec{AP} = \frac{m}{m+n} \vec{AB}$.

The vector $\vec{AP}$ can be written as $\vec{p} - \vec{a}$.

The vector $\vec{AB}$ can be written as $\vec{b} - \vec{a}$.

Substituting these into the equation: $\vec{p} - \vec{a} = \frac{m}{m+n} (\vec{b} - \vec{a})$

$\vec{p} = \vec{a} + \frac{m}{m+n} (\vec{b} - \vec{a})$

$\vec{p} = \frac{(m+n)\vec{a} + m(\vec{b} - \vec{a})}{m+n}$

$\vec{p} = \frac{m\vec{a} + n\vec{a} + m\vec{b} - m\vec{a}}{m+n}$

$\vec{p} = \frac{n\vec{a} + m\vec{b}}{m+n}$

Now, substituting the components of the position vectors: $\hat{i}x + \hat{j}y + \hat{k}z = \frac{n(\hat{i}x_1 + \hat{j}y_1 + \hat{k}z_1) + m(\hat{i}x_2 + \hat{j}y_2 + \hat{k}z_2)}{m+n}$

$\hat{i}x + \hat{j}y + \hat{k}z = \frac{\hat{i}(nx_1 + mx_2) + \hat{j}(ny_1 + my_2) + \hat{k}(nz_1 + mz_2)}{m+n}$

By equating the coefficients of $\hat{i}$, $\hat{j}$, and $\hat{k}$ on both sides, we get the coordinates of point $P$:

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

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

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

So, the coordinates of point $P$ that divides the line segment joining $A(x_1, y_1, z_1)$ and $B(x_2, y_2, z_2)$ internally in the ratio $m:n$ are: $P \left( \frac{mx_2 + nx_1}{m+n}, \frac{my_2 + ny_1}{m+n}, \frac{mz_2 + nz_1}{m+n} \right)$

Memory Trick for Internal Division: Think of it as a weighted average. The ratio $m:n$ means $P$ is closer to $A$ if $n$ is larger, and closer to $B$ if $m$ is larger. The coordinate corresponding to $A$ ($x_1$) is multiplied by $n$ (the part of the ratio associated with $B$), and the coordinate corresponding to $B$ ($x_2$) is multiplied by $m$ (the part of the ratio associated with $A$). The sum of these weighted coordinates is then divided by the sum of the ratio parts ($m+n$).

Example 1:

Find the coordinates of the point that divides the line segment joining the points $A(1, 2, 3)$ and $B(4, 5, 6)$ internally in the ratio $2:1$.

Here, $(x_1, y_1, z_1) = (1, 2, 3)$, $(x_2, y_2, z_2) = (4, 5, 6)$, $m=2$, and $n=1$.

Using the section formula: $x = \frac{m x_2 + n x_1}{m+n} = \frac{2(4) + 1(1)}{2+1} = \frac{8+1}{3} = \frac{9}{3} = 3$

$y = \frac{m y_2 + n y_1}{m+n} = \frac{2(5) + 1(2)}{2+1} = \frac{10+2}{3} = \frac{12}{3} = 4$

$z = \frac{m z_2 + n z_1}{m+n} = \frac{2(6) + 1(3)}{2+1} = \frac{12+3}{3} = \frac{15}{3} = 5$

So, the coordinates of the point are $P(3, 4, 5)$.

External Division

Now, consider a point $P(x, y, z)$ that divides the line segment $AB$ externally in the ratio $m:n$. This means that $P$ lies on the line containing $A$ and $B$, but outside the segment $AB$, such that the ratio of the distance $AP$ to the distance $PB$ is $m/n$. For external division, we assume $m > n$ if $P$ is on the side of $B$, or $n > m$ if $P$ is on the side of $A$.

Derivation using Vectors:

Let $\vec{a}$ and $\vec{b}$ be the position vectors of points $A$ and $B$ respectively. Let $\vec{p}$ be the position vector of point $P$.

If $P$ divides $AB$ externally in the ratio $m:n$, then $\vec{AP} = \frac{m}{n} \vec{BP}$ is not the correct vector relationship. The correct way to express this is that $P$ divides $AB$ in the ratio $m:(-n)$ internally.

So, we can use the internal division formula by replacing $n$ with $-n$.

The position vector $\vec{p}$ is given by: $\vec{p} = \frac{n\vec{a} + m\vec{b}}{m+n}$ (from internal division)

Replacing $n$ with $-n$: $\vec{p} = \frac{(-n)\vec{a} + m\vec{b}}{m+(-n)}$

$\vec{p} = \frac{m\vec{b} - n\vec{a}}{m-n}$

Substituting the components of the position vectors: $\hat{i}x + \hat{j}y + \hat{k}z = \frac{m(\hat{i}x_2 + \hat{j}y_2 + \hat{k}z_2) - n(\hat{i}x_1 + \hat{j}y_1 + \hat{k}z_1)}{m-n}$

$\hat{i}x + \hat{j}y + \hat{k}z = \frac{\hat{i}(mx_2 - nx_1) + \hat{j}(my_2 - ny_1) + \hat{k}(mz_2 - nz_1)}{m-n}$

Equating the coefficients of $\hat{i}$, $\hat{j}$, and $\hat{k}$ gives the coordinates of point $P$:

$x = \frac{mx_2 - nx_1}{m-n}$

$y = \frac{my_2 - ny_1}{m-n}$

$z = \frac{mz_2 - nz_1}{m-n}$

So, the coordinates of point $P$ that divides the line segment joining $A(x_1, y_1, z_1)$ and $B(x_2, y_2, z_2)$ externally in the ratio $m:n$ are: $P \left( \frac{mx_2 - nx_1}{m-n}, \frac{my_2 - ny_1}{m-n}, \frac{mz_2 - nz_1}{m-n} \right)$

Memory Trick for External Division: Notice the similarity to the internal division formula. The only difference is the minus sign in the numerator and the denominator. The coordinate corresponding to $A$ ($x_1$) is multiplied by $n$ (and subtracted), and the coordinate corresponding to $B$ ($x_2$) is multiplied by $m$ (and remains positive). The sum of these weighted coordinates is then divided by the difference of the ratio parts ($m-n$). This formula is valid only if $m \neq n$. If $m=n$, the denominator becomes zero, which signifies that the point is at infinity, meaning the line is parallel to the line segment.

Example 2:

Find the coordinates of the point that divides the line segment joining the points $A(1, 2, 3)$ and $B(4, 5, 6)$ externally in the ratio $2:1$.

Here, $(x_1, y_1, z_1) = (1, 2, 3)$, $(x_2, y_2, z_2) = (4, 5, 6)$, $m=2$, and $n=1$.

Using the external division formula: $x = \frac{m x_2 - n x_1}{m-n} = \frac{2(4) - 1(1)}{2-1} = \frac{8-1}{1} = 7$

$y = \frac{m y_2 - n y_1}{m-n} = \frac{2(5) - 1(2)}{2-1} = \frac{10-2}{1} = 8$

$z = \frac{m z_2 - n z_1}{m-n} = \frac{2(6) - 1(3)}{2-1} = \frac{12-3}{1} = 9$

So, the coordinates of the point are $P(7, 8, 9)$.

Special Cases

Midpoint Formula in 3D

The midpoint of a line segment is a special case of internal division where the ratio is $1:1$. If $P(x, y, z)$ is the midpoint of the line segment joining $A(x_1, y_1, z_1)$ and $B(x_2, y_2, z_2)$, then $m=1$ and $n=1$.

Using the internal division formula: $x = \frac{1 \cdot x_2 + 1 \cdot x_1}{1+1} = \frac{x_1 + x_2}{2}$ $y = \frac{1 \cdot y_2 + 1 \cdot y_1}{1+1} = \frac{y_1 + y_2}{2}$ $z = \frac{1 \cdot z_2 + 1 \cdot z_1}{1+1} = \frac{z_1 + z_2}{2}$

Thus, the midpoint $P$ has coordinates: $P \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}, \frac{z_1 + z_2}{2} \right)$

Example 3:

Find the midpoint of the line segment joining $A(2, -1, 4)$ and $B(6, 3, 0)$.

Using the midpoint formula: $x = \frac{2+6}{2} = \frac{8}{2} = 4$ $y = \frac{-1+3}{2} = \frac{2}{2} = 1$ $z = \frac{4+0}{2} = \frac{4}{2} = 2$

The midpoint is $(4, 1, 2)$.

Centroid of a Triangle in 3D

Consider a triangle with vertices $A(x_1, y_1, z_1)$, $B(x_2, y_2, z_2)$, and $C(x_3, y_3, z_3)$. The centroid $G$ is the point of intersection of the medians. A median connects a vertex to the midpoint of the opposite side.

Let $D$ be the midpoint of side $BC$. The coordinates of $D$ are: $D \left( \frac{x_2 + x_3}{2}, \frac{y_2 + y_3}{2}, \frac{z_2 + z_3}{2} \right)$

The centroid $G$ divides the median $AD$ internally in the ratio $2:1$.

Using the section formula with $m=2$ and $n=1$, and points $A(x_1, y_1, z_1)$ and $D \left( \frac{x_2 + x_3}{2}, \frac{y_2 + y_3}{2}, \frac{z_2 + z_3}{2} \right)$:

$x_G = \frac{2 \left(\frac{x_2 + x_3}{2}\right) + 1(x_1)}{2+1} = \frac{x_2 + x_3 + x_1}{3} = \frac{x_1 + x_2 + x_3}{3}$

$y_G = \frac{2 \left(\frac{y_2 + y_3}{2}\right) + 1(y_1)}{2+1} = \frac{y_2 + y_3 + y_1}{3} = \frac{y_1 + y_2 + y_3}{3}$

$z_G = \frac{2 \left(\frac{z_2 + z_3}{2}\right) + 1(z_1)}{2+1} = \frac{z_2 + z_3 + z_1}{3} = \frac{z_1 + z_2 + z_3}{3}$

So, the coordinates of the centroid $G$ are: $G \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3}, \frac{z_1 + z_2 + z_3}{3} \right)$

Shortcut for Centroid: The coordinates of the centroid are simply the average of the corresponding coordinates of the vertices of the triangle.

Example 4:

Find the centroid of the triangle with vertices $A(1, 2, 3)$, $B(0, 4, -1)$, and $C(2, -1, 5)$.

Using the centroid formula: $x_G = \frac{1 + 0 + 2}{3} = \frac{3}{3} = 1$ $y_G = \frac{2 + 4 + (-1)}{3} = \frac{5}{3}$ $z_G = \frac{3 + (-1) + 5}{3} = \frac{7}{3}$

The centroid is $G \left( 1, \frac{5}{3}, \frac{7}{3} \right)$.

Applications and Significance

The section formula in three dimensions is a cornerstone for solving various problems in analytical geometry and physics.

Vector Analysis: It forms the basis for vector operations involving points in space, such as finding the resultant of forces or determining the center of mass of systems of particles.
Physics: In mechanics, it's used to find the center of mass of rigid bodies or systems of particles. In electromagnetism, it can be applied in calculating electric and magnetic fields.
Computer Graphics: It's fundamental in computer graphics for transformations, object manipulation, and defining spatial relationships between points and lines.
Engineering: Used in structural analysis, robotics, and path planning where precise spatial calculations are required.

Understanding the section formula, both for internal and external division, is crucial for mastering three-dimensional geometry. It provides a direct method to locate points based on ratios and is a stepping stone to more complex concepts like planes, lines in 3D, and surfaces of revolution. The ability to switch between vector notation and coordinate notation makes this formula versatile.