Section Formula in 3D Geometry – Class 11 Mathematics
Written by Neeraj Anand
Published by ANAND TECHNICAL PUBLISHERS
For CBSE Board Exams and JEE Mains & Advanced Preparation
📘 Introduction
In three-dimensional geometry, the section formula helps determine the coordinates of a point that divides a line segment joining two points in a given ratio. This formula is an extension of the two-dimensional section formula and plays a vital role in solving geometry problems related to space.
This concept is essential for students preparing for CBSE board exams and JEE Mains & Advanced as it forms the foundation for more complex topics in 3D geometry.
In order to locate the position of a point in space, we require a coordinate system. Section Formula is a useful tool in 3D coordinate geometry, which helps us find the coordinate of any point on a space which is dividing the plane into some known ratio.
Suppose a point divides a plane segment into two parts which may be equal or not, with the help of the section formula we can find the coordinates of that point.
✏️ Definition of Section Formula in 3D
If a point P(x,y,z) divides the line segment joining two points
A(x1,y1,z1) and B(x2,y2,z2) in the ratio m:n, then the coordinates of P can be found using the section formula.
There are two types of division:
- Internal Division: The point divides the line segment between A and B.
- External Division: The point divides the line segment outside the segment AB.
Derivation of Section Formula in 3D
Let us consider two points A (x1, y1, z1) and B(x2, y2, z2). Consider a point P(x, y, z) dividing AB in the ratio m:n as shown in the figure given below.
To find the coordinates of the point P, the following steps are followed:
- Draw AL, PN, and BM perpendicular to XY plane such that AL || PN || BM as shown above in the Figure.
- The points L, M and N lie on the straight line formed due to the intersection of a plane containing AL, PN and BM and XY- plane.
- From point P, a line segment ST is drawn such that it is parallel to LM.
- ST intersects AL externally at S, and it intersects BM at T internally.
Since ST is parallel to LM and AL || PN || BM, therefore, the quadrilaterals LNPS and NMTP qualify as parallelograms.
Also, ∆ASP ~∆BTP therefore,
\(\begin{array}{l}\large \frac{m}{n} = \frac{AP}{BP} = \frac{AS}{BT} = \frac{SL – AL}{BM – TM} = \frac{NP – AL}{BM – PN} = \frac{z – z_{1}}{z_{2} – z}\end{array} \)
After Rearranging the above equation we get,
\(\begin{array}{l}z=\large \frac{mz_{2}+ nz_{1}}{m + n}\end{array} \)
The above procedure can be repeated by drawing perpendiculars to XZ and YZ- planes to get the x and y coordinates of the point P that divides the line segment AB in the ratio m:n internally.
\(\begin{array}{l}\large x = \frac{mx_{2}+ nx_{1}}{m+n}, y = \frac{my_{2}+ ny_{1}}{m+n}\end{array} \)
Sectional Formula (Internally) in 3D
The coordinates of the point P(x, y, z) dividing the line segment joining the points A(x1, y1, z1) and B(x2, y2, z2) in the ratio m:n internally are given by:
\(\begin{array}{l}\large \left ( \frac{mx_{2}+ nx_{1}}{m+n}, \frac{my_{2}+ ny_{1}}{m+n}, \frac{mz_{2}+ nz_{1}}{m+n} \right )\end{array} \)
Sectional Formula (Externally) in 3D
If the given point P divides the line segment joining the points A(x1, y1, z1) and B(x2, y2, z2) externally in the ratio m:n, then the coordinates of P are given by replacing n with –n as:
\(\begin{array}{l}\large \left ( \frac{mx_{2}- nx_{1}}{m-n}, \frac{my_{2}- ny_{1}}{m-n}, \frac{mz_{2}- nz_{1}}{m-n} \right )\end{array} \)
This represents the section formula for three dimension geometry.
If the point P divides the line segment joining points A and B internally in the ratio k:1, then the coordinates of point P will be
\(\begin{array}{l}\large \left ( \frac{kx_{2}+ x_{1}}{k+1}, \frac{ky_{2}+ y_{1}}{k+1}, \frac{kz_{2}+ z_{1}}{k+1} \right )\end{array} \)
Sectional Formula in 3D For Mid Point
In this case the point P dividing the line segment joining points A(x1, y1, z1) and B(x2, y2, z2) is the midpoint of line segment AB. Here, point P is the midpoint of line segment AB, then P divides the line segment AB in the ratio 1:1, i.e. m=n=1.
Coordinates of mid point P will be given as:
\(\begin{array}{l}\large \left ( \frac{1 \times x_{2}+ 1 \times x_{1}}{1+1}, \frac{1 \times y_{2}+ 1 \times y_{1}}{1+1}, \frac{1 \times z_{2}+ 1 \times z_{1}}{1+1} \right )\end{array} \)
Therefore, the coordinates of the midpoint of line segment joining points A(x1, y1, z1) and B(x2, y2, z2) are given by,
\(\begin{array}{l}\large \left ( \frac{x_{2}+ x_{1}}{2}, \frac{y_{2}+ y_{1}}{2}, \frac{z_{2}+ z_{1}}{2} \right )\end{array} \)
Solved Examples of Section Formula in 3D
Example 1: Determine the coordinate points which divide the line segment that joins the point (1, -2, 3) and (3, 4, -5) in the ratio of 2:3 internally and externally.
Solution:
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,
⇒ [(mx2+nx1)/m+n, (my2+ny1)/m+n, (mz2+nz1)/m+n ]
⇒ [(2(3)+3(1))/2+3,(2(4)+3(-2))/2+3, (2(-5)+3(3))/2+3 ]
⇒[(9/5), (⅖), (-⅕)]
Hence, the required point is [(9/5), (⅖), (-⅕)]
Let the point P(x, y, z) divides the line segment that joins A(1, -2, 3) and B(3, 4, -5) externally in the ratio of 2:3, then
⇒ [(mx2-nx1)/m-n, (my2-ny1)/m-n, (mz2-nz1)/m-n ]
⇒ [(2(3)-3(1))/2-3,(2(4)-3(-2))/2-3, (2(-5)-3(3))/2-3 ]
⇒[-3, -14, 19]
Hence, the required point is (-3, -14, 19).
Example 2: Show that the three points (– 4, 6, 10), (2, 4, 6) and (14, 0, –2) are collinear using the section formula.
Solution:
Assume that the three points are A(– 4, 6, 10), B(2, 4, 6), C(14, 0, –2). Now, the point P divides the line segment AB in the ratio of k:1, then the coordinates of p are:
⇒[(2k-4)/k+1, (4k+6)/k+1, (6k+10)/k+1]
Now, let’s check whether some value of k, the point P coincides with the point C.
By equating (2k-4)/k+1 = 14, we get the value of k as -3/2.
If k=-3/2, then
⇒(4k+6)/k+1
⇒(4(-3/2)+6)/(-3/2)+1
⇒ 0
Similarly,
⇒ (6k+10)/k+1
⇒(6(-3/2)+10)/(-3/2)+1
⇒-2
Hence, the point C(14, 0, -2) which divides the line segment externally in the ratio of 3:2, which is the same as the point P.
Therefore, the points A, B and C are collinear.
Practice Problems
Solve the problems given below:
- Show that the points P(2, –3, 4), Q(–1, 2, 1), R(0, ⅓, 2) are collinear using section formula.
- Calculate the ratio in which the YZ-plane divides the line segment formed by joining the points (-2, 4, 7) and (3, -5, 8).
- Find the coordinate point that divides the line segment that joins the points (– 2, 3, 5) and (1, – 4, 6) in the ratio of 2:3 externally.
✅ Applications of Section Formula in 3D Geometry
- Finding coordinates of specific points dividing a segment in a given ratio
- Calculating centroids of triangles or tetrahedrons
- Useful for solving geometrical problems involving ratios and proportions in space
- Essential for advanced topics like vectors and spatial analysis in physics
📝 Conclusion
Mastering the section formula in 3D geometry is crucial for students aiming for high scores in board exams and competitive exams like JEE Mains and Advanced. It builds the foundation for solving complex spatial problems and enhances understanding of geometric relationships in three dimensions.
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