3D Coordinate System & Distance Formula

Master the three-dimensional coordinate system, distance formula, section formula, and coordinate transformations for JEE Main and Advanced

7 min read

Real-World Hook: GPS Navigation

When your phone shows “You are 150m away” from a destination, it’s calculating the distance formula in 3D space - considering latitude, longitude, and altitude! GPS satellites use 3D coordinates to pinpoint your exact location on Earth. Similarly, video game engines use 3D coordinate systems to render characters and objects in virtual worlds.

The 3D Coordinate System

3D Coordinate System with Point P(x, y, z)XYZP(x, y, z)P'(x, y, 0)(x, 0, 0)(0, y, 0)(x, 0, z)(0, y, z)OPO(0,0,0)zXY plane (z=0)YZ plane(x=0)XZ plane (y=0)Right-Hand Rule:Thumb: +XIndex: +YMiddle: +ZDistance from Origin:OP = sqrt(x^2 + y^2 + z^2)Extension of 2D formulaDistance Between Points:AB = sqrt((x2-x1)^2 + (y2-y1)^2+ (z2-z1)^2)A(x1,y1,z1), B(x2,y2,z2)8 Octants:I: (+,+,+)V: (+,+,-)II: (-,+,+)VI: (-,+,-)III: (-,-,+)VII: (-,-,-)IV: (+,-,+)VIII:(+,-,-)Projections:XY: (x, y, 0)YZ: (0, y, z)XZ: (x, 0, z)

Understanding the Three Axes

In 3D space, any point P is represented by an ordered triplet (x, y, z) where:

  • x-coordinate: Distance from the YZ-plane (left/right)
  • y-coordinate: Distance from the XZ-plane (forward/backward)
  • z-coordinate: Distance from the XY-plane (up/down)

Key Convention: We use the right-hand coordinate system:

  • Point your right thumb along positive X-axis
  • Index finger along positive Y-axis
  • Middle finger points along positive Z-axis

The Eight Octants

Just like 2D has 4 quadrants, 3D space is divided into 8 octants:

OctantxyzExample Point
I+++(2, 3, 4)
II-++(-2, 3, 4)
III--+(-2, -3, 4)
IV+-+(2, -3, 4)
V++-(2, 3, -4)
VI-+-(-2, 3, -4)
VII---(-2, -3, -4)
VIII+--(2, -3, -4)

Distance Formula in 3D

Formula

The distance between two points A(x₁, y₁, z₁) and B(x₂, y₂, z₂) is:

$$\boxed{AB = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}}$$

Interactive Demo: Visualize 3D Distance

Explore points and distances in three-dimensional space with interactive visualization.

Memory Trick 🧠

“XYZ Differences Squared and Rooted”

  • Think of it as the 2D distance formula plus the z-component
  • 2D: √[(Δx)² + (Δy)²]
  • 3D: √[(Δx)² + (Δy)² + (Δz)²]

Special Case: Distance from Origin

Distance of point P(x, y, z) from origin O(0, 0, 0):

$$\boxed{OP = \sqrt{x^2 + y^2 + z^2}}$$

Common Mistake ⚠️

Wrong: Taking absolute differences instead of squares

❌ AB = √(|x₂ - x₁| + |y₂ - y₁| + |z₂ - z₁|)
✓  AB = √[(x₂ - x₁)² + (y₂ - y₁)² + (z₂ - z₁)²]

Sign Awareness: The order (x₂ - x₁) vs (x₁ - x₂) doesn’t matter because of squaring!

Section Formula

Internal Division

If point P(x, y, z) divides the line segment joining A(x₁, y₁, z₁) and B(x₂, y₂, z₂) internally in the ratio m:n, then:

$$\boxed{P = \left(\frac{mx_2 + nx_1}{m+n}, \frac{my_2 + ny_1}{m+n}, \frac{mz_2 + nz_1}{m+n}\right)}$$

Memory Trick 🧠

“M loves 2, N loves 1”

  • m goes with the second point (x₂, y₂, z₂)
  • n goes with the first point (x₁, y₁, z₁)
  • Denominator is always (m + n) for internal division

External Division

If P divides AB externally in ratio m:n, then:

$$\boxed{P = \left(\frac{mx_2 - nx_1}{m-n}, \frac{my_2 - ny_1}{m-n}, \frac{mz_2 - nz_1}{m-n}\right)}$$

Key Difference:

  • Internal: m + n in denominator, + in numerator
  • External: m - n in denominator, - in numerator (second term)

Midpoint Formula (Special Case: m = n = 1)

$$\boxed{\text{Midpoint} = \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}, \frac{z_1 + z_2}{2}\right)}$$

Common Mistake ⚠️

Mixing up internal and external formulas:

Internal division (2:1) from A(1,2,3) to B(4,5,6):
❌ P = ((2·4 - 1·1)/(2-1), ...) = (7, ...)  [Used external formula!]
✓  P = ((2·4 + 1·1)/(2+1), ...) = (3, ...)  [Correct internal formula]

Centroid of a Triangle and Tetrahedron

Centroid of Triangle

For triangle with vertices A(x₁, y₁, z₁), B(x₂, y₂, z₂), C(x₃, y₃, z₃):

$$\boxed{G = \left(\frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3}, \frac{z_1 + z_2 + z_3}{3}\right)}$$

Centroid of Tetrahedron

For tetrahedron with vertices A, B, C, D:

$$\boxed{G = \left(\frac{x_1 + x_2 + x_3 + x_4}{4}, \frac{y_1 + y_2 + y_3 + y_4}{4}, \frac{z_1 + z_2 + z_3 + z_4}{4}\right)}$$

Memory Trick 🧠

“Average all coordinates” - Just take the arithmetic mean of all x’s, y’s, and z’s separately!

Worked Examples

Example 1: Distance Calculation (JEE Main Level)

Problem: Find the distance between A(2, -1, 3) and B(-1, 0, 4).

Solution: Using distance formula:

$$AB = \sqrt{(-1-2)^2 + (0-(-1))^2 + (4-3)^2}$$ $$= \sqrt{(-3)^2 + (1)^2 + (1)^2}$$ $$= \sqrt{9 + 1 + 1} = \sqrt{11}$$

Answer: √11 units

Example 2: Section Formula (JEE Main Level)

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

Solution: Using internal section formula with m = 2, n = 1:

$$x = \frac{2(4) + 1(1)}{2+1} = \frac{9}{3} = 3$$ $$y = \frac{2(5) + 1(2)}{2+1} = \frac{12}{3} = 4$$ $$z = \frac{2(6) + 1(3)}{2+1} = \frac{15}{3} = 5$$

Answer: P(3, 4, 5)

Verification: P is closer to B than A (ratio 2:1 means P divides closer to B).

Example 3: Collinear Points (JEE Advanced Level)

Problem: Show that points A(1, 2, 3), B(2, 3, 4), and C(3, 4, 5) are collinear.

Solution: For collinear points, AB + BC = AC (or any similar combination).

$$AB = \sqrt{(2-1)^2 + (3-2)^2 + (4-3)^2} = \sqrt{1+1+1} = \sqrt{3}$$ $$BC = \sqrt{(3-2)^2 + (4-3)^2 + (5-4)^2} = \sqrt{1+1+1} = \sqrt{3}$$ $$AC = \sqrt{(3-1)^2 + (4-2)^2 + (5-3)^2} = \sqrt{4+4+4} = 2\sqrt{3}$$

Since AB + BC = √3 + √3 = 2√3 = AC, the points are collinear.

Alternative Method: Check if the coordinates follow a linear pattern:

  • x: 1, 2, 3 (common difference = 1)
  • y: 2, 3, 4 (common difference = 1)
  • z: 3, 4, 5 (common difference = 1)

All coordinates have the same pattern → Points are collinear.

Practice Problems

Level 1: JEE Main Basics

  1. Find the distance between points P(3, 4, 5) and Q(6, 8, 10).

  2. Find the midpoint of the line segment joining A(-2, 3, 5) and B(4, -1, 7).

  3. In which octant does the point P(-3, 4, -2) lie?

  4. Find the distance of point (2, 3, 6) from the origin.

Level 2: JEE Main Standard

  1. Point P divides the line joining A(2, 3, 4) and B(5, 6, 7) in ratio 2:1 internally. Find the coordinates of P.

  2. The vertices of a triangle are A(1, 2, 3), B(4, 5, 6), C(7, 8, 9). Find the centroid.

  3. If the distance between points (3, y, 5) and (1, 2, 3) is √12, find the value(s) of y.

  4. Find the ratio in which the YZ-plane divides the line joining (2, 4, 5) and (-3, 5, 6).

Level 3: JEE Advanced

  1. Show that the points A(1, 2, 3), B(2, 4, 5), C(3, 6, 7) are collinear and find the ratio in which B divides AC.

  2. Find the locus of a point which is equidistant from the points A(1, 2, 3) and B(3, 2, 1).

  3. The vertices of a tetrahedron are (0, 0, 0), (4, 0, 0), (0, 4, 0), (0, 0, 4). Find:

    • (a) Centroid
    • (b) Distance from centroid to each vertex
    • (c) Volume using determinant formula

  4. Point P moves such that its distance from A(1, 0, 0) is twice its distance from B(0, 0, 0). Find the equation of the locus.

Quick Revision Formulas

ConceptFormula
Distance AB√[(x₂-x₁)² + (y₂-y₁)² + (z₂-z₁)²]
Distance from origin√(x² + y² + z²)
Internal division (m:n)((mx₂+nx₁)/(m+n), …)
External division (m:n)((mx₂-nx₁)/(m-n), …)
Midpoint((x₁+x₂)/2, (y₁+y₂)/2, (z₁+z₂)/2)
Centroid (triangle)((x₁+x₂+x₃)/3, …)
Centroid (tetrahedron)((x₁+x₂+x₃+x₄)/4, …)

Common Exam Patterns

JEE Main typically asks:

  • Direct distance calculations (1-2 marks)
  • Section formula applications (2-3 marks)
  • Centroid and midpoint problems (2 marks)
  • Combined with vectors or lines (3-4 marks)

JEE Advanced patterns:

  • Locus problems in 3D
  • Proving collinearity/coplanarity
  • Optimization (max/min distance)
  • Integration with vector algebra

Time-Saving Tip: For exam speed, memorize the midpoint formula separately rather than substituting m=n=1 in section formula each time!

Last updated: November 2025