Equation of a Plane in 3D Space

Master all forms of plane equations - normal form, intercept form, general form, and plane through three points for JEE Main and Advanced

10 min read

Real-World Hook: Flat Surfaces Everywhere

Every flat wall in your room, every table top, every floor represents a plane in 3D space! Architects use plane equations to design buildings, aerospace engineers use them to model aircraft wings, and geologists use them to describe fault lines in Earth’s crust. Even your phone screen is mathematically a plane!

What is a Plane?

A plane is a flat, two-dimensional surface extending infinitely in 3D space. It requires:

  1. A point on the plane, OR
  2. A normal vector (perpendicular to the plane), OR
  3. Three non-collinear points, OR
  4. Intercepts on the axes

Form 1: General Equation of a Plane

Equation

$$\boxed{ax + by + cz + d = 0}$$

Interactive Demo: Visualize Planes in 3D

See how planes are oriented in three-dimensional space.

where:

  • a, b, c are constants (not all zero)
  • (a, b, c) represents the normal vector to the plane
  • d is a constant
  • (x, y, z) is any point on the plane

Key Insight

The coefficients (a, b, c) give the direction ratios of the normal (perpendicular) to the plane!

Memory Trick ๐Ÿง 

“ABC is the Normal Direction”

  • Plane: ax + by + cz + d = 0
  • Normal vector: n = ai + bj + ck
  • This is analogous to 2D: ax + by + c = 0 has normal (a, b)

Form 2: Normal Form (Point-Normal Form)

Equation

If plane passes through point (xโ‚, yโ‚, zโ‚) with normal vector (a, b, c):

$$\boxed{a(x - x_1) + b(y - y_1) + c(z - z_1) = 0}$$

Vector Form

$$\boxed{(\vec{r} - \vec{a}) \cdot \vec{n} = 0}$$

where:

  • r = position vector of any point on plane = xi + yj + zk
  • a = position vector of given point = xโ‚i + yโ‚j + zโ‚k
  • n = normal vector = ai + bj + ck

Geometric Interpretation

Vector from given point to any point on plane (r - a) is perpendicular to normal n. Hence their dot product = 0!

Form 3: Intercept Form

Equation

Plane making intercepts a, b, c on X, Y, Z axes:

$$\boxed{\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1}$$

where:

  • a = x-intercept (plane cuts X-axis at (a, 0, 0))
  • b = y-intercept (plane cuts Y-axis at (0, b, 0))
  • c = z-intercept (plane cuts Z-axis at (0, 0, c))

Memory Trick ๐Ÿง 

“Sum of Ratios Equals 1”

  • Exactly like 2D line intercept form: x/a + y/b = 1
  • Extended to 3D: x/a + y/b + z/c = 1

Converting to General Form

Multiply throughout by abc:

$$bcx + acy + abz = abc$$

Or: bcx + acy + abz - abc = 0

Common Mistake โš ๏ธ

Intercept vs coefficient confusion:

Plane: x/2 + y/3 + z/4 = 1

โŒ x-intercept = 1/2  [Taking reciprocal!]
โœ“  x-intercept = 2   [Denominator itself]

The plane cuts X-axis at (2, 0, 0), Y-axis at (0, 3, 0), Z-axis at (0, 0, 4)

Form 4: Plane Through Three Points

Method

Given three non-collinear points A(xโ‚, yโ‚, zโ‚), B(xโ‚‚, yโ‚‚, zโ‚‚), C(xโ‚ƒ, yโ‚ƒ, zโ‚ƒ), the plane equation is:

$$\boxed{\begin{vmatrix} x - x_1 & y - y_1 & z - z_1 \\ x_2 - x_1 & y_2 - y_1 & z_2 - z_1 \\ x_3 - x_1 & y_3 - y_1 & z_3 - z_1 \end{vmatrix} = 0}$$

Alternative Method (Easier!)

Step 1: Let plane be ax + by + cz + d = 0

Step 2: Substitute all three points to get three equations

Step 3: Solve for a, b, c in terms of d (or vice versa)

Step 4: Write final equation

Memory Trick ๐Ÿง 

“Determinant with Differences”

  • First row: (x - xโ‚, y - yโ‚, z - zโ‚) [general point minus first point]
  • Second row: (xโ‚‚ - xโ‚, yโ‚‚ - yโ‚, zโ‚‚ - zโ‚) [second point minus first]
  • Third row: (xโ‚ƒ - xโ‚, yโ‚ƒ - yโ‚, zโ‚ƒ - zโ‚) [third point minus first]

Form 5: Vector Form of Plane

Equation

$$\boxed{\vec{r} = \vec{a} + \lambda\vec{b} + \mu\vec{c}}$$

where:

  • r = position vector of any point on plane
  • a = position vector of a given point on plane
  • b, c = two non-parallel vectors parallel to the plane
  • ฮป, ฮผ = parameters (scalars)

This is the parametric form of a plane.

Converting to Cartesian Form

If r = xi + yj + zk, a = aโ‚i + aโ‚‚j + aโ‚ƒk, b = bโ‚i + bโ‚‚j + bโ‚ƒk, c = cโ‚i + cโ‚‚j + cโ‚ƒk:

$$x = a_1 + \lambda b_1 + \mu c_1$$ $$y = a_2 + \lambda b_2 + \mu c_2$$ $$z = a_3 + \lambda b_3 + \mu c_3$$

Eliminate ฮป and ฮผ to get Cartesian form.

Special Planes

Plane Parallel to XY-plane

z = k (constant)

General form: 0x + 0y + 1z - k = 0 Normal: (0, 0, 1) - parallel to Z-axis

Plane Parallel to YZ-plane

x = k (constant)

Normal: (1, 0, 0) - parallel to X-axis

Plane Parallel to XZ-plane

y = k (constant)

Normal: (0, 1, 0) - parallel to Y-axis

Plane Through Origin

If plane passes through origin (0, 0, 0):

$$\boxed{ax + by + cz = 0}$$

(The constant term d = 0)

Finding Intercepts from General Form

Given: ax + by + cz + d = 0

Step 1: Rearrange to: ax + by + cz = -d

Step 2: Divide by -d:

$$\frac{ax}{-d} + \frac{by}{-d} + \frac{cz}{-d} = 1$$

Step 3: Rewrite as:

$$\frac{x}{-d/a} + \frac{y}{-d/b} + \frac{z}{-d/c} = 1$$

Intercepts:

  • x-intercept = -d/a
  • y-intercept = -d/b
  • z-intercept = -d/c

Worked Examples

Example 1: Point-Normal Form (JEE Main Level)

Problem: Find equation of plane passing through (1, 2, 3) with normal (2, 3, 4).

Solution: Using normal form:

$$a(x - x_1) + b(y - y_1) + c(z - z_1) = 0$$ $$2(x - 1) + 3(y - 2) + 4(z - 3) = 0$$ $$2x - 2 + 3y - 6 + 4z - 12 = 0$$ $$\boxed{2x + 3y + 4z = 20}$$

Or: 2x + 3y + 4z - 20 = 0

Example 2: Three Points (JEE Main Level)

Problem: Find plane through A(1, 1, 1), B(2, 3, 1), C(3, 2, 2).

Solution: Let plane be ax + by + cz = d (assuming it doesn’t pass through origin)

Substituting points:

  • A: a + b + c = d … (i)
  • B: 2a + 3b + c = d … (ii)
  • C: 3a + 2b + 2c = d … (iii)

From (i): d = a + b + c

Substitute in (ii): 2a + 3b + c = a + b + c โ†’ a + 2b = 0 โ†’ a = -2b

Substitute in (iii): 3a + 2b + 2c = a + b + c โ†’ 2a + b + c = 0 โ†’ 2(-2b) + b + c = 0 โ†’ -4b + b + c = 0 โ†’ c = 3b

Let b = 1, then a = -2, c = 3 From (i): d = -2 + 1 + 3 = 2

Equation: -2x + y + 3z = 2

Or: 2x - y - 3z + 2 = 0

Example 3: Intercept Form (JEE Main Level)

Problem: Find equation of plane with intercepts 2, 3, 4 on X, Y, Z axes respectively.

Solution: Using intercept form:

$$\frac{x}{2} + \frac{y}{3} + \frac{z}{4} = 1$$

Multiplying by LCM(2, 3, 4) = 12:

$$\boxed{6x + 4y + 3z = 12}$$

Or: 6x + 4y + 3z - 12 = 0

Example 4: Finding Intercepts (JEE Main Level)

Problem: Find intercepts made by plane 2x + 3y + 6z = 12 on the axes.

Solution: Method 1: Convert to intercept form

Divide equation by 12:

$$\frac{2x}{12} + \frac{3y}{12} + \frac{6z}{12} = 1$$ $$\frac{x}{6} + \frac{y}{4} + \frac{z}{2} = 1$$

Intercepts: x = 6, y = 4, z = 2

Method 2: Direct calculation

  • x-intercept: Put y = z = 0 โ†’ 2x = 12 โ†’ x = 6
  • y-intercept: Put x = z = 0 โ†’ 3y = 12 โ†’ y = 4
  • z-intercept: Put x = y = 0 โ†’ 6z = 12 โ†’ z = 2

Example 5: Plane Through Origin (JEE Advanced Level)

Problem: Find plane through origin and points (1, 2, 3), (3, 2, 1).

Solution: Since plane passes through origin, equation is: ax + by + cz = 0

Substituting (1, 2, 3): a + 2b + 3c = 0 … (i) Substituting (3, 2, 1): 3a + 2b + c = 0 … (ii)

From (ii) - (i): 2a - 2c = 0 โ†’ a = c

Substitute in (i): a + 2b + 3a = 0 โ†’ 4a + 2b = 0 โ†’ b = -2a

Let a = 1, then c = 1, b = -2

Equation: x - 2y + z = 0

Practice Problems

Level 1: JEE Main Basics

  1. Find the equation of plane passing through (2, 3, 1) with normal vector (1, 2, 3).

  2. Write the equation x + 2y + 3z = 6 in intercept form.

  3. Find intercepts made by plane 3x + 4y + 6z = 12.

  4. Find the equation of YZ-plane.

Level 2: JEE Main Standard

  1. Find equation of plane through points (1, 0, 0), (0, 2, 0), (0, 0, 3).

  2. Find equation of plane passing through (1, 1, 1), (2, 3, 4), and parallel to Z-axis.

  3. Find the value of k if plane 2x + 3y + kz = 6 passes through (1, 1, 1).

  4. Find equation of plane through (1, 2, 3) and perpendicular to line joining (1, 2, 3) and (3, 4, 5).

Level 3: JEE Advanced

  1. Find equation of plane equidistant from points (1, 2, 3) and (3, 4, 5).

  2. Find plane through line (x-1)/2 = (y-2)/3 = (z-3)/4 and point (1, 1, 1).

  3. Find equation of plane through intersection of planes x + 2y + 3z = 4 and 2x + y - z = 5, and passing through (1, 1, 1).

  4. Find the locus of a point equidistant from the planes x + 2y + 3z = 0 and 3x + 2y + z = 0.

Quick Reference: All Forms

FormEquationWhen to Use
Generalax + by + cz + d = 0Standard form, (a,b,c) is normal
Normala(x-xโ‚) + b(y-yโ‚) + c(z-zโ‚) = 0Given point and normal
Interceptx/a + y/b + z/c = 1Given intercepts
Vector(r-a)ยทn = 0Vector problems
Three pointsUse determinant or substitutionGiven 3 points

Algorithm: Finding Plane Equation

Given: Information about the plane

Case 1 - Point and Normal: Use a(x-xโ‚) + b(y-yโ‚) + c(z-zโ‚) = 0

Case 2 - Intercepts: Use x/a + y/b + z/c = 1

Case 3 - Three Points:

  • Substitute in ax + by + cz = d
  • Solve for a, b, c (in terms of d)
  • Write final equation

Case 4 - Two Points + Parallel to Line/Plane:

  • Find normal using cross product
  • Use point-normal form

Common Exam Patterns

JEE Main typically asks:

  • Finding plane from point and normal (2-3 marks)
  • Intercept form conversions (2 marks)
  • Plane through three points (3-4 marks)
  • Finding intercepts from general form (2 marks)

JEE Advanced patterns:

  • Plane through line and point
  • Plane through intersection of two planes
  • Family of planes
  • Locus problems with planes

Pro Tip:

  • For three points, substitution method is often faster than determinant
  • Always verify by substituting points back into final equation
  • Check if plane passes through origin - simplifies to ax + by + cz = 0

Last updated: November 2025