Maths · JEE Advanced only

Equation of a Plane in 3D Geometry (JEE Advanced)

Not in the JEE Main syllabus, but it is in JEE Advanced. Study time: about 8 to 10 hours. Spend 2 to 3 hours on plane equations and normal vectors, 2 hours on distance and angle formulas, 2 hours on line-plane relationships, and 2 to 3 hours on coplanarity and mixed problems..

Last updated: 26 October 2026

Direct answer

A plane in 3D is a flat surface that extends infinitely in two dimensions.

A plane in 3D is a flat surface that extends infinitely in two dimensions. You can describe it using a point on the plane and a normal vector perpendicular to it, or using three non-collinear points. The equation of a plane lets you work with distances, angles, and intersections in three-dimensional space.

Official syllabus

What the JEE Advanced syllabus says

“equation of a plane, distance of a point from a plane, angle between two planes, angle between a line and the plane, coplanar lines”

Source: JEE Advanced 2026 Information Bulletin, Annexure-I. Closest JEE Main topic: Equation of a line; Skew lines, the shortest distance between them and its equation.

JEE Main focuses on lines in 3D and skew lines, but Advanced adds planes, which means you need to handle plane equations, point-to-plane distances, and angles involving planes. These concepts appear frequently in Advanced and require comfort with vector operations you may not have practiced much.

Key ideas

What you must understand

  • A plane with normal vector (a, b, c) passing through point (x₀, y₀, z₀) has equation a(x − x₀) + b(y − y₀) + c(z − z₀) = 0
  • The general form ax + by + cz + d = 0 tells you the normal vector is (a, b, c)
  • Three non-collinear points determine a unique plane. Find two vectors in the plane, take their cross product to get the normal
  • The angle between two planes equals the angle between their normal vectors
  • A line is parallel to a plane if the direction vector of the line is perpendicular to the normal of the plane
  • Two lines are coplanar if they either intersect or are parallel. Use the scalar triple product to test this.
Formulas

Formulas and key facts

  • General equation of plane: ax + by + cz + d = 0
  • Vector equation: r⃗ · n⃗ = d, where n⃗ is normal vector
  • Distance from point (x₁, y₁, z₁) to plane ax + by + cz + d = 0: |ax₁ + by₁ + cz₁ + d| / √(a² + b² + c²)
  • Angle between two planes with normals n⃗₁ and n⃗₂: cos θ = |n⃗₁ · n⃗₂| / (|n⃗₁| |n⃗₂|)
  • Angle between line (direction d⃗) and plane (normal n⃗): sin θ = |d⃗ · n⃗| / (|d⃗| |n⃗|)
  • Plane through three points A, B, C: use n⃗ = AB⃗ × AC⃗ as normal
  • Coplanarity test for lines: if lines pass through points a⃗, b⃗ with directions p⃗, q⃗, then [b⃗ − a⃗, p⃗, q⃗] = 0 for coplanar lines
  • Equation of plane through intersection of planes P₁ and P₂: P₁ + λP₂ = 0
Avoid

Common mistakes

  • Forgetting to take absolute value in the distance formula, which can give negative results
  • Confusing the angle formula for two planes (uses cosine) with the angle formula for a line and plane (uses sine)
  • Using the direction vector of a line as if it were the normal vector of a plane containing that line
  • In coplanarity problems, mixing up which vectors go into the scalar triple product
Worked example

A JEE Advanced-level example

Question. Find the distance from the point (1, 2, 3) to the plane passing through the points A(1, 0, 0), B(0, 2, 0), and C(0, 0, 3).

Show solution
  1. Find vectors in the plane: AB⃗ = (−1, 2, 0) and AC⃗ = (−1, 0, 3)
  2. Compute normal vector n⃗ = AB⃗ × AC⃗ = (2×3 − 0×0, 0×(−1) − (−1)×3, (−1)×0 − 2×(−1)) = (6, 3, 2)
  3. Plane equation using point A(1, 0, 0): 6(x − 1) + 3(y − 0) + 2(z − 0) = 0, which gives 6x + 3y + 2z − 6 = 0
  4. Apply distance formula: d = |6(1) + 3(2) + 2(3) − 6| / √(36 + 9 + 4) = |6 + 6 + 6 − 6| / √49 = 12/7

Answer: 12/7 units

Solved twice independently; both solutions agree.

FAQs

Frequently asked questions

Is Equation of a Plane in 3D Geometry in the JEE Main syllabus?

No. It is in the JEE Advanced 2026 syllabus only. The closest JEE Main topic is Equation of a line; Skew lines, the shortest distance between them and its equation.

How long does Equation of a Plane in 3D Geometry take to prepare for JEE Advanced?

About 8 to 10 hours. Spend 2 to 3 hours on plane equations and normal vectors, 2 hours on distance and angle formulas, 2 hours on line-plane relationships, and 2 to 3 hours on coplanarity and mixed problems., if you already know the related JEE Main topic.

WA✎
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