# Angle Between Two Planes

> Angle between planes in JEE Mathematics: cosine of normals formula, parallel and perpendicular conditions, line-plane angle and the plane family trap.

- Canonical URL: https://prepelephant.com/topics/jee/mathematics/angle-between-planes
- Exam / course: JEE · Subject: Mathematics
- Publisher: PrepElephant (https://prepelephant.com) — Prepared and reviewed by the PrepElephant Academic Review Team
- First published: 2026-10-02
- Last updated: 2026-10-02
- How to cite: "Angle Between Two Planes", PrepElephant, https://prepelephant.com/topics/jee/mathematics/angle-between-planes

## Direct answer

The angle between two planes is defined as the angle between their normals. For planes a1x + b1y + c1z = d1 and a2x + b2y + c2z = d2, cos θ = |a1a2 + b1b2 + c1c2| / (√(a1^2 + b1^2 + c1^2) × √(a2^2 + b2^2 + c2^2)), the modulus keeping θ acute by convention. The planes are parallel when a1/a2 = b1/b2 = c1/c2 and perpendicular when a1a2 + b1b2 + c1c2 = 0. For a line with direction vector b meeting a plane with normal n, the angle uses sine instead: sin θ = |b·n| / (|b||n|), because the angle a line makes with the normal and with the plane are complementary.

## What you must remember

- **Plane-angle formula:** cos θ = |n1·n2| / (|n1||n2|); the angle between planes equals the angle between normals, taken acute via the modulus.
- **Parallel planes:** normals proportional, a1/a2 = b1/b2 = c1/c2; the constant terms decide whether the planes coincide or stay distinct.
- **Perpendicular planes:** dot product of normals zero, a1a2 + b1b2 + c1c2 = 0.
- **Line-plane angle:** sin θ = |b·n| / (|b||n|) — a line parallel to the plane has b·n = 0, a line normal to the plane has |sin θ| = 1.
- **Vector form:** for r·n1 = d1 and r·n2 = d2 the same machinery runs on n1, n2 directly; a plane through the line of intersection forms the family P1 + λP2 = 0.
- **Perpendicular member of a family:** to find the plane in P1 + λP2 = 0 perpendicular to a given plane with normal n3, solve (n1 + λn2)·n3 = 0 for λ.
- **Cosine without modulus** gives angles in [0°, 180°]; JEE answer options almost always want the acute value, so the modulus is not decoration.

## Computing one angle end to end

Take the planes 2x - y + z = 3 and x + y + 2z = 1. Normals are n1 = (2, -1, 1) and n2 = (1, 1, 2), each of length √6. The dot product is 2 - 1 + 2 = 3, so cos θ = |3|/(√6 × √6) = 1/2 and θ = 60°. Now watch the convention at work: flipping the sign of n2 to (-1, -1, -2) — the same plane — makes the raw dot product -3 and the raw cosine -1/2, suggesting 120°. Both 60° and 120° describe the same geometric pair, one for the acute and one for the obtuse angle between the planes; the modulus selects the acute one that exams expect. The same reasoning explains the line-plane switch: a line with direction b making angle φ with the normal makes 90° - φ with the plane itself, and sin(90° - φ) = cos φ converts the dot product into the correct sine formula without any new geometry.

## Where students slip

JEE Main tests this as a direct formula evaluation with clean numbers — normals like (2, -1, 1) against (1, 1, 2) yielding cosines of 1/2, 1/√2, 0 — often as numerical-value questions. Advanced wraps the same computation inside the family P1 + λP2 = 0, asking for the member perpendicular or at a given angle to a third plane, or combines it with distance-from-point and image-of-point work. The predictable errors: dropping the modulus and marking 120° where 60° is expected; treating direction ratios of a line lying in a plane as if they were the plane's normal; and in the family question, forgetting that the coefficient of λ runs on the left-hand sides only (constants combine as d1 + λd2, and mixing that up shifts the whole plane). Line-plane questions that give the angle with a plane and ask for the angle with its normal require the complement — a favourite single-step trap. Three-dimensional geometry carries reliable weight in both papers, and this is its most formulaic corner.

## Frequently asked questions

### How is the angle between two planes computed?

As the angle between their normals: cos θ = |a1a2 + b1b2 + c1c2| / (√(a1^2 + b1^2 + c1^2) × √(a2^2 + b2^2 + c2^2)), with the modulus enforcing the acute angle.

### When are two planes perpendicular?

When their normals are: a1a2 + b1b2 + c1c2 = 0, regardless of the constant terms d1 and d2.

### Why does the line-plane angle use sine rather than cosine?

Because the angle with the normal and the angle with the plane are complementary, so sin θ = |b·n|/(|b||n|) measures the smaller angle the line makes with the plane itself.

### What does it mean when the raw cosine is negative?

The normals form an obtuse angle; the planes' acute angle is its supplement, obtained by taking the absolute value of the dot product.

### How do you find a plane through the line of intersection of two planes, perpendicular to a third?

Write the family P1 + λP2 = 0 with normal n1 + λn2, then solve (n1 + λn2)·n3 = 0 for λ and substitute back.