Angle Between Two Planes
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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.