IB Maths AA HL Topic 3 — Geometry & Trigonometry Paper 1 & 2 ~6 min read HL only

Angles Between Two Planes

The angle between two intersecting planes equals the angle between their normal vectors — same dot-product formula as for two lines or vectors, no subtraction step. Use absolute value to get the acute angle.

📘 What you need to know

Formula: cos θ = |n₁ · n₂| / (|n₁| |n|₂) (in the formula booklet under angle between two vectors).
Angle between two planes = angle between their normals.
Two pairs of equal angles form when planes intersect — the acute and obtuse pair sum to 180°.
Use absolute value on the dot product to guarantee the acute angle.
For vector form: get the normal via n = b × c (cross product of the two direction vectors).
Special cases: n₁ · n₂ = 0 → planes perpendicular (θ = 90°); n₁ ∥ n₂ → planes parallel (θ = 0°).
No “subtract from 90°” step — that’s only for line-and-plane.

The formula

Acute angle between two planes cos θ = |n₁ · n₂||n₁| |n₂|

The two planes’ surfaces stand at the same angle as their normal vectors — both perpendicular to their respective surfaces, so they tilt in lockstep. Just compute the standard dot-product angle between the two normals.

Cartesian → normal

n = (a, b, c)

read off coefficients of x, y, z

Vector form → normal

n = b × c

cross product of the two direction vectors

Special cases

Condition	Geometric meaning	Angle θ
n₁ · n₂ = 0	normals perpendicular	θ = 90° (planes perpendicular)
n₁ parallel to n₂	normals scalar multiples	θ = 0° (planes parallel)
otherwise	normals at some angle	0° < θ < 90°

🧭 Recipe — angle between two planes

Identify n₁ and n₂: read off coefficients (Cartesian) or take b × c (vector form).
Compute |n₁ · n₂|, |n₁|, |n₂|.
Apply: cos θ = |n₁ · n₂| / (|n₁| |n₂|).
Take cos⁻¹ to get the acute angle.
Convert to radians if the question requires it (multiply by π/180).

Worked examples

WE 1

Acute angle between two planes (in degrees)

Find the acute angle between the planes Π₁: x + 2y + 2z = 6 and Π₂: 2x − y + 2z = 4. Give your answer to 3 s.f.

Step 1: Normals n₁ = (1, 2, 2); n₂ = (2, −1, 2) Step 2: Components n₁ · n₂ = (1)(2) + (2)(−1) + (2)(2) = 4 |n₁| = √(1+4+4) = 3 |n₂| = √(4+1+4) = 3 Step 3: cos θ = |4|/(3 × 3) = 4/9 θ = cos⁻¹(4/9) ≈ 63.61° θ ≈ 63.6° no subtraction step — the formula gives the angle directly

WE 2

Acute angle between two planes (in radians)

Find the acute angle in radians between the planes Π₁: x + y + z = 4 and Π₂: 2x + 2y − z = 5. Give your answer to 3 s.f.

Step 1: Normals n₁ = (1, 1, 1); n₂ = (2, 2, −1) Step 2: Components n₁ · n₂ = 2 + 2 − 1 = 3 |n₁| = √3; |n₂| = 3 Step 3: cos θ = |3|/(√3 × 3) = 1/√3 θ = cos⁻¹(1/√3) ≈ 0.9553 rad θ ≈ 0.955 rad in degrees, this is ≈ 54.7°

WE 3

Show two planes are perpendicular

Show that the planes Π₁: 2x + y − 2z = 5 and Π₂: x − 2y = 3 are perpendicular.

Step 1: Identify normals n₁ = (2, 1, −2); n₂ = (1, −2, 0) Step 2: Test n₁ · n₂ n₁ · n₂ = (2)(1) + (1)(−2) + (−2)(0) = 2 − 2 + 0 = 0 Dot product zero → normals perpendicular → planes perpendicular θ = 90° (or π/2 rad) whenever n₁ · n₂ = 0, the planes meet at a right angle

WE 4

Show two planes are parallel

Show that the planes Π₁: 3x − y + 2z = 5 and Π₂: 6x − 2y + 4z = 7 are parallel and find the angle between them.

Step 1: Compare normals n₁ = (3, −1, 2); n₂ = (6, −2, 4) = 2 × n₁ ✓ → normals are scalar multiples → planes are parallel Step 2: Check if same plane RHS: for same plane would need 2 × 5 = 10 ≠ 7 → parallel but distinct planes Planes parallel; θ = 0° the angle between any two parallel planes is 0° regardless of the gap between them

WE 5

Find a value to make two planes perpendicular

Find the value of k for which the planes Π₁: 2x + ky − z = 5 and Π₂: x + y + 3z = 4 are perpendicular.

Step 1: For perpendicular, n₁ · n₂ = 0 (2)(1) + (k)(1) + (−1)(3) = 0 2 + k − 3 = 0 k = 1 verify with k=1: n₁ = (2, 1, −1); n₁ · n₂ = 2 + 1 − 3 = 0 ✓

WE 6

Angle between a plane in vector form and a plane in Cartesian form

Find the acute angle between the planes Π₁: r = (1, 0, 2) + λ(1, 1, 0) + μ(0, 1, 2) and Π₂: 2x − y + z = 5. Give your answer in degrees to 3 s.f.

Step 1: Get n₁ from Π₁ via cross product of directions n₁ = (1, 1, 0) × (0, 1, 2) i: (1)(2) − (0)(1) = 2 j: −[(1)(2) − (0)(0)] = −2 k: (1)(1) − (1)(0) = 1 n₁ = (2, −2, 1) Step 2: n₂ from Π₂ Cartesian n₂ = (2, −1, 1) Step 3: Apply formula n₁ · n₂ = 4 + 2 + 1 = 7 |n₁| = √(4+4+1) = 3; |n₂| = √(4+1+1) = √6 cos θ = 7/(3√6) ≈ 0.9526 θ = cos⁻¹(7/(3√6)) ≈ 17.72° θ ≈ 17.7° in radians: ≈ 0.309 rad

💡 Top tips

Don’t subtract from 90° — that step is only for line-and-plane, not plane-and-plane.
Always use absolute value on the dot product — guarantees an acute angle.
For vector-form planes, take n = b × c first; then proceed as if Cartesian.
Quick perpendicular check: if n₁ · n₂ = 0, planes are perpendicular — no further calculation.
Quick parallel check: if normals are scalar multiples, planes are parallel — angle is 0°.

⚠ Common mistakes

Mistakenly subtracting from 90° — this only applies to line-and-plane, not plane-and-plane.
Forgetting absolute value — a negative dot product would give the obtuse angle (≥ 90°).
Computing the cross product backwards — j-component flips sign in the determinant expansion.
Using direction vectors instead of the normal for vector-form planes — must take the cross product first.
Forgetting to convert when the question asks for radians but you’ve calculated in degrees (or vice versa).

Final note in this section: Shortest Distances with Planes. Three sub-cases — perpendicular distance from a point to a plane, point on a line to a plane, and parallel-plane separation. The trick: build a line through the external point in the direction of the plane’s normal, find where it hits the plane, then measure |λn|.

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