IB Maths AA HL Topic 3 — Geometry & Trigonometry Paper 1 & 2 ~7 min read HL only

Angles Between a Line & a Plane

The angle between a line and a plane is the angle between the line and its projection onto the plane — the smallest angle between the two. The trick: compute the angle between the line’s direction and the plane’s normal first, then subtract from 90°.

📘 What you need to know

Definition: angle between a line and its projection onto the plane (the smallest angle).
Formula: cos α = |b · n| / (|b| |n|), then θ = 90° − α.
α is the angle between the direction b and the normal n.
θ is the angle between the line and the plane (what the question asks for).
Right-triangle picture: line = hypotenuse, projection on plane = adjacent (to θ), normal = opposite.
Use absolute value in the dot product to guarantee an acute α (and so a valid θ between 0° and 90°).
Special cases: b · n = 0 → line parallel to plane (θ = 0°); b parallel to n → line perpendicular (θ = 90°).
Radians or degrees: subtract from π/2 instead of 90° if the question is in radians.

The formula

Angle between line direction and plane normal cos α = |b · n||b| |n|

Angle between line and plane θ = 90° − α (or π/2 − α in radians)

The line and the normal are perpendicular complements: if you know the angle to the normal, the angle to the plane is what’s left to make a right angle. The absolute value bars on the dot product guarantee α is acute, so θ is positive.

The sin shortcut

Since cos α = sin(90° − α) = sin θ, you can skip the subtraction entirely:

Direct formula for θ sin θ = |b · n||b| |n|

Both methods give the same answer. The cos⁻¹-then-subtract approach is what most mark schemes show; the direct sin⁻¹ is faster.

Special cases

Condition	Geometric meaning	Angle θ
b · n = 0	direction perpendicular to normal	θ = 0° (line parallel to plane)
b parallel to n	direction parallel to normal	θ = 90° (line perpendicular to plane)
otherwise	line crosses the plane at one point	0° < θ < 90°

🧭 Recipe — angle between a line and a plane

Identify b and n: line direction and plane normal (the coefficients of x, y, z in the Cartesian form).
Compute |b · n|, |b|, |n|.
Apply the formula: cos α = |b · n| / (|b| |n|).
Find α via cos⁻¹.
Subtract from 90° (or π/2) to get the angle between line and plane.

Worked examples

WE 1

Angle between a line and a plane (in degrees)

Find the acute angle between the line r = (1, 2, 0) + λ(2, 1, 2) and the plane 3x − 2y + z = 4. Give your answer to 3 s.f.

Step 1: b = (2, 1, 2); n = (3, −2, 1) Step 2: Compute components b · n = (2)(3) + (1)(−2) + (2)(1) = 6 |b| = √(4+1+4) = 3; |n| = √(9+4+1) = √14 Step 3: cos α cos α = |6|/(3√14) = 2/√14 ≈ 0.5345 α = cos⁻¹(2/√14) ≈ 57.69° Step 4: θ = 90° − α θ ≈ 32.3° positive value < 90° → line crosses plane at one point

WE 2

Angle between a line and a plane (in radians)

Find the acute angle in radians between the line r = (1, −2, 3) + λ(2, −1, 2) and the plane x + y + z = 6. Give your answer to 3 s.f.

Step 1: b = (2, −1, 2); n = (1, 1, 1) Step 2: Components b · n = 2 − 1 + 2 = 3 |b| = 3; |n| = √3 Step 3: cos α = 3/(3√3) = 1/√3 α = cos⁻¹(1/√3) ≈ 0.9553 rad Step 4: θ = π/2 − α θ ≈ 1.5708 − 0.9553 = 0.6155 rad θ ≈ 0.616 rad in degrees, this is ≈ 35.3°

WE 3

Angle for a different line and plane

The line l has equation r = (2, 1, −1) + λ(1, 1, 0). Find the acute angle between l and the plane x + y + z = 1. Give your answer in radians to 3 s.f.

Step 1: b = (1, 1, 0); n = (1, 1, 1) Step 2: Components b · n = 1 + 1 + 0 = 2 |b| = √2; |n| = √3 Step 3: cos α = 2/(√2 · √3) = 2/√6 α = cos⁻¹(2/√6) ≈ 0.6155 rad Step 4: θ = π/2 − α θ ≈ 1.5708 − 0.6155 = 0.9553 rad θ ≈ 0.955 rad θ ≈ 54.7° — quite steep, since b is almost aligned with n

WE 4

Show a line is perpendicular to a plane

Show that the line r = (1, 2, 3) + λ(2, −1, 1) is perpendicular to the plane 4x − 2y + 2z = 6.

Step 1: Compare direction and normal b = (2, −1, 1); n = (4, −2, 2) n = 2 × b → b parallel to n ✓ Step 2: When direction parallel to normal, line is perpendicular to plane Verify with formula: b · n = 8 + 2 + 2 = 12; |b| = √6; |n| = 2√6 cos α = 12/(√6 × 2√6) = 12/12 = 1 → α = 0 θ = 90° − 0° = 90° Line is perpendicular to plane (θ = 90°) whenever direction is a scalar multiple of normal, the line is perpendicular

WE 5

Find a value to make a line parallel to a plane

Find the value of k for which the line r = (1, 0, 2) + λ(2, k, 1) is parallel to the plane x + y − z = 4. State the angle between the line and the plane for this value.

Step 1: Line parallel to plane ⟺ b · n = 0 (2)(1) + (k)(1) + (1)(−1) = 0 2 + k − 1 = 0 → k = −1 Step 2: Angle between line and plane b · n = 0 → cos α = 0 → α = 90° θ = 90° − 90° = 0° k = −1; angle = 0° (line parallel to plane) b · n = 0 means direction is perpendicular to normal → direction lies in the plane

WE 6

Angle for a line defined by two points

The line l passes through A(1, 0, 2) and B(3, 1, 4). Find the acute angle between l and the plane x − 2y + 2z = 7. Give your answer in degrees to 3 s.f.

Step 1: Direction AB b = AB = (2, 1, 2); n = (1, −2, 2) Step 2: Components b · n = 2 − 2 + 4 = 4 |b| = √(4+1+4) = 3; |n| = √(1+4+4) = 3 Step 3: cos α = 4/(3 × 3) = 4/9 α = cos⁻¹(4/9) ≈ 63.61° Step 4: θ = 90° − α θ ≈ 26.4° in radians: θ ≈ 0.461 rad

💡 Top tips

Always use absolute value on the dot product — guarantees an acute α and a valid θ.
Match units: subtract α from 90° in degrees, π/2 in radians.
Use sin⁻¹ shortcut: sin θ = |b · n| / (|b| |n|) — saves the subtraction step.
Sanity check: if direction is parallel to normal, line should be perpendicular (θ = 90°); if direction is perpendicular to normal, line should be parallel (θ = 0°).
Read the question: “angle between line and plane” ≠ “angle between direction and normal” — they’re complementary.

⚠ Common mistakes

Stopping at α instead of subtracting from 90° (or π/2) — α is the angle to the normal, not the plane.
Forgetting absolute value — a negative dot product gives an obtuse α, leading to a negative θ.
Mixing radians and degrees — keep one mode throughout.
Treating “b · n = 0″ as line in plane — it means parallel, but you also need the anchor on the plane to be in it.
Using sin θ = (b · n) / (|b||n|) without the absolute value — same sign issue as above.

Next: Angles Between Two Planes. The angle between two planes equals the angle between their normals — same dot-product formula, no subtraction needed. The geometry is symmetric: each plane’s normal stands perpendicular to its own surface, and the angle between those normals matches the dihedral angle of the two planes.

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