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

Angle Between Two Vectors & Perpendicular Vectors

Rearrange the scalar product formula and you get the angle between any two vectors: cos θ = (v · w) / (|v| |w|). The cleanest case is perpendicular: v · w = 0 ⟺ they meet at 90°.

📘 What you need to know

Angle formula: cos θ = v · w|v| |w| — in the formula booklet.
Process: scalar product on top, magnitudes on bottom, arccos to finish.
Sign of dot product = sign of cos θ. Positive → acute; zero → 90°; negative → obtuse.
Perpendicular condition: v · w = 0 (assuming neither is the zero vector).
Converse holds: if v · w = 0 for non-zero v and w, they’re perpendicular.
For unknowns with “perpendicular” in the question: set up the dot product, equate to 0, solve.
Round angles to 3 s.f. (e.g. 68.5°) unless an exact value comes out.

The angle formula

Angle between two vectors cos θ = v · w|v| |w|

This comes from rearranging v · w = |v||w| cos θ. Compute the three numbers on the right, divide, then take arccos.

Sign of dot product reveals the angle: v·w > 0 → acute (θ < 90°); = 0 → right angle (θ = 90°); < 0 → obtuse (θ > 90°). Quick classifier without arccos.

The perpendicular condition

Perpendicular ⟺ dot product is zero v ⊥ w ⟺ v · w = 0 (neither is 0)

The angle between perpendicular vectors is 90°, and cos 90° = 0. So the dot product collapses to zero exactly when the two vectors meet at right angles.

“Perpendicular” in a question = “set the dot product to zero”. It’s the most common move to find an unknown component when two vectors are guaranteed to meet at right angles.

🧭 Recipe — find the angle between two vectors

Compute the scalar product using the component formula.
Compute each magnitude, |v| and |w|.
Plug into cos θ = (v·w) / (|v||w|).
Take arccos to get θ.
Round to 3 s.f. in degrees, or give the exact angle if cos θ is a special value (1/2, √2/2, etc.).

Worked examples

WE 1

Find the angle between two column vectors

Find the angle between a = (1, 2, −2) and b = (3, 0, 4). Give your answer to 3 s.f.

Step 1: Scalar product a · b = (1)(3) + (2)(0) + (−2)(4) = 3 + 0 − 8 = −5 Step 2: Magnitudes |a| = √(1 + 4 + 4) = √9 = 3 |b| = √(9 + 0 + 16) = √25 = 5 Step 3: Apply the formula cos θ = −5 / (3 × 5) = −1/3 θ = arccos(−1/3) ≈ 109.471° θ ≈ 109° (3 s.f.) negative cos → obtuse angle, just as expected

WE 2

Angle with mixed notation

Find the angle between u = 2i + 4j + 4k and v = (−1, 2, 2). Give your answer to 3 s.f.

Step 1: Scalar product (treat both as columns) u · v = (2)(−1) + (4)(2) + (4)(2) = −2 + 8 + 8 = 14 Step 2: Magnitudes |u| = √(4 + 16 + 16) = √36 = 6 |v| = √(1 + 4 + 4) = √9 = 3 Step 3: Apply the formula cos θ = 14 / (6 × 3) = 14/18 = 7/9 θ = arccos(7/9) ≈ 38.942° θ ≈ 38.9° (3 s.f.)

WE 3

Test if two vectors are perpendicular

Determine whether a = (3, 5, −1) and b = (2, −1, 1) are perpendicular.

Compute the scalar product — if it’s zero, they’re perpendicular a · b = (3)(2) + (5)(−1) + (−1)(1) = 6 − 5 − 1 = 0 a · b = 0 → a and b are perpendicular ✓ no need to compute angles or magnitudes — zero dot product seals it

WE 4

Find unknown given perpendicularity

Given that a = (2, k, 3) and b = (4, −1, 2) are perpendicular, find the value of k.

Step 1: Set the scalar product to zero a · b = (2)(4) + (k)(−1) + (3)(2) = 0 Step 2: Simplify and solve 8 − k + 6 = 0 14 − k = 0 k = 14 “perpendicular” in the question → set dot product = 0, then solve

WE 5

Find an exact angle

Find the exact angle between a = i + j and b = i + k.

Step 1: Scalar product a · b = (1)(1) + (1)(0) + (0)(1) = 1 Step 2: Magnitudes |a| = √(1 + 1 + 0) = √2; |b| = √(1 + 0 + 1) = √2 Step 3: Apply the formula cos θ = 1 / (√2 × √2) = 1/2 θ = arccos(1/2) θ = 60° (or π/3) cos θ = 1/2 is an exact value — give the angle exactly, not as a decimal

WE 6

Find an angle in a triangle

The points A, B, and C have coordinates (1, 2, 3), (3, 4, 4), and (2, 1, 5). Find the angle BAC, giving your answer to 3 s.f.

Step 1: Form vectors from A — AB and AC AB = B − A = (2, 2, 1) AC = C − A = (1, −1, 2) Step 2: Scalar product AB · AC = (2)(1) + (2)(−1) + (1)(2) = 2 − 2 + 2 = 2 Step 3: Magnitudes |AB| = √(4 + 4 + 1) = √9 = 3 |AC| = √(1 + 1 + 4) = √6 Step 4: Apply the formula cos(BAC) = 2 / (3√6) BAC = arccos(2/(3√6)) ≈ 74.207° angle BAC ≈ 74.2° (3 s.f.) always form vectors emanating FROM the vertex named in the angle (BAC = vectors from A)

💡 Top tips

Sign of dot product = sign of cos θ. Positive → acute; negative → obtuse; zero → right angle.
For triangle angles like BAC, always form both vectors starting at the named vertex (here A).
Watch for exact angles: cos θ = 1/2 → 60°; 1/√2 → 45°; √3/2 → 30°. Give exact, not decimal.
“Perpendicular” = set dot product to zero — fastest route for unknown components.
Verify your answer with the dot product sign before computing arccos.

⚠ Common mistakes

Forgetting to take arccos. cos θ = 0.5 means θ = 60°, NOT θ = 0.5°.
Using BA and AC for angle BAC instead of AB and AC. The vectors must both point AWAY from A.
Computing arccos of values outside [−1, 1] — usually a sign or magnitude error.
Skipping the dot product = 0 step for “perpendicular” questions.
Wrong calculator mode — degrees vs radians. Default to degrees unless the question is in radians.

Next note: The Vector Product. Multiplying two vectors to give another vector (not a scalar) — perpendicular to both inputs and useful for normals, areas, and 3D geometry.

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