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

The Scalar Product

The scalar product (or dot product) takes two vectors and returns a single number. It has two equivalent formulas — a quick component sum and a geometric magnitude × cos(angle) — both in the formula booklet. Sets up perpendicularity tests, angle calculations, and physics applications like work done.

📘 What you need to know

Component formula: v · w = v₁w₁ + v₂w₂ + v₃w₃ — multiply matching components, add.
Angle formula: v · w = |v||w| cos θ, where θ is the angle between the vectors.
The output is a scalar (a number) — not a vector. That’s why it’s called the “scalar” product.
Commutative: v · w = w · v.
Distributive: u · (v + w) = u · v + u · w.
Self-product: v · v = |v|² (a vector dotted with itself = its magnitude squared).
Sign reveals the angle: positive → acute; zero → perpendicular; negative → obtuse.

Two equivalent formulas

Component formula

v · w = v₁w₁ + v₂w₂ + v₃w₃

use when you have components

Angle formula

v · w = |v||w| cos θ

use when you have magnitudes & angle

Both give the same answer. Pick whichever matches the data you’ve got.

Properties worth remembering

Property	What it says
Commutative	v · w = w · v
Distributive	u · (v + w) = u · v + u · w
Scalar associativity	(kv) · w = k(v · w)
Self-product	v · v = \|v\|²
Perpendicular	v · w = 0 (when neither is the zero vector)
Parallel	\|v · w\| = \|v\|\|w\|

Sign trick: cos θ has the sign of the dot product. So v · w > 0 → angle acute; = 0 → perpendicular; < 0 → angle obtuse. Quick angle classification without computing θ.

🧭 Recipe — compute a scalar product

Match notation — write both vectors in column or base form.
Multiply matching components: x×x, y×y, z×z (in that order).
Add the three products — that’s the scalar product.
If you only have magnitudes and angle, use |v||w|cos θ instead.
Watch signs when components are negative.

Worked examples

WE 1

Scalar product using the component formula

Calculate the scalar product of a = (3, −2, 5) and b = (1, 4, −2).

Multiply matching components and add a · b = (3)(1) + (−2)(4) + (5)(−2) = 3 − 8 − 10 a · b = −15 negative → angle between a and b is obtuse

WE 2

Scalar product with mixed notation

Calculate the scalar product of u = 4i − j + 3k and v = (2, −5, 1).

Step 1: Convert u to column form u = (4, −1, 3) Step 2: Multiply components and add u · v = (4)(2) + (−1)(−5) + (3)(1) = 8 + 5 + 3 u · v = 16

WE 3

Scalar product using the angle formula

Two vectors a and b have magnitudes |a| = 4 and |b| = 3. The angle between them is 60°. Find a · b.

Use a · b = |a||b| cos θ a · b = 4 × 3 × cos 60° = 12 × 1/2 a · b = 6 use this formula when you don’t have components, just magnitudes & angle

WE 4

Use v · v = |v|² to find an unknown

The vector v = (2, p, −3) satisfies v · v = 22. Find the possible values of p.

Step 1: v · v = sum of squared components = |v|² v · v = 2² + p² + (−3)² = 4 + p² + 9 = 13 + p² Step 2: Set equal to 22 13 + p² = 22 → p² = 9 Step 3: Solve — both signs work p = ±3 v · v always gives |v|², so the dot product with itself is just the magnitude squared

WE 5

Use the distributive property

Given that a · b = 5 and a · c = −2, find the value of a · (2b − 3c).

Step 1: Expand using the distributive property a · (2b − 3c) = 2(a · b) − 3(a · c) Step 2: Substitute the known values = 2(5) − 3(−2) = 10 + 6 a · (2b − 3c) = 16 treat the dot product like a number multiplication when expanding brackets

WE 6

Work done by a force (real-world application)

A force F = 5i + 2j − 3k N moves an object through a displacement s = 4i − j + 2k m. Find the work done by the force, given that work = F · s.

Step 1: Apply the component formula F · s = (5)(4) + (2)(−1) + (−3)(2) = 20 − 2 − 6 work = 12 J work is a scalar — the dot product naturally gives a number, not a direction

💡 Top tips

Use the formula that fits your data — components → component formula; magnitudes/angle → angle formula.
The sign tells you the angle type — positive (acute), zero (perpendicular), negative (obtuse).
v · v = |v|² is a quick way to convert between dot product and magnitude.
Treat dot product like number multiplication when expanding (distributive law works just like algebra).
Keep components in the same order when multiplying — x×x, y×y, z×z.

⚠ Common mistakes

Treating the result as a vector. The dot product always returns a number.
Sign errors when components are negative: (−2)(4) = −8, but (−2)(−5) = +10.
Adding instead of multiplying components — it’s component-wise multiplication, then sum.
Mixing up the formulas: |v||w|cos θ uses the angle between, not the angle of either vector with the axes.
Forgetting that v · v = |v|², not |v|. The square root only kicks in for magnitude itself.

Next note: Angle Between Two Vectors & Perpendicular Vectors. Rearrange the angle formula to find θ directly, plus the cleanest test for perpendicularity in vector geometry: just check if the scalar product is zero.

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