IB Maths AI HL Vector Properties Paper 1 & 2 v · w = v₁w₁ + v₂w₂ + v₃w₃ ~7 min read

The Scalar Product

The scalar product (also called the dot product) takes two vectors and returns a single number — not another vector. There are two equivalent formulas: a component formula for direct computation from coordinates, and a geometric formula involving the magnitudes and the angle between the vectors. Both appear in the IB formula booklet, and both give the same answer; switching between them is the foundation for the angle-finding and perpendicularity questions in the next sub-topic.

📘 What you need to know

Component formula: v · w = v₁w₁ + v₂w₂ + v₃w₃ — multiply matching components, then sum.
Geometric formula: v · w = |v||w| cos θ, where θ is the angle between the vectors when drawn from a common start.
Output is a scalar: the answer is a single number (positive, negative, or zero), not a vector.
Perpendicular vectors have v · w = 0 (because cos 90° = 0). The converse also holds for non-zero vectors.
Parallel vectors satisfy |v · w| = |v||w| — the scalar product equals ±|v||w| depending on whether they point the same way (+) or opposite (−).
Sign tells direction: v · w > 0 when the angle is acute; v · w < 0 when the angle is obtuse; v · w = 0 when exactly perpendicular.

Two formulas, one number

The scalar product can be computed two ways, and both give the same number. Use the component formula when you have coordinates; use the geometric formula when you’re given the magnitudes and the angle directly. Either way the result is a scalar — positive if the angle between the vectors is acute, negative if obtuse, and exactly zero when they’re perpendicular. That sign-based interpretation is what makes the dot product such a useful diagnostic for vector geometry.

The angle θ between v and w drives both formulas: the component formula adds matching coordinate products; the geometric formula multiplies magnitudes by cos θ. The sign of the result diagnoses whether the angle is acute, right, or obtuse.

Two formulas for v · w v · w = v₁w₁ + v₂w₂ + v₃w₃ = |v||w| cos θ both formulas in the formula booklet (geometry & trigonometry section)

Choosing the right formula

The two formulas trade information for information. The component formula needs you to know the coordinates of both vectors; the geometric formula needs the magnitudes and the angle. If you have both, either formula works and you can use one to verify the other. If you have neither all of one nor all of the other, you’ll typically be asked to combine them — for instance, computing the dot product via components, then equating it with |v||w| cos θ to solve for θ. That combined use is exactly what the next sub-topic exploits.

Symmetry: v · w = w · v — the order doesn’t matter (unlike the vector product, which is anti-commutative). And v · v = |v|², since the angle between any vector and itself is 0°.

🧭 Recipe — compute v · w

Check what’s given: coordinates → component formula; magnitudes + angle → geometric formula.
For the component formula: write both vectors in column form, multiply matching components, then add the three products.
For the geometric formula: compute |v||w| first, then multiply by cos θ.
Mind the angle: match the calculator mode (degrees vs radians) to how θ is stated.
Interpret the sign: positive → acute angle; zero → perpendicular; negative → obtuse.

Worked examples

WE 1

Component formula in 3D

Given a = (4, −2, 3)^T and b = (1, 5, −2)^T, find a · b.

multiply matching components a · b = 4·1 + (−2)·5 + 3·(−2) simplify each term = 4 + (−10) + (−6) = 4 − 10 − 6 a · b = −12 negative result ⇒ the angle between a and b is obtuse.

WE 2

2D scalar product

Given p = 5i − 2j and q = 3i + 4j, find p · q.

2D — only two component products p · q = 5·3 + (−2)·4 = 15 − 8 p · q = 7 positive ⇒ acute angle. In 2D the formula has just two terms; no k-components.

WE 3

Mixed notation

Given u = (2, −1, 4)^T and v = 3i + 5j + 2k, find u · v.

convert v to column form first v = (3, 5, 2)^T apply the component formula u · v = 2·3 + (−1)·5 + 4·2 = 6 − 5 + 8 u · v = 9 always line up i, j, k coefficients before multiplying — order of the original terms doesn’t matter.

WE 4

Geometric formula — acute angle

Two vectors a and b have magnitudes |a| = 4 and |b| = 7, and the angle between them is 60°. Find a · b.

geometric formula a · b = |a||b|cos θ a · b = 4 · 7 · cos(60°) cos(60°) = 1/2 = 28 · ½ a · b = 14 no coordinates given — geometric formula is the only option.

WE 5

Geometric formula — obtuse angle

Given that |v| = 3, |w| = 5, and the angle between them is 120°, find v · w.

geometric formula v · w = 3 · 5 · cos(120°) cos(120°) = −1/2 (Q2 value) = 15 · (−½) v · w = −7.5 obtuse angle ⇒ negative dot product, as expected.

WE 6

Special case — parallel vectors

The vectors c and d are parallel, with |c| = 6 and |d| = 4. Find c · d if they point in (a) the same direction, (b) opposite directions.

(a) same direction ⇒ θ = 0° c · d = 6 · 4 · cos(0°) = 24 · 1 = 24 (b) opposite directions ⇒ θ = 180° c · d = 6 · 4 · cos(180°) = 24 · (−1) = −24 (a) c · d = 24; (b) c · d = −24 |c · d| = 24 = |c||d| in both cases — the parallel-vector property.

💡 Top tips

Read the question for clues: coordinates given → component formula; magnitudes + angle → geometric formula.
Convert mixed forms first: write everything as columns before multiplying — keeps the components lined up.
Don’t forget signs: a negative component multiplied by a positive component gives a negative product.
Use the sign as a sanity check: positive ⇒ acute, zero ⇒ perpendicular, negative ⇒ obtuse. Mismatches signal an arithmetic slip.
Calculator mode: the geometric formula needs cos θ — make sure your calculator matches degrees (60°) or radians (π/3).

⚠ Common mistakes

Returning a vector instead of a scalar: the dot product yields a single number, not (v₁w₁, v₂w₂, v₃w₃)^T.
Sign errors with two negatives: (−2)(−5) = +10, not −10.
Multiplying out of order: pair v₁ with w₁, v₂ with w₂ — never v₁ with w₂ across components.
Wrong angle convention: θ is measured between the vectors when drawn from a common start — not the supplementary angle on the other side.
Confusing dot and cross: v · w outputs a scalar; v × w outputs a vector. Different operations, different results.

Next up — Angle Between Two Vectors & Perpendicular Vectors. Combining both formulas gives cos θ = (v · w) / (|v||w|), turning the scalar product into a direct angle-finder. The perpendicularity test — v · w = 0 — becomes a one-line check for any “are these perpendicular?” question.

Need help with Vectors?

Get 1-on-1 help from an IB examiner who knows exactly what Paper 1 & 2 are looking for.

Book Free Session →