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

The Vector Product

The vector product (or cross product) takes two vectors and produces a third vector — perpendicular to both inputs. Two formulas: a component formula for the full vector, and |v||w|sin θ for its magnitude. Used to find normals to planes, perpendicular directions, and (next note) areas.

📘 What you need to know

Component formula (in the formula booklet): v × w = (v₂w₃ − v₃w₂, v₃w₁ − v₁w₃, v₁w₂ − v₂w₁).
Magnitude formula: |v × w| = |v||w| sin θ (also in the formula booklet).
The output is a vector, perpendicular to both v and w. Direction follows the right-hand rule.
NOT commutative: v × w = −(w × v) — order matters.
Distributive: u × (v + w) = u × v + u × w.
Self-product is zero: v × v = 0 (any vector parallel to itself).
Parallel vectors: v × w = 0 ⟺ v and w are parallel (or one is zero).
Perpendicular vectors: |v × w| = |v||w| (since sin 90° = 1).

Two formulas to know

Component formula

v × w = v₂w₃ − v₃w₂v₃w₁ − v₁w₃v₁w₂ − v₂w₁

gives the full vector — direction and magnitude

Magnitude formula

|v × w| = |v||w| sin θ

gives only the length

Pattern for the components: each entry is a 2×2 “cross” — top × bottom minus bottom × top, cycling through (i, j, k) but skipping the row’s index. Practice a few times and it sticks.

Properties — different from dot product

Property	Cross product	Dot product (for comparison)
Output	vector	scalar
Commutative?	NO; v × w = −w × v	yes; v · w = w · v
Self-product	v × v = 0	v · v = \\|v\\|²
Parallel vectors	v × w = 0	\\|v · w\\| = \\|v\\| \\|w\\|
Perpendicular vectors	\\|v × w\\| = \\|v\\| \\|w\\|	v · w = 0

The vector product gives a vector that is automatically perpendicular to both inputs — making it the fastest way to find a “normal” direction in 3D. The dot product can’t do that; it only spits out a number.

🧭 Recipe — compute v × w with the component formula

Write both vectors as columns with components (v₁, v₂, v₃) and (w₁, w₂, w₃).
First entry: v₂w₃ − v₃w₂ (skip the first row).
Second entry: v₃w₁ − v₁w₃ (skip the second row, swap order).
Third entry: v₁w₂ − v₂w₁ (skip the third row).
Sanity check: dot the result with v and w separately — both should be 0.

Worked examples

WE 1

Compute the vector product

Find u × v for u = (3, −1, 2) and v = (2, 4, −1).

Apply the component formula i-comp: u₂v₃ − u₃v₂ = (−1)(−1) − (2)(4) = 1 − 8 = −7 j-comp: u₃v₁ − u₁v₃ = (2)(2) − (3)(−1) = 4 + 3 = 7 k-comp: u₁v₂ − u₂v₁ = (3)(4) − (−1)(2) = 12 + 2 = 14 u × v = (−7, 7, 14) Check perpendicularity (sanity check) u · (u×v) = (3)(−7) + (−1)(7) + (2)(14) = −21 − 7 + 28 = 0 ✓ v · (u×v) = (2)(−7) + (4)(7) + (−1)(14) = −14 + 28 − 14 = 0 ✓

WE 2

Vector product with mixed notation

Find a × b for a = 2i − 3j + k and b = (1, 2, −2).

Step 1: Convert a to column form a = (2, −3, 1) Step 2: Apply the formula i: (−3)(−2) − (1)(2) = 6 − 2 = 4 j: (1)(1) − (2)(−2) = 1 + 4 = 5 k: (2)(2) − (−3)(1) = 4 + 3 = 7 a × b = 4i + 5j + 7k

WE 3

Magnitude using the sin formula

Two vectors a and b have magnitudes |a| = 5 and |b| = 4. The angle between them is 30°. Find |a × b|.

Use |a × b| = |a||b| sin θ |a × b| = 5 × 4 × sin 30° = 20 × 1/2 |a × b| = 10 use this when you have magnitudes & angle but not components

WE 4

Verify u × v is perpendicular to both u and v

Given u = (2, −1, 3) and v = (1, 4, 2), find u × v and verify it is perpendicular to both u and v.

Step 1: Compute u × v i: (−1)(2) − (3)(4) = −2 − 12 = −14 j: (3)(1) − (2)(2) = 3 − 4 = −1 k: (2)(4) − (−1)(1) = 8 + 1 = 9 u × v = (−14, −1, 9) Step 2: Check via dot products = 0 u · (u × v) = (2)(−14) + (−1)(−1) + (3)(9) = −28 + 1 + 27 = 0 ✓ v · (u × v) = (1)(−14) + (4)(−1) + (2)(9) = −14 − 4 + 18 = 0 ✓ u × v ⊥ u and u × v ⊥ v this perpendicularity is the defining feature of the vector product

WE 5

Test if two vectors are parallel

Use the vector product to determine whether a = (2, 4, −2) and b = (−1, −2, 1) are parallel.

Compute a × b — if it’s the zero vector, a and b are parallel i: (4)(1) − (−2)(−2) = 4 − 4 = 0 j: (−2)(−1) − (2)(1) = 2 − 2 = 0 k: (2)(−2) − (4)(−1) = −4 + 4 = 0 a × b = (0, 0, 0) a and b are parallel notice b = −1/2 × a — they’re scalar multiples, so this matches

WE 6

Find a normal to a plane through three points

The points A, B, and C have coordinates (1, 0, 2), (3, 1, 4), and (2, −1, 1) respectively. Find a vector perpendicular to the plane ABC.

Step 1: Form two vectors in the plane (from A) AB = B − A = (2, 1, 2) AC = C − A = (1, −1, −1) Step 2: A normal is AB × AC i: (1)(−1) − (2)(−1) = −1 + 2 = 1 j: (2)(1) − (2)(−1) = 2 + 2 = 4 k: (2)(−1) − (1)(1) = −2 − 1 = −3 normal = i + 4j − 3k any non-zero scalar multiple of (1, 4, −3) is also a valid normal

💡 Top tips

Don’t memorise the wrong sign on the j-component — it’s v₃w₁ − v₁w₃, with the order swapped relative to the others.
Sanity check by dotting: u · (u × v) and v · (u × v) should both equal 0.
For “find a normal” or “perpendicular to both” questions, the cross product is the move.
Order matters: u × v and v × u point in opposite directions.
Cross product = 0 vector is the fastest test for parallel vectors in 3D.

⚠ Common mistakes

Wrong sign on the j-component. Watch the cyclic pattern carefully.
Treating the result as a scalar. Vector product → vector. (Magnitude is a scalar — but |v × w| is the size of the vector, not the vector itself.)
Computing v × w when the question asks for w × v — they’re negatives of each other.
Forgetting v × v = 0 (the zero vector, not the scalar 0).
Confusing properties with dot product: cross is NOT commutative, has no |v|² self-product, etc.

Next note: Areas using the Vector Product. The magnitude |v × w| is the area of the parallelogram with v and w as adjacent sides — and half that gives the area of a triangle.

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