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 1Compute 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 2Vector 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 3Magnitude 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 4Verify 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 5Test 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 6Find 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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