IB Maths AI HL Vector Properties Paper 1 & 2 v × w (cross product) ~8 min read

The Vector Product

The vector product (also called the cross product) takes two vectors and returns a new vector that’s perpendicular to both. Its direction follows the right-hand rule; its magnitude is |v||w| sin θ. Where the scalar product detects perpendicularity (giving zero), the vector product detects parallel vectors (giving the zero vector). The component formula looks heavy at first — nine multiplications — but it follows a clean cyclic pattern.

📘 What you need to know

Component formula: v × w = (v₂w₃ − v₃w₂, v₃w₁ − v₁w₃, v₁w₂ − v₂w₁)^T — given in the formula booklet.
Geometric magnitude: |v × w| = |v||w| sin θ — also in the formula booklet (only the magnitude version).
Output is a vector (not a scalar) and is perpendicular to both inputs, so (v × w) · v = 0 and (v × w) · w = 0.
Parallel test: non-zero vectors are parallel iff v × w = 0 (the zero vector). The converse holds too.
Perpendicular case: when v and w are perpendicular, sin θ = 1, so |v × w| = |v||w|.
Order matters: v × w = −(w × v). Swapping the order flips the direction (reverses every component); this is unlike the scalar product, which is symmetric.

Direction perpendicular to the plane

If you draw v and w from a common starting point, they define a plane. The cross product v × w is a vector that points straight out of this plane — perpendicular to both v and w. The right-hand rule fixes which of the two possible perpendicular directions is correct: with your right hand, index finger along v and middle finger along w, the thumb points in the direction of v × w. The magnitude is |v||w| sin θ — equal to the area of the parallelogram spanned by v and w (a fact used heavily in the next sub-topic on areas).

Vectors v (teal) and w (orange) lie in the grey parallelogram (their plane). The cross product v × w (blue) shoots perpendicular to this plane, with its direction set by the right-hand rule. Its magnitude equals the parallelogram’s area.

Vector product formulas v × w = (v₂w₃ − v₃w₂, v₃w₁ − v₁w₃, v₁w₂ − v₂w₁)^T |v × w| = |v||w| sin θ component formula is in the booklet; only the magnitude version of the geometric formula is given

Component formula — the cyclic pattern

The component formula has a memorable structure: each output component skips its own index and uses the other two in a cyclic order. The first component (1) uses indices 2 and 3; the second (2) uses 3 and 1; the third (3) uses 1 and 2. Within each component, multiply the diagonal one way, subtract the diagonal the other way. Practising on a few small examples cements the pattern quickly. Many students prefer to memorise it as a 3×3 determinant, but the index-based version above gives the same answer and avoids requiring linear algebra notation.

Verification trick: after computing v × w, check (v × w) · v and (v × w) · w. Both should equal zero. If either isn’t zero, you’ve made an arithmetic slip in the cross product itself.

🧭 Recipe — compute v × w

Line up the two vectors as columns: write v = (v₁, v₂, v₃)^T and w = (w₁, w₂, w₃)^T.
First component: v₂w₃ − v₃w₂ — multiply row 2 by row 3 then subtract row 3 by row 2.
Second component: v₃w₁ − v₁w₃ — cycle: 3, 1.
Third component: v₁w₂ − v₂w₁ — cycle: 1, 2.
Verify perpendicularity (optional but recommended): the cross product should give zero when dotted with either v or w.

Worked examples

WE 1

Compute v × w from components

Find v × w where v = (2, 3, −1)^T and w = (4, −1, 2)^T.

first component: v₂w₃ − v₃w₂ = 3·2 − (−1)·(−1) = 6 − 1 = 5 second component: v₃w₁ − v₁w₃ = (−1)·4 − 2·2 = −4 − 4 = −8 third component: v₁w₂ − v₂w₁ = 2·(−1) − 3·4 = −2 − 12 = −14 v × w = (5, −8, −14)^T double-negative trap in component 1: −(−1)·(−1) = −1, not +1.

WE 2

Verify perpendicularity to both inputs

Given a = (1, 2, 3)^T and b = (2, −1, 1)^T, find a × b, then verify that the result is perpendicular to both a and b.

compute a × b comp 1: 2·1 − 3·(−1) = 2 + 3 = 5 comp 2: 3·2 − 1·1 = 6 − 1 = 5 comp 3: 1·(−1) − 2·2 = −1 − 4 = −5 a × b = (5, 5, −5)^T verify (a × b) · a = 0 = 5·1 + 5·2 + (−5)·3 = 5 + 10 − 15 = 0 ✓ verify (a × b) · b = 0 = 5·2 + 5·(−1) + (−5)·1 = 10 − 5 − 5 = 0 ✓ a × b = (5, 5, −5)^T; perpendicular to both ✓ always a useful cross-check; both dot products MUST be exactly 0.

WE 3

Magnitude from geometric formula

The vectors v and w have magnitudes |v| = 6 and |w| = 4, and the angle between them is 30°. Find |v × w|.

geometric magnitude formula |v × w| = |v| |w| sin θ = 6 · 4 · sin(30°) sin(30°) = 1/2 = 24 · 1/2 |v × w| = 12 the geometric formula gives only the magnitude, not the direction — that requires components or the right-hand rule.

WE 4

Parallel vectors give the zero vector

Find p × q where p = 2i + 4j − 6k and q = 3i + 6j − 9k.

check for parallel: q = (3/2) p 3/2 = 3/2 = (−9)/(−6) ⇒ parallel compute components anyway comp 1: 4·(−9) − (−6)·6 = −36 + 36 = 0 comp 2: (−6)·3 − 2·(−9) = −18 + 18 = 0 comp 3: 2·6 − 4·3 = 12 − 12 = 0 p × q = (0, 0, 0)^T = 0 spotting q = (3/2)p first saves time — the answer is automatically the zero vector for parallel inputs.

WE 5

Order matters — v × w vs w × v

Given v = (3, 1, 2)^T and w = (−1, 2, 0)^T, find v × w and w × v. Comment on the result.

compute v × w comp 1: 1·0 − 2·2 = −4 comp 2: 2·(−1) − 3·0 = −2 comp 3: 3·2 − 1·(−1) = 7 v × w = (−4, −2, 7)^T compute w × v (swap and recompute) comp 1: 2·2 − 0·1 = 4 comp 2: 0·3 − (−1)·2 = 2 comp 3: (−1)·1 − 2·3 = −7 w × v = (4, 2, −7)^T w × v = −(v × w) every component is the negative — swapping the order reverses the cross product’s direction.

WE 6

Find a vector perpendicular to two given vectors

Find a vector that is perpendicular to both a = (1, 1, 0)^T and b = (1, 0, 1)^T.

a × b gives a perpendicular vector comp 1: 1·1 − 0·0 = 1 comp 2: 0·1 − 1·1 = −1 comp 3: 1·0 − 1·1 = −1 verify perpendicular to both (1, −1, −1) · a = 1 − 1 + 0 = 0 ✓ (1, −1, −1) · b = 1 + 0 − 1 = 0 ✓ (1, −1, −1)^T (any scalar multiple also works) the cross product is the standard way to construct a perpendicular vector to two given inputs.

💡 Top tips

Memorise the cyclic pattern: 23, 31, 12 — that’s the order for the three components.
Verify with dot products: (v × w) · v = 0 and (v × w) · w = 0 are quick post-checks.
Spot parallel inputs: if v and w are scalar multiples of each other, the answer is the zero vector — no component calculation needed.
For “perpendicular to two vectors” questions, compute v × w and you’re done.
Watch the order: v × w and w × v differ by a sign — always note which the question wants.

⚠ Common mistakes

Mixing up the order in each component: v₂w₃ − v₃w₂, not v₃w₂ − v₂w₃ — the difference is a sign flip.
Treating the cross product as a scalar: the answer is a vector with three components, not a single number.
Sign errors with double negatives: e.g. −(−1)·(−1) = −1, not +1.
Forgetting that parallel ⇒ zero vector: not zero scalar — 0 = (0, 0, 0)^T.
Using cos instead of sin in the geometric formula: the cross product uses sin θ (compare with dot product’s cos θ).

Next up — Areas using the Vector Product. The magnitude |v × w| equals the area of the parallelogram with v and w as adjacent sides; halve it for the area of a triangle. This single fact handles every “find the area of triangle ABC” or “parallelogram on three vertices” question on the syllabus — just compute the cross product and take its magnitude.

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