IB Maths AI HLVector PropertiesPaper 1 & 2v × 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 formulasv × w = (v2w3 − v3w2, v3w1 − v1w3, v1w2 − v2w1)T|v × w| = |v||w| sin θcomponent 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.
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 × bcomp 1: 2·1 − 3·(−1) = 2 + 3 = 5comp 2: 3·2 − 1·1 = 6 − 1 = 5comp 3: 1·(−1) − 2·2 = −1 − 4 = −5a × b = (5, 5, −5)Tverify (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| = 12the 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) p3/2 = 3/2 = (−9)/(−6) ⇒ parallelcompute components anywaycomp 1: 4·(−9) − (−6)·6 = −36 + 36 = 0comp 2: (−6)·3 − 2·(−9) = −18 + 18 = 0comp 3: 2·6 − 4·3 = 12 − 12 = 0p × q = (0, 0, 0)T = 0spotting 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 × wcomp 1: 1·0 − 2·2 = −4comp 2: 2·(−1) − 3·0 = −2comp 3: 3·2 − 1·(−1) = 7v × w = (−4, −2, 7)Tcompute w × v (swap and recompute)comp 1: 2·2 − 0·1 = 4comp 2: 0·3 − (−1)·2 = 2comp 3: (−1)·1 − 2·3 = −7w × v = (4, 2, −7)Tw × 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 vectorcomp 1: 1·1 − 0·0 = 1comp 2: 0·1 − 1·1 = −1comp 3: 1·0 − 1·1 = −1verify 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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