Cross product

A spanner turns a bolt only because the push has both a size and a sense of rotation — clockwise or anticlockwise, out of the page or into it. Ordinary multiplication cannot carry that extra piece of information; the cross product (vector product) can, because unlike the dot product it hands back a whole new vector, perpendicular to both inputs, whose direction encodes exactly that sense of turning. This article builds the cross product from its definition a⃗×b⃗=(absin⁡θ)n^\vec a\times\vec b=(ab\sin\theta)\hat n, shows how the right-hand rule fixes the direction of n^\hat n, derives the component (unit-vector) formula used to actually compute one, and applies it to torque, τ⃗=r⃗×F⃗\vec\tau=\vec r\times\vec F — the motivating physical example this syllabus pairs it with. Two worked numericals, one on torque and one on components, carry every number through in full.

The core insight

The cross product a⃗×b⃗=(absin⁡θ)n^\vec a\times\vec b=(ab\sin\theta)\hat n is a vector of magnitude absin⁡θab\sin\theta, perpendicular to both a⃗\vec a and b⃗\vec b with direction fixed by the right-hand rule, and it is anti-commutative: a⃗×b⃗=−(b⃗×a⃗)\vec a\times\vec b=-(\vec b\times\vec a).

The core idea

The cross product (or vector product) of two vectors a⃗\vec a and b⃗\vec b, inclined at angle θ\theta to each other, is defined as

a⃗×b⃗=(absin⁡θ) n^,\vec a \times \vec b = (ab\sin\theta)\,\hat n,

where: a=∣a⃗∣a=|\vec a| and b=∣b⃗∣b=|\vec b| are the magnitudes of the two vectors, θ\theta is the angle between them (measured with both tails together), and n^\hat n is a unit vector perpendicular to both a⃗\vec a and b⃗\vec b, its direction fixed by the right-hand rule: curl the fingers of the right hand from a⃗\vec a towards b⃗\vec b, through the smaller angle, and the extended thumb points along n^\hat n. Unlike the dot product, whose output is a plain number, the cross product's output is itself a vector — it carries both a size (absin⁡θab\sin\theta) and an orientation in space.

Properties that follow directly from the definition:

  • Not commutative. In fact a⃗×b⃗=−(b⃗×a⃗)\vec a\times\vec b=-(\vec b\times\vec a) — swapping the order reverses the direction of the result while leaving its magnitude unchanged. This is called anti-commutativity, and it is the single most important difference from ordinary multiplication.
  • Parallel vectors give zero. If a⃗∥b⃗\vec a\parallel\vec b (so θ=0°\theta=0° or 180°180°), then sin⁡θ=0\sin\theta=0 and a⃗×b⃗=0⃗\vec a\times\vec b=\vec 0, whatever the magnitudes involved. In particular a⃗×a⃗=0⃗\vec a\times\vec a=\vec 0 for every vector, since the angle with itself is zero.
  • Perpendicular vectors give the maximum. If a⃗⊥b⃗\vec a\perp\vec b (θ=90°\theta=90°), then sin⁡θ=1\sin\theta=1 and the magnitude is largest possible for those two lengths: ∣a⃗×b⃗∣=ab|\vec a\times\vec b|=ab.
  • Unit-vector table. Applying the right-hand rule to the standard axes in cyclic order gives ı^×ȷ^=k^\hat\imath\times\hat\jmath=\hat k, ȷ^×k^=ı^\hat\jmath\times\hat k=\hat\imath, k^×ı^=ȷ^\hat k\times\hat\imath=\hat\jmath; reversing any pair flips the sign, e.g. ȷ^×ı^=−k^\hat\jmath\times\hat\imath=-\hat k. This short table is all that is needed to expand a cross product given in component form, using the distributive property a⃗×(b⃗+c⃗)=a⃗×b⃗+a⃗×c⃗\vec a\times(\vec b+\vec c)=\vec a\times\vec b+\vec a\times\vec c.
  • Magnitude bound. Since sin⁡θ\sin\theta never exceeds 11, the magnitude of a cross product is always bounded by the product of the two lengths: ∣a⃗×b⃗∣≤ab|\vec a\times\vec b|\le ab, with equality exactly when the two vectors are perpendicular. This is a quick sanity check — a computed magnitude larger than abab signals an arithmetic slip.

Where the dot product measures "how parallel" two vectors are (largest when θ=0°\theta=0°), the cross product measures "how perpendicular" they are (largest when θ=90°\theta=90°) — the two products are complementary tools, and problems that combine both usually ask for one quantity built from cos⁡θ\cos\theta and another from sin⁡θ\sin\theta.

Figure
Right hand rule for the cross product a cross b showing vector a along one direction vector b at an angle to it and the result n hat pointing straight up out of the plane of a and b
Right-hand rule — curl from a towards b, thumb gives n-hat a b θ n-hat = a × b plane containing a and b

Physical example — torque. The moment of a force about a point, or torque, is τ⃗=r⃗×F⃗\vec\tau=\vec r\times\vec F, where r⃗\vec r is the position vector from the point to where the force acts. Its magnitude is τ=rFsin⁡θ\tau=rF\sin\theta: largest when the force is applied perpendicular to r⃗\vec r (why a door handle sits far from the hinge, and is pushed at right angles for maximum turning effect) and zero when the force acts along r⃗\vec r itself (pushing straight toward or away from the hinge turns nothing).

Derivation AExpanding a cross product into unit-vector (component) form

The goal is to turn the geometric definition a⃗×b⃗=(absin⁡θ)n^\vec a\times\vec b=(ab\sin\theta)\hat n into a purely algebraic recipe for two vectors given by their components, using nothing beyond the distributive law and the unit-vector cross-product table ı^×ȷ^=k^\hat\imath\times\hat\jmath=\hat k, ȷ^×k^=ı^\hat\jmath\times\hat k=\hat\imath, k^×ı^=ȷ^\hat k\times\hat\imath=\hat\jmath, so that any cross product can be computed without ever measuring an angle by protractor.

Step 1. Write both vectors in component form: a⃗=axı^+ayȷ^+azk^\vec a=a_x\hat\imath+a_y\hat\jmath+a_z\hat k and b⃗=bxı^+byȷ^+bzk^\vec b=b_x\hat\imath+b_y\hat\jmath+b_z\hat k. The cross product a⃗×b⃗\vec a\times\vec b is taken to distribute over addition, exactly like ordinary multiplication distributes over a sum — this distributive property is assumed, not re-derived, at this syllabus level.

Step 2. Multiply out term by term, pairing every component of a⃗\vec a with every component of b⃗\vec b. There are nine such terms, each of the form aibj(e^i×e^j)a_ib_j(\hat e_i\times\hat e_j), where e^i,e^j\hat e_i,\hat e_j range over ı^,ȷ^,k^\hat\imath,\hat\jmath,\hat k.

Step 3. Simplify with the unit-vector table. Three terms vanish immediately because a unit vector crossed with itself is zero: ı^×ı^=ȷ^×ȷ^=k^×k^=0⃗\hat\imath\times\hat\imath=\hat\jmath\times\hat\jmath=\hat k\times\hat k=\vec 0. The remaining six terms use ı^×ȷ^=k^\hat\imath\times\hat\jmath=\hat k, ȷ^×k^=ı^\hat\jmath\times\hat k=\hat\imath, k^×ı^=ȷ^\hat k\times\hat\imath=\hat\jmath, and their sign-reversed partners ȷ^×ı^=−k^\hat\jmath\times\hat\imath=-\hat k, k^×ȷ^=−ı^\hat k\times\hat\jmath=-\hat\imath, ı^×k^=−ȷ^\hat\imath\times\hat k=-\hat\jmath.

Step 4. Collect terms by which unit vector they multiply. Gathering everything gives a⃗×b⃗=(aybz−azby)ı^−(axbz−azbx)ȷ^+(axby−aybx)k^\vec a\times\vec b=(a_yb_z-a_zb_y)\hat\imath-(a_xb_z-a_zb_x)\hat\jmath+(a_xb_y-a_yb_x)\hat k. For vectors confined to the xy-plane (az=bz=0a_z=b_z=0, the case this syllabus mainly uses), the first two brackets vanish and only the last term survives: a⃗×b⃗=(axby−aybx)k^\vec a\times\vec b=(a_xb_y-a_yb_x)\hat k — a single number times k^\hat k, which is exactly the pattern used in the worked example below.

Worked example

Part A — torque. A force of 20 N20\ \text{N} is applied perpendicular to a wrench at a distance of 0.3 m0.3\ \text{m} from the bolt. Find the magnitude of the torque produced.

Step 1. Identify the values: r=0.3 mr=0.3\ \text{m}, F=20 NF=20\ \text{N}, and since the force is perpendicular to the wrench, θ=90°\theta=90°, so sin⁡θ=1\sin\theta=1.

Step 2. Apply τ=rFsin⁡θ=(0.3)(20)(1)=6 N ⋅ m\tau=rF\sin\theta=(0.3)(20)(1)=6\ \text{N}\,\cdot\,\text{m}. The torque has magnitude 6 N·m, and by the right-hand rule (curling from r⃗\vec r towards F⃗\vec F) it points out of the plane of rotation, indicating the anticlockwise sense in which the wrench tends to turn the bolt.

Part B — components. Given a⃗=2ı^+3ȷ^\vec a=2\hat\imath+3\hat\jmath and b⃗=4ı^−1ȷ^\vec b=4\hat\imath-1\hat\jmath (both in the xy-plane), find a⃗×b⃗\vec a\times\vec b.

Step 3. Since neither vector has a k^\hat k component, use the xy-plane shortcut from the derivation, a⃗×b⃗=(axby−aybx)k^\vec a\times\vec b=(a_xb_y-a_yb_x)\hat k, with ax=2a_x=2, ay=3a_y=3, bx=4b_x=4, by=−1b_y=-1: a⃗×b⃗=(2(−1)−3(4))k^=(−2−12)k^=−14k^\vec a\times\vec b=\big(2(-1)-3(4)\big)\hat k=(-2-12)\hat k=-14\hat k.

Step 4. Sanity check by full expansion: 2(4)(ı^×ı^)+2(−1)(ı^×ȷ^)+3(4)(ȷ^×ı^)+3(−1)(ȷ^×ȷ^)=0+(−2)k^+12(−k^)+0=−14k^2(4)(\hat\imath\times\hat\imath)+2(-1)(\hat\imath\times\hat\jmath)+3(4)(\hat\jmath\times\hat\imath)+3(-1)(\hat\jmath\times\hat\jmath)=0+(-2)\hat k+12(-\hat k)+0=-14\hat k, matching Step 3. a⃗×b⃗=−14k^\vec a\times\vec b=-14\hat k, a vector of magnitude 14 pointing into the page — consistent with the rule that two vectors confined to the xy-plane always produce a cross product purely along ±k^\pm\hat k.

Common traps

  • Treating the cross product as commutative like ordinary multiplication — it is anti-commutative, a⃗×b⃗=−b⃗×a⃗\vec a\times\vec b=-\vec b\times\vec a, so the order of the two vectors matters.
  • Forgetting that the cross product of parallel (or anti-parallel) vectors is zero — a common error in torque problems where a force is applied directly along the position vector, e.g. pulling straight along a rod produces no torque about its own axis.
  • Confusing sin⁡θ\sin\theta (cross product magnitude) with cos⁡θ\cos\theta (dot product) — mixing these up gives a wrong numeric answer even when the method is otherwise correct.
  • Getting the right-hand-rule direction backwards — always curl from the first vector to the second in a⃗×b⃗\vec a\times\vec b, never the reverse, and never use the left hand.
Why is the cross product a vector when the dot product is just a number?

The two products encode different physical questions. The dot product a⃗⋅b⃗=abcos⁡θ\vec a\cdot\vec b=ab\cos\theta asks "how much do the two vectors point the same way," and that is captured by a single number. The cross product asks "what new direction and turning sense do these two vectors together define," and a plane has infinitely many perpendicular directions unless you also specify which one — so the answer has to carry direction information, which only a vector can do. The right-hand rule is simply the convention that picks one of the two possible perpendicular directions consistently.

How do I remember the sign in the unit-vector table without memorising all six products?

Write ı^,ȷ^,k^\hat\imath,\hat\jmath,\hat k around a small circle in that cyclic order. Going with the arrows (ı^→ȷ^→k^→ı^\hat\imath\to\hat\jmath\to\hat k\to\hat\imath) gives a plus sign, e.g. ı^×ȷ^=+k^\hat\imath\times\hat\jmath=+\hat k and k^×ı^=+ȷ^\hat k\times\hat\imath=+\hat\jmath. Going against the arrows gives a minus sign, e.g. ȷ^×ı^=−k^\hat\jmath\times\hat\imath=-\hat k. You only ever need to memorise the cycle, not six separate facts, and ı^×ı^=ȷ^×ȷ^=k^×k^=0⃗\hat\imath\times\hat\imath=\hat\jmath\times\hat\jmath=\hat k\times\hat k=\vec 0 always, since no vector is perpendicular to itself.

Why does a longer wrench handle produce more torque for the same push?

Because torque magnitude is τ=rFsin⁡θ\tau=rF\sin\theta, and increasing rr — the distance from the bolt to where you push — increases τ\tau directly, for the same force FF applied at the same angle. Physically, the same push creates a bigger "turning moment" the farther it acts from the pivot, which is exactly why spanners and door handles are made long and why you instinctively push near the far end of a lever rather than close to its hinge.

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Where this leads