Cross Product

In rotational physics, many quantities — torque, angular momentum, and magnetic force — require a special multiplication of two vectors that produces a third vector perpendicular to both. This operation is called the cross product (or vector product). Unlike the dot product, which yields a scalar, the cross product yields a new vector whose direction encodes the axis of rotation and whose magnitude encodes how much two vectors 'push against' each other at an angle. The cross product is essential for rotational mechanics because rotation itself is inherently three-dimensional. When a force is applied to a wrench, the twisting effect (torque) depends not only on the size of the force and the length of the arm, but also on the angle between them and the axis about which rotation occurs — all of which are captured simultaneously by the cross product. Without this tool, describing angular momentum or torque in 3D would be extremely cumbersome. Geometrically, the cross product A×B\mathbf{A} \times \mathbf{B} produces a vector whose magnitude equals the area of the parallelogram spanned by A\mathbf{A} and B\mathbf{B}, and whose direction is given by the right-hand rule: curl the fingers of the right hand from A\mathbf{A} toward B\mathbf{B}, and the thumb points in the direction of A×B\mathbf{A} \times \mathbf{B}. This makes the cross product a precise, coordinate-free tool for describing rotation.

Key takeaways
  • Magnitude Definition
  • Direction and the Right-Hand Rule
  • Component Form Using the Determinant
  • Standard Basis Vector Cross Products
  • Physical Application: Torque

Magnitude Definition

The magnitude of the cross product of two vectors A\mathbf{A} and B\mathbf{B} is defined as |\mathbf{A} \times \mathbf{B}| = |\mathbf{A}||\mathbf{B}|\sin\theta where θ\theta is the angle between the vectors measured from A\mathbf{A} to B\mathbf{B} (with 0≤θ≤180°0 \leq \theta \leq 180°). When θ=0°\theta = 0° or 180°180° the vectors are parallel and the cross product is zero — parallel vectors have no perpendicular component between them. The cross product is maximum when θ=90°\theta = 90°, i.e., when the vectors are perpendicular.

Direction and the Right-Hand Rule

The direction of A×B\mathbf{A} \times \mathbf{B} is perpendicular to the plane containing both A\mathbf{A} and B\mathbf{B}. To find which perpendicular direction: point the fingers of your right hand along A\mathbf{A}, then curl them toward B\mathbf{B} through the smaller angle θ\theta. Your extended thumb points in the direction of A×B\mathbf{A} \times \mathbf{B}. Critically, the cross product is anti-commutative: B×A=−(A×B)\mathbf{B} \times \mathbf{A} = -(\mathbf{A} \times \mathbf{B}), pointing in the exactly opposite direction. This anti-commutativity is not a quirk — it reflects the physical fact that reversing the sense of rotation reverses the direction of the angular axis.

Component Form Using the Determinant

If A=(Ax,Ay,Az)\mathbf{A} = (A_x, A_y, A_z) and B=(Bx,By,Bz)\mathbf{B} = (B_x, B_y, B_z), the cross product is computed using the formal determinant expansion: \mathbf{A} \times \mathbf{B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ A_x & A_y & A_z \\ B_x & B_y & B_z \end{vmatrix} Expanding this determinant along the first row gives: \mathbf{A} \times \mathbf{B} = (A_y B_z - A_z B_y)\hat{i} - (A_x B_z - A_z B_x)\hat{j} + (A_x B_y - A_y B_x)\hat{k} A useful memory aid: for each component, cover the row and column of that unit vector in the 3×33 \times 3 array, then take the difference of the diagonal products, with alternating signs (+,−,++, -, +).

Standard Basis Vector Cross Products

The unit vectors i^\hat{i}, j^\hat{j}, k^\hat{k} along the xx-, yy-, zz-axes obey cyclic cross product rules: i^×j^=k^\hat{i} \times \hat{j} = \hat{k}, j^×k^=i^\hat{j} \times \hat{k} = \hat{i}, k^×i^=j^\hat{k} \times \hat{i} = \hat{j}, and reversing the order negates the result: j^×i^=−k^\hat{j} \times \hat{i} = -\hat{k}, etc. A parallel vector crossed with itself is always zero: i^×i^=0\hat{i} \times \hat{i} = \mathbf{0}. These rules are the building blocks for all component-wise cross product calculations.

Physical Application: Torque

The torque τ\boldsymbol{\tau} produced by a force F\mathbf{F} applied at position r\mathbf{r} from the pivot is τ=r×F\boldsymbol{\tau} = \mathbf{r} \times \mathbf{F}. Its magnitude is τ=rFsin⁡θ\tau = rF\sin\theta, where θ\theta is the angle between r\mathbf{r} and F\mathbf{F}. Only the component of force perpendicular to the lever arm r\mathbf{r} contributes to rotation — a force directed straight toward the pivot (θ=0\theta = 0) produces no torque. The direction of τ\boldsymbol{\tau} points along the rotation axis according to the right-hand rule, with counterclockwise rotation (when viewed from the tip of τ\boldsymbol{\tau}) being the positive convention.

The Key Equation

A×B=∣A∣∣B∣sin⁡θ  n^\mathbf{A} \times \mathbf{B} = |\mathbf{A}||\mathbf{B}|\sin\theta\;\hat{n}
  • \mathbf{A},\, \mathbf{B} — Vector operands (any (e.g., m, N)): The two vectors being multiplied; their units determine the unit of the result.
  • |\mathbf{A}|,\, |\mathbf{B}| — Magnitudes (same as respective vector): The lengths (magnitudes) of vectors A and B.
  • \theta — Included angle (radians or degrees): The angle between vectors A and B, measured in the plane they define; must satisfy 0 ≤ θ ≤ π.
  • \hat{n} — Unit normal vector (dimensionless): A unit vector perpendicular to both A and B, whose direction is determined by the right-hand rule.

Worked Example

Problem. A force F=(3i^+0j^+0k^)\mathbf{F} = (3\hat{i} + 0\hat{j} + 0\hat{k}) N is applied at position r=(0i^+2j^+0k^)\mathbf{r} = (0\hat{i} + 2\hat{j} + 0\hat{k}) m from a pivot. Calculate the torque vector τ=r×F\boldsymbol{\tau} = \mathbf{r} \times \mathbf{F} and its magnitude.

τ=−6k^N ⋅ m,∣τ∣=6N ⋅ m\boldsymbol{\tau} = -6\hat{k} \text{N}\,\cdot\,\text{m}, \quad |\boldsymbol{\tau}| = 6 \text{N}\,\cdot\,\text{m}

Answer. The torque vector is −6k^-6\hat{k} N·m, with magnitude 6 N·m directed into the page.

Figure
Diagram
Cross Product: A × BArea = |A||B|sinθABθA × B(out of plane)ABA×B outof pageRight-Hand Rule1. Point fingers along A 2. Curl toward B 3. Thumb → direction of A × BAnti-commutative: B × A = −(A × B)
Origins · 1 minJosiah Willard Gibbs and how this was discovered
JW
Josiah Willard Gibbs
1839–1903 · American

Gibbs developed the modern notation and formalism of vector calculus in the 1880s, including the dot product and cross product as distinct, useful operations. His pamphlet 'Elements of Vector Analysis' (1881–84) established much of the notation physicists and engineers still use today, directly enabling compact expressions like τ=r×F\boldsymbol{\tau} = \mathbf{r} \times \mathbf{F} for torque. Gibbs spent his entire academic career at Yale, where he worked for nine years without a salary — he had inherited enough money to live on. Yale eventually paid him when Johns Hopkins University offered him a position to lure him away.

  • Torque and Rotational Force — Torque is defined as τ=r×F\boldsymbol{\tau} = \mathbf{r} \times \mathbf{F}, making it the most direct physical application of the cross product in rotational mechanics.
  • Angular Momentum — Angular momentum is L=r×p\mathbf{L} = \mathbf{r} \times \mathbf{p}, another fundamental cross product in rotational dynamics.
  • Dot Product (Scalar Product) — The dot product is the complementary vector multiplication operation; comparing A⋅B=ABcos⁡θ\mathbf{A} \cdot \mathbf{B} = AB\cos\theta with the cross product deepens understanding of both.
  • 3D Vector Components and Unit Vectors — Computing cross products in component form requires fluency with i^\hat{i}, j^\hat{j}, k^\hat{k} notation and 3D coordinate systems.
  • The Right-Hand Rule in Rotational Motion — The right-hand rule is the geometric convention that gives the cross product its direction and connects it to physical rotation axes.