Cross Product in Rotation

When we move from straight-line motion to rotation, the familiar scalar tools of algebra are no longer enough. Rotation happens in three-dimensional space, and both the magnitude and the direction of a rotational effect matter. The cross product is a special operation between two vectors that produces a third vector perpendicular to both of them — exactly the geometric relationship we need to describe how forces create rotation and how rotating objects carry momentum. Two of the most important quantities in rotational dynamics — torque and angular momentum — are defined using the cross product. Torque τ⃗\vec{\tau} tells us how effectively a force rotates an object about a pivot, while angular momentum L⃗\vec{L} describes the rotational inertia and spin of a moving system. Both are vector quantities with magnitude and a direction given by the right-hand rule, and both are zero unless there is a genuine rotational effect present. Without the cross product, we could not capture this directionality in a compact, mathematically useful way. Understanding the cross product in the context of rotation also reveals deep symmetries: Newton's second law for rotation, τ⃗=dL⃗/dt\vec{\tau} = d\vec{L}/dt, mirrors its linear counterpart F⃗=dp⃗/dt\vec{F} = d\vec{p}/dt, and angular momentum is conserved whenever net torque is zero — the principle behind a spinning figure skater pulling in their arms. This article develops the cross product from scratch and shows exactly how it enters the definitions of torque and angular momentum.

Key takeaways
  • Definition of the Cross Product
  • Component Form of the Cross Product
  • Torque as a Cross Product
  • Angular Momentum as a Cross Product
  • Newton's Second Law for Rotation

Definition of the Cross Product

For two vectors A⃗\vec{A} and B⃗\vec{B} in three-dimensional space, the cross product A⃗×B⃗\vec{A} \times \vec{B} is a new vector whose magnitude is ∣A⃗∣∣B⃗∣sin⁡θ|\vec{A}||\vec{B}|\sin\theta, where θ\theta is the angle between the two vectors (taken between 0°0° and 180°180°). Its direction is perpendicular to the plane containing A⃗\vec{A} and B⃗\vec{B}, found using the right-hand rule: point your fingers along A⃗\vec{A}, curl them toward B⃗\vec{B}, and your thumb points in the direction of A⃗×B⃗\vec{A} \times \vec{B}. Crucially, the cross product is anti-commutative: A⃗×B⃗=−(B⃗×A⃗)\vec{A} \times \vec{B} = -(\vec{B} \times \vec{A}), so order matters.

Component Form of the Cross Product

If A⃗=Axi^+Ayj^+Azk^\vec{A} = A_x\hat{i} + A_y\hat{j} + A_z\hat{k} and B⃗=Bxi^+Byj^+Bzk^\vec{B} = B_x\hat{i} + B_y\hat{j} + B_z\hat{k}, then the cross product is computed via a determinant: \vec{A} \times \vec{B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ A_x & A_y & A_z \\ B_x & B_y & B_z \end{vmatrix} = (A_yB_z - A_zB_y)\hat{i} - (A_xB_z - A_zB_x)\hat{j} + (A_xB_y - A_yB_x)\hat{k}. This formula is essential for solving problems where vectors are given in component form. Note that 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 any vector crossed with itself is zero.

Torque as a Cross Product

Torque is defined as τ⃗=r⃗×F⃗\vec{\tau} = \vec{r} \times \vec{F}, where r⃗\vec{r} is the position vector from the pivot (axis of rotation) to the point where force F⃗\vec{F} is applied. The magnitude is ∣τ⃗∣=rFsin⁡θ|\vec{\tau}| = rF\sin\theta, where θ\theta is the angle between r⃗\vec{r} and F⃗\vec{F}. This means only the component of force perpendicular to the position vector creates torque — a force directed straight toward or away from the pivot creates zero torque. The direction of τ⃗\vec{\tau} (along the axis of rotation) is given by the right-hand rule: if τ⃗\vec{\tau} points out of a page, the rotation is counter-clockwise when viewed from that direction.

Angular Momentum as a Cross Product

For a single particle of mass mm moving with velocity v⃗\vec{v}, the angular momentum about a point is L⃗=r⃗×p⃗=r⃗×mv⃗\vec{L} = \vec{r} \times \vec{p} = \vec{r} \times m\vec{v}, where r⃗\vec{r} is the position vector from the reference point to the particle. Its magnitude is L=rmvsin⁡θL = rmv\sin\theta, and its direction again follows the right-hand rule. For a rigid body rotating about a fixed axis with angular velocity ω⃗\vec{\omega}, this generalises to L⃗=Iω⃗\vec{L} = I\vec{\omega}, where II is the moment of inertia — but the cross product definition r⃗×mv⃗\vec{r}\times m\vec{v} remains the fundamental starting point from which this result is derived by summing over all mass elements.

Newton's Second Law for Rotation

Just as F⃗=dp⃗/dt\vec{F} = d\vec{p}/dt in linear motion, the rotational analogue is τ⃗net=dL⃗/dt\vec{\tau}_{\text{net}} = d\vec{L}/dt. We can verify this starting from L⃗=r⃗×mv⃗\vec{L} = \vec{r} \times m\vec{v}: differentiating gives dL⃗/dt=r⃗˙×mv⃗+r⃗×mv⃗˙d\vec{L}/dt = \dot{\vec{r}}\times m\vec{v} + \vec{r}\times m\dot{\vec{v}}. The first term is mv⃗×v⃗=0⃗m\vec{v}\times\vec{v} = \vec{0} (any vector crossed with itself is zero), and the second term is r⃗×F⃗=τ⃗\vec{r}\times\vec{F} = \vec{\tau}. Therefore τ⃗=dL⃗/dt\vec{\tau} = d\vec{L}/dt, confirming that the cross-product definition of angular momentum is fully consistent with Newton's laws. When τ⃗net=0⃗\vec{\tau}_{\text{net}} = \vec{0}, angular momentum is conserved.

The Key Equation

$$\vec{\tau} = \vec{r} \times \vec{F}, \quad |\vec{\tau}| = rF\sin\theta
- **\vec{\tau}** — Torque (N\cdot m): Vector quantity representing the rotational effect of a force about a pivot point; direction gives the axis and sense of rotation. - **\vec{r}** — Position vector (m): Vector from the pivot (axis of rotation) to the point of force application. - **\vec{F}** — Applied force (N): The force whose rotational effect is being calculated. - **\theta** — Angle between vectors (degrees or radians): The angle between $\vec{r}$ and $\vec{F}$; torque is maximum when $\theta = 90°$ and zero when $\theta = 0°$ or $180°$. ## Worked Example **Problem.** A mechanic applies a force $\vec{F} = 20\hat{j}$ N at the end of a wrench. The position vector from the bolt (pivot) to the point of application is $\vec{r} = 0.30\hat{i}$ m. (a) Find the torque vector $\vec{\tau}$. (b) Find its magnitude and describe the direction of rotation.

\vec{\tau} = 6.0\hat{k} \text{ N·m}

∗∗Answer.∗∗Thetorquevectoris6.0N⋅minthe+zdirection,correspondingtocounter−clockwiserotationaboutthebolt.**Answer.** The torque vector is 6.0 N\cdot m in the +z direction, corresponding to counter-clockwise rotation about the bolt.
Figure
Diagram
Cross Product: Torque = r × FPivot Or⃗F⃗90°τ⃗(out of page)Counter-clockwise rotation|τ| = rF sinθ = rF (θ=90°)
Origins · 1 minLeonhard Euler and how this was discovered
LE
Leonhard Euler
1707–1783 · Swiss

Euler formalized the mathematical framework of rigid-body rotation, introducing the concept of the moment of inertia and deriving the equations of rotational motion (Euler's equations) that relate torques to angular accelerations. His notation and vector methods laid the groundwork for the modern treatment of angular momentum and torque as cross-product quantities. Euler lost sight in his right eye around age 31 and became totally blind by age 59, yet he produced nearly half of his total lifetime mathematical output after losing his sight — dictating papers to his sons.

  • Torque and Rotational Equilibrium — Torque is defined directly by the cross product τ⃗=r⃗×F⃗\vec{\tau} = \vec{r}\times\vec{F}; this article shows how net torque determines whether a rigid body is in rotational equilibrium.
  • Angular Momentum and Its Conservation — Angular momentum L⃗=r⃗×mv⃗\vec{L} = \vec{r}\times m\vec{v} is defined by the cross product, and its conservation law dL⃗/dt=τ⃗netd\vec{L}/dt = \vec{\tau}_{\text{net}} is derived from it.
  • Dot Product and Work — The dot product is the companion vector operation to the cross product; contrasting them shows why work (a scalar) uses the dot product while torque (a vector) uses the cross product.
  • Moment of Inertia — For a rigid body, L⃗=Iω⃗\vec{L} = I\vec{\omega} connects the cross-product definition of angular momentum to the scalar moment of inertia, completing the rotational dynamics picture.
  • Vectors in Three Dimensions — A solid understanding of 3D vector components and the right-hand rule is essential prerequisite knowledge for working with cross products in rotation.