1.3 - The Vector (Cross) Product

Introduction

The vector product (also called the cross product or outer product) is a way to multiply two vectors together to obtain another vector. Unlike the scalar product, which produces a scalar, the cross product produces a vector that is perpendicular to both of the original vectors. This operation is fundamental in physics for calculating torques, angular momentum, magnetic forces, and many other applications.

Definition of the Vector Product:

The vector product of two vectors $\mathbf{a}$ and $\mathbf{b}$ is denoted by $\mathbf{a} \times \mathbf{b}$ and is a vector with the following properties:

Magnitude: The magnitude of the cross product is given by: $|\mathbf{a} \times \mathbf{b}| = ab\sin\theta$ where $\theta$ is the angle between the two vectors when they are placed tail-to-tail, and $a$ and $b$ are the magnitudes of vectors $\mathbf{a}$ and $\mathbf{b}$ respectively.
Direction: The direction of $\mathbf{a} \times \mathbf{b}$ is perpendicular to both $\mathbf{a}$ and $\mathbf{b}$, following the right-hand rule: point your right-hand fingers in the direction of $\mathbf{a}$, curl them toward $\mathbf{b}$, and your thumb points in the direction of $\mathbf{a} \times \mathbf{b}$.

Geometric Interpretation

The cross product has the following geometric meanings:

Area of a parallelogram: The magnitude $|\mathbf{a} \times \mathbf{b}|$ equals the area of the parallelogram formed by vectors $\mathbf{a}$ and $\mathbf{b}$: $\text{Area} = |\mathbf{a} \times \mathbf{b}| = ab\sin\theta$
Perpendicular direction: The cross product points in the direction perpendicular to the plane containing $\mathbf{a}$ and $\mathbf{b}$, with direction determined by the right-hand rule.
Parallel or anti-parallel vectors ($\theta = 0°$ or $180°$): When two vectors are parallel or anti-parallel, $\sin(0°) = \sin(180°) = 0$, so the cross product is the zero vector: $\mathbf{a} \times \mathbf{b} = \mathbf{0}$ This is a useful test for parallel vectors: if $\mathbf{a} \times \mathbf{b} = \mathbf{0}$ (and neither vector is zero), then the vectors are parallel.

Component Form

We can use the components of the vectors to calculate the cross product as follows:

Vector Product in Component Form (Cartesian Coordinates):

For two vectors $\mathbf{a} = a_x\hat{x} + a_y\hat{y} + a_z\hat{z}$ and $\mathbf{b} = b_x\hat{x} + b_y\hat{y} + b_z\hat{z}$, the cross product is given by the determinant of the following matrix (see The 3×3 Determinant for more info):

\begin{aligned}\mathbf{a} \times \mathbf{b} &= \begin{vmatrix} \hat{x} & \hat{y} & \hat{z} \\ a_x & a_y & a_z \\ b_x & b_y & b_z \end{vmatrix}\\ \\ &= \begin{vmatrix} a_y & a_z \\ b_y & b_z \end{vmatrix} \hat{x} - \begin{vmatrix} a_x & a_z \\ b_x & b_z \end{vmatrix} \hat{y} + \begin{vmatrix} a_x & a_y \\ b_x & b_y \end{vmatrix} \hat{z} \\ \\ &= (a_y b_z - a_z b_y)\hat{x} - (a_x b_z - a_z b_x)\hat{y} + (a_x b_y - a_y b_x)\hat{z}. \end{aligned}

Equivalently, we can calculate the cross product using the Levi-Civita symbol (summation notation):

(\mathbf{a} \times \mathbf{b})_i = \sum_{j=1}^3 \sum_{k=1}^3 \epsilon_{ijk} a_j b_k

where $\epsilon_{ijk}$ is the Levi-Civita symbol (1 for even permutations, -1 for odd permutations, 0 otherwise).

Example 1: Basic Calculation

Calculate the cross product of $\mathbf{a} = 3\hat{x} + 4\hat{y}$ and $\mathbf{b} = 2\hat{x} - \hat{y} + 5\hat{z}$.

Solution:

Using the component form:

\begin{aligned}\mathbf{a} \times \mathbf{b} &= \begin{vmatrix} \hat{x} & \hat{y} & \hat{z} \\ a_x & a_y & a_z \\ b_x & b_y & b_z \end{vmatrix}\\ &= (a_y b_z - a_z b_y)\hat{x} - (a_x b_z - a_z b_x )\hat{y} + (a_x b_y - a_y b_x)\hat{z} \end{aligned}

Substitute the vector components into the determinant:

\begin{aligned} \mathbf{a} \times \mathbf{b} &= [(4)(5) - (0)(-1)]\hat{x} - [(3)(5) - (0)(2) ]\hat{y} + [(3)(-1) - (4)(2)]\hat{z} \\ &= (20 - 0)\hat{x} - (15 - 0)\hat{y} + (-3 - 8)\hat{z} \\ &= 20\hat{x} - 15\hat{y} - 11\hat{z} \end{aligned}

Example 2

Show that the cross product calculated in Example 1 is perpendicular to both $\mathbf{a}$ and $\mathbf{b}$.

Solution:

We test for orthogonality by calculating the dot product of each vector with the cross product. Here are the vectors:

\begin{aligned} \mathbf{a} &= 3\hat{x} + 4\hat{y}\\ \mathbf{b} &= 2\hat{x} - \hat{y} + 5\hat{z}\\ \mathbf{a} \times \mathbf{b} &= 20\hat{x} - 15\hat{y} - 11\hat{z} \end{aligned}

First, verify that $(\mathbf{a} \times \mathbf{b}) \cdot \mathbf{a} = 0$:

\begin{aligned} (\mathbf{a} \times \mathbf{b}) \cdot \mathbf{a} &= (20\hat{x} - 15\hat{y} - 11\hat{z}) \cdot(3\hat{x} + 4\hat{y})\\ &= 20(3) + (-15)(4) + (-11)(0)\\ &= 60 - 60 + 0\\ &= 0 \end{aligned}

Next, verify that $(\mathbf{a} \times \mathbf{b}) \cdot \mathbf{b} = 0$:

\begin{aligned} (\mathbf{a} \times \mathbf{b}) \cdot \mathbf{b} &= (20\hat{x} - 15\hat{y} - 11\hat{z}) \cdot(2\hat{x} - \hat{y} + 5\hat{z})\\ &= 20(2) + (-15)(-1) + (-11)(5)\\ &= 40 + 15 - 55\\ &= 0 \end{aligned}

Thus, the cross product is perpendicular to both $\mathbf{a}$ and $\mathbf{b}$.

Key Properties

Properties of the Vector Product:

Anti-commutative: $\mathbf{a} \times \mathbf{b} = -\mathbf{b} \times \mathbf{a}$ (order matters!)
Distributive: $\mathbf{a} \times (\mathbf{b} + \mathbf{c}) = \mathbf{a} \times \mathbf{b} + \mathbf{a} \times \mathbf{c}$
Scalar multiplication: $(k\mathbf{a}) \times \mathbf{b} = k(\mathbf{a} \times \mathbf{b})$ for any scalar $k$
Cross product with itself: $\mathbf{a} \times \mathbf{a} = \mathbf{0}$ (the zero vector)
Parallel vectors: If $\mathbf{a} \times \mathbf{b} = \mathbf{0}$, then the vectors are co-linear (parallel or anti-parallel), or one or both are zero
Not associative: In general, $\mathbf{a} \times (\mathbf{b} \times \mathbf{c}) \neq (\mathbf{a} \times \mathbf{b}) \times \mathbf{c}$

Example 2: Cross Product with Unit Vectors

Calculate $\hat{x} \times \hat{y}$.

Solution:

Using the component form with $\hat{x} = \hat{x}$ and $\hat{y} = \hat{y}$:

\begin{aligned} \hat{x} \times \hat{y} &= \begin{vmatrix} \hat{x} & \hat{y} & \hat{z} \\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{vmatrix}\\ &= [(0)(0) - (0)(1)]\hat{x} + [(0)(0) - (1)(0)]\hat{y} + [(1)(1) - (0)(0)]\hat{z} \\ &= 0\hat{x} + 0\hat{y} + 1\hat{z} \\&= \hat{z} \end{aligned}

This confirms the right-hand rule: $\hat{x} \times \hat{y} = \hat{z}$. Similarly, $\hat{y} \times \hat{z} = \hat{x}$ and $\hat{z} \times \hat{x} = \hat{y}$.

Example 3: Parallel Vectors

Show that $\mathbf{a} = 2\hat{x} + 4\hat{y} + 6\hat{z}$ and $\mathbf{b} = \hat{x} + 2\hat{y} + 3\hat{z}$ are colinear (parallel or anti-parallel) by computing their cross product.

Solution:

Calculate the cross product:

\begin{aligned} \mathbf{a} \times \mathbf{b} &= \begin{vmatrix} \hat{x} & \hat{y} & \hat{z} \\ 2 & 4 & 6 \\ 1 & 2 & 3 \end{vmatrix}\\ &= [(4)(3) - (6)(2)]\hat{x} -[(2)(3) - (6)(1) ]\hat{y} + [(2)(2) - (4)(1)]\hat{z} \\ &= (12 - 12)\hat{x} + (6 - 6)\hat{y} + (4 - 4)\hat{z} \\ &= \mathbf{0} \end{aligned}

Since $\mathbf{a} \times \mathbf{b} = \mathbf{0}$ and neither vector is zero, the vectors are parallel. Note that a simpler way to show this is to observe that $\mathbf{b} = \frac{1}{2}\mathbf{a}$, confirming they point along the same line.

Example 4: Finding the Magnitude

Find the magnitude of the cross product of $\mathbf{u} = \hat{x} + 2\hat{y} + 2\hat{z}$ and $\mathbf{v} = 3\hat{x} + 4\hat{z}$.

Solution:

First, calculate the cross product:

\begin{aligned} \mathbf{u} \times \mathbf{v} &= \begin{vmatrix} \hat{x} & \hat{y} & \hat{z} \\ 1 & 2 & 2 \\ 3 & 0 & 4 \end{vmatrix}\\ &= [(2)(4) - (2)(0)]\hat{x} - [(1)(4) - (2)(3) ]\hat{y} + [(1)(0) - (2)(3)]\hat{z}\\ &= (8 - 0)\hat{x} - (4-6)\hat{y} + (0 - 6)\hat{z} \\ &= 8\hat{x} + 2\hat{y} - 6\hat{z} \end{aligned}

The magnitude is:

|\mathbf{u} \times \mathbf{v}| = \sqrt{8^2 + 2^2 + (-6)^2} = \sqrt{64 + 4 + 36} = \sqrt{104} = 2\sqrt{26}

Alternatively, we can use the formula $|\mathbf{u} \times \mathbf{v}| = |\mathbf{u}||\mathbf{v}|\sin\theta$. First find $|\mathbf{u}| = \sqrt{1^2 + 2^2 + 2^2} = 3$ and $|\mathbf{v}| = \sqrt{3^2 + 0^2 + 4^2} = 5$. From Example 2 in the scalar product page, we know $\cos\theta = 11/15$, so $\sin\theta = \sqrt{1 - (11/15)^2} = \sqrt{104}/15$. Therefore:

|\mathbf{u} \times \mathbf{v}| = (3)(5)\frac{\sqrt{104}}{15} = \sqrt{104} = 2\sqrt{26}

Example 5: Anti-commutative Property

Verify that $\mathbf{a} \times \mathbf{b} = -\mathbf{b} \times \mathbf{a}$ for $\mathbf{a} = \hat{x} + 2\hat{z}$ and $\mathbf{b} = 3\hat{x} + \hat{y}$.

Solution:

First compute $\mathbf{a} \times \mathbf{b}$:

\mathbf{a} \times \mathbf{b} = [(0)(0) - (2)(1)]\hat{x} - [(1)(0) - (2)(3)]\hat{y} + [(1)(1) - (0)(3)]\hat{z}

\mathbf{a} \times \mathbf{b} = -2\hat{x} + 6\hat{y} + \hat{z}

Now compute $\mathbf{b} \times \mathbf{a}$:

\mathbf{b} \times \mathbf{a} = [(1)(2) - (0)(0)]\hat{x} - [(3)(2) - (0)(1) ]\hat{y} + [(3)(0) - (1)(1)]\hat{z}

\mathbf{b} \times \mathbf{a} = 2\hat{x} - 6\hat{y} - \hat{z}

Indeed, $\mathbf{b} \times \mathbf{a} = 2\hat{x} - 6\hat{y} - \hat{z} = -(-2\hat{x} + 6\hat{y} + \hat{z}) = -\mathbf{a} \times \mathbf{b}$, confirming the anti-commutative property.

Applications in Physics

The cross product appears frequently in physics:

Torque: $\boldsymbol{\tau} = \mathbf{r} \times \mathbf{F}$, where torque is the cross product of position and force
Angular momentum: $\mathbf{L} = \mathbf{r} \times \mathbf{p}$, where angular momentum is the cross product of position and linear momentum
Magnetic force: $\mathbf{F} = q\mathbf{v} \times \mathbf{B}$, where the force on a moving charge is the cross product of velocity and magnetic field
Area calculations: The magnitude of the cross product gives the area of parallelograms and triangles
Normal vectors: Cross products are used to find vectors perpendicular to surfaces