# Cross Product of Two Vectors: Formula, Rule, Examples

TL;DR

The cross product of two vectors \(\mathbf{a}\) and \(\mathbf{b}\) produces a third vector perpendicular to both, written \(\mathbf{a} \times \mathbf{b}\), with magnitude \(\|\mathbf{a}\|\|\mathbf{b}\|\sin\theta\) and direction set by the right-hand rule. This article covers the formula, the determinant method of computing it, the right-hand rule, the area-of-a-parallelogram link, and how it differs from the dot product.

## What Is the Cross Product of Two Vectors?

The **cross product of two vectors** \(\mathbf{a}\) and \(\mathbf{b}\), written \(\mathbf{a} \times \mathbf{b}\), is a **vector** that is perpendicular to both \(\mathbf{a}\) and \(\mathbf{b}\), and therefore perpendicular to the plane they lie in. It is defined only in **three dimensions**. Its defining formula gives both its size and its direction at once:

$$\mathbf{a} \times \mathbf{b} = \|\mathbf{a}\|\|\mathbf{b}\|\sin\theta \hat{\mathbf{n}},$$

where \(\|\mathbf{a}\|\) and \(\|\mathbf{b}\|\) are the magnitudes (lengths) of the two vectors, \(\theta\) is the angle between them, and \(\hat{\mathbf{n}}\) is the **unit vector** perpendicular to both, pointing in the direction the right-hand rule selects.

## The Cross Product Formula in Determinant Form

When the vectors are given by components, the cross product is computed as a **determinant**. For

$$\mathbf{a} = a_1\hat{\mathbf{i}} + a_2\hat{\mathbf{j}} + a_3\hat{\mathbf{k}}, \qquad \mathbf{b} = b_1\hat{\mathbf{i}} + b_2\hat{\mathbf{j}} + b_3\hat{\mathbf{k}},$$

the cross product is the symbolic \(3 \times 3\) determinant:

$$\mathbf{a} \times \mathbf{b} = \begin{vmatrix} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \end{vmatrix}.$$

Expanding along the top row gives the component formula:

$$\mathbf{a} \times \mathbf{b} = (a_2 b_3 - a_3 b_2)\hat{\mathbf{i}} - (a_1 b_3 - a_3 b_1)\hat{\mathbf{j}} + (a_1 b_2 - a_2 b_1)\hat{\mathbf{k}}.$$

## The Right-Hand Rule: Reading the Direction

The formula gives a vector, but there are always two perpendicular directions to a plane. The **right-hand rule** picks which one. Point the fingers of your right hand along \(\mathbf{a}\), curl them toward \(\mathbf{b}\) through the angle \(\theta\), and your thumb points along \(\mathbf{a} \times \mathbf{b}\).

This is why **order matters**. Swapping the inputs reverses the curl, so it flips the thumb:

$$\mathbf{b} \times \mathbf{a} = -(\mathbf{a} \times \mathbf{b}).$$

The cross product is **anti-commutative**, with vectors \(\mathbf{a} \times \mathbf{b}\) and \(\mathbf{b} \times \mathbf{a}\) having the same length but pointing in opposite directions.

## Cross Product and the Area of a Parallelogram

The magnitude \(\|\mathbf{a}\|\|\mathbf{b}\|\sin\theta\) is the **area of the parallelogram** spanned by \(\mathbf{a}\) and \(\mathbf{b}\).

Lay \(\mathbf{a}\) and \(\mathbf{b}\) tail-to-tail; they span a parallelogram whose area is base times height. Thus:

$$\text{Area} = \text{base} \times \text{height} = \|\mathbf{a}\|\|\mathbf{b}\|\sin\theta = \|\mathbf{a} \times \mathbf{b}\|.$$

Half of that, \(\tfrac{1}{2}\|\mathbf{a} \times \mathbf{b}\|\), is the area of the **triangle** with sides \(\mathbf{a}\) and \(\mathbf{b}\).

## Cross Product vs Dot Product

The two ways to multiply vectors answer different questions:

| Feature | Dot product \(\mathbf{a} \cdot \mathbf{b}\) | Cross product \(\mathbf{a} \times \mathbf{b}\) |
| --- | --- | --- |
| Result | A **scalar** (number) | A **vector** |
| Zero when | Vectors are perpendicular (\(\theta = 90°\)) | Vectors are parallel (\(\theta = 0°\) or \(180°\)) |
| Order | Commutative: \(\mathbf{a} \cdot \mathbf{b} = \mathbf{b} \cdot \mathbf{a}\) | Anti-commutative: \(\mathbf{b} \times \mathbf{a} = -(\mathbf{a} \times \mathbf{b})\) |
| Measures | How **aligned** two vectors are | How **perpendicular** they are (and the area they span) |

## Where the Cross Product of Two Vectors Shows Up

The cross product is the natural language of anything that turns, twists, or needs a perpendicular direction. Examples include:

- **Torque:** \(\boldsymbol{\tau} = \mathbf{r} \times \mathbf{F}\)
- **Angular momentum and magnetic force:** \(\mathbf{F} = q\mathbf{v} \times \mathbf{B}\)
- **Surface normals in graphics:** Every lit 3D surface needs a **normal vector**.
- **Navigation and structures:** Finding a direction perpendicular to a measured plane.

## Examples of the Cross Product of Two Vectors

### Example 1
**Find \(\mathbf{a} \times \mathbf{b}\) for \(\mathbf{a} = 2\hat{\mathbf{i}} + 3\hat{\mathbf{j}} + \hat{\mathbf{k}}\) and \(\mathbf{b} = \hat{\mathbf{i}} + 4\hat{\mathbf{j}} + 5\hat{\mathbf{k}}\).**

Set up the determinant:

$$\mathbf{a} \times \mathbf{b} = \begin{vmatrix} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ 2 & 3 & 1 \\ 1 & 4 & 5 \end{vmatrix}.$$

Expanding gives:
$$= 11\hat{\mathbf{i}} - 9\hat{\mathbf{j}} + 5\hat{\mathbf{k}}.$$

### Example 2
**Find \(\mathbf{a} \times \mathbf{b}\) for \(\mathbf{a} = \hat{\mathbf{i}} + 2\hat{\mathbf{j}} + 3\hat{\mathbf{k}}\) and \(\mathbf{b} = 4\hat{\mathbf{i}} + 5\hat{\mathbf{j}} + 6\hat{\mathbf{k}}\).**

Using the determinant gives:
$$\mathbf{a} \times \mathbf{b} = \begin{vmatrix} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ 1 & 2 & 3 \\ 4 & 5 & 6 \end{vmatrix} = -3\hat{\mathbf{i}} + 6\hat{\mathbf{j}} - 3\hat{\mathbf{k}}.$$

### Example 3
**Find the magnitude of \(\mathbf{a} \times \mathbf{b}\) if \(\|\mathbf{a}\| = 4\), \(\|\mathbf{b}\| = 5\), and the angle between them is \(30°\).**

Use the magnitude form:
$$\|\mathbf{a} \times \mathbf{b}\| = 4 \cdot 5 \cdot \sin 30° = 10.$$

### Example 4
**Find the area of the parallelogram whose adjacent sides are \(\mathbf{a} = 3\hat{\mathbf{i}} + \hat{\mathbf{j}}\) and \(\mathbf{b} = \hat{\mathbf{i}} + 2\hat{\mathbf{j}}\).**

$$\mathbf{a} \times \mathbf{b} = \begin{vmatrix} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ 3 & 1 & 0 \\ 1 & 2 & 0 \end{vmatrix} = 5\hat{\mathbf{k}}.$$

### Example 5
**Find the cross product of the parallel vectors \(\mathbf{a} = 2\hat{\mathbf{i}} + 4\hat{\mathbf{j}} + 6\hat{\mathbf{k}}\) and \(\mathbf{b} = \hat{\mathbf{i}} + 2\hat{\mathbf{j}} + 3\hat{\mathbf{k}}\).**

Computing the determinant gives the zero vector:
$$\mathbf{a} \times \mathbf{b} = \begin{vmatrix} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ 2 & 4 & 6 \\ 1 & 2 & 3 \end{vmatrix} = \mathbf{0}.$$

### Example 6
**Find a unit vector perpendicular to both \(\mathbf{a} = \hat{\mathbf{i}} + \hat{\mathbf{j}}\) and \(\mathbf{b} = \hat{\mathbf{j}} + \hat{\mathbf{k}}\).**

Compute the cross product:
$$\mathbf{a} \times \mathbf{b} = \begin{vmatrix} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ 1 & 1 & 0 \\ 0 & 1 & 1 \end{vmatrix} = \hat{\mathbf{i}} - \hat{\mathbf{j}} + \hat{\mathbf{k}}.$$

## Key Takeaways
- The **cross product of two vectors** gives a vector perpendicular to both, with magnitude \(\|\mathbf{a}\|\|\mathbf{b}\|\sin\theta\).
- Compute it using the $3 \times 3$ determinant while keeping the middle term negative.
- The right-hand rule helps fix the direction, and the order matters.