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:

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.