The Cross Product

Quick answer

The cross product u × v of two vectors in ℝ³ is a vector perpendicular to both, pointing by the right-hand rule, with length |u||v| sin θ, the area of the parallelogram the two vectors span. It is computed from a 3 × 3 determinant, which also explains both facts. The cross product reverses sign when the order is swapped and is zero exactly for parallel vectors. The scalar triple product u · (v × w) gives the volume of a box, and the cross product of a position and a force gives torque.

What you'll learn

Compute a cross product with the determinant pattern
Use the cross product to find a vector perpendicular to two others
Find areas of parallelograms and triangles in space
Find volumes with the scalar triple product

A product that makes a new direction

The dot product turns two vectors into a number. In three dimensions there is a second product that turns them into a vector, and a useful one: perpendicular to both. Write $\mathbf{i}$ , $\mathbf{j}$ , $\mathbf{k}$ for the unit vectors along the axes. Then

\mathbf{u} \times \mathbf{v} = \det\begin{bmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k}\\ u_1 & u_2 & u_3\\ v_1 & v_2 & v_3 \end{bmatrix} = (u_2v_3 - u_3v_2)\,\mathbf{i} - (u_1v_3 - u_3v_1)\,\mathbf{j} + (u_1v_2 - u_2v_1)\,\mathbf{k}

The “determinant” is a memory device, expanded along the first row as in cofactor expansion. Its three components are $2 \times 2$ determinants, and the middle one carries the minus sign.

Why it is perpendicular, and why its length is an area

Dot $\mathbf{u}$ with $\mathbf{u} \times \mathbf{v}$ : the result is the same determinant with $\mathbf{u}$ written into the first row in place of $\mathbf{i}, \mathbf{j}, \mathbf{k}$ . A determinant with two equal rows is zero, so $\mathbf{u} \cdot (\mathbf{u} \times \mathbf{v}) = 0$ , and the same holds for $\mathbf{v}$ . For the length, expanding both sides shows the identity

\|\mathbf{u} \times \mathbf{v}\|^2 = \|\mathbf{u}\|^2\|\mathbf{v}\|^2 - (\mathbf{u} \cdot \mathbf{v})^2 = \|\mathbf{u}\|^2\|\mathbf{v}\|^2\sin^2\theta

The cross product points straight out of the plane of $\mathbf{u}$ and $\mathbf{v}$ , and its length $\|\mathbf{u}\|\|\mathbf{v}\|\sin\theta$ is the area of the parallelogram they span: base $\|\mathbf{u}\|$ times height $\|\mathbf{v}\|\sin\theta$ . Of the two perpendicular directions, it takes the one given by the right-hand rule: fingers curling from $\mathbf{u}$ toward $\mathbf{v}$ , thumb along $\mathbf{u} \times \mathbf{v}$ .

The parallelogram on u = (3, 1, 0) and v = (1, 2, 0)

Worked examples

Common mistakes

Practice problems

Find $(2, 0, 1) \times (1, 3, 0)$ .

Answer

$(-3, 1, 6)$

Full solution

$(0 \cdot 0 - 1 \cdot 3,\ 1 \cdot 1 - 2 \cdot 0,\ 2 \cdot 3 - 0 \cdot 1) = (-3, 1, 6)$ . Check: $(2, 0, 1) \cdot (-3, 1, 6) = -6 + 0 + 6 = 0$ .
Find a vector perpendicular to both $(1, 1, 0)$ and $(0, 1, 1)$ .

Answer

$(1, -1, 1)$

Full solution

Their cross product is $(1 \cdot 1 - 0 \cdot 1,\ 0 \cdot 0 - 1 \cdot 1,\ 1 \cdot 1 - 1 \cdot 0) = (1, -1, 1)$ .
Find $(2, 4, 6) \times (1, 2, 3)$ . What does the answer say?

Answer

$(0, 0, 0)$ ; the vectors are parallel.

Full solution

$(2, 4, 6) = 2(1, 2, 3)$ , so $\sin\theta = 0$ and the parallelogram is flat.
Find the area of the parallelogram spanned by $(1, 2, 2)$ and $(0, 1, 0)$ .

Answer

$\sqrt{5}$

Full solution

The cross product is $(2 \cdot 0 - 2 \cdot 1,\ 2 \cdot 0 - 1 \cdot 0,\ 1 \cdot 1 - 2 \cdot 0) = (-2, 0, 1)$ , of length $\sqrt{5}$ .
Find the volume of the box spanned by $(1, 0, 0)$ , $(0, 1, 0)$ and $(0, 0, 4)$ .

Answer

$4$

Full solution

$(0, 1, 0) \times (0, 0, 4) = (4, 0, 0)$ , and its dot product with $(1, 0, 0)$ is $4$ . The box is $1 \times 1 \times 4$ .
Do $(1, 1, 0)$ , $(1, 0, 1)$ and $(2, 1, 1)$ lie in one plane through the origin?

Answer

Yes

Full solution

$(1, 0, 1) \times (2, 1, 1) = (-1, 1, 1)$ , and $(1, 1, 0) \cdot (-1, 1, 1) = 0$ . A zero triple product means zero volume. Indeed the third vector is the sum of the first two.
Find $\mathbf{k} \times \mathbf{j}$ .

Answer

$-\mathbf{i}$

Full solution

$\mathbf{j} \times \mathbf{k} = \mathbf{i}$ , and reversing the order flips the sign.
A force $(0, 0, 50)$ N acts at the end of a lever $(0.3, 0, 0)$ m. Find the torque.

Answer

$(0, -15, 0)$ N·m

Full solution

$(0 \cdot 50 - 0 \cdot 0,\ 0 \cdot 0 - 0.3 \cdot 50,\ 0.3 \cdot 0 - 0 \cdot 0) = (0, -15, 0)$ .
$\|\mathbf{u}\| = 3$ , $\|\mathbf{v}\| = 4$ and the angle between them is $30°$ . Find $\|\mathbf{u} \times \mathbf{v}\|$ .

Answer

$6$

Full solution

$3 \cdot 4 \cdot \sin 30° = 12 \cdot \tfrac{1}{2}$ .
A student computes $(1, 2, 3) \times (4, 5, 6)$ as $(-3, -6, -3)$ . What went wrong?

Hint

Dot the answer with $(1, 2, 3)$ .

Answer

The middle component lost its minus sign. The answer is $(-3, 6, -3)$ .

Full solution

$(1, 2, 3) \cdot (-3, -6, -3) = -3 - 12 - 9 = -24 \ne 0$ , so the student’s vector is not perpendicular. The $\mathbf{j}$ component is $-(1 \cdot 6 - 3 \cdot 4) = 6$ .

Frequently asked questions

How do you compute a cross product?

For u = (u₁, u₂, u₃) and v = (v₁, v₂, v₃), u × v = (u₂v₃ − u₃v₂, u₃v₁ − u₁v₃, u₁v₂ − u₂v₁), the expansion of a determinant with i, j, k in the first row.

What direction does u × v point?

Perpendicular to both u and v, on the side given by the right-hand rule: fingers from u toward v, thumb along u × v.

What is the length of the cross product?

|u||v| sin θ, where θ is the angle between the vectors. That is the area of the parallelogram they span.

Is the cross product commutative?

No. v × u = −(u × v): swapping the order flips the direction.

What does a zero cross product mean?

The vectors are parallel, or one of them is zero. Then sin θ = 0 and the parallelogram has no area.

What to learn next

Lines and Planes in SpaceDescribe a line in space with a point and a direction, a plane with a point and a normal vector, and find intersections, angles and distances between them.

For teachersPrint problems onlyTick to print a blank worksheet; leave it clear to print the lesson with the full answer key.