Math Core

The dot product turns two vectors into a number. The cross product turns two vectors into a new vector, one that is perpendicular to both of them. That makes it the natural tool for building normal vectors to planes, computing areas and volumes, and describing rotation (torque and angular momentum). Unlike the dot product, it only works in three dimensions.

Definition

Cross product

For $\mathbf{a} = \langle a_1, a_2, a_3 \rangle$ and $\mathbf{b} = \langle b_1, b_2, b_3 \rangle$ , the cross product is

\mathbf{a} \times \mathbf{b} = \langle a_2 b_3 - a_3 b_2,\ a_3 b_1 - a_1 b_3,\ a_1 b_2 - a_2 b_1 \rangle.

Nobody memorizes this formula as written. Instead, use a $3 \times 3$ determinant, expanded along the first row:

\mathbf{a} \times \mathbf{b} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \end{vmatrix} = \begin{vmatrix} a_2 & a_3 \\ b_2 & b_3 \end{vmatrix}\mathbf{i} - \begin{vmatrix} a_1 & a_3 \\ b_1 & b_3 \end{vmatrix}\mathbf{j} + \begin{vmatrix} a_1 & a_2 \\ b_1 & b_2 \end{vmatrix}\mathbf{k},

where $\begin{vmatrix} p & q \\ r & s \end{vmatrix} = ps - qr$ . (The first row contains vectors, so this is a memory device rather than a true determinant, but it gives exactly the right answer.)

Common mistake

The middle term has a minus sign. Forgetting it is the most common cross-product error. Always check your answer by dotting it with $\mathbf{a}$ and with $\mathbf{b}$ : both results must be $0$ .

Worked example: Computing a cross product

Find $\mathbf{a} \times \mathbf{b}$ for $\mathbf{a} = \langle 2, -1, 3 \rangle$ and $\mathbf{b} = \langle 1, 4, -2 \rangle$ .

Solution.

\begin{aligned} \mathbf{a} \times \mathbf{b} &= \big((-1)(-2) - (3)(4)\big)\mathbf{i} - \big((2)(-2) - (3)(1)\big)\mathbf{j} + \big((2)(4) - (-1)(1)\big)\mathbf{k} \\ &= (2 - 12)\mathbf{i} - (-4 - 3)\mathbf{j} + (8 + 1)\mathbf{k} = \langle -10, 7, 9 \rangle. \end{aligned}

Check: $\langle -10, 7, 9 \rangle \cdot \mathbf{a} = -20 - 7 + 27 = 0$ and $\langle -10, 7, 9 \rangle \cdot \mathbf{b} = -10 + 28 - 18 = 0$ .

Geometry: direction and length

The check in the example is no accident. Expanding $(\mathbf{a} \times \mathbf{b}) \cdot \mathbf{a}$ from the definition, every term cancels, and the same happens for $\mathbf{b}$ . So $\mathbf{a} \times \mathbf{b}$ is orthogonal to both $\mathbf{a}$ and $\mathbf{b}$ . Of the two perpendicular directions, it points the way given by the right-hand rule: curl the fingers of your right hand from $\mathbf{a}$ toward $\mathbf{b}$ (through the smaller angle), and your thumb points along $\mathbf{a} \times \mathbf{b}$ .

a × b is perpendicular to the plane of a and b, pointing by the right-hand rule. Its length equals the area of the parallelogram spanned by a and b.

For the length, a direct (if tedious) algebraic computation gives Lagrange's identity $|\mathbf{a} \times \mathbf{b}|^2 = |\mathbf{a}|^2|\mathbf{b}|^2 - (\mathbf{a} \cdot \mathbf{b})^2$ . Substituting $\mathbf{a} \cdot \mathbf{b} = |\mathbf{a}|\,|\mathbf{b}|\cos\theta$ gives $|\mathbf{a}|^2|\mathbf{b}|^2(1 - \cos^2\theta) = |\mathbf{a}|^2|\mathbf{b}|^2\sin^2\theta$ . Since $\sin\theta \ge 0$ for $0 \le \theta \le \pi$ , we get the following.

Length of the cross product

If $\theta$ is the angle between $\mathbf{a}$ and $\mathbf{b}$ , then

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

which is the area of the parallelogram with sides $\mathbf{a}$ and $\mathbf{b}$ . In particular, two nonzero vectors are parallel if and only if $\mathbf{a} \times \mathbf{b} = \mathbf{0}$ .

The parallelogram has base $|\mathbf{a}|$ and height $|\mathbf{b}|\sin\theta$ , which is why its area appears. A triangle with two sides $\mathbf{a}$ and $\mathbf{b}$ has half that area.

Worked example: Area of a triangle in space

Find the area of the triangle with vertices $P(1, 0, 1)$ , $Q(3, 2, 1)$ and $R(1, 3, 4)$ .

Solution. Two sides from $P$ are $\overrightarrow{PQ} = \langle 2, 2, 0 \rangle$ and $\overrightarrow{PR} = \langle 0, 3, 3 \rangle$ . Then

\overrightarrow{PQ} \times \overrightarrow{PR} = \langle 2 \cdot 3 - 0 \cdot 3,\ 0 \cdot 0 - 2 \cdot 3,\ 2 \cdot 3 - 2 \cdot 0 \rangle = \langle 6, -6, 6 \rangle,

with length $\sqrt{36 \cdot 3} = 6\sqrt3$ . The triangle's area is half of that: $3\sqrt3$ .

Algebraic properties

The basis vectors multiply cyclically: $\mathbf{i} \times \mathbf{j} = \mathbf{k}$ , $\mathbf{j} \times \mathbf{k} = \mathbf{i}$ , $\mathbf{k} \times \mathbf{i} = \mathbf{j}$ , and reversing any order flips the sign. In general:

$\mathbf{b} \times \mathbf{a} = -(\mathbf{a} \times \mathbf{b})$ (anticommutative), and so $\mathbf{a} \times \mathbf{a} = \mathbf{0}$ .
$(c\mathbf{a}) \times \mathbf{b} = c(\mathbf{a} \times \mathbf{b}) = \mathbf{a} \times (c\mathbf{b})$ .
$\mathbf{a} \times (\mathbf{b} + \mathbf{c}) = \mathbf{a} \times \mathbf{b} + \mathbf{a} \times \mathbf{c}$ , and similarly on the other side.
The cross product is not associative: $(\mathbf{i} \times \mathbf{i}) \times \mathbf{j} = \mathbf{0}$ , but $\mathbf{i} \times (\mathbf{i} \times \mathbf{j}) = \mathbf{i} \times \mathbf{k} = -\mathbf{j}$ .

The scalar triple product

Combining both products gives the scalar triple product $\mathbf{a} \cdot (\mathbf{b} \times \mathbf{c})$ , which is the ordinary determinant

\mathbf{a} \cdot (\mathbf{b} \times \mathbf{c}) = \begin{vmatrix} a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3 \end{vmatrix}.

Geometrically, $|\mathbf{b} \times \mathbf{c}|$ is the area of the base parallelogram, and $|\mathbf{a}|\,|\cos\varphi|$ (where $\varphi$ is the angle between $\mathbf{a}$ and $\mathbf{b} \times \mathbf{c}$ ) is the height of the slanted box above it.

Volume of a parallelepiped

The parallelepiped with edges $\mathbf{a}$ , $\mathbf{b}$ , $\mathbf{c}$ from one vertex has volume $V = |\mathbf{a} \cdot (\mathbf{b} \times \mathbf{c})|$ . The three vectors are coplanar exactly when this triple product is $0$ .

Worked example: Volume and a torque

(a) Find the volume of the parallelepiped with edges $\mathbf{a} = \langle 1, 2, 0 \rangle$ , $\mathbf{b} = \langle 0, 1, 3 \rangle$ , $\mathbf{c} = \langle 2, 0, 1 \rangle$ . (b) A wrench handle points along $\mathbf{r} = \langle 0, 0.25, 0 \rangle$ meters from the bolt, and a force $\mathbf{F} = \langle 0, 0, -40 \rangle$ newtons is applied at its end. Find the torque $\boldsymbol{\tau} = \mathbf{r} \times \mathbf{F}$ .

Solution. (a) Expand along the first row:

\begin{vmatrix} 1 & 2 & 0 \\ 0 & 1 & 3 \\ 2 & 0 & 1 \end{vmatrix} = 1(1 - 0) - 2(0 - 6) + 0 = 13,

so $V = 13$ cubic units.

(b) $\mathbf{r} \times \mathbf{F} = \langle (0.25)(-40) - 0 \cdot 0,\ 0 \cdot 0 - 0 \cdot (-40),\ 0 \cdot 0 - 0.25 \cdot 0 \rangle = \langle -10, 0, 0 \rangle$ . The torque has magnitude $10$ N·m and points along the negative $x$ -axis, the axis the bolt turns about. Here $\mathbf{r} \perp \mathbf{F}$ , so the magnitude is simply $|\mathbf{r}|\,|\mathbf{F}| = 0.25 \cdot 40$ .

Tip

To get a unit vector perpendicular to two given vectors, compute their cross product and divide by its length. Its negative is the other answer.

Practice

Practice 1

Compute $\langle 1, 2, 3 \rangle \times \langle 4, 5, 6 \rangle$ .

Enter a point like (2, -3)

Practice 2

Simplify $(\mathbf{i} \times \mathbf{j}) \times \mathbf{j}$ .

Practice 3

Find the area of the parallelogram spanned by $\langle 3, 0, 1 \rangle$ and $\langle -1, 2, 2 \rangle$ .