The Cross Product

The dot product multiplies two vectors and gives a number. The cross product multiplies two vectors in 3-space and gives a vector, one that is perpendicular to both of them. It’s the tool for finding a direction at right angles to two given directions, and its length measures area. That makes it the key to torque, areas and volumes, and (in the next unit) the equations of planes.

Key ideas

The definition

For $\vec{a} = [a_1, a_2, a_3]$ and $\vec{b} = [b_1, b_2, b_3]$, the cross product $\vec{a} \times \vec{b}$ (read “a cross b”) is

$$\vec{a} \times \vec{b} = [\,a_2 b_3 - a_3 b_2,\ \ a_3 b_1 - a_1 b_3,\ \ a_1 b_2 - a_2 b_1\,]$$

The cross product is only defined for vectors in 3-space. (To use it with 2-D vectors, give them a $z$-component of $0$.)

Two ways to remember the formula

The “write it twice” method. Write the components of $\vec{a}$ in a row, and those of $\vec{b}$ underneath, then repeat the first two columns at the right:

$$\begin{array}{ccccc} a_1 & a_2 & a_3 & a_1 & a_2 \\ b_1 & b_2 & b_3 & b_1 & b_2 \end{array}$$

Cover the first column. Each component comes from a $2 \times 2$ block, read as “down-right product minus up-right product”: columns 2–3 give $a_2 b_3 - a_3 b_2$, columns 3–4 give $a_3 b_1 - a_1 b_3$, and columns 4–5 give $a_1 b_2 - a_2 b_1$.

The determinant method. A $2 \times 2$ determinant is $\begin{vmatrix} p & q \\ r & s \end{vmatrix} = ps - qr$. The cross product can be written as a $3 \times 3$ determinant with $\vec{i}$, $\vec{j}$, $\vec{k}$ in the top row, expanded along that row:

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

Watch the minus sign in front of the $\vec{j}$ term. It’s the most common slip.

It’s perpendicular to both vectors

The whole point of the cross product: $\vec{a} \times \vec{b}$ is perpendicular to $\vec{a}$ and to $\vec{b}$. You can prove it with dot products:

$$\begin{aligned} \vec{a} \cdot (\vec{a} \times \vec{b}) &= a_1(a_2 b_3 - a_3 b_2) + a_2(a_3 b_1 - a_1 b_3) + a_3(a_1 b_2 - a_2 b_1) \\ &= a_1 a_2 b_3 - a_1 a_3 b_2 + a_2 a_3 b_1 - a_1 a_2 b_3 + a_1 a_3 b_2 - a_2 a_3 b_1 \\ &= 0 \end{aligned}$$

Every term cancels with another one. The same thing happens for $\vec{b} \cdot (\vec{a} \times \vec{b})$ (Practice question 9). This also gives you a built-in check: after computing a cross product, dot it with both vectors. If you don’t get $0$ twice, there’s an arithmetic error.

Direction: the right-hand rule

There are two directions perpendicular to both $\vec{a}$ and $\vec{b}$ (one is the opposite of the other). The right-hand rule tells you which one $\vec{a} \times \vec{b}$ uses: point the fingers of your right hand along $\vec{a}$, then curl them toward $\vec{b}$ (through the smaller angle). Your thumb points along $\vec{a} \times \vec{b}$.

For example, $\vec{i} \times \vec{j} = \vec{k}$: fingers along the $x$-axis, curl toward the $y$-axis, and your thumb points up the $z$-axis. Going around the cycle $\vec{i} \to \vec{j} \to \vec{k} \to \vec{i}$ gives

$$\vec{i} \times \vec{j} = \vec{k}, \qquad \vec{j} \times \vec{k} = \vec{i}, \qquad \vec{k} \times \vec{i} = \vec{j}$$

and going backwards gives the negatives: $\vec{j} \times \vec{i} = -\vec{k}$, $\vec{k} \times \vec{j} = -\vec{i}$, $\vec{i} \times \vec{k} = -\vec{j}$.

Magnitude

If $\theta$ is the angle between $\vec{a}$ and $\vec{b}$ ($0^\circ \le \theta \le 180^\circ$), then

$$|\vec{a} \times \vec{b}| = |\vec{a}|\,|\vec{b}|\sin\theta$$

This is the area of the parallelogram with sides $\vec{a}$ and $\vec{b}$: base $|\vec{a}|$ times height $|\vec{b}|\sin\theta$. The applications page uses this for areas, volumes, and torque.

$\vec{a} \times \vec{b}$ is perpendicular to the plane of $\vec{a}$ and $\vec{b}$, and $\vec{b} \times \vec{a}$ points the opposite way. Its length is the parallelogram’s area.

Properties

Property	Rule
Anticommutative	$\vec{b} \times \vec{a} = -(\vec{a} \times \vec{b})$
Distributive	$\vec{a} \times (\vec{b} + \vec{c}) = \vec{a} \times \vec{b} + \vec{a} \times \vec{c}$
Scalar multiples	$(k\vec{a}) \times \vec{b} = k(\vec{a} \times \vec{b}) = \vec{a} \times (k\vec{b})$
Parallel vectors	if $\vec{a} \parallel \vec{b}$, then $\vec{a} \times \vec{b} = \vec{0}$; in particular $\vec{a} \times \vec{a} = \vec{0}$

• Anticommutative: swapping the order reverses the direction (your right hand curls the other way). Unlike numbers and dot products, order matters.
• Parallel vectors: $\theta = 0^\circ$ or $180^\circ$, so $\sin\theta = 0$. This gives a quick test: two non-zero vectors are parallel exactly when their cross product is $\vec{0}$.
• Not associative: in general $(\vec{a} \times \vec{b}) \times \vec{c} \ne \vec{a} \times (\vec{b} \times \vec{c})$. A quick counterexample with unit vectors:

$$(\vec{i} \times \vec{i}) \times \vec{j} = \vec{0} \times \vec{j} = \vec{0} \qquad\text{but}\qquad \vec{i} \times (\vec{i} \times \vec{j}) = \vec{i} \times \vec{k} = -\vec{j}$$

So with cross products, brackets matter, and you must keep both the order and the grouping.

Worked examples

Example 1: Computing a cross product

Find $\vec{a} \times \vec{b}$ for $\vec{a} = [2, -1, 3]$ and $\vec{b} = [1, 4, -2]$, and check that it’s perpendicular to both.

Solution. Use the formula component by component:

$$\begin{aligned} \text{first: } & a_2 b_3 - a_3 b_2 = (-1)(-2) - (3)(4) = 2 - 12 = -10 \\ \text{second: } & a_3 b_1 - a_1 b_3 = (3)(1) - (2)(-2) = 3 + 4 = 7 \\ \text{third: } & a_1 b_2 - a_2 b_1 = (2)(4) - (-1)(1) = 8 + 1 = 9 \end{aligned}$$

So $\vec{a} \times \vec{b} = [-10, 7, 9]$.

Check:

$$\begin{aligned} [-10, 7, 9] \cdot [2, -1, 3] &= -20 - 7 + 27 = 0 \ \checkmark \\ [-10, 7, 9] \cdot [1, 4, -2] &= -10 + 28 - 18 = 0 \ \checkmark \end{aligned}$$

Example 2: The determinant method, and reversing the order

Let $\vec{a} = [1, 0, 2]$ and $\vec{b} = [3, -2, 1]$. Find $\vec{a} \times \vec{b}$ and $\vec{b} \times \vec{a}$.

Solution.

$$\begin{aligned} \vec{a} \times \vec{b} &= \begin{vmatrix} \vec{i} & \vec{j} & \vec{k} \\ 1 & 0 & 2 \\ 3 & -2 & 1 \end{vmatrix} \\ &= \big(0(1) - 2(-2)\big)\vec{i} - \big(1(1) - 2(3)\big)\vec{j} + \big(1(-2) - 0(3)\big)\vec{k} \\ &= 4\vec{i} - (-5)\vec{j} + (-2)\vec{k} \\ &= [4, 5, -2] \end{aligned}$$

By the anticommutative property, $\vec{b} \times \vec{a} = -[4, 5, -2] = [-4, -5, 2]$, with no extra work.

Check: $[4, 5, -2] \cdot [1, 0, 2] = 4 + 0 - 4 = 0$ ✓ and $[4, 5, -2] \cdot [3, -2, 1] = 12 - 10 - 2 = 0$ ✓

Example 3: Magnitude and angle

Let $\vec{a} = [1, 1, 0]$ and $\vec{b} = [0, 1, 1]$. Find $|\vec{a} \times \vec{b}|$ and use it to find the angle between the vectors.

Solution.

$$\vec{a} \times \vec{b} = [\,1(1) - 0(1),\ \ 0(0) - 1(1),\ \ 1(1) - 1(0)\,] = [1, -1, 1]$$

So $|\vec{a} \times \vec{b}| = \sqrt{3}$. Both vectors have magnitude $\sqrt{2}$, so

$$\sin\theta = \frac{|\vec{a} \times \vec{b}|}{|\vec{a}||\vec{b}|} = \frac{\sqrt{3}}{\sqrt{2}\,\sqrt{2}} = \frac{\sqrt{3}}{2}$$

Careful: $\sin\theta = \dfrac{\sqrt{3}}{2}$ has two answers between $0^\circ$ and $180^\circ$, namely $60^\circ$ and $120^\circ$. The sine can’t tell acute from obtuse. Use the dot product to decide: $\vec{a} \cdot \vec{b} = 0 + 1 + 0 = 1 \gt 0$, so the angle is acute and $\theta = 60^\circ$.

(Lesson: to find an angle, the dot product is the better tool. Use the cross product when you need a perpendicular vector or an area.)

Example 4: Scalar multiples

Let $\vec{a} = [1, 2, 0]$ and $\vec{b} = [0, 1, -1]$. Find $\vec{a} \times \vec{b}$ and $(2\vec{a}) \times (3\vec{b})$, and describe the relationship. Does it hold for any two vectors?

Solution.

$$\vec{a} \times \vec{b} = [\,2(-1) - 0(1),\ \ 0(0) - 1(-1),\ \ 1(1) - 2(0)\,] = [-2, 1, 1]$$

With $2\vec{a} = [2, 4, 0]$ and $3\vec{b} = [0, 3, -3]$:

$$(2\vec{a}) \times (3\vec{b}) = [\,4(-3) - 0(3),\ \ 0(0) - 2(-3),\ \ 2(3) - 4(0)\,] = [-12, 6, 6]$$

That’s $6[-2, 1, 1] = 6(\vec{a} \times \vec{b})$. It holds for any vectors: every component of the cross product is a difference of products with one factor from each vector, so scaling $\vec{a}$ by $2$ and $\vec{b}$ by $3$ scales every term by $6$. In general, $(m\vec{a}) \times (n\vec{b}) = mn(\vec{a} \times \vec{b})$.

Geometrically: the new parallelogram has sides $2$ and $3$ times as long, so $6$ times the area, and the perpendicular direction hasn’t changed.

Common mistakes

Forgetting the minus sign on the middle term. With the determinant method, the $\vec{j}$ term is subtracted. With the component formula, the middle component is $a_3 b_1 - a_1 b_3$ (not $a_1 b_3 - a_3 b_1$). Always check by dotting your answer with both vectors.

Swapping the order. $\vec{a} \times \vec{b}$ and $\vec{b} \times \vec{a}$ point in opposite directions. If a question asks for $\vec{b} \times \vec{a}$, put $\vec{b}$‘s components in the first row.

Treating the answer as a number. The cross product is a vector. Its magnitude $|\vec{a} \times \vec{b}|$ is the number that equals $|\vec{a}||\vec{b}|\sin\theta$.

Finding an angle from the sine alone. $\sin\theta$ is the same for $\theta$ and $180^\circ - \theta$. If you use the cross product to find an angle, check the sign of the dot product to choose between them, or just use the dot product instead.

Using the left hand (or curling the wrong way). The right-hand rule needs your right hand, with fingers along the first vector curling toward the second. Check with $\vec{i} \times \vec{j} = \vec{k}$.

Assuming you can regroup. $(\vec{a} \times \vec{b}) \times \vec{c}$ and $\vec{a} \times (\vec{b} \times \vec{c})$ are usually different vectors (see Practice question 8). Work out the inner bracket first, exactly as written.

Practice

1. (Warm-up) Find each cross product.

• (a) $[3, 0, 0] \times [0, 2, 0]$
• (b) $\vec{i} \times \vec{k}$
• (c) $\vec{k} \times \vec{j}$

Solution

(a) $[\,0(0) - 0(2),\ \ 0(0) - 3(0),\ \ 3(2) - 0(0)\,] = [0, 0, 6]$. (It’s $6(\vec{i} \times \vec{j}) = 6\vec{k}$.)

(b) $\vec{k} \times \vec{i} = \vec{j}$, so $\vec{i} \times \vec{k} = -\vec{j} = [0, -1, 0]$.

(c) $\vec{j} \times \vec{k} = \vec{i}$, so $\vec{k} \times \vec{j} = -\vec{i} = [-1, 0, 0]$.

2. (Warm-up) Find $[1, 2, 3] \times [4, 5, 6]$.

Solution

$$[\,2(6) - 3(5),\ \ 3(4) - 1(6),\ \ 1(5) - 2(4)\,] = [12 - 15,\ 12 - 6,\ 5 - 8] = [-3, 6, -3]$$

Check: $[-3, 6, -3] \cdot [1, 2, 3] = -3 + 12 - 9 = 0$ ✓ and $[-3, 6, -3] \cdot [4, 5, 6] = -12 + 30 - 18 = 0$ ✓

3. (Core) Let $\vec{a} = [2, 3, -1]$ and $\vec{b} = [-1, 1, 4]$. Find $\vec{a} \times \vec{b}$ and verify that it’s perpendicular to both vectors.

Solution

$$\begin{aligned} \vec{a} \times \vec{b} &= [\,3(4) - (-1)(1),\ \ (-1)(-1) - 2(4),\ \ 2(1) - 3(-1)\,] \\ &= [12 + 1,\ 1 - 8,\ 2 + 3] \\ &= [13, -7, 5] \end{aligned}$$

Check: $[13, -7, 5] \cdot [2, 3, -1] = 26 - 21 - 5 = 0$ ✓ and $[13, -7, 5] \cdot [-1, 1, 4] = -13 - 7 + 20 = 0$ ✓

4. (Core) Suppose $\vec{a} \times \vec{b} = [2, -3, 6]$. Without knowing $\vec{a}$ or $\vec{b}$, find:

• (a) $\vec{b} \times \vec{a}$
• (b) $(3\vec{a}) \times \vec{b}$
• (c) $(2\vec{a}) \times (5\vec{b})$
• (d) $|\vec{a} \times \vec{b}|$

Solution

(a) Anticommutative: $\vec{b} \times \vec{a} = [-2, 3, -6]$.

(b) $3(\vec{a} \times \vec{b}) = [6, -9, 18]$.

(c) $2 \times 5 = 10$, so $10(\vec{a} \times \vec{b}) = [20, -30, 60]$.

(d) $\sqrt{4 + 9 + 36} = \sqrt{49} = 7$.

5. (Core) $|\vec{a}| = 4$, $|\vec{b}| = 6$, and the angle between them is $30^\circ$.

• (a) Find $|\vec{a} \times \vec{b}|$.
• (b) Would the answer change if the angle were $150^\circ$? Would $\vec{a} \cdot \vec{b}$ change?

Solution

(a) $|\vec{a} \times \vec{b}| = (4)(6)\sin 30^\circ = 24 \times \dfrac{1}{2} = 12$.

(b) No: $\sin 150^\circ = \sin 30^\circ = \dfrac{1}{2}$, so the magnitude is still $12$. But the dot product changes sign: $24\cos 30^\circ = 12\sqrt{3}$, while $24\cos 150^\circ = -12\sqrt{3}$.

6. (Core) Use the cross product to show that $\vec{u} = [2, -4, 6]$ and $\vec{v} = [-3, 6, -9]$ are parallel.

Solution

$$\begin{aligned} \vec{u} \times \vec{v} &= [\,(-4)(-9) - 6(6),\ \ 6(-3) - 2(-9),\ \ 2(6) - (-4)(-3)\,] \\ &= [36 - 36,\ -18 + 18,\ 12 - 12] \\ &= [0, 0, 0] \end{aligned}$$

The cross product is the zero vector, so the vectors are parallel. (Indeed, $\vec{v} = -\dfrac{3}{2}\vec{u}$, so they point in opposite directions.)

7. (Core) Let $\vec{a} = [1, 2, 2]$ and $\vec{b} = [2, -1, 2]$. Find $\vec{a} \cdot \vec{b}$ and $|\vec{a} \times \vec{b}|$, then show that $\cos^2\theta + \sin^2\theta = 1$ for the angle between them. Find $\theta$ to one decimal place.

Solution

$\vec{a} \cdot \vec{b} = 2 - 2 + 4 = 4$, and $|\vec{a}| = |\vec{b}| = 3$.

$$\vec{a} \times \vec{b} = [\,2(2) - 2(-1),\ \ 2(2) - 1(2),\ \ 1(-1) - 2(2)\,] = [6, 2, -5], \qquad |\vec{a} \times \vec{b}| = \sqrt{36 + 4 + 25} = \sqrt{65}$$

So $\cos\theta = \dfrac{4}{9}$ and $\sin\theta = \dfrac{\sqrt{65}}{9}$, and

$$\cos^2\theta + \sin^2\theta = \frac{16}{81} + \frac{65}{81} = \frac{81}{81} = 1 \ \checkmark$$

The cosine is positive, so $\theta$ is acute: $\theta = \cos^{-1}\left(\dfrac{4}{9}\right) \approx 63.6^\circ$.

8. (Challenge) Let $\vec{a} = [1, 1, 0]$, $\vec{b} = [0, 1, 0]$, and $\vec{c} = [0, 0, 1]$. Compute $(\vec{a} \times \vec{b}) \times \vec{c}$ and $\vec{a} \times (\vec{b} \times \vec{c})$. What does this show?

Solution

$\vec{a} \times \vec{b} = [\,1(0) - 0(1),\ \ 0(0) - 1(0),\ \ 1(1) - 1(0)\,] = [0, 0, 1]$, so

$$(\vec{a} \times \vec{b}) \times \vec{c} = [0, 0, 1] \times [0, 0, 1] = [0, 0, 0]$$

(a vector crossed with itself is $\vec{0}$).

$\vec{b} \times \vec{c} = \vec{j} \times \vec{k} = [1, 0, 0]$, so

$$\vec{a} \times (\vec{b} \times \vec{c}) = [1, 1, 0] \times [1, 0, 0] = [\,1(0) - 0(0),\ \ 0(1) - 1(0),\ \ 1(0) - 1(1)\,] = [0, 0, -1]$$

The results are different, so the cross product is not associative.

9. (Challenge) Use components to prove that $\vec{b} \cdot (\vec{a} \times \vec{b}) = 0$ for any vectors $\vec{a}$ and $\vec{b}$ in 3-space. Together with the same fact for $\vec{a}$, explain what this tells you about the direction of $\vec{a} \times \vec{b}$.

Solution

$$\begin{aligned} \vec{b} \cdot (\vec{a} \times \vec{b}) &= b_1(a_2 b_3 - a_3 b_2) + b_2(a_3 b_1 - a_1 b_3) + b_3(a_1 b_2 - a_2 b_1) \\ &= a_2 b_1 b_3 - a_3 b_1 b_2 + a_3 b_1 b_2 - a_1 b_2 b_3 + a_1 b_2 b_3 - a_2 b_1 b_3 \\ &= 0 \end{aligned}$$

Each term cancels with another. A zero dot product means $\vec{a} \times \vec{b}$ is perpendicular to $\vec{b}$. Together with $\vec{a} \cdot (\vec{a} \times \vec{b}) = 0$ (shown in Key ideas), this verifies that the cross product is perpendicular to both of the original vectors.