The Cross Product of Two Vectors Explained

What the cross product produces

The dot product turns two vectors into a number. The cross product, also called the vector product, turns two vectors into a third vector. It is defined only in three dimensions, and the result has two defining features. Its direction is perpendicular to both original vectors, and its length equals the area of the parallelogram they form:

Here $\theta$ is the angle between the vectors, from 0 to 180 degrees. Because the sine of 0 and 180 degrees is zero, parallel vectors always have a cross product of zero, and perpendicular vectors give the largest possible length for their sizes.

The component formula

For $\mathbf{a} = (a_1, a_2, a_3)$ and $\mathbf{b} = (b_1, b_2, b_3)$, the components are:

That formula is hard to memorise, so most people use a symbolic determinant instead. Put the unit vectors $\mathbf{i}$, $\mathbf{j}$ and $\mathbf{k}$ in the first row, then the components of $\mathbf{a}$ and $\mathbf{b}$, and expand along the first row:

Always finish with the two dot product checks. They take seconds and catch nearly every sign error, especially in the middle component.

The right-hand rule and order

Two opposite directions are perpendicular to both vectors, and the right-hand rule picks one. Point the fingers of your right hand along $\mathbf{a}$, curl them towards $\mathbf{b}$, and your thumb points along $\mathbf{a} \times \mathbf{b}$. For the standard unit vectors this gives $\mathbf{i} \times \mathbf{j} = \mathbf{k}$, $\mathbf{j} \times \mathbf{k} = \mathbf{i}$ and $\mathbf{k} \times \mathbf{i} = \mathbf{j}$.

Reversing the order points your thumb the other way, so the cross product is anticommutative:

It is also not associative: $(\mathbf{i} \times \mathbf{i}) \times \mathbf{j} = \mathbf{0}$, while $\mathbf{i} \times (\mathbf{i} \times \mathbf{j}) = \mathbf{i} \times \mathbf{k} = -\mathbf{j}$. It does distribute over addition, and scalars can be factored out, so $(k\mathbf{a}) \times \mathbf{b} = k(\mathbf{a} \times \mathbf{b})$.

Areas of parallelograms and triangles

Because its length is a parallelogram's area, the cross product gives areas directly from coordinates. The vectors in Example 1 span a parallelogram of area $|(-3, 6, -3)| = \sqrt{54} \approx 7.35$, and the triangle with those two sides has half that area, about 3.67.

Normal vectors and equations of planes

A plane is determined by a point on it and a normal vector perpendicular to it. Since the cross product of two vectors lying in a plane is perpendicular to both, it is exactly such a normal. In Example 2, the vector $(6, 3, 2)$ is normal to the plane through $P$, $Q$ and $R$, so the plane has the equation $6x + 3y + 2z = d$. Substituting $P(1, 0, 0)$ gives $d = 6$.

The plane is $6x + 3y + 2z = 6$, and $Q$ and $R$ also satisfy it: $3 \cdot 2 = 6$ and $2 \cdot 3 = 6$. The same idea is used constantly in 3D graphics, where every triangle of a model needs a normal vector to decide how it is lit and which side faces the camera.

Torque and other physics

Torque, the turning effect of a force, is a cross product: $\boldsymbol{\tau} = \mathbf{r} \times \mathbf{F}$, where $\mathbf{r}$ runs from the pivot to the point where the force acts. Its magnitude is $|\mathbf{r}|\,|\mathbf{F}|\sin\theta$.

Angular momentum and the magnetic force on a moving charge are cross products too, and in each case the right-hand rule gives the direction of the result.

The scalar triple product and volume

Combining both products gives the scalar triple product $\mathbf{a} \cdot (\mathbf{b} \times \mathbf{c})$, which equals the $3 \times 3$ determinant with rows $\mathbf{a}$, $\mathbf{b}$ and $\mathbf{c}$. Its absolute value is the volume of the slanted box the three vectors span. For $(1, 0, 0)$, $(0, 2, 0)$ and $(0, 0, 3)$, the box is an ordinary rectangular box and the triple product is $1 \cdot 2 \cdot 3 = 6$. A triple product of zero means the three vectors lie in one plane.

The cross product calculator computes cross products with every determinant step shown, and also reports areas and the angle between the vectors. Pair it with the dot product calculator to verify perpendicularity.

Distance from a point to a line

The area interpretation also solves a classic distance problem. Take a line through a point $A$ with direction vector $\mathbf{u}$, and a point $P$ off the line. The parallelogram spanned by $\overrightarrow{AP}$ and $\mathbf{u}$ has area $|\overrightarrow{AP} \times \mathbf{u}|$, and since its base is $|\mathbf{u}|$, its height, which is the distance from $P$ to the line, is the area divided by the base.

For example, take the line through the origin in the direction $(1, 2, 2)$ and the point $P(2, 1, 3)$. The cross product $(2, 1, 3) \times (1, 2, 2) = (-4, -1, 3)$ has length $\sqrt{26}$, and the direction vector has length 3, so the distance is $\frac{\sqrt{26}}{3} \approx 1.70$. You can confirm this with a projection: the point is $\sqrt{14}$ from the origin and its projection along the line has length $\frac{10}{3}$, and the Pythagorean theorem gives the same distance.

Common mistakes and practice

• Forgetting the minus sign on the middle component when expanding the determinant.
• Reversing the order of the vectors, which flips the direction of the answer.
• Using the cross product's length as the triangle's area instead of half of it.
• Trying to take the cross product of two-dimensional vectors without first adding a zero third component.

For practice, check that $(1, 0, 0) \times (0, 1, 0) = (0, 0, 1)$, which is the rule $\mathbf{i} \times \mathbf{j} = \mathbf{k}$. Then confirm that $(1, 2, 3) \times (2, 4, 6)$ is the zero vector, since the second vector is twice the first. Finally, compute $(1, 1, 0) \times (0, 1, 1)$; you should get $(1, -1, 1)$, and both dot product checks should come out to zero.

Why the formula gives a perpendicular vector

It is worth seeing why the component formula works rather than just trusting it. Dot the result with $\mathbf{a}$ itself: you get $a_1(a_2b_3 - a_3b_2) + a_2(a_3b_1 - a_1b_3) + a_3(a_1b_2 - a_2b_1)$. Expand the brackets and every term appears twice with opposite signs, so the total is exactly zero. The same cancellation happens when you dot with $\mathbf{b}$. In determinant language, dotting with $\mathbf{a}$ replaces the row of unit vectors by $\mathbf{a}$, producing a determinant with two equal rows, and such a determinant is always zero. Either way, the perpendicularity is built into the formula rather than being a coincidence of particular examples.

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