2.4 The Cross Product

Big idea

The dot product answers a question about alignment and returns a number. This section answers a different question and returns a vector. Given two vectors that are not parallel, there is a direction perpendicular to both of them, and there is a natural size to attach to that direction. The cross product packages both into one object, and it exists only in three dimensions, because only there does a plane have a single perpendicular direction up to sign.

The formula looks like a determinant, and that is not a coincidence or a memory aid dressed up as mathematics. The symbolic determinant with the standard unit vectors in the top row, the components of the first vector in the second row and the components of the second vector in the third row expands exactly into the three components of the cross product. Writing it that way also explains the properties for free: swapping two rows of a determinant flips its sign, which is why the cross product reverses when you reverse the order of the factors, and a determinant with two equal rows is zero, which is why a vector crossed with itself vanishes.

The size of the cross product is where the geometry lives. It equals $\|\mathbf{u}\| \, \|\mathbf{v}\| \sin\theta$, which is the area of the parallelogram the two vectors span. So the cross product measures how much two vectors fail to be parallel, in the same way that the dot product measures how much they fail to be perpendicular. Combining the two products gives the volume of the box built on three vectors, and a volume of zero is the test for three vectors lying in a common plane.

Physics uses this operation constantly. Torque, angular momentum and the force on a moving charge in a magnetic field are all cross products, and they all share the feature that the answer points out of the plane of the inputs. The right-hand rule is what fixes the sign, and it is the reason the coordinate system was declared right-handed in the first place.

Decoder

The cross product of two vectors is the vector perpendicular to both of them, with length equal to the area of the parallelogram they span, and direction given by the right-hand rule.

Each clause is doing work. Perpendicular to both settles the line the answer lies on. The area settles how long it is. The right-hand rule settles which of the two opposite directions along that line is chosen: point your right fingers along the first vector, curl them toward the second, and your thumb gives $\mathbf{u} \times \mathbf{v}$. Reverse the order of the factors and your thumb reverses too.

The clause about area also explains the degenerate case. If $\mathbf{u}$ and $\mathbf{v}$ are parallel, the parallelogram they span is flat and has no area, so the cross product is the zero vector. That gives a parallelism test to sit alongside the orthogonality test from the previous section: $\mathbf{u} \cdot \mathbf{v} = 0$ detects perpendicular vectors and $\mathbf{u} \times \mathbf{v} = \mathbf{0}$ detects parallel ones.

The classic mistake is assuming that the cross product behaves like multiplication. It does not commute, and it is not associative: grouping three factors differently generally changes the answer. Nor can you cancel: $\mathbf{u} \times \mathbf{v} = \mathbf{u} \times \mathbf{w}$ does not force $\mathbf{v} = \mathbf{w}$. The safe habit is to verify every cross product you compute by dotting it against both original vectors and confirming that you get zero twice.

Definitions and results

Definition. For $\mathbf{u} = \langle u_1, u_2, u_3 \rangle$ and $\mathbf{v} = \langle v_1, v_2, v_3 \rangle$,

$$ \mathbf{u} \times \mathbf{v} = \langle u_2 v_3 - u_3 v_2, \; u_3 v_1 - u_1 v_3, \; u_1 v_2 - u_2 v_1 \rangle $$

The result is a vector, not a scalar, and both inputs must be three-dimensional.

Determinant form. The same computation is the expansion of

$$ \mathbf{u} \times \mathbf{v} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ u_1 & u_2 & u_3 \\ v_1 & v_2 & v_3 \end{vmatrix} $$

along the top row. The middle term carries a minus sign from the expansion, which is where sign errors usually enter.

Orthogonality. The cross product is orthogonal to both factors: $\mathbf{u} \cdot (\mathbf{u} \times \mathbf{v}) = 0$ and $\mathbf{v} \cdot (\mathbf{u} \times \mathbf{v}) = 0$. This is the cheapest available check on a computed answer.

Algebraic properties. The operation is anticommutative, $\mathbf{v} \times \mathbf{u} = -(\mathbf{u} \times \mathbf{v})$, and $\mathbf{u} \times \mathbf{u} = \mathbf{0}$. It distributes over addition and scalars pull out, but it is neither commutative nor associative.

The standard unit vectors. In a right-handed system, $\mathbf{i} \times \mathbf{j} = \mathbf{k}$, $\mathbf{j} \times \mathbf{k} = \mathbf{i}$ and $\mathbf{k} \times \mathbf{i} = \mathbf{j}$, while reversing any of these pairs flips the sign. The cycle $\mathbf{i} \to \mathbf{j} \to \mathbf{k} \to \mathbf{i}$ is worth memorizing.

Magnitude and area. If $\theta$ is the angle between the vectors, then

$$ \|\mathbf{u} \times \mathbf{v}\| = \|\mathbf{u}\| \, \|\mathbf{v}\| \sin\theta $$

which is the area of the parallelogram with sides $\mathbf{u}$ and $\mathbf{v}$. Half of it is the area of the triangle with those two sides. Two nonzero vectors are parallel exactly when the cross product is $\mathbf{0}$.

Scalar triple product and volume. For three vectors, $\mathbf{u} \cdot (\mathbf{v} \times \mathbf{w})$ is a scalar equal to the determinant whose rows are the three vectors. Its absolute value is the volume of the parallelepiped they span:

$$ V = |\mathbf{u} \cdot (\mathbf{v} \times \mathbf{w})| $$

The value is zero exactly when the three vectors are coplanar, since a flat box has no volume.

Torque. A force $\mathbf{F}$ applied at a position $\mathbf{r}$ measured from a pivot produces torque $\boldsymbol{\tau} = \mathbf{r} \times \mathbf{F}$. Its magnitude is $\|\mathbf{r}\| \, \|\mathbf{F}\| \sin\theta$, so a force directed straight along the lever arm produces no turning effect, and a perpendicular force produces the most.

Worked examples

Computing a cross product and checking it

Let $\mathbf{u} = \langle 1, 2, -1 \rangle$ and $\mathbf{v} = \langle 3, 0, 2 \rangle$. Expanding the determinant along the top row:

$$ \mathbf{u} \times \mathbf{v} = \mathbf{i}\big(2 \cdot 2 - (-1) \cdot 0\big) - \mathbf{j}\big(1 \cdot 2 - (-1) \cdot 3\big) + \mathbf{k}\big(1 \cdot 0 - 2 \cdot 3\big) $$

$$ \mathbf{u} \times \mathbf{v} = \langle 4, -5, -6 \rangle $$

Check both orthogonality conditions: $\langle 1,2,-1\rangle \cdot \langle 4,-5,-6\rangle = 4 - 10 + 6 = 0$ and $\langle 3,0,2\rangle \cdot \langle 4,-5,-6\rangle = 12 + 0 - 12 = 0$. Reversing the order would give $\mathbf{v} \times \mathbf{u} = \langle -4, 5, 6 \rangle$, the same line with the opposite heading.

Area of a triangle in space

Find the area of the triangle with vertices $A(1,1,0)$, $B(3,2,1)$ and $C(2,4,3)$.

Two sides emanating from $A$ are $\overrightarrow{AB} = \langle 2,1,1 \rangle$ and $\overrightarrow{AC} = \langle 1,3,3 \rangle$. Their cross product is

$$ \overrightarrow{AB} \times \overrightarrow{AC} = \mathbf{i}(1 \cdot 3 - 1 \cdot 3) - \mathbf{j}(2 \cdot 3 - 1 \cdot 1) + \mathbf{k}(2 \cdot 3 - 1 \cdot 1) = \langle 0, -5, 5 \rangle $$

Its magnitude is $\sqrt{0 + 25 + 25} = 5\sqrt{2}$, which is the area of the parallelogram. The triangle is half of it:

$$ \text{Area} = \frac{5\sqrt{2}}{2} \approx 3.54 $$

The first component came out zero, which says the normal vector has no $x$-part, so the plane of the triangle is parallel to the $x$-axis. Dotting $\langle 0,-5,5\rangle$ into each side gives $0$ both times, confirming the computation.

Volume, and a test for coplanarity

Using the same $\mathbf{u} = \langle 1,2,-1 \rangle$ and $\mathbf{v} = \langle 3,0,2 \rangle$, with $\mathbf{u} \times \mathbf{v} = \langle 4,-5,-6 \rangle$, take $\mathbf{w} = \langle 1,1,1 \rangle$. The scalar triple product is

$$ \mathbf{w} \cdot (\mathbf{u} \times \mathbf{v}) = 4 - 5 - 6 = -7 $$

so the parallelepiped built on the three vectors has volume $7$. The negative sign only records the orientation of the triple.

Now replace $\mathbf{w}$ by $\mathbf{w}' = \langle 4, 2, 1 \rangle$. Then $\mathbf{w}' \cdot (\mathbf{u} \times \mathbf{v}) = 16 - 10 - 6 = 0$, so these three vectors are coplanar. The reason is visible on inspection: $\mathbf{u} + \mathbf{v} = \langle 4, 2, 1 \rangle$, so $\mathbf{w}'$ already lies in the plane spanned by the other two and adds no thickness to the box.

Torque on a wrench

A wrench extends $0.25$ meters from a bolt along the positive $x$-axis, and you push straight down with $60$ newtons, so $\mathbf{r} = \langle 0.25, 0, 0 \rangle$ and $\mathbf{F} = \langle 0, 0, -60 \rangle$. Then

$$ \boldsymbol{\tau} = \mathbf{r} \times \mathbf{F} = \mathbf{i}(0 - 0) - \mathbf{j}\big(0.25(-60) - 0\big) + \mathbf{k}(0 - 0) = \langle 0, 15, 0 \rangle $$

The torque has magnitude $15$ newton-meters and points along the positive $y$-axis, which is the axis the bolt turns about. Since the force is perpendicular to the wrench, $\sin\theta = 1$ and the magnitude is just $0.25 \times 60$. Pushing along the handle instead, with $\mathbf{F}$ parallel to $\mathbf{r}$, would give a cross product of $\mathbf{0}$: no turning at all, which matches experience.

Practice

These problems cover the whole operation: expanding the determinant, using the right-hand rule and the standard unit vectors, finding a vector orthogonal to two given vectors, computing areas and volumes, and applying the triple product as a coplanarity test.

Quiz

Five items on computing cross products, orthogonality and the right-hand rule, areas of parallelograms and triangles, and the scalar triple product.