F5 - The vector product

Cross-product components

Given two directions in space, how can you find a direction perpendicular to both? For example, a vector along the positive horizontal axis and one along the positive vertical axis in a flat page have a common perpendicular pointing out of the page.

You already know that a scalar product produces a number. The vector product, also called the cross product, produces a vector. For the standard unit vectors in a right-handed coordinate system, $\mathbf{i}\times\mathbf{j}=\mathbf{k}$. Its direction and size are both important.

The first component uses the second and third coordinates; the second uses the third and first; the third uses the first and second. Each is a difference of two products. A determinant-style memory aid is

The minus sign before the middle minor is essential. The column-vector formula has already incorporated it.

Now change only the last component of $\mathbf{b}$. Notice which components of the product respond to that change.

These scalar-product checks detect many slips. They do not distinguish a correct cross product from its negative; the order in the component rule determines the sign.

Use the same calculation with a fresh pair.

Orientation, magnitude and noncommutativity

The size of a cross product measures the area spanned by two vectors, while its direction records their order.

Take $\mathbf{a}=(2,0,0)$ and $\mathbf{b}=(1,3,0)$. They span a parallelogram of base $2$ and perpendicular height $3$, so its area is $6$. Their cross product is $(0,0,6)$, perpendicular to the parallelogram. Swapping the order gives $(0,0,-6)$: the area stays the same, but the normal reverses.

[DIAGRAM: asset_name: Area and orientation of the vector product; asset_slug: aqa7367-cross-orientation; caption: The parallelogram has area six while reversing the vector order reverses the normal; description: Two equal-scale views of the same xy-plane parallelogram spanned by a equals (2,0,0) and b equals (1,3,0). Blue a points right and clay b points up and right. The base is two and perpendicular height three. The left panel labels a cross b as six k pointing out of the page with a circle-dot symbol; the right labels b cross a as minus six k pointing into the page with a circle-cross symbol. A note explains the two normal symbols.]

The component rule fixes this orientation, rather than allowing either perpendicular direction.

The magnitude follows from base times height: the perpendicular height of the parallelogram is $|\mathbf{b}|\sin\theta$. Since sine is nonnegative on this angle interval, the magnitude is nonnegative. Parallel or antiparallel vectors give zero; if either input is zero, the component rule also gives zero. A zero cross product has no direction. For two nonzero vectors, a zero cross product therefore tests whether they are parallel.

Keep the lengths fixed and change only the angle. An obtuse angle remains a valid vector angle.

There is a useful connection to the scalar product. Squaring its cosine formula and the cross-product sine formula gives

because $\cos^2\theta+\sin^2\theta=1$. Thus dot and cross products together recover the product of the lengths without finding the angle. The identity also holds when an input is zero.

Cross products distribute over addition, and scalar factors can be taken outside:

Each rule follows by expanding the component formula. They hold for real scalars, including negative numbers and zero. Also, $\mathbf{a}\times\mathbf{a}=\mathbf{0}$. These properties simplify expressions even when you are not given the original vector components. Cross products are not generally associative: keep the brackets in a product of three vectors.