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On the web I’ve often seen people ask where the cross product comes from. Often they get back the formula with no explanation, or a proof of the determinant form having the desired properties. This might be useful for mathematicians or physicists, but the value of a 3×3 determinant is something I need to manually compute even less often than cross products.

Some might be happy to take the definition as is and commit it to memory, but I prefer to save my brain space for more complicated topics. Instead I have a derivation I use when I need to pull out the formula for the cross product, which only uses things I already know.

One place the cross product appears is in quaternion multiplication so we’ll start from there. Don’t fret if you don’t know anything about quaternions since not much is needed for this purpose. Everything you need to know about them for this should be covered in this post.

One common method of writing a quaternion is

where is a scalar value, and is a 3D vector. Given this representation, the multiplication of two quaternions can be written as

Where and are the regular vector dot and cross products.

Using this we can “promote” a vector to a quaternion by writing it as

Multiplying two vectors as quaternions in this form then looks like

So the vector part of the product of two vectors “promoted” to quaternions is the cross product of the vectors. Well how does that get us any closer to knowing the formula for the cross product?

The derivation of the quaternion product is actually fairly straight forward as well, but we need a (simple) formula given to us by Hamilton. We can write our quaternion as a real number part plus products of three imaginary values , , with the property .

Then if we have two quaternions and written as

We can multiply these in the same fashion as regular polynomials except treating , , as not commutative under multiplication. To be completely verbose this looks like

Next distribute, and rearrange constants to a more pleasing form. As noted above it is important to not treat , , as commutative.

This looks like a complete train wreck, but at this point we can make use of our quaternion equation from before to clean things up. Again being quite explicit this changes the above into

Now rearrange so the product is in the same form as and were above

which is the normal quaternion product that was given in its scalar + vector form earlier.

Remembering back to the first part of this post, the cross product is just the vector part of our quaternion multiplication if we take . So taking the w values as zero, the vector part of the last formula is

which is the definition of the cross product as expected. So there you have it a derivation of the cross product that doesn’t treat it as a rabbit from a hat.