A few weeks ago, my colleague who teaches Physics asked me about the derivation and justification of the **cross-product** formula.

The cross product for two vectors will find a third vector that is perpendicular to the original two vectors given.

The given vectors are assumed to be perpendicular (orthogonal) to the vector that will result from the cross product. This means that the **dot** product of each of the original vectors with the new vector will be zero.

So, given two vectors

and

we want to find a third vector

so that is perpendicular to both and

As I mentioned above this means that we want the dot product of with each of the two original vectors to be zero.

and

This gives us two equations to work with. Since we have three variables to solve for , we’ll need another equation to work with.

The website *Heaven’s in the backyard* introduces a third constraint that the modulus of the cross product vector be equal to 1.

This creates a third equation and allows us to solve for and in terms of and

As mentioned above the web page *Heaven’s in the backyard* does a nice job with the derivation of the values for and and ends up with the formula .

I don’t see this mentioned in the derivation, but it appears that the term that is factored out and defined to make the derivation work more smoothly is actually the modulus of the cross product vector .

Assuming that the cross product vector has a length of results in an answer that is multiplied by , because the actual perpendicular has a modulus equal to which is .

After working through some of these ideas, I became interested in where the cross-product came from.

At Wikipedia, they mention that Joseph-Louis Lagrange, the French/Italian mathematician of the late 18th century provided a formula for this in a paper from 1773 that was focused on the properties of a tetrahedron. The calculations related to the cross-product appear in the first few pages of the paper. Lagrange posits 9 “quantités” and then proceeds through a blizzard of calculations based on these quantities.

If each triplet of values , and is considered as the coordinates of a vector, then the first calculations are the cross-products of each vector with each of the others. In Lagrange’s notation, we could identify each of the cross products as

, ,

On the following page Lagrange identifies the square of the modulus for each of the cross-product vectors as , and

Several pages later Lagrange notes that the dot product of each original vector with the appropriate cross-product produces a zero result.

There are two things that fascinate me about this – (1) the depth of this seemingly simple question – how do you justify the cross-product formula? and (2) what was Lagrange up to in this paper? – what is the purpose of the multitude of calculations that he completes in the paper from 1773 (*Solutions analytiques de quelques problèmes sur les pyramides triangulaires*)?

on August 5, 2018 at 1:08 AM |LekdhenIs it possible if you can attach some screenshots from Heaven’s in the backyard? If there are no copyright issues because that blog is invite-only.