We have always been told that vector division is impossible (or better, not defined), but Sunday night I came up with an idea on how to divide vectors to each other. I will do my usual brainstorm thing in English this time, for this fourth post of what apparently became a series of articles on vector calculus and trigonometry.
To quickly put English readers in context, in the preceding posts I was encouraging people to use vector operations instead of specific coordinate systems, dimension and trigonometric transcendental functions (much more error prone, abstract to understand and less generic). Indeed I showed how to solve school exercises with then new tools (like computing the area of triangles or calculating the intersection of a ray with a cylinder). I introduced the three laws of trigonometry ( “law of projections” (a·b) = ½(|a|² + |b|² – |c|²), the “law of areas” |axb| = |bxc| = |cxa| and the “law of the products” (a·b)² + |axb|² = |a|²|b|² that naturally replace the cosinus and sinus laws plus the basic trigonometric identity), plus some other ideas like projections. Also, in other posts, I arbitrarily used vectors or functions and signals showing that way that actually correlation, Fourier transforms and so forth have some sort of geometric interpretation. Well.
So yes, I decided to invent the division of vectors and then investigate it’s properties. I proceed like this:
We first realize that somehow vector division is actually kind-of naturally defined for vectors that are aligned, since if a and b are this two aligned vectors then a = b·k, so it seems tempting to assert that the division a/b is indeed k, a scalar. Now, what happens when a is not aligned to b ?
The big trick is to decompose a into two vectors, one that is parallel to b and another one that is perpendicular. Let’s call them ap (for parallel) and ao (for orthogonal) such that a = ap+ao. So, a/b = (ap+ao)/b = ap/b + ao/b. By projection we get ap=b(a.b)/|b|², and therefore ao=a-b(a·b)/|b|². The first part of the division is easy to solve, cause ap and b are parallel by definition and so the division is just a scalar. Actually it’s value is ap/b = b(a·b)/|b|²/b = (a·b)/|b|² . Hey, so far so good!
We now need to decide what to do with the division of perpendicular vectors. Well, my proposal is that we just do as if nothing happened and they were parallel, but just taking care to keep track of the fact that they actually are perpendicular. Let’s add an exclamation mark (“!“) in front of their scalar division as a remainder. Then we first need the length of ao, |ao|² = |a-b(a·b)/|b|²|² . We expand it as normal binomial to get |ao|² = (|a|²-(a·b)²)/|b|². Now we note that by the third law of vector trigonometry |ao|² = |axb|²/|b|², meaning that ao/b = ! |axb|/|b|². Putting parallel and orthogonal parts together we get:
what pretty much looks like a complex scalar, with it’s real part somehow saying about the parallel nature of the division and the imaginary part about the perpendicular part. In fact, the modulo of this scalar gives the ratio of the lengths of a to b. See: |k|² = ((a·b)² + |axb|²)/|b|^4 = |a|²|b|^4/|b|² = |a|²/|b|², so |k|=|a|/|b|=|a/b|. And even more, check for it, the argument of this complex number is nothing but the angle between our vectors a and b !!
Basically, in this game, whatever the amount of dimensions, is seems that the division of two vectors is a complex scalar (real if the vectors are aligned). For those going too fast, caution!, a complex number is NOT a vector (and you should not think as if it was), it’s just a scalar (yes, with more than one components). All this also means of course we can rotate vector b into vector a by just multiplying by a scalar k by doing a = b·k. Now we will have to think for what we can use the division of vectors. Anyone has ideas?
Note: all this little discoveries suspuciusly start to point to much in the direction of that Geometry Algebra thingy?