What's behind dot product and cross product?

Question

I would like to know where the dot product and the cross product came from.

Why they exist, why they are in the way they are?

$A\times B \\A\cdot B$

Thank you

Answer 1

The cross product has been "created" almost out of scratch by the great american physicist Gibbs, around 1895, for writing under a more convenient form Maxwell equations (that had already been very simplified by Heaviside). He had to fight the "quaternionists" camarilla who advocated the use of quaternions for writing these equations. The mathematicians only slowly adopted cross product (say it was around the 1940-1950s that it was universally accepted), whereas...

The dot product had made its way in mathematics at the eve of XXth century. One can say that it has a very humble origin. It was something that you could not get rid off in some computations; in particular, it was the supplementary sum of terms when expanding the "norm" (a name invented in the XXth century) of the sum $V+V'$ of two vectors $V=(a,b)$ and $V'(a',b')$ (think to physicists who had to understand the intensity of two non collinear forces):

$$\underbrace{(a+a')^2+(b+b')^2}_{\text{square of the length of V+V'}} \ \ \ =\underbrace{a^2+b^2}_{\text{square of the length of V}}+\underbrace{a'^2+b'^2}_{\text{square of the length of V'}}+2\underbrace{(aa'+bb')}_{\text{something...}}$$

A "something" that was looking as a kind of double product, as if one could write the preceding equation under the form:

$$(V+V')^2=V^2+V'^2+2V.V'...$$

People who have dared to use this analogy with "ordinary algebra" have, for some of them, endured criticism, and in particular the founding father of vector algebra, Grassmann, who had ideas such ahead of his time (circa 1840-1850) that his works were only understood after his death...

Hilbert (circa 1910) has played an important role in imposing the concept of dot product, by enlarging it: what a giant mental step to say that two functions $f$ and $g$ are orthogonal is their 'dot product" $\int_a^b f(t)g(t)dt=0$...

Remark: for understanding the interplay between quaternions and cross product, have a look at (https://math.stackexchange.com/q/1917093).

Answer 2

Since this is a mathematics forum (not physics), I'll respond with a mathematician's answer:

The dot product comes from the only non-trivial rotation- and scale-invariant scalar function of two vectors that is symmetric in arguments (i.e., $f({\bf A}, {\bf B}) = f({\bf B}, {\bf A})$) and is linear in each vector, that is, obeys:

$f(2{\bf A},{\bf B}) = f({\bf A}, 2 {\bf B}) = 2 f({\bf A}, {\bf B})$

That is, first think of "all" possible functions of two vectors that leads to a scalar. You might try the (scalar) length of the sum of ${\bf A}$ and ${\bf B}$:

$g({\bf A}, {\bf B}) = |{\bf A} + {\bf B}|$

This function simply doesn't have the linearity property as described above:

$g(2 {\bf A}, {\bf B}) = |2 {\bf A} + {\bf B}| \neq 2 g({\bf A}, {\bf B})$.

As you can see, this length is not linear in each vector: If you double ${\bf A}$ or ${\bf B}$, you generally will not get double the length of ${\bf A} + {\bf B}$. So that doesn't work. Try other functions. You'll find that the only scalar function that is linear in both vectors is ${\bf A} \cdot {\bf B}$:

$h({\bf A}, {\bf B}) = {\bf A} \cdot {\bf B} = |{\bf A}| |{\bf B}| \cos \theta$.

Now check linearity:

$h(2 {\bf A}, {\bf B}) = |2 {\bf A}| |{\bf B}| \cos \theta = 2 |{\bf A}| |{\bf B}| \cos \theta = 2 h({\bf A}, {\bf B})$,

as desired!

The cross product comes from the only non-trivial linear function of the two component input vectors (up to sign... right-hand or left-hand convention). Try to think of a function of ${\bf A}$ and ${\bf B}$ that gives a vector. The simplest function that comes to mind is: ${\bf A} + {\bf B}$. We want a function that is linear in each vector, i.e., if we double ${\bf A}$ or ${\bf B}$ we get double the output. The simple sum does not work, as you can check for yourself. Only the cross product will work!

Try it!

Answer 3

Two examples from physics:

The dot product $\vec F\cdot\vec d$ gives the work performed by the force $\vec F$ acting upon a body that had a "small enough" displacement $\vec d$.

The cross product $\vec v \times \vec B$ gives us the Lorentz force (per unit charge) acting upon an electrically charged particle that moves with velocity $\vec v$ in a constant magnetic field $\vec B$.