Are Dot and Cross products fundamental ideas?

Question

Are Dot and Cross Products fundamental ideas of Vector Algebra? Isn't there some intuition they were born from? From the looks of the Cross Product, especially, it's hard to believe there isn't one.

Answer 1

They're very fundamental.

You can show the dot product is invariant under rotations, and is the unique bilinear generalisation of squared length.

The cross product $a\times b$ in $3$ dimensions can be considered in any of several ways: it's the determinant of a matrix with $a,\,b$ as two rows and $(i,\,j,\,k)$ as the third; $a,\,b,\,a\times b$ form a right-hand set of axes, with the third orthogonal to the plane the others span; $(a\cdot b)^2+(a\times b)^2=a^2b^2$; and it makes even more sense once you learn about the Levi-Civita symbol.

There's also a geometric connection: a parallelepiped with edges parallel to $a,\,b,\,c$ had volume $a\cdot b\times c$, which can also be written as a matrix determinant.

Answer 2

Geometrically, it is quite clear that the (signed) volume of a parallelepiped is a linear function in each of the three side vectors $\vec a,\vec b,\vec c$. This means that if we fix $\vec a$ and $\vec b$, then the signed volume is a linear function of $\vec c$. If you already know that every linear function $\Bbb R^3\to \Bbb R$ can be written as $\vec x\mapsto \vec x\cdot \vec n$ for a suitable vector $\vec n$, this means that there is a map $(\vec a, \vec b)\mapsto \vec n(\vec a, \vec b)$, which we however shall write $\vec a\times \vec b$ instead of $\vec n(\vec a,\vec b)$. From the above, it follows that $\vec a\times \vec b$ is linear in each factor and that $\vec b\times \vec a=-\vec a\times \vec b$. So much for the existence of this interesting product - in order to actually compute it, you may first investigate how a volume can be computed in terms of determinnants ...

Answer 3

A question that arises when talking about objects with multiple pieces is that is there a sensible way to describe a product involving each of the pieces individually?

There are two natural answers to that question $-$ we can take the product symmetrically or antisymmetrically.

Take the $2$D case for example. If I had two vectors $v=(a,b)$ and $w=(c,d)$, how could I describe a product such that $v \star w = w \star v$ or $v \star' w = -w \star' v$ ? If we only consider products that involve linear combinations of all of the possible products of the components, we get

$$ac+bd$$

$$ad-bc$$

These are fundamental in that these choices were mostly unique, mod taking arbitrary linear combinations of the first and mod a minus sign for the second. But we can get around both issues by declaring something "natural" about just adding the first one evenly with no weights because the second one cannot have unequal weights, and choosing a convention for the sign of the second.

Now consider the $3$D case. For $v=(a,b,c)$ and $w=(d,e,f)$ we have the following products:

$$ad+be+cf$$ $$ae-bd$$ $$af-cd$$ $$bf-ce$$

Notice that if we demand that our product formula must include equal weights of all components of the vectors, the first product is our unique choice for the symmetric product, the dot product.

All three of the following products are pairwise antisymmetric so any linear combination of the three will do for the antisymmetric product. But notice that we happen to have three independent terms, and we are in a $3$D space. By pure coincidence we can choose to put the product with some sign convention into a vector because we happened to have three components. And this is where the cross product comes from and why it only exists in $3$D (and some direct analogue in $7$D).

The fundamental idea with a few assumptions was our search for a symmetry and an antisymmetry which gives rise the products. This line of thinking is very powerful and leads to many ideas in differential geometry, for example.