The dot product is the special case of a more general concept, the inner product. If you have a vector space $ V $ over the reals or the complex numbers, then an inner product is a map $ f : V \times V \to \mathbb{C} $ or $ f : V \times V \to \mathbb{R} $ which is conjugate symmetric, positive definite, and linear in its first argument. We usually write $ f(u, v) = \langle u, v \rangle $, in which case these properties can be summed up as follows:
- Conjugate symmetry: $ \overline{\langle u, v \rangle} = \langle v, u \rangle $, where $ \bar{z} $ denotes complex conjugation. Note that this implies $ \langle u, u \rangle $ is always real for any vector $ u $.
- Positive definiteness: $ \langle v, v \rangle \geq 0 $ for any $ v \in V $, with equality holding iff $ v = 0 $.
- Linearity in the first argument: $ \langle \alpha u + \beta v, w \rangle = \alpha \langle u, w \rangle + \beta \langle v, w \rangle $ where $ u, v, w \in V $ and $ \alpha, \beta $ are in the field of scalars.
If $ V = \mathbb{R}^n $, then we can fix a basis $ B = \{ b_i \in \mathbb{R}, 1 \leq i \leq n \} $ and define $ \langle b_i, b_i \rangle = 1 $ and $ \langle b_i, b_j \rangle = 0 $ for $ i \neq j $. Extending this to all of $ \mathbb{R}^n $ by linearity gives us
$$ \left \langle \sum_{k=1}^{n} c_k b_k, \sum_{j=1}^{n} d_j b_j \right \rangle = \sum_{1 \leq k, j \leq n} d_k c_j \langle b_i, b_j \rangle = \sum_{i=1}^{n} c_i d_i $$
where positive definiteness is readily verified. You will recognize this expression as the definition of the dot product. Indeed, if we take our basis $ B $ to be the standard basis of $ \mathbb{R}^n $, then this inner product is the dot product.
Why is this formalism more powerful? A result about the inner product is the Cauchy-Schwarz inequality, which says that $ |\langle u, v \rangle| \leq |u| |v| $ where $ |u| = \sqrt{\langle u, u \rangle} $. This tells us that
$$ -1 \leq \frac{\langle u, v \rangle}{|u| |v|} \leq 1 $$
assuming that our field of scalars is $ \mathbb{R} $. We then see that the arccosine of this expression is well-defined, so we can define the angle between nonzero vectors $ u $ and $ v $ as
$$ \theta = \arccos \left( \frac{\langle u, v \rangle}{|u| |v|} \right) $$
The properties we expect to be true are then easily verified. This notion extends to infinite dimensional vector spaces over $ \mathbb{R} $, where defining angle is not at all obvious. It is then trivially true that we have $ \langle u, v \rangle = |u| |v| \cos(\theta) $, since that is how $ \theta $ was defined.
The cross product is an entirely separate concept which allows us to find a vector orthogonal to two given vectors in $ \mathbb{R}^3 $. In addition, its magnitude also gives the area of the parallelogram spanned by the vectors. These properties can be taken as the definition of the cross product (with appropriate care for orientation), or they can be derived as theorems starting from the algebraic definition.
Best Answer
Here's an explanation in terms of the Hodge dual and the exterior (wedge) product.
Let ${e_1, e_2, e_3}$ be the standard orthonormal basis for $\mathbb{R}^3$. Consider the two vectors $a = a_1 e_1 + a_2 e_2 + a_3 e_3$ and $b = b_1 e_1 + b_2 e_2 + b_3 e_3$. From the matrix computation we obtain the familiar formula
$a\times b = (a_2 b_3 - a_3 b_2) e_1 + (a_3 b_1 - a_1 b_3) e_2 + (a_1 b_2 - a_2 b_1) e_3$.
But (see note at the bottom)
$a \wedge b = (a_1 b_2 - a_2 b_1) e_1 \wedge e_2 + (a_2 b_3 - a_3 b_2) e_2 \wedge e_3 + (a_3 b_1 - a_1 b_3) e_3 \wedge e_1$,
where the wedge $\wedge$ represents the exterior product. One can now compute the dual of this latter expression using that the left contraction of $(e_1 \wedge e_2)$ onto $(e_3 \wedge e_2 \wedge e_1)$ is $e_3$ (and similar relations). The result is that
$a \times b = (a \wedge b)^*$,
that is, the cross product of $a$ and $b$ is the dual of their exterior product.
Geometrically, this is an incredible picture. The exterior product is the plane element spanned by both $a$ and $b$, and the dual is the vector orthogonal to that plane.
This is my favorite interpretation of the cross product, but it's only helpful, of course, if you're familiar with exterior algebra and the Hodge dual.
Note: The wedge product can be found by formally computing
$(a_1 e_1 + a_2 e_2 + a_3 e_3) \wedge (b_1 e_1 + b_2 e_2 + b_3 e_3)$
using the distributivity and anticommutation relations of the exterior product.