Properties:
These properties imply:
$$ |\bold a\cdot\bold b|\le|\bold a||\bold b|\\~\\ (\bold a+\lambda\bold b)\cdot(\bold a+\lambda\bold b)\ge0~~\forall\lambda\\ \bold a\cdot\bold a+2\lambda\bold a\cdot\bold b+\lambda^2\bold b\cdot\bold b\ge0~~\forall\lambda\\ 4\lambda^2(\bold a\cdot\bold b)^2-4\lambda^2(\bold a\cdot\bold a)(\bold b\cdot\bold b) \le0\\ |\bold a\cdot\bold b|\le|\bold a||\bold b| $$
Where in the third line we use the discriminant of a quadratic in $\lambda$. This quadratic either has one solution or no solutions if the entire quadratic is positive which produces our $\le0$ condition. This is the Cauchy Inequality
Using the Cauchy Inequality we can define angles using the following definition:
$$ \bold a\cdot\bold b=|\bold a||\bold b|\cos\theta $$
Thus, vectors are orthogonal if $\bold a\cdot\bold b=0$
We can also show that $|\bold a+\bold b|\le|\bold a|+|\bold b|$, the triangle inequality
$$ |\bold a||\bold b|\cos\theta\le|\bold a||\bold b|\\ 2\bold a\cdot\bold b\le2|\bold a||\bold b|\\ \bold a\cdot\bold a+2\bold a\cdot\bold b+\bold b\cdot\bold b\le\bold a\cdot\bold a+2|\bold a||\bold b|+\bold b\cdot\bold b\\ (\bold a+\bold b)\cdot(\bold a+\bold b)\le\Big(\sqrt{(\bold a\cdot\bold a)}+\sqrt{(\bold b\cdot\bold b)}\Big)^2\\ |\bold a+\bold b|\le|\bold a|+|\bold b|
$$
The dot products between a set of orthonormal basis vectors $\bold e_i$ follows
$$ \bold e_i\cdot\bold e_j=\delta_{ij} $$
Where $\delta_{ij}$ is the delta kronecker symbol equal to 1 if $i= j$ and equal to 0 if $i\ne j$
Using these facts we can develop a formula to calculate the dot product in Euclidean Space. By writing vectors as components of the $n$ orthonoromal basis vectors in $\R^n$:
$$ \bold a\cdot\bold b=\sum^n_{i=1}a_ib_i $$
Only defined in $\R^3$
Its biggest limitation is that the Cross Product can only be defined for three dimensions
In two dimensions the orthogonal direction is outside the plane
In more than 3 directions, an orthogonal vector to two vectors by the right hand rule is not unique
Properties