Argument of Feynman for equivalence of dot product definitions The following argument from the Feynman Lectures on Physics (Vol I, Lecture 11), which relates to the equivalence of the algebraic and geometric definitions, does not particularly convince me.

Also, there is a simple geometrical way to calculate $\vec{a} \cdot \vec{b}$, without
having to calculate the components of $\vec{a}$ and $\vec{b}$: $\vec{a} \cdot \vec{b}$ is the product of
the length of $\vec{a}$ and the length of $\vec{b}$ times the cosine of the angle
between them. Why? Suppose that we choose a special coordinate system
in which the x-axis lies along $\vec{a}$; in those circumstances, the only
component of $\vec{a}$ that will be there is $a_x$, which is of course the whole
length of $\vec{a}$. Thus Eq. (11.19) reduces to $a \cdot b = a_x b_x$ for this case, and
this is the length of $\vec{a}$ times the component of $\vec{b}$ in the direction of
$\vec{a}$, that is, $b \cos \theta$: $a \cdot b = a b \cos \theta$. Therefore, in that special coordinate
system, we have proved that $\vec{a} \cdot \vec{b}$ is the length of $\vec{a}$ times the length of
$\vec{b}$ times $\cos \theta$. But if it is true in one coordinate system, it is true
in all, because $\vec{a} \cdot \vec{b}$ is independent of the coordinate system; that is
our argument.

In fact, most of this argument seems just fine, but it seems like Feynman is casually asserting a priori that the dot product should be independent of the coordinate system. This is something I do not like, since I can't see an obvious justification for it. (Indeed, if by "coordinate system" he means basis, then there are clearly bases for which this is not true, e.g., ${2\hat{i}, 2\hat{j}, 2\hat{k}}$.)
Could someone who is better at reading between the lines of Feynman please clarify this for me?
 A: This does seem to be a gap in the argument. Maybe Feynman filled in the gap elsewhere, or maybe it's a true gap. He's a physicist so he's not aiming for full mathematical rigor, he just wants great insight.
Here's how I would fill in the gap. Suppose that $\beta = (u_1, u_2, u_3)$ is an orthonormal basis for $\mathbb R^3$. Let $Q = \begin{bmatrix} u_1 & u_2 & u_3 \end{bmatrix}$ (so the $i$th column of $Q$ is the column vector $u_i$). The change of basis matrix from the standard basis to $\beta$ is $Q^{-1} = Q^T$.
Now suppose that $x$ and $y$ are vectors in $\mathbb R^3$. Notice that
\begin{align}
 (Q^T x) \cdot (Q^T y ) &= (Q^T x)^T Q^T y \\
&= x^T Q Q^T y \\
&= x^T y \\
&= x \cdot y.
\end{align}
This shows that changing basis from the standard basis to the basis $\beta$ does not change the dot product.
A: If I read correctly that you're looking for a justification that the two computations are equivalent, I came up with a way to demonstrate this a few months ago. Myself, I always found the geometric $\vec{a} \cdot \vec{b} = ||\vec{a}|| \cos(θ) ||\vec{b}||$ computation to be the more intuitive, so I'll start with that definition and justify the element-based definition. This won't require anything more than basic trig and some geometric intuitions.
Say we have 2 vectors, $ \vec{u} $ and $ \vec{v} $, in $ \mathbb{R}^n$, and let's assume that these vectors are non-collinear-- one isn't a scaled up version of the other (if they are collinear, then cos(θ) = 1 which simplifies the problem considerably). Explicitly,
$$\vec{u}=<u_1,u_2,u_3,\cdots,u_n>$$
$$\vec{v}=<v_1,v_2,v_3,\cdots,v_n>$$
If they aren't collinear, and if their tails lie at the origin of our n-dimensional space, then we can use their positions two define two lines: one line going between  the origin and $\vec{u}$, and another line going through the origin and $\vec{v}$. The great thing about this setup is that now we're dealing with 3 points in space. It doesn't matter how many dimensions those points may exist in according to our axes, because 3 points is enough to define a unique plane in that space. We don't know how to label those points in that plane, but we know those points exist in it. Turning to GeoGebra,

I've labeled the coordinate points in context of our n-dimensional space-- we're not using the 2D coordinates of the plane itself. Now, we know we want to connect this element computation to geometry. To do that, I've labeled θ, the angle between the two vectors in the plane that contains them, and I've defined a vector $$\vec{w} = \vec{v}-\vec{u}$$
$$\vec{w} = <v_1 - u_1, v_2 - u_2, v_3 - u_3, \cdots, v_n - u_n>$$
We notice that this diagram draws out a triangle, and, if we assume we know θ as well as our two vectors, we get the triangle

And now we do some math.
It's not guranteed that our vectors will produce a right triangle, so to solve this thing we'll need the Law of Cosines, as well as the magnitudes of the sides of our triangle.
$$\text{Law of Cosines:} \: \: c^2 = a^2 +b^2 - 2ab\cos(\angle C)$$
$$ ||\vec{u}|| = \sqrt{u^2_1 + u^2_2 + u^2_3 + \cdots + u^2_n}$$
$$ ||\vec{v}|| = \sqrt{v^2_1 + v^2_2 + v^2_3 + \cdots + v^2_n}$$
$$ ||\vec{w}|| = \sqrt{(v_1 - u_1)^2 + (v_2 - u_2)^2 + (v_3 - u_3)^2 + \cdots + (v_n - u_n)^2}$$
If our side $a$ is represented by $\vec{u}$, and side $b$ is represented by $\vec{v}$, then side $c$ and $\angle C$ are given by $\vec{w}$ and $θ$. Using these values in the Law of Cosines formula and solving this for $\cos(θ)$ gives
$$\cos(\angle C) = -\frac{c^2 - a^2 - b^2}{2ab}$$
$$\cos(θ) = -\frac{||\vec{w}||^2 - ||\vec{u}||^2 - ||\vec{v}||^2}{2||\vec{u}|||\vec{v}||}$$
$$\cos(θ) = -\frac{[(v_1 - u_1)^2 + (v_2 - u_2)^2 + \cdots + (v_n - u_n)^2] - [u^2_1 + u^2_2 + \cdots + u^2_n] - [v^2_1 + v^2_2 + \cdots + v^2_n]}{2||\vec{u}|||\vec{v}||}$$
Expanding out our terms for $||\vec{w}||^2$,
$$\cos(θ) = -\frac{
[v_1^2 - 2 u_1 v_1 + u_1^2 + v_2^2 - 2 u_2 v_2 + u_2^2 + \cdots + v_n^2 - 2 u_n v_n + u_n^2]
 - [u^2_1 + u^2_2 + \cdots + u^2_n] - [v^2_1 + v^2_2 + \cdots + v^2_n]}{2||\vec{u}|||\vec{v}||}$$
It is at this point we notice that every $u_i^2$ and $v_i^2$ in the numerator has a corresponding $-u_i^2$ and $-v_i^2$, allowing us to reduce all those terms to $0$. Doing this, along with distributing the negatives across, gives us
$$\cos(θ) = \frac{2 u_1 v_1 + 2 u_2 v_2 + \cdots + 2 u_n v_n}{2||\vec{u}|||\vec{v}||}$$
$$\cos(θ) = \frac{ u_1 v_1 +  u_2 v_2 + \cdots +  u_n v_n}{||\vec{u}|||\vec{v}||}$$
Tying all of this back to our given definition for the dot product
$$\vec{u} \cdot \vec{v} = ||\vec{u}||\cos(θ)||\vec{v}||$$
$$\vec{u} \cdot \vec{v} = ||\vec{u}||||\vec{v}||\frac{ u_1 v_1 +  u_2 v_2 + \cdots +  u_n v_n}{||\vec{u}|||\vec{v}||}$$
$$\vec{u} \cdot \vec{v} = u_1 v_1 +  u_2 v_2 + \cdots +  u_n v_n$$
Hopefully that gives an intuitive path for how we would go about connecting those two computations.
