The Euclidean n space $(\mathbb{E}^n,d)$ is a metric space with metric $d$ for which there exists a bijective map $\mathbb{E}^n\rightarrow\mathbb{R}^n$, such that if P,Q $\in \mathbb{E}^n$ are mapped to $\mathbf{x,y}\in\mathbb{R}^n$ then $d(P,Q)=\lvert{\mathbf{y}-\mathbf{x}}\rvert$
If $P,Q,R\in\mathbb{R}^n$ are mapped to $\mathbf{x},\mathbf{y},\mathbf{z}\in\mathbb{R}^n$, then the angle spanned by P,Q,R is equal to $\angle\mathbf{xyz}$ and we can evaluate it using
$\cos(\angle\mathbf{xyz})=\frac{(\mathbf{x}-\mathbf{y})\cdot(\mathbf{z}-\mathbf{y})}{\lvert{\mathbf{x}-\mathbf{y}}\rvert\lvert{\mathbf{z}-\mathbf{y}}\rvert}$
The angle is independent of this choice, because the inner product is determined by the quadratic form.
The above is what is written in the book.
I can't understand why the angle is independent of the choice of coordinates(bijection) and the meaning of 'inner product is determined by the quadratic form.'.
Could you explain them?
Suppose you change the coordinate system, then
$ x = O' + R x' $
$ y = O'+ R y' $
$ z = O' + R z' $
where $O'$ is the position of the origin of the $O'x'y'z'$ reference frame when expressed in the $Oxyz$ system, and $R$ is a rotation matrix. it follows that
$ x- y = R (x' - y') $
$ z - y = R (z' - y') $
Then
$ (x - y) \cdot (z - y) = (x - y)^T (z - y) \\= (x' - y')^T R^T R (z' - y') = (x' - y')^T (z' - y') = (x' - y') \cdot (z' - y') $
And for the same reason,
$ \| x - y \| = \sqrt{ (x - y) \cdot (x - y) } = \sqrt{ (x' - y') \cdot (x' - y') } $
and
$ \| z - y \| = \sqrt{ (z - y) \cdot (z - y) } = \sqrt{ (z' - y') \cdot (z' - y') } $
Therefore,
$ \dfrac{ (x - y) \cdot (z - y) }{\| x - y \| \| z- y\|} = \dfrac{ (x' - y') \cdot (z' - y') }{ \| x' - y' \| \| z' - y' \| } $
Hence, the cosine of the angle $\angle xyz $ will be the same when computed using any coordinate system.