"Centered" Rigid Motions Preserve Dot Product

Suppose that we have two vectors $\vec{AB}$ and $\vec{CD}$, and that the angle between them is $\theta$.

We know that $\vec{AB}\cdot\vec{CD}=|\vec{AB}||\vec{CD}|\cos\theta$.

Suppose that we map space by an isometry $F$, such that $A$ is mapped to $A'$, $B$ is mapped to $B'$, and so on.

Since isometries preserve distance by definition, we have: $$A'B'=AB$$ $$C'D'=CD$$ Moreover, isometries also preserve angles. If you are unfamiliar with this, you can look at this article: All Isometries Preserve Angles. However, it is pretty simple: the idea boils down to the fact that a triangle may be determined up to congruence by its side lengths, even if you don't know the angles.

Thus, we have the following, if $\theta'$ is the angle between $\vec{A'B'}$ and $\vec{C'D'}$: $$\theta'=\theta$$ Combining these statements, $$|\vec{AB}||\vec{CD}|\cos\theta=|\vec{A'B'}||\vec{C'D'}|\cos\theta'$$ Therefore, we can conclude: $$\vec{AB}\cdot\vec{CD}=\vec{A'B'}\cdot\vec{C'D'}$$ Since the points $A$, $B$, $C$, and $D$ were purely arbitrary choices, we may conclude that the same holds for any choice of points. Thus, all isometries preserve dot products of vectors.