In J.Rotman's "A First Course in Abstract Algebra", page 135.
Definition. An isometry of the plane is a function $\varphi:\mathbb{R}^2 \to \mathbb{R}^2$ that is distance preserving: for all points $P=(a,b)$ and $Q=(c,d)$ in $\mathbb{R}^2$, $$\parallel\varphi(P)-\varphi(Q)\parallel=\parallel{P-Q}\parallel,$$ where $\parallel{P-Q}\parallel=\sqrt{(a-c)^2+(b-d)^2}$ is the distance from P to Q.
Let $P\cdot Q=ac+bd.$
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It follows that every isometry $\varphi$ preserves dot products: $$\varphi(P)\cdot\varphi(Q)=P\cdot Q,$$ because $$\varphi(P)\cdot\varphi(Q)=\parallel\varphi(P)-\varphi(Q)\parallel^2=\parallel{P-Q}\parallel^2=P\cdot Q.$$
but this seems not right, because $$2P \cdot Q = \parallel P\parallel^2+\parallel Q\parallel^2 - \parallel P-Q\parallel^2$$ and $$2\varphi(P) \cdot \varphi(Q) = \parallel \varphi(P)\parallel^2+\parallel \varphi(Q)\parallel^2 - \parallel \varphi(P)-\varphi(Q)\parallel^2,$$ the last terms are equal from isometry, but if I want $\varphi(P)\cdot\varphi(Q)=P\cdot Q$, it seems $\parallel \varphi(P)\parallel^2=\parallel P\parallel^2$ should hold for any point $P$, which may needs $\varphi(O)=O$.
But then Isom$(\mathbb{R}^2)$ is defined as all isometries of the plane, its subset consisting of all those isometries with $\varphi(O)=O$ is called orthogonal group of the plane denoted by $O_2(\mathbb{R})$.
So it seems there should be some isometrics preseving dot product but $O$ not fixed.
Am I doing wrong?
Well, I just find this error has been fixed in later version, it is indeed those isometry fixing $O$, and the strange equation is also corrected.