All lines of $\mathbb R^3$ are isometric to $\mathbb R$

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I have just started reading Metric Spaces by Michael Searcoid. The first Chapter states a result :

Suppose $n \in \mathbb N~\forall~ i \in \mathbb N_n,(X_i,\tau_i) $ is a non empty metric space. Let $e$ be a conserving metric on $P = \Pi_{i=1}^{n} X_i.$

For each $j \in \mathbb N_n, a \in P:$ let $ X_{j,a}= \{x \in P~|~x_i=a_i~\forall~i \in \mathbb N_n-\{j\} \}$.

Then, the map $x\rightarrow x_j$ is an isometry from $X_{j,a}$ to $X$.

By the above result, we get that certain lines of $\mathbb R^3$ parallel to an axis are isometric to $\mathbb R $.

However, the text goes on to say that all lines of $\mathbb R^3$ are isometric to $\mathbb R $ and gives the following reasoning :

because for $a,b \in \mathbb R^3, b \ne 0, $ the map $t \rightarrow a + tb / \delta_{(0,0,0)}(b)$ from $\mathbb R$ onto $\{a + tb~|~t \in \mathbb R \}$ is an isometry.

I do not understand the immediate above reasoning ( especially, the term written as : $a + tb / \delta_{(0,0,0)}(b)$ ). Could somebody please help me understand this. Thank you very much for your help in this regard.

Definitions used :

Two spaces are isometric when an isometry between them exists.

On a given metric space :$~(X,d) , \delta_x (y) = \{ d(x,y)~\forall~y \in X ; x \in X \} $

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$a+tb$, where $t$ ranges through $\mathbb R$, is a line. But perhaps the map $t \mapsto a+tb$ is not an isometry. Normalize by a certain factor to make it isometric. That factor is the distance from the origin to $b$.