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I have a question from a book.

First we subtract these two vectors \begin{align} \mathbf{r}&=x\hat{e}_x+y\hat{e}_y+z\hat{e}_z\\ \mathbf{r'}&=x'\hat{e}_x+y'\hat{e}_y+z'\hat{e}_z \end{align} so we get $$ \mathbf{R}=\mathbf{r}-\mathbf{r'}=(x-x')\hat{e}_x+(y-y')\hat{e}_y+(z-z')\hat{e}_z $$ Now, let $f$ be a scalar function of $\mathbf{R}$, such that $f(\mathbf{R})$.

$f$ is a multivariable function, right? But of which variables?

Does it mean $f(\mathbf{R})=f(x,y,z)$?

Or $f(\mathbf{R})=f(x',y',z')$?

Or maybe $f(\mathbf{R})=f(x,x',y,y',z,z')$?

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  • $\begingroup$ It should say $f(\mathbb{R}) =$ something. Otherwise it's not a complete sentence. Anyway in this case it would be a function of three variables: $x-x'$, $y-y'$, and $z-z'$. $\endgroup$ – Grad student Jun 23 '17 at 20:02
  • $\begingroup$ I mean, it could be. Is there any other context from the book, or is that the whole problem as stated by the book? In any case, it would be a scalar function that takes vectors as its input. $\endgroup$ – Johnq Jun 23 '17 at 20:04
  • $\begingroup$ your conclusion from the information given that it must be a scalar function is just wrong. $\endgroup$ – Max Jun 23 '17 at 20:09
  • $\begingroup$ @Max I updated my post. $\endgroup$ – JDoeDoe Jun 23 '17 at 20:15
  • $\begingroup$ $\mathbf R$ is a function of $\mathbf r$ and $\mathbf r'$, or of $x, y, z, x', y', z'$, whichever you prefer. It's common, especially in physics, to not be explicit with such "obvious" things but to leave it open so that any variable in an expression can be seen as an argument of a function. $\endgroup$ – md2perpe Jun 23 '17 at 21:41

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