Intuitive understanding of relationship between unit vectors and position vector

Question

On Wolfram Mathworld they give a unit vector in the $\textbf{x}_n $ direction as: $$ \hat{\textbf{x}_n} \equiv \dfrac{\frac{\partial \textbf{r}}{\partial x_n}}{\lvert \frac{\partial \textbf{r}}{\partial x_n} \rvert} $$

Can anyone give an intuitive interpretation of this? Alternatively, seeing a proof would help explain where this equation comes from, and help me out.

My current understanding is as follows. For an example in polar coordinates, $\dfrac{\partial \textbf{r}}{\partial \theta}$ is the change in $\textbf{r}$ generated by a change of $\theta \rightarrow \theta+d\theta$. That change is a movement along the $\theta$ direction, as shown in the picture below.

I'm not too sure, however, if this is a valid interpretation.

Answer 1

That is overly complicated. A unit vector is a simple thing.

A unit vector is just a vector with size 1. You can get a unit vector just by dividing a vector by its norm:

$$\hat{x}=\frac{x}{\|x\|}.$$

Now $\hat{x}$ is a unit vector pointing in the same direction as $x$.

Wolfram Alpha is giving an example of that.

Answer 2

Your interpretation is correct. Consider the spherical coordinate system, $(r,\theta,\phi)$. Then the unit vectors $\hat{\mathbf{r}}$, $\hat{\boldsymbol{\theta}}$, and $\hat{\boldsymbol{\phi}}$ are defined as the direction that an arbitrary vector $\mathbf{R}(r,\theta,\phi)$ moves in when the coordinates $r$, $\theta$, and $\phi$ are infinitesimally tweaked.

To be concrete, we can write $\mathbf{R}(r,\theta,\phi)$ in Cartesian coordinates:

$$\mathbf{R}(r,\theta,\phi) = r\sin\phi\cos\theta \,\hat{\mathbf{i}} + r\sin\phi\sin\theta \,\hat{\mathbf{j}} + r\cos\phi\,\hat{\mathbf{k}}.$$

Then $\hat{\mathbf{r}}$ is defined to be along the direction that $r$ changes along: $$ \hat{\mathbf{r}}\propto \frac{d\mathbf{R}}{dr} = \sin\phi\cos\theta \,\hat{\mathbf{i}} + \sin\phi\sin\theta \,\hat{\mathbf{j}} + \cos\phi\, \hat{\mathbf{k}}, $$ and similarly $$ \hat{\boldsymbol{\theta}}\propto \frac{d\mathbf{R}}{d\theta} = -r\sin\phi\sin\theta\,\hat{\mathbf{i}} + r\sin\phi\cos\theta \,\hat{\mathbf{j}} $$ and $$ \hat{\boldsymbol{\phi}}\propto \frac{d\mathbf{R}}{d\phi} = r\cos\phi\cos\theta \,\hat{\mathbf{i}} + r\cos\phi\sin\theta \,\hat{\mathbf{j}} - r\sin\phi\, \hat{\mathbf{k}}. $$

We also have the condition that $|\hat{\mathbf{r}}| = |\hat{\boldsymbol{\theta}}| = |\hat{\boldsymbol{\phi}}| = 1$, which means we have to normalize the vectors: this where the division by the magnitude comes from. For example, $$\hat{\boldsymbol{\theta}} = \frac{\frac{d\mathbf{R}}{d\theta}}{|\frac{d\mathbf{R}}{d\theta}|} = \frac{-r\sin\phi\sin\theta\,\hat{\mathbf{i}} + r\sin\phi\cos\theta \,\hat{\mathbf{j}}}{\sqrt{(-r\sin\phi\sin\theta)^2 + (r\sin\phi\cos\theta)^2}} = -\sin\theta \hat{\mathbf{i}}+\cos\theta \hat{\mathbf{j}}.$$