In this post, I find an expression for the radial unit vector, $\vec{e}_r$. The three unit vectors in the following digram form a right-handed spherical coordinate system.

This unit vector is easier to find than the other two unit vectors because all that is needed is vector addition.

### The Radial Unit Vector in Terms of Spherical Coordinates

Suppose $r=1$. Using vector addition,

$\vec{r} = r’ \cos\phi \vec{e}_x + r’ \sin\phi \vec{e}_y + \cos \theta \vec{e}_z$.

Since $r=1$, the expression on the right is equal to $\vec{e}_r$:

$\vec{e}_r = r’ \cos\phi \vec{e}_x + r’ \sin\phi \vec{e}_y + \cos \theta \vec{e}_z$.

From trigonometry, $r’ = r \sin\theta$. Replacing $r’$ by $\sin \theta$ since $r=1$,

$ \boxed{ \vec{e}_r = \sin \theta \cos\phi \vec{e}_x + \sin \theta \sin\phi \vec{e}_y + \cos \theta \vec{e}_z }$.

### The Radial Unit Vector in Terms of Cartesian Coordinates

The previous equation can be written in terms of Cartesian coordinates, by using the following equations.

$\cos\theta = \frac{z}{\sqrt{x^2 + y^2 + z^2}}$

$\sin\theta = \frac{\sqrt{x^2 + y^2}}{\sqrt{x^2 + y^2 + z^2}}$

$\cos\phi = \frac{x}{\sqrt{x^2 + y^2}}$

$\sin\phi = \frac{y}{\sqrt{x^2 + y^2}}$.

$ \boxed{ \vec{e}_r = \frac{x}{\sqrt{x^2 + y^2 + z^2}} \vec{e}_x + \frac{ y }{\sqrt{x^2 + y^2 + z^2}} \vec{e}_y + \frac{z}{\sqrt{x^2 + y^2 + z^2}} \vec{e}_z }$.