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