Gradient of distance vector length Suppose that there is a surface described by:
$\phi(x,y,z)=c$
And suppose that there is a fixed point A:
$\vec{r_A}=(x_A,y_A,z_A)$
Let $\vec r$ be position vector of any point on the surface so that:
$R=|\vec r-\vec r_A|$ 
Show that $\nabla R$ is a unit vector whose direction is along $\vec r-\vec r_A$
I tried to write $z=f(x,y)$ and then calculate what is $\nabla R$ but I got that it has no $z$ component which leads to a contradiction...

$$R=|\vec{r}-\vec{r}_a|=|(x-x_a)\mathbf{i}+(y-y_a)\mathbf{j}+(z-z_a)\mathbf{k}|=\sqrt{(x-x_a)^2+(y-y_a)^2+(z-z_a)^2}$$
Now take $\nabla R$
