 Solution

Use the ambient position coordinates $\mathbf{r}$ and a fixed point $\mathbf{a}$. The distance is $R=|\mathbf{r}-\mathbf{a}|$, so $R^2=(x-a_x{)}^2+(y-a_y{)}^2+(z-a_z{)}^2$. Taking its gradient gives $2R\nabla R=2(\mathbf{r}-\mathbf{a})$. Therefore, away from $P=A$,It points from $A$ to $P$. The gradient is taken in the surrounding space and then evaluated on the surface; the intrinsic surface gradient of restricted distance would instead be its tangential projection and need not have unit magnitude.