Unit normal of sphere in cartesian coordinates

Question

I'm trying to find the unit normal of the following sphere: $g(x,y,z)=x^2+y^2+z^2-1$ \begin{equation} \vec{n}=\frac{\nabla \vec{g}}{|\nabla \vec{g}|}\\ \nabla \vec{g}=2x\hat{i}+2y\hat{j}+2\hat{z}\\ |\nabla\vec{g}|=\sqrt{4x^2+4y^2+4z^2}\\ \text{and finally } \vec{n}=\frac{2x\hat{i}+2y\hat{j}+2\hat{z}}{\sqrt{4x^2+4y^2+4z^2}} \end{equation} However the equality book give me at a point $S(x_0,y_0,z_0)$; \begin{equation} \vec{n}=(x_0\hat{i}+y_0\hat{j}+z_0\hat{k}) \end{equation}

Answer 1

Note that in this case since $x^2+y^2+z^2=1$

$$\vec{n_1}=\frac{x_0\hat{i}+y_0\hat{j}+\hat{z_0}}{\sqrt{x_0^2+y_0^2+z_0^2}}\,=\,\vec{n_2}=x_0\hat{i}+y_0\hat{j}+z_0\hat{k}$$

In more general case for $x^2+y^2+z^2=R^2$ we have $n_1\parallel n_2$ are parallel and both are two normal vectors to the surface with the difference that $\vec {n_1}$ is normalized (length=1).