which one is the right way for finding the normal vector to the surface

Question

In wolfram website it is mentioned that if $z=f(x,y)$ then the normal vector to the surface $z$ is $[f_x(x,y) \quad f_y(x,y) \quad -1]^T$ where $f_x$ and $f_y$ are the partial derivatives with respect to $x$ and $y$ respectively. Further, it is written that if $ax+by+cz+d=0$ then the normal vector is $[a \quad b \quad c]. $ Now if I write $ax+by+cz+d=0$ as $z=-\left(d/c+ax/c+by/c\right)$ and from this expression the normal vector turns out to be $[-a/c \quad -b/c \quad -1]$ which is not exactly $[a \quad b \quad c]$. Why we get two different answers here for normal vector?

Answer 1

Both vectors are correct. The point to note here is that they are indeed parallel vectors, since we can write: $$v_2=\alpha*v_1$$ where $\alpha$ in this case is equal to $-1/c$.

Since they're linearly dependent, they're both normal to the given plane.

Answer 2

They both are valid normal vectors since they are multiple and then parallel , indeed

$$[a \, b \, c]=-c \cdot [-a/c \, -b/c \, -1]$$

Answer 3

Consider $g(x,y,z) = z-f(x,y) = 0$ instead.

Now

$$ \nabla g(x,y,z) \ \mbox{is the normal vector to the surface}\ \ z-f(x,y)=0 $$