Computing a normal to a plane using cross product

Question

I have a plane $ ax + by +cz = 0 $. I know that the normal to the plan is the 3 dimensional vector $(a,b,c)^T $ because I define $ f(x,y,z) = ax + by +cz $ and the normal vector is $ \nabla f $

Now I want to do it in another way - I parameterized the plane in this way $$ \vec {s}(x,y) =(ax,by,-(ax +by)/c )^T $$

$\vec{s}_x = (a,0,-a/c)^T $

$\vec{s}_y = (0,b,-b/c)^T $

From what I read the normal vector should be $ \vec{s}_x \times \vec{s}_y $ but when I compute it I don't get the vector $(a,b,c)^T$

Any idea where is my mistake ?

Answer 1

The normal vector of a plane is not unique. There is a whole line of normal vectors. The one you get by taking the cross product and the one given by one implicit equation must be proportional in any case.