Understanding Normal One-Forms for Plane x=0

[Mmmm]

I'm reading through Schutz's first course in relativity book and am finding question 12 on page 83 a bit problematic.

If I understand it correctly an normal one-form to a plane is a one-form that, when operating on a normal vector to the plane, will give the result 0. This seems fairly straight forward to me.
The question is talking about the plane x=0.
So all vectors normal to this must be of the form (a,0,0) (ie parallel to the x axis)
In that case, the normal one form must have components (0,b,c) then
\tilde{n}(\vec{V})= 0*a+b*0+c*0=0

Part (c) of the question says

Show that any normal to S is a multiple of \tilde{n}

and the answer provided is:

On the Cartesian basis, the components of \tilde{n} are (0,0,\beta) for some \beta. Thus any \tilde{n} is a multiple of any other.

But my understanding is that (0,0,\beta) is just a subset of all possible normal one forms to this plane, and I'd agree that of this subset any \tilde{n} is a multiple of any other. But this isn't true for all \tilde{n}, as surley (0,\alpha, \beta) is also valid.

Obviously I'm missing something fairly fundamental here, and I just have to understand this before I move on... Please help :)

 

[Mmmm]

Oh... I am being stupid.. Just realized that the one-form has to operate on a vector tangent to the surface, not the vector normal... I really should read more carefully...
now the one-form is actually perpendicular to the plane and so calling it a normal one-form to the plane makes much more sense!