Tangent plane of a parametric function

[BiGyElLoWhAt]

Ok, so I'm really hoping someone can help me logic my way through this.

I have a function to the effect of: $r(u,v)=f(u,v)\hat{i} + g(u,v)\hat{j} +h(u,v)\hat{k}$

I need to find an equation of a tangent plane at a point $(u_{0},v_{0})$
and quite frankly I'm at a loss on how to do this.

So what I ultimately want is 2 vectors tangent to the surface at that point, I'm not sure that with what I have I can simply chain some derivatives together and cross them...

If I had z(x,y) I would simply take $\frac{\partial z}{\partial x} (cross) \frac{\partial z}{\partial y}$ as that would give me 2 (presumably different) vectors and crossing them will give me my normal to use in my tangent function...

Now how to apply this to a parametric function in 3-space...

Ok... I think I just answered my question as I was typing this... (It happens that way quite a bit)
$\frac{\partial r}{\partial u} (cross) \frac{\partial r}{\partial v}$

But what if I had $r(u,v,w)$ ?
In my head , whether I chose to cross
$\frac{\partial r}{\partial u} (cross) \frac{\partial r}{\partial v}$
or
$\frac{\partial r}{\partial u} (cross) \frac{\partial r}{\partial w}$
or
$\frac{\partial r}{\partial w} (cross) \frac{\partial r}{\partial v}$
It really shouldn't matter, I guess at this point I just want to hear somebody's input other than my own. All of these will give me a normal, no?

(once I plug in my point that is...)

 

[LCKurtz]

A surface only has two parameters. So you will only run into the $r_u\times r_v$ case for a surface. And you are correct, that is how you find a normal.