Finding normal vector to a surface

[Kuma]

Homework Statement

Given a parameterized surface:

C(u,v) = (3 cos u sin v, 2 sin u sin v, cos v) 0<u<2pi, 0<v<pi

I have to find a normal vector to that surface.

Homework Equations

The Attempt at a Solution

So tangent vectors can be Tu = (dx/du, dy/du, dz/du) and Tv = (dx/dv, dy/dv, dz/dv)

And I can take the cross of those to find a normal vector. But what points of u and v do i use? The cross product gave me:

(-2sin^2 v cos u, 3sin^2 v sin u, 6sin^2 u sin^2 v - 6 cos^2 u sin^2 v)

 

[Simon Bridge]

But what points of u and v do i use?

What difference does it make? (And why?)

 

[Kuma]

It shouldn't make a difference. Do i just plug in the endpoints of u and v? ie would (0,0) work? I get (0,0,0) if I use that point. Not a vector...

 

[Simon Bridge]

Why shouldn't it make a difference?
What happens if the surface is curved?

 

[Kuma]

Right. When the surface is curved the cross product of the tangents shouldn't be 0.

 

[Simon Bridge]

You are saying that the cross product of the tangents to a plane (flat) surface are zero?
Then how would you find the normal vector to a plane surface?