Finding normal vector to a surface

Kuma
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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)
 
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But what points of u and v do i use?
What difference does it make? (And why?)
 
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...
 
Why shouldn't it make a difference?
What happens if the surface is curved?
 
Right. When the surface is curved the cross product of the tangents shouldn't be 0.
 
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?
 
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