Quickie: vector normal to surface

ilyas.h
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Homework Statement


Find vector normal to z = x^2 + y^2 - 3 at point r = (2, -1, 2)

Homework Equations

The Attempt at a Solution


normal.png


here is the markscheme. I understand how to find the gradient, but i don't understand how they calculated the magnitude.

thanks
 
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## z = x^2 +y^2 - 3 ## is satisfied by points that form a three dimensional surface, and it's helpful to view that surface as the level surface of some function. How about viewing it as the level surface ## f(x,y,z) = 0 ## for ## f(x,y,z) = z - x^2 - y^2 + 3 ##? Does that clarify things?
 
Geofleur said:
## z = x^2 +y^2 - 3 ## is satisfied by points that form a three dimensional surface, and it's helpful to view that surface as the level surface of some function. How about viewing it as the level surface ## f(x,y,z) = 0 ## for ## f(x,y,z) = z - x^2 - y^2 + 3 ##? Does that clarify things?

nope. Take the magnitude of the llevel surface?
 
ilyas.h said:
nope. Take the magnitude of the llevel surface?
What's the magnitude of the vector -4i + 2j + k ? This is the gradient vector at (2, -1, 2)
 
SteamKing said:
What's the magnitude of the vector -4i + 2j + k ? This is the gradient vector at (2, -1, 2)

how does that help? i need to know how to compute the magnitude.
 
ilyas.h said:
how does that help? i need to know how to compute the magnitude.
Good Lord, you're working gradient problems, but you've skipped basic vector arithmetic.

If V = ai + bj + ck, then |V| = (a2 + b2 + c2) 1/2

Also |V| = (VV)1/2, where ⋅ signifies the dot product of two vectors.
 
SteamKing said:
Good Lord, you're working gradient problems, but you've skipped basic vector arithmetic.

If V = ai + bj + ck, then |V| = (a2 + b2 + c2) 1/2

Also |V| = (VV)1/2, where ⋅ signifies the dot product of two vectors.

My fault all along. I was trying to calculate the magnitude of the surface instead of the magnitude of the gradient of the surface.

Yes i know how to calculate the magnitude of a vector thank you very much.

answered.
 
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