# Quickie: vector normal to surface

## Homework Statement

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

## The Attempt at a Solution here is the markscheme. I understand how to find the gradient, but i dont understand how they calculated the magnitude.

thanks

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Geofleur
Gold Member
$z = x^2 +y^2 - 3$ 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?

$z = x^2 +y^2 - 3$ 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?

SteamKing
Staff Emeritus
Homework Helper
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)

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.

SteamKing
Staff Emeritus
Homework Helper
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.

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.