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In theory, if I have a certain function I can get his normal unit vector by using the gradient of it.

$$\hat{f} = \dfrac{\nabla f}{|| \nabla f ||}$$

Example (correction from answer):

$$ z = 2 -x -y$$ $$ f(x,y,z)= z + x + y -2 $$ $$ \nabla f(x,y,z)= \hat{i} + \hat{j} + \hat{k}$$ $$ \dfrac{\nabla z}{|| \nabla z ||}= \dfrac{1}{\sqrt{3}} (\hat{i} + \hat{j} + \hat{k})$$

Is that correct?

what about this example: $$ z = \sqrt{x^2+y^2} $$ $$ \nabla f(x,y,z)= \dfrac{x}{\sqrt{x^2+y^2}} \hat{i} + \dfrac{y}{\sqrt{x^2+y^2}} \hat{j} + -\hat{k}$$ $$ \dfrac{\nabla f}{|| \nabla f ||}= \dfrac{\dfrac{x}{\sqrt{x^2+y^2}} \hat{i} + \dfrac{y}{\sqrt{x^2+y^2}} \hat{j} + -\hat{k}}{\sqrt{ (\dfrac{x}{\sqrt{x^2+y^2}} )^2 + (\dfrac{y}{\sqrt{x^2+y^2}} )^2 + (-1)^2 }}$$

$$ \dfrac{\nabla f}{|| \nabla f ||}= \dfrac{\dfrac{x}{\sqrt{x^2+y^2}} \hat{i} + \dfrac{y}{\sqrt{x^2+y^2}} \hat{j} + -\hat{k}}{\sqrt{2}}$$

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Is $z$ a function name, or are you describing a surface? –  copper.hat Jul 10 '12 at 17:20
    
It's a surface. –  fneron Jul 10 '12 at 17:23
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1 Answer

up vote 2 down vote accepted

No, this is not correct. You must first write $f(x,y,z)=x+y+z-2$. Then, calculate $\nabla f=\vec i+\vec j+\vec k$ which gives $$\hat{f} = \dfrac{\nabla f}{|| \nabla f ||}=\frac{1}{\sqrt 3}(\vec i+\vec j+\vec k)$$ This is the required normal unit vector.

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Thank you. When I have a surface, I have to put everything on one side? –  fneron Jul 10 '12 at 17:26
    
look at my new edit... –  fneron Jul 10 '12 at 17:34
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In your new problem,in calculation of $\nabla f$,there is an error; it should be $\nabla f=\frac{x}{\sqrt{x^2+y^2}}\vec i+\frac{y}{\sqrt{x^2+y^2}}\vec j-\vec k$. Check it.Everything else is fine. –  Aang Jul 10 '12 at 18:33
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