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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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    It's a surface.2012-07-10

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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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    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.2012-07-10