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How would I find the vector perpendicular to the surface $\phi(x,y,z)=0 $?

My initial thoughts are to calculate grad $\phi$? But would this not just give me zero? Thanks

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    $\begingroup$ If $\phi$ is differentiable it is correct to calculate $\operatorname{grad}\phi$ If it's 0 or not depends on $\phi$ of course.. $\endgroup$ – user127.0.0.1 Jan 15 '14 at 16:05
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The gradient of $\phi$ would work quite nicely, as long as it's not, as you say, equal to $0$. For most calculation purposes you would need to normalize it as well, though.

Example: $\phi(x, y, z) = x^2 + y^2 + z^2 - 1$ (this gives the unit sphere). The gradient is $$ (2x, 2y, 2z) $$ Of course, you would only be interested in this at the points of the unit sphere, which means that in this case the length of the gradient will always be $2$, so to normalize it, we divide by $2$ and get the normal vector $\vec n(x, y, z) = (x, y, z)$, pointing directly outwards from the origin as expected.

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  • $\begingroup$ Ok but what about the case where phi = 0 ? $\endgroup$ – user1887919 Jan 16 '14 at 20:15
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    $\begingroup$ @user1887919 Then, with some luck, the limit of $\frac{\operatorname{grad}\phi}{|\operatorname{grad}\phi|}$ exist at that point (possibly with a sign change from one side of the surface to another, but that's OK). If that doesn't work, I'm afraid you gust have to be clever. $\endgroup$ – Arthur Jan 16 '14 at 20:26

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