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

  • 2
    $\begingroup$ 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

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.

  • $\begingroup$ Ok but what about the case where phi = 0 ? $\endgroup$ – user1887919 Jan 16 '14 at 20:15
  • 1
    $\begingroup$ @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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.