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Analytic Geometry - finding the normal vector at a point

  1. Jun 9, 2010 #1
    I'm struggling to find the method to use when trying to get the normal vector n to a (flat) surface at a specific point. Every textbook I've researched has the same scenerio, where they show how to get a plane given a point and normal vector, but I have the opposite problem, where I know the plane and know the point but want the normal. And since that method uses a dot product I can't simply "undo" it to find n.

    I know how to get the general normal to the surface by taking the gradient of the plane equation, but I need the specific one that goes through a known coordinate.

    And sorry to make it more difficult, but this is in cylindrical coordinates... if there is a way to do this without converting to Cartesian that would be some excellent information.

  2. jcsd
  3. Jun 9, 2010 #2
    I don't understand... that answer should just be that normal vector you have already found rooted at p. The normal vector has the same coordinates everywhere on the plane.

    Your surface is just a mapping from (r, theta, z) to (x,y,z). If you cross-prod your two coordinate curves mapped into your set of xyz coordinates defining your surface, you get a normal vector in cartesian coordinates.

    Maybe you can post the whole problem statement?
  4. Jun 9, 2010 #3
    Thanks 7thSon

    I understand the idea that the normal is the same everywhere and that I just have to root it at the point p to obtain the equation of the line perpendicular to the surface passing through p, I guess I just don't know how to do that..? Let's say I have the general normal, how do I plug in p to make it a specific line?

    Here's the scenario:

    Dealing with ray optics, I have a beam incident on a prism. And to use Snell's Law I need to define the incident plane. To define that plane I need two vectors, one is the vector of the incident beam, and the other is the vector that's perpendicular to the surface of the prism at the point where the beam hits it.

    Finding that vector is my priority, doing it in cylindrical is secondary, though it would be extremely useful, because there is more than one prism and they are all rotating and it's a big mess in Cartesian.
  5. Jun 9, 2010 #4
    I can define a vector field on all of [tex] R^3 [/tex] that is constant, e.g. F(x,y,z) = ( 1, 2 , -1). Therefore at any location in R3 you have an associated vector (1 , 2, -1). You can either claim a vector exists somewhere in a manifold like R3 or define a vector field on some set of space.

    What you really want is a line, not a vector. More precisely what you want is a line(contour) whose tangent is a constant equal to that vector. There are infinite formulas/parameterizations for that line. One of them is to let [tex] t [/tex] parameterize your line and define C := (x0 + n1*t, y0 + n2*t, z0 + n3*t), where n = (n1, n2, n3) is your normal vector and P = (x0, y0, z0) is the coordinate of your point of interest. You can let t run from negative infinity to infinity or whatever interval you want. You can also pick another point (x1, y1,z1) such that the curve passes through that point at t = 0.

    You could parameterize that curve in cylindrical coordinates but I doubt you would really want to.
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