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Normal vectors on tangent spaces.

  1. Feb 10, 2010 #1
    1. The problem statement, all variables and given/known data
    Given a surface parameterized by the function [itex] f(x) [/itex] and a point p on that surface, assume that [itex] P [/itex] is a point on the tangent space of f at p. Find the normal vector to the hyperplane at [itex] P [/itex].

    3. The attempt at a solution

    The tangent hyperplane to f at p is given by the equation
    [tex] \nabla f(p) \cdot ( x- p) = 0 [/tex]
    Since we know that [itex] P [/itex] is on the tangent space, we must have that [itex] \nabla f(p) \cdot ( P - p ) = 0 [/itex]. However, here is where I am stuck. I'm not sure how to use this to calculate the normal vector. Any ideas?
     
  2. jcsd
  3. Feb 10, 2010 #2

    Dick

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    Hmm. The normal vector to a plane is the vector n such that n.(a-b)=0 for all a and b in the plane, right?
     
  4. Feb 10, 2010 #3
    If I'm not mistaken, in the case of n.(a-b) = 0 the normal is very specifically the point passing through b. The set of all a for which this is satisfied help to define the plane.

    In this case, the planes should be the same, so it seems that I should be able to do something like equation the equations

    [tex] \nabla f(p)\cdot(x-p) = n\cdot (x-P) [/itex]

    In this case, solving for n will give the desired result. I'm playing around with this right now, but haven't gotten anywhere.
     
  5. Feb 10, 2010 #4

    Dick

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    So if n.(p-a)=0 and n.(p-b)=0 isn't n.(a-b)=0? Just subtract the first two equations. But let me rephrase the hint. Isn't grad(f)(p) a normal vector?
     
  6. Feb 10, 2010 #5
    Yes, [itex] \nabla f(p) [/itex] is normal, and I see what you're saying as far determining it being invariant under choice of internal coordinates.

    This yields some important insight into my problem and answers my posed question, so thank you. Now perhaps you can help me with my real question.

    Suppose that [itex]f:\mathbb R^n \to \mathbb R^{m\times m} [/itex] is a multivariate, matrix-valued function so that [itex] \nabla f(p) [/itex] is a rank-3 tensor defining the tangent manifold to the orientable surface defined by by [itex] f(x) = 0 [/itex]. Namely, [itex] \nabla f(p) [/itex] generates [itex] T_p M [/itex] where [itex] M = \{ x \in \mathbb R^n : f(x) = 0 \} [/itex].

    Now through some magic I get a point P (solving a linear programming problem under the constraint that P lies on the tangent manifold) and I want to project P back onto M. Numerically, this is best done by considering vectors orthogonal to [itex] T_p M[/itex] at P. This being said, what is the ``best direction'' to consider? Analogously, I would take the normal vector to P. Should I just use orthogonal vectors at p to define a search direction? Or is there a frame in which the unique ``vector'' passing through P defines the best search direction?
     
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