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Normal vectors

  1. Apr 19, 2007 #1
    I need to learn how to find normal unit vectors to surfaces. Is there a trick or a formula that I can use?

    F. ex. the normal unit vector to the surface x^2+y^2+z^2=4 is 0,5(x,y,z), but why?

    What can I do to find normal vectors in other cases?
     
    Last edited: Apr 19, 2007
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  3. Apr 19, 2007 #2

    Dick

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    For surfaces defined in the form f(x,y,z)=0, the gradient vector is normal to the surfaces. My question is why do you think 0.5*(x,y,z) is a unit vector?
     
  4. Apr 19, 2007 #3
    I don't know, but my book says so.
     
  5. Apr 19, 2007 #4

    Dick

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    At a particular point, it may be, but the length of that vector is 0.5*sqrt(x^2+y^2+z^2). Not usually equal to one.
     
  6. Apr 19, 2007 #5
    So the gradient vector is the normal vector to f(x,y,z)=x^2+y^2+z^2 - 4 =0.

    Grad f(x,y,z) = (2x, 2y, 2z) = 2(x,y,z). This is a normal vector with length 2 according to my book. Why?

    I'm going to calculate a flux problem, so I need the unit tangent vector.
     
  7. Apr 19, 2007 #6

    Dick

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    To convert a normal vector to a unit normal vector, just divide the vector by its length. You know how to do that right? 2(x,y,z) actually has length 4. How did I figure that out?
     
    Last edited: Apr 19, 2007
  8. Apr 19, 2007 #7

    HallsofIvy

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    Are you sure your book says that? I presume that you know that the length of vector (a, b, c) is [itex]\sqrt{a^2+ b^3+ c^2}[/itex] so that the length of (2x,2y,2z) is [itex]2\sqrt{x^2+ y^2+ z^2}[/itex]. That is equal to two if and only if the length of (x,y,z) itself is 1- that would mean your point is on the unit circle.
     
  9. Apr 19, 2007 #8
    No idea :blushing:

    Edit: yes, my mistake, the book says that the length is 4.
     
  10. Apr 19, 2007 #9

    Dick

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    Ah. You edited your problem post. Now 0.5*(x,y,z) IS a unit normal. Why?
     
  11. Apr 19, 2007 #10
    Because x^2+y^2+z^2=r^2=4 and sqrt(4)=2?
     
    Last edited: Apr 19, 2007
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