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Spherical, Cylindrical Coordinates - quick easy questions

  1. Sep 20, 2006 #1
    EDIT: I figured it out.

    I'm sick, tired, and just confusing myself. I have a test in a few hours, and would like to clear some easy stuff up so I'm not thinking about it during the test.

    Vector subtraction works in spherical and cylindrical coordinates right?

    That is to say, [itex] \vec {AB} [/itex], points from point [itex] A [/itex] to point [itex] B [/itex] where [itex] \vec {AB} = \vec B - \vec A [/itex] in spherical, cylindrical, and cartesian?

    For some reason I'm thinking that I read somewhere that we first need to convert to cartesian, but that might actually be related to my second question.

    Is the modulus for a vector the same in cartesian, cylindrical, and spherical coordinates? I mean if I am trying to find the length in spherical.
    [tex] \vec V_{sph} = (R, \theta, \phi) [/tex]
    Isn't [itex] R [/itex] just the length from orgin. It doesn't make sense to say that, [itex] | \vec V_{sph} | = \sqrt{(R)^2+(\theta)^2+(\phi)^2} [/itex].

    The same for cylindrical, is [itex] | \vec V_{cyl} | = \sqrt{(r)^2 + (z)^2} [/itex] ?


    EDIT: I figured it out
    Last edited: Sep 20, 2006
  2. jcsd
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