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Calc iii question/analytic geometry

  1. Feb 9, 2009 #1
    Given vertices of regular tetrahedron OABC, prove that vector OA + OB + OC is perpendicular to plane ABC...

    I've been racking my brain on this one, can't figure it out... would appreciate some help

    I thought I'd center it on the origin:
    (0, 0, 0)
    (1, 0, 0)
    (0, 1, 0)
    (0, 0, 1)

    A + B + C = (1, 0, 0) + (0, 1, 0) +(0, 0, 1) = (1, 1, 1)
    (not sure if that's right - seems like it would be outside the tetrahedron)

    No idea where to go from there.
     
  2. jcsd
  3. Feb 9, 2009 #2

    Tom Mattson

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    The problem here is that your tetrahedron isn't regular. Consider triangle OAB. Two of the sides have length 1, but the hypoenuse has length [itex]\sqrt{2}[/itex]. You'll have to choose different vectors.
     
  4. Feb 9, 2009 #3
    Is there an easier way to do it? Showing that this is true for one regular tetrahedron doesn't necessarily/formally mean it holds for all of them, does it?

    Also, even when I switch coordinates, it doesn't work. For example, a regular tetrahedron (... at least wikipedia says this one is regular)

    O (1, 1, 1)
    A (-1, -1, 1)
    B (-1, 1, -1)
    C (1, -1, -1)

    vector OA = A - O = (-2, -2, 0)
    Similarly, vector OB = (-2, 0, -2)
    OC = (0, -2, -2)

    AB = (0, 2, -2)
    AC = (2, 0, -2)

    AB x AC = (-4, -4, -4)
    OA + OB + OC = (-4, -4, -4)

    And when you take the dot product of those it should be 0 (if they are perpendicular, as they are supposed to be)... but it isn't 0

    What's wrong? Also, again is showing it holds for one regular tetrahedron sufficient?
     
    Last edited: Feb 9, 2009
  5. Feb 9, 2009 #4

    Tom Mattson

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    If there is, I don't see it.

    No, but you're trying to do hear is easily generalized. Instead of trying to come up with a tetrahedron whose sides are of unit length just let the length be some variable, say [itex]a[/itex]. Then *presto* you're working with any possible regular tetrahedron.
     
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