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3 Vectors from a Common Point onto a Plane

  1. Nov 3, 2012 #1
    1. The problem statement, all variables and given/known data

    Three vectors, [itex]\vec A, \vec B, \vec C[/itex] originiate from a common point, and have their heads in a plane.
    Show that [itex]\vec A \times \vec B + \vec B \times \vec C + \vec C \times \vec A[/itex] is perpendicular to the plane.

    2. Relevant equations

    I'm not exactly sure, but probably solvable with the definition of a cross product (or even without it?)


    3. The attempt at a solution

    First, I attempted to simplify the problem by assuming all three vectors were in the same plane, and that they all pointed to a line (simplified version of a plane). However, it turned out that the sum of the three cross products seem to be parallel, not perpendicular to the plane/line.

    Then, I tried solving the problem using the mathematical definition of cross products, after assigning angles between each pair of vectors. However I'm not sure how to relate the three.

    I'm guessing this may need to be solved by breaking the vectors down into components, or there is a much simpler method that I'm just not seeing.. I'd really appreciate it if someone can give me a hint or point me in the right direction! :)
     
  2. jcsd
  3. Nov 3, 2012 #2

    mfb

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    Can you find vectors which are parallel to the plane?
    What is the angle between those vectors and a vector perpendicular to the plane?
    Which operation is useful to check that?

    No, don't do that.
     
  4. Nov 3, 2012 #3

    LCKurtz

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    And, adding to mfb's advice, no, don't break it into components. Do you have the theorem that you can interchange dot and cross in a triple scalar product?
     
  5. Nov 4, 2012 #4
    Yes, we learned the triple scalar product.
     
  6. Nov 4, 2012 #5
    I was thinking about this, but wasn't sure how to go about this.. since we're not given information about the plane itself. I guess maybe by using the 3 vectors I can piece out the plane?

    EDIT: As in, [itex]\vec A - \vec B[/itex], etc. would be parallel to the plane?
     
  7. Nov 4, 2012 #6

    Dick

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    Yes, it would.
     
  8. Nov 4, 2012 #7
    Ah, so you would you take dot products of these differences with the given expression and prove that it equals zero?
     
  9. Nov 4, 2012 #8

    Dick

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    That's the idea. How many difference dot products would you need to show are zero?
     
  10. Nov 4, 2012 #9
    3, then. Thanks so much! :)
     
  11. Nov 4, 2012 #10

    Dick

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    You could do 3, but you only need two vectors to span a plane. 1 wouldn't be enough.
     
  12. Nov 4, 2012 #11

    mfb

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    The third one is trivial anyway:

    If ##X \cdot (A-B)=0## and ##X \cdot (B-C)=0## , then ##X \cdot ((A-B)+(B-C))=X \cdot (A-C)=0##
     
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