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Linear algebra area with vectors

  1. Sep 20, 2007 #1
    The figure below shows the tetrahedron determined by a,b,c E R^3. The area vector of a face is a vector perpendicular to the face, pointing outward, whose magnitude is the area of the face. Show that the sum of the four outward pointing area vectors of the faces equals the zero vector. I have attached the image.


    Since the area of each face is identical can I jsut say it is some variable say c. So i am trying to look for the directional vector of each side.. but i am having trouble doing this. I think that vec b is one and vec (b-c) is another, but i am having trouble picturing the other two vectors especially the one pointing downwards.
    Is this even the right approach to this problem?
     

    Attached Files:

  2. jcsd
  3. Sep 20, 2007 #2

    HallsofIvy

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    Are you told that the area of each face is identical? That is NOT true for a general tetrahedron. (And, even if they were, it would be really bad idea to use the same symbol, c, for the area as for one of the vectors!)

    Do you know the "cross product"? In particular, do you know that the area of a parallelogram having vectors [itex]\vec{a}[/itex] and [itex]\vec{b}[/itex] as adjacent sides is [itex]|\vec{a}\times\vec{b}|[/itex]? If you know that then the vectors referred to are just the cross products of two vectors each. Using the notation and order of edges that you have in your picture, the outward 'area vectors' are:
    bottom face: [itex]\vec{c}\times\vec{b}[/itex]
    front face: [itex]\vec{b}\times\vec{a}[/itex]
    back face: [itex]\vec{a}\times\vec{c}[/itex]
    right face: [itex](\vec{b}-\vec{c})\times\(\vec{b}-\vec{a})[/itex]
    Using the distributive law and the fact that the cross product is anti-commutative will give you the result.

    If you can't use the cross product- sorry!
     
  4. Sep 20, 2007 #3
    hi i'm just wondering why for the right face it's b-c x b-a and not b-a x c-a
     
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