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Math Help - Vectors in R3 ---- how to approach this problem

  1. #1
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    Vectors in R3 ---- how to approach this problem

    (a) Given a general (not necessarily rectangular) tetrahedron, let V_{1}, V_{2}, V_{3}, V_{4} denote vectors whose lengths are equal to the areas of the four faces, and whose directions are perpendicular to the these faces and point outward. Show that V_{1} + V_{2} + V_{3} + V_{4} = 0.


    (b) Now consider a rectangular tetrahedron, i.e., one which has one vertex at which all three angles are right angles. Let D denote the area of the face opposite this rectangular vertex (i.e., the analog of a hypothenuse), and let A;B;C denote the areas of the other three faces. Prove that D^2 = A^2 + B^2 + C^2.
    Last edited by nikcs123; October 10th 2010 at 06:15 PM.
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  2. #2
    Junior Member bondesan's Avatar
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    The approach I'm thinking for a) would be associate each edge with a vector. If you cross product two adjacent vectors, suppose A and B, you will get a vector, lets call it C, which is perpendicular to the face supported by those two vectors. You can find the area by taking |C|/2. Other edges will be a subtraction of two other known vectors, so maybe this will help to prove the asked identity.
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  3. #3
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    Thanks bondesan.



    So I have tetrahedron ABCD (ignore PQSR in the picture, only decent picture i could find).

    So in order to define vectors V_{1}, V_{2}, V_{3}, V_{4}, I start with one of the sides, lets use side ABC.

    To find a vector normal to side ABC, I take the cross product (AB x AC) of 2 edges. That produces a vector with a direction normal and a magnitude of 2x the area of ABC (the area of a parallelogram with sides AB and AC). So then V_{1} = \frac{AB \times AC}{2}.

    Analogously for the other 3 sides, it can be found that:
    V_{2} = \frac{AB \times AD}{2}.
    V_{3} = \frac{AD \times AC}{2}.
    V_{4} = \frac{BC \times BD}{2}.

    Is that correct? At this point I'm still confused as to how to prove that the sum of those 4 vectors is equal to 0.

    EDIT: from here the denominators can be removed and the equation can be simplified to: (AB x AC) + (AB x AD) + (AD x AC) + (BC x BD) = 0.
    Last edited by nikcs123; October 10th 2010 at 07:12 PM.
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    As for part (b), I took a different tetrahedron (one with B as the "rectangular vertex") so D is the area of side ACD.

    Using similar properties, I found that D = \frac{\parallel AC \times AD\parallel}{2}, and then A = \frac{\parallel AB \times AC\parallel}{2}, B = \frac{\parallel AB \times AD\parallel}{2}, C = \frac{\parallel BC \times BD\parallel}{2}.

    So then from here I have to prove that:
    (\frac{\parallel AC \times AD\parallel}{2})^2 = (\frac{\parallel AB \times AC\parallel}{2})^2 + (\frac{\parallel AB \times AD\parallel}{2})^2 + (\frac{\parallel BC \times BD\parallel}{2})^2

    Simplifying, (\parallel AC \times AD\parallel)^2 = (\parallel AB \times AC\parallel)^2 + (\parallel AB \times AD\parallel)^2 + (\parallel BC \times BD\parallel)^2

    stuck at this point
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  5. #5
    Junior Member bondesan's Avatar
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    Quote Originally Posted by nikcs123 View Post



    Is that correct? At this point I'm still confused as to how to prove that the sum of those 4 vectors is equal to 0.

    EDIT: from here the denominators can be removed and the equation can be simplified to: (AB x AC) + (AB x AD) + (AD x AC) + (BC x BD) = 0.
    I think you are correct. You must notice that, for instance, BC can be written in terms of other vectors: BC = AC - AB - and this will help you to simplify the whole thing, remembering some properties of the cross product ( AxB = -BxA, ...) and so on.

    For b) maybe you should use ||AxB|| = |A|.|B|.sin t where t is the angle between A and B. You will have a lots of sin(pi/2) = 1.

    Regards
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  6. #6
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    Thanks again.

    I've managed to completely go through and prove part a.

    As for part b, I redefined my vectors as D = \frac{\parallel AC \times AD\parallel}{2}, A = \frac{\parallel AB \times BC\parallel}{2}, B = \frac{\parallel AB \times BD\parallel}{2}, C = \frac{\parallel BC \times BD\parallel}{2}.

    So then:
    (\frac{\parallel AC \times AD\parallel}{2})^2 = (\frac{\parallel AB \times BC\parallel}{2})^2 + (\frac{\parallel AB \times BD\parallel}{2})^2 + (\frac{\parallel BC \times BD\parallel}{2})^2

    (\parallel AC \times AD\parallel)^2 = (\parallel AB \times BC\parallel)^2 + (\parallel AB \times BD\parallel)^2 + (\parallel BC \times BD\parallel)^2

    Using the fact that magnitude of a cross product is magnitude of the vectors times the angle in between, I then find that:

    (\parallel AC \parallel \parallel AD \parallel sin(CAD))^2 = (\parallel AB \parallel \parallel BC \parallel sin(ABC))^2 + (\parallel AB \parallel \parallel BD \parallel sin(ABD))^2 + (\parallel BC \parallel \parallel BD \parallel sin(DBC))^2

    Where CAD, ABC, ABD, DBC are the angles between their respective vectors. Based on the properties of a rectangular tetrahedron, angles ABC, ABD, DBC are all pi/2, so the sin of those is equal to 1. Leaves me with:

    (\parallel AC \parallel \parallel AD \parallel sin(CAD))^2 = \parallel AB \parallel ^2 \parallel BC \parallel ^2 + \parallel AB \parallel ^2 \parallel BD \parallel ^2 + \parallel BC \parallel ^2 \parallel BD \parallel ^2

    Kind of stuck at this point, don't know where to take it from here.
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  7. #7
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    a) let A, B, C, D be the 4 vertex. For the face ABC, its area vector is
    (AB x AC)/2, so the question is to show that 0 = \sum (AB x AC), iterating all 4 faces.
    Note that AB x AC = (A-B) x (A-C) = B x C - B x A - A x C = A x B + B x C + C x A
    adding all the 4 equations up, you will get a 0, since for each edge, like AB, there are two faces( e.g ABC and ABD) adjointed, each face contributes a term A x B, but with opposite signs.

    b) is a consequence of a) and the Pythagoras theorem
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    Quote Originally Posted by xxp9 View Post
    b) is a consequence of a) and the Pythagoras theorem
    Makes sense, but how does one prove that?
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  9. #9
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    Quote Originally Posted by nikcs123 View Post
    Makes sense, but how does one prove that?
    from a) V1 + V2 + V3 + V4 = 0, but now V1, V2 and V3 are orthogonal now, according to the Pythagoras theorem, we have |V1+V2+V3|^2 = |V1|^2+|V2|^2+|V3|^2. DONE
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