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Math Help - "z" graph of a tetrahedron

  1. #1
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    "z" graph of a tetrahedron

    How do you find the "z" graph of a tetrahedron when you are only given the vertices?

    For example....

    What's the "z" graph of a tetrahedron when the xy-plane vertices are (0,2), (0,0), and (1,0) and when z=3 as long as x=0 and y=0?
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  2. #2
    Member Ruun's Avatar
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    What do you mean by graph? What I've done is just plotting the points and join them with lines, first the triangle in the x-y plane and then, with the last point at (0,0,3)
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  3. #3
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    Every tetrahedron is bounded by a 3D plane, akin to a z-based graph z(x,y). I was wondering how you would find out the exact z-equation based on the vertices. We know the vertices and the shape, but what is the equation of the 3D plane?
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  4. #4
    Member Ruun's Avatar
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    Ok.

    For example take the vectors (0-0,0-2,3-0) and (0-1,0-0,3-0), do the cross product and you will have a normal vector, then the plane can be written as Ax+By+Cz=0 where (A,B,C) is a normal vector and (x,y,z) it's a point on the plane.

    I think this should work
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  5. #5
    Member Ruun's Avatar
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    Well, my answer is that a normal vector is (6,3,2). Then 6x+3y+2z=0 and so  z= -2x-y

    Hope it will be right!
    Last edited by Ruun; June 8th 2009 at 01:46 PM. Reason: latex fail lol
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  6. #6
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    No that's not the right one. I used my graphing software and it doesn't fit in with the vertices.
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  7. #7
    Member Ruun's Avatar
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    In fact it's wrong: I divided by 3 instead of 2. z=-3x - \frac{3}{2}y, check this one.
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  8. #8
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    Nope lol.

    Keep trying.

    Fyi.... I have a very limited knowledge of vectors (cross product, dot product, etc.) right now so I kinda can't figure out how you're deriving the equations but that's for another time.
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  9. #9
    Member Ruun's Avatar
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    Ok no problem! Then a brief explanation

    You gave some points in the space, (0,0,3), (0,2,0) and (0,1,0). it's possible to form vectors by substracting the coordinates of two given points, with three points we form the vectors (0,-2,3) and (-1,0,3) As those vectors are the sides of tetrahedron, they sall belong to the plane we are looking for

    From cross product theory, the vector u \times v = w its perpendicular to both u, and v, that's it, to the plane that belong to those vectors.

    Now from dot product theory, if we take some vector with coordinates (x,y,z) wich has its end (thinking in vectors as oriented segments) in the point (x,y,z) and another vector (A,B,C) wich is perpendicular to the plane, the dot product is zero. Explicitly: (A,B,C) \cdot (x,y,z) = 0 . Then we have Ax + By + Cz = 0. Our (A,B,C) is (6,3,2) so this is were my equation comes from.

    You're saying that there's something wrong, so there should be some fail in derivation or computation, but at least we're trying!
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