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Math Help - Coordinates, vectors and cross products

  1. #1
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    Question Coordinates, vectors and cross products

    Hi

    I have 4 3D points and I have been told I need to create 2 vecors out of them. I then need to cross product these vecotrs to then get another vector.

    Now with my terrible grasp of maths, I'm ahving a hard time working out this stuff.

    Can anybody help me work out a solution for solving this problem?
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  2. #2
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    Quote Originally Posted by yellowFattyBEan View Post
    Hi

    I have 4 3D points and I have been told I need to create 2 vecors out of them. I then need to cross product these vecotrs to then get another vector.

    Now with my terrible grasp of maths, I'm ahving a hard time working out this stuff.

    Can anybody help me work out a solution for solving this problem?
    Which problem?

    Please post the complete text of the question and your work. Otherwise we can't help you very much.
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  3. #3
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    ok

    I'm working with a 3d graphics package so it's kinda hard to relate everything.

    I have a 2d plane, and I've tried to mark the corners of this. (sorry for the crudeness)

    (-6,0,-6) (6,0,6)
    _________
    / /
    / /
    / /
    /________/

    (-6,0,6) (6,0,-6)


    Basically, I need to take these points, convert them to vectors and find the cross product so I can then go on to prove that this is a 2d plane ranther than a 3d object.
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  4. #4
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    Quote Originally Posted by yellowFattyBEan View Post
    ok

    I'm working with a 3d graphics package so it's kinda hard to relate everything.

    I have a 2d plane, and I've tried to mark the corners of this. (sorry for the crudeness)

    (-6,0,-6) (6,0,6)
    _________
    / /
    / /
    / /
    /________/

    (-6,0,6) (6,0,-6)


    Basically, I need to take these points, convert them to vectors and find the cross product so I can then go on to prove that this is a 2d plane ranther than a 3d object.
    If I understand your question correctly you want to show that these 4 points are placed in a plane (?).

    If so:

    1. Three points which don't lie on a straight line allways lie in a plane. Take 3 points and determine the equation of this plane:
    A(-6, 0 ,-6), B(6, 0, 6), C(-6, 0, 6)

    p: \vec r = (-6, 0, -6) + s \cdot ((6,0,6)-(-6,0,-6))+t \cdot ((-6,0,6)-(-6,0,-6))

    p:\vec r = (-6,0-6)+s(12,0,12)+t(0,0,12)

    2. Now check if the point D(6,0,-6) belongs to the plane p:

    (6,0,-6)=(-6,0-6)+s(12,0,12)+t(0,0,12)

    This equation has the only solution s = 1 and t = -1. Therefore D\in p

    3. Since the second coordinate of all points is zero, all points are situated in the x-z-plane.
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