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Math Help - 6 lengths define the 3D coordinate of a Tetrahedron

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    6 lengths define the 3D coordinate of a Tetrahedron

    Given 6 lengths l1..l6 in R,
    assuming that the Tetrahedron has a face lies on the x-y plane;
    and there is an edge of the Tetrahedron lies on the x-axis.

    Find the 4 3D coordinates v1..v4 that define the Tetrahedron.

    Thank you
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    Quote Originally Posted by williamlai3a View Post
    Given 6 lengths l1..l6 in R,
    assuming that the Tetrahedron has a face lies on the x-y plane;
    and there is an edge of the Tetrahedron lies on the x-axis.

    Find the 4 3D coordinates v1..v4 that define the Tetrahedron.
    We can take v_1 = (0,0,0) and v_2 = (l_1,0,0). Let v_3 = (p,q,0), with l_2,\ l_3 being the lengths of the edges v_1v_3,\ v_2v_3. Then
    (1)\qquad p^2+q^2 = l_2^2,
    (2)\qquad (p-l_1)^2+q^2 = l_3^2.

    Multiply out the bracket in (2), and substitute (1) into (2), to get
    (3)\qquad l_2^2 - 2l_1p + l_1^2 = l_3^2.

    From (3) you can find p, and then from (1) you can find q. So that determines the coordinates of v_3.

    The procedure to find the location of v_4 is similar, except that this time we have to work in three dimensions rather than two. Let v_4 = (a,b,c), with l_4,\ l_5,\ l_6 being the lengths of the edges v_1v_4,\ v_2v_4,\ v_3v_4. Then
    (4)\qquad a^2+b^2+c^2 = l_4^2,
    (5)\qquad (a-l_1)^2+b^2+c^2 = l_5^2,
    (6)\qquad (a-p)^2+(b-q)^2+c^2 = l_6^2.

    Multiply out the brackets in (5) and (6), and substitute (4) into those equations to replace a^2+b^2+c^2 by l_4^2. That will leave you with two linear equations for a and b, which you can solve. Finally, substitute these values for a and b into (4) to find c.
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