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Thread: 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 $\displaystyle v_1 = (0,0,0)$ and $\displaystyle v_2 = (l_1,0,0)$. Let $\displaystyle v_3 = (p,q,0)$, with $\displaystyle l_2,\ l_3$ being the lengths of the edges $\displaystyle v_1v_3,\ v_2v_3$. Then
    $\displaystyle (1)\qquad p^2+q^2 = l_2^2,$
    $\displaystyle (2)\qquad (p-l_1)^2+q^2 = l_3^2.$

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

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

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

    Multiply out the brackets in (5) and (6), and substitute (4) into those equations to replace $\displaystyle a^2+b^2+c^2$ by $\displaystyle l_4^2$. 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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