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Ax = b when b = [0; 0; 0; 0]

  1. Jul 30, 2008 #1
    1. The problem statement, all variables and given/known data
    given z = (0, 0, 0, 1), and v = (0, x, y, z),

    and the following properties hold

    [tex] v \times u = L [/tex]
    [tex] L \times v = u [/tex]
    [tex] u \times L = v [/tex]

    v, u, l, and z are unit quaternions

    In other words; we define forward (v), up (u) and right(L), find a unit quaternion that when you apply it on z, you get v, in the proper orientation.

    solve for q = (a, b, c, d)

    2. Relevant equations
    (equation to solve for q)
    [tex]qzq^{-1} = v[/tex]
    Left side applys q on z and right side is v

    3. The attempt at a solution
    [tex] qz = vq [/tex]
    [tex] dk = (xi + yj + zk)(a + bi + cj + dk) [/tex]

    after expanding and grouping the terms on the same axis together. In a matrix form I got

    [tex] \left(
    \begin{array}{cccc}
    x & y & z & 0 \\
    0 & z & y & x \\
    z & 0 & -x & y \\
    -y & x & -1 & z
    \end{array}
    \right)

    \left(
    \begin{array}{c}
    a \\
    b \\
    c \\
    d
    \end{array}
    \right)
    =
    \left(
    \begin{array} {c}
    0 \\
    0 \\
    0 \\
    0
    \end{array}
    \right)
    [/tex]

    [a, b, c, d] = [0, 0, 0, 0] ? That doesn't seem right?

    Shouldn't I need to use atleast the vector u?

    Thank you for your help
     
  2. jcsd
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