Given the parameters p 1 , p 2 , p 3 does the following system can be solved:This system actually discribes a Rotation Matrix p 1 a b c p 2 d e f p 3 where only p 1 , p 2 and p 3 are know. For example p 1 = p 2 = p 3 = 1 would result in a = b = c = d = e = f = 0.Why is this system (unabigiously) solvable / not solvable?If it is solvable, what is the solution?

Question

Given the parameters       p    1    ,      p    2    ,      p    3   does the following system can be solved:This system actually discribes a Rotation Matrix      p    1          a              b    c                  p    2          d    e              f                  p    3  where only       p    1  ,       p    2   and       p    3   are know. For example       p    1    =      p    2    =      p    3    =  1 would result in   a  =  b  =  c  =  d  =  e  =  f  =  0.Why is this system (unabigiously) solvable / not solvable?If it is solvable, what is the solution?

Dalton Erickson · Accepted Answer

Let  M  =      (                                        p            1                                    a                          b                                      c                                      p            2                                    d                                      e                          f                                      p            3                                )    .If   M is a rotation matrix, then       M          −      1        =      M    T    . This implies that   M and       M          −      1       have the same entries on the main diagonal. But if   M is not the identity,   M  ≠      M          −      1        , so the rotation matrix is not completely determined by the entries on its main diagonal.On the other hand, if   [      u    1           u    2           u    3        ]    T   is a unit vector on the axis of rotation of   M and if 𝜃 is the angle of rotation about that axis, then                    (1)                              p          i                =        (        1        −        cos        ⁡        θ        )                  u          i          2                +        cos        ⁡        θ            for   i  =  1  ,  2  ,  3. Moreover,      p    1    +      p    2    +      p    3    =  1  +  2  cos  ⁡  θ  .Therefore we can express cos𝜃 in terms of       p    1    .,       p    2    ., and       p    3    .. Plug that value of cos𝜃 into equation (1) for each   i  ; this either gives       u    1    =  0 or gives two possible values of       u    1   which differ only by a sign change.We can safely assume that   0  ≤  θ  ≤  π  ,, because the rotation described by angle   −  θ and unit vector   [      u    1           u    2           u    3        ]    T   is the same as the rotation described by angle   θ and unit vector   [      −          u      1               −          u      2               −          u      3            ]    T    .. That means that in general there are eight possible ways to fill in the matrix   M (one for each choice of the signs of each of the       u    i    )  ,), therefore eight possible solutions to the given set of equations. (For   0  &amp;lt;  θ  &amp;lt;  π  ,, there are four solutions if exactly one of the       u    i   is zero, two solutions if two of the 𝑢𝑖 are equal to zero. There are half as many solutions if   θ  =  π  ,, and of course only one solution if   θ  =  0.  ).) Moreover, by computing the rotation matrix for the rotation by angle   θ around the axis given by   [      u    1           u    2           u    3        ]    T    ,, we can compute all the unknown entries   a  ,  b  ,  c  ,  d  ,  e  , and   f in the rotation matrix for a specific choice of   [      u    1           u    2           u    3        ]    T    .

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