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Transforming a 3D vector using matrix

  1. Hi Friends,

    I have a problem in transforming a vector in 3d using a matrix.

    I have two points A, B. Assume origin is O, here A,B ,O are in 3d.

    First construct a vector AB:

    OB = OA + AB
    AB = OB - OA
    AB = OB + AO

    I have a 4*4 matrix that gives the transformation matrix(rotation+translation)

    I need to transform vector AB using this matrix.

    Currently what I do is transform individual points A , B first and the reconstruct the vector A'B' again (after tranforming)
    What I want to do is to without doing the above steps, transform the vector as it is (without transforming the individual points). Is this possible?

    How could I do this computation?

    I am looking for your reply!

    thank you.

    /Gajan
     
  2. jcsd
  3. HallsofIvy

    HallsofIvy 40,413
    Staff Emeritus
    Science Advisor

    Using a single 4x4 matrix to do both rotation and translation, you have to use a "projective" space. That means you are representing the point (x,y,z) as the (column) vector [x y z 1] with the provision that [a b c d] is the same as [a/d b/d c/d 1] (d can never be 0). In that case the matrix that rotates, say, [itex]\theta[/itex] degrees about the y-axis and translates by (tx,ty, tz) is
    [tex]\begin{bmatrix} cos(\theta) & 0 & -sin(\theta) & tx \\ 0 & 1 & 0 & ty \\ sin(\theta) & 0 & cos(\theta) & tz \\ 0 & 0 & 0 & 1\end{bmatrix}[/tex]
    Notice that in the particular case of [itex]\theta= 0[/itex] where there is no rotation and so a pure translation, this becomes
    [tex]\begin{bmatrix} 1 & 0 & 0 & tx \\ 0 & 1 & 0 & ty \\ 0 & 0 & 1 & tz \\ 0 & 0 & 0 & 1\end{bmatrix}\begin{bmatrix} x \\ y \\ z \\ 1\end{bmatrix}= \begin{bmatrix} x+ tx \\ y+ ty \\ z+ tz \\ 1\end{bmatrix}[/tex]

    While if tx= ty= tz= 0 so there is a pure rotation and no translation it is
    [tex]\begin{bmatrix} cos(\theta) & 0 & -sin(\theta) & 0 \\ 0 & 1 & 0 & 0 \\ sin(\theta) & 0 & cos(\theta) & 0 \\ 0 & 0 & 0 & 1\end{bmatrix}\begin{bmatrix} x \\ y \\ z \\ 1\end{bmatrix}= \begin{bmatrix} xcos(\theta)- zsin(\theta) \\ y \\ xsin(\theta)+ zcos(\theta)\end{bmatrix}[/tex]

    I would handle a general rotation as the product of two rotations around coordinate axes.
     
  4. Hi,

    First of all thank you for your reply.

    In my case , I have the matrix in the following form:

    Transpose matrix of the transformation : M
    vector :V

    V * M

    still is it the same way the matrix product is done?
     
    Last edited: Feb 18, 2009
  5. HallsofIvy

    HallsofIvy 40,413
    Staff Emeritus
    Science Advisor

    Assuming that you are writing V as a row matrix, yes, swapping "row" and "column" is purely a matter of convention.
     
  6. Thanking you. I understand it now.
     
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