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Lin algebra rotation difficault stuff

  1. Mar 28, 2009 #1
    1. The problem statement, all variables and given/known data

    1) Find a a matrix for counterclockwise rotation 70° about the point (2,5) and use homogeneous coordinates.
    2) Then, use that matrix to find out where the point (8,7) projects.

    2. Relevant equations



    3. The attempt at a solution
    1) [cos() -sin() 0 ; sin() cos() 0 ; 0 0 1] * [1 0 -2; 0 1 -5] = .....

    2) Multiply what I get in one by [8, 7, ???]

    or what?
     
  2. jcsd
  3. Mar 28, 2009 #2
    how can we use 3x3 matrices acting on augmented vectors in R^3 to represent transformations in R^2? your rotation matrix is correct, but you'll need to fix your translation matrix and fix the vector these act on. hint:think consistency! perform your (corrected) translation matrix on an arbitrary vector in R^3, what form do you need on this vector to get a good result?
     
  4. Mar 29, 2009 #3

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    You really need to do three things:
    1) translate the point (2,5) to (0,0)
    2) rotate about (0,0) 70 degrees
    3) translate (0,0) back to 2,5).

    In homogeneous coordinates, the point (a, b) is represented by the array [a, b, 1] with the understanding that any multiple, [sa, sb, s], is equivalent to the point (a,b). That is, if the third component is not 1, divide the entire array by that component to get the point.

    The advantage of homogenous coordinates (they are used a lot in computer graphics) is that you can represent translation as a matrix multiplication. For example:
    [tex]\begin{bmatrix}1 & 0 & a \\ 0 & 1 & b\\ 0 & 0 & 1\end{bmatrix}\begin{bmatrix}x \\ y \\ 1\end{bmatrix}= \begin{bmatrix}x+ a \\ y+ b \\ 1\end{bmatrix}[/tex].

    Of course, rotation around the origin, by [itex]\theta[/itex] degrees, is given by
    [tex]\begin{bmatrix}cos(\theta) & -sin(\theta) & 1 \\ sin(\theta) & cos(\theta) & 0 \\ 0 & 0 & 1\end{bmatrix}[/tex].

    To solve this problem you need to construct three matrices: the matrix, A, that translates (2, 5) to (0, 0), the matrix, B, that rotates 70 degrees around (0,0), and the matrix, C, that translates (0, 0) to (2, 5) (It will, of course, be the inverse of the matrix A). Finally multiply the matrices in order CBA.

    The point (8, 7) is represented by the array [8, 7, 1].
     
    Last edited: Mar 29, 2009
  5. Mar 29, 2009 #4
    Are you sure there´s supposed to be 1 at the top in the right?
    not zero?


    I already did the translation and have [x+2; y+5; 1]

    What do I do next?
     
  6. Mar 31, 2009 #5
    Its also useful to think about rotations rationally, instead of with cos and sin, which don't usually give exact values. Use the fact that a point on the circle can be better coordinatized often with [(1-t^2)/(1+t^2),2t/(1+t^2)].

    This also connects with Rational Trigonometry.
     
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