Results 1 to 4 of 4
Like Tree1Thanks
  • 1 Post By SlipEternal

Math Help - Visualization of rotation

  1. #1
    Junior Member
    Joined
    Oct 2013
    From
    Moscow, Russia
    Posts
    41

    Visualization of rotation

    I am trying to visualize the following rotation of R^3, but it is very difficult. I want to get the answer by intuition, and not by using the Rodrigues rotation formula or conjugation of matrices, etc.


    Help please.


    Problem statement: Determine the matrix that represents the following rotation of R^3: an angle of pi/2 about the fixed axis containing the vector (1,1,0)^t


    Here is what I have tried in my diagram:


    ![Coordinate axes][1]


    Should I find a 3x3 rotation matrix A such that A(1,1,0)^t=(-1,1,0)^t?


    [1]: http://i.stack.imgur.com/88MKh.png
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Nov 2010
    Posts
    1,879
    Thanks
    742

    Re: Visualization of rotation

    Consider a rotation by an angle of pi/2 about the fixed axis containing the vector e_1. That would be a rotation that would take e_2\mapsto e_3 and e_3 \mapsto -e_2. So, find a basis for \mathbb{R}^3 containing (1,1,0)[/tex]. Then, you want a 3x3 rotation matrix that will take (1,1,0) to itself, the second basis vector to the third, and the third basis vector to negative the first.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Oct 2013
    From
    Moscow, Russia
    Posts
    41

    Re: Visualization of rotation

    Would the basis {[1,0,0],[0,1,0]} work? For the rotation matrix, could you show me an example?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Joined
    Nov 2010
    Posts
    1,879
    Thanks
    742

    Re: Visualization of rotation

    Oh, I should have mentioned that you want an orthonormal basis. For example: (1,1,0), (-1,1,0), (0,0,1) are a set of three orthogonal vectors. So, an orthonormal basis would be: \dfrac{1}{\sqrt{2}}\begin{pmatrix}1 \\ 1 \\ 0\end{pmatrix}, \dfrac{1}{\sqrt{2}}\begin{pmatrix}-1 \\ 1 \\ 0\end{pmatrix}, \begin{pmatrix}0 \\ 0 \\ 1\end{pmatrix}. Also, you can always use the Gram-Schmidt process if you cannot find a simple orthogonal basis to normalize.

    So, you want:

    A\begin{pmatrix}\dfrac{1}{\sqrt{2}} & -\dfrac{1}{\sqrt{2}} & 0 \\ \dfrac{1}{\sqrt{2}} & \dfrac{1}{\sqrt{2}} & 0 \\ 0 & 0 & 1\end{pmatrix} = \begin{pmatrix}\dfrac{1}{\sqrt{2}} & 0 & \dfrac{1}{\sqrt{2}} \\ \dfrac{1}{\sqrt{2}} & 0 & -\dfrac{1}{\sqrt{2}} \\ 0 & 1 & 0\end{pmatrix}

    That means A = \begin{pmatrix}\dfrac{1}{\sqrt{2}} & 0 & \dfrac{1}{\sqrt{2}} \\ \dfrac{1}{\sqrt{2}} & 0 & -\dfrac{1}{\sqrt{2}} \\ 0 & 1 & 0\end{pmatrix}\begin{pmatrix}\dfrac{1}{\sqrt{2}} & -\dfrac{1}{\sqrt{2}} & 0 \\ \dfrac{1}{\sqrt{2}} & \dfrac{1}{\sqrt{2}} & 0 \\ 0 & 0 & 1\end{pmatrix}^{-1} = \dfrac{1}{2}\begin{pmatrix}1 & 1 & \sqrt{2} \\ 1 & 1 & -\sqrt{2} \\ -\sqrt{2} & \sqrt{2} & 0\end{pmatrix}
    Last edited by SlipEternal; November 10th 2013 at 12:05 PM.
    Thanks from abscissa
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. translating world rotation values to local rotation axis
    Posted in the Advanced Algebra Forum
    Replies: 0
    Last Post: April 12th 2013, 07:24 AM
  2. r formula visualization
    Posted in the Trigonometry Forum
    Replies: 18
    Last Post: November 20th 2010, 04:37 AM
  3. 3-d visualization problem
    Posted in the Geometry Forum
    Replies: 2
    Last Post: September 17th 2009, 04:49 AM
  4. Replies: 3
    Last Post: March 14th 2009, 07:37 AM
  5. Tensor Visualization
    Posted in the LaTeX Help Forum
    Replies: 10
    Last Post: September 22nd 2007, 07:29 PM

Search Tags


/mathhelpforum @mathhelpforum