Results 1 to 3 of 3

Math Help - Rotating Figures

  1. #1
    Newbie
    Joined
    Apr 2005
    Posts
    12

    HELP: Rotating Figures

    Hi

    These four points A(3,1), B(5/2,2), C(9/2,3), D(5,2)

    Is turned 90 degrees clockwise around the center of the figure.

    I'm supposed to write a transformation that does this.

    I get the following:

    (x,y) \rightarrow \left[ \begin{array}{cc} -cos( 2 \theta ) & sin(2 \theta) \\ sin(2 \theta) & cos(2 \theta) \end{array} \right] \left[ \begin{array}{c} x \\ y \end{array} \right]

    My teacher is says this is wrong cause any transformations that isn't rotated around (0,0) is not a linear transformation.

    If thats true what should the transformation then look like ???

    /Fred
    Follow Math Help Forum on Facebook and Google+

  2. #2
    hpe
    hpe is offline
    Member hpe's Avatar
    Joined
    Apr 2005
    Posts
    158
    Quote Originally Posted by Mathman24
    [These four points are to be rotated] 90 degrees clockwise around the center of the figure.
    Let's say a rotation of the desired form abou the origin is of the form \bold{y} = A \bold{x} where A is a matrix and x and y are 2-vectors. You'll have to determine the matrix A - your solution isn't correct since it doesn't pay attention to the 90 degrees. Such a rotation is linear.

    Suppose now you want to rotate about a different point, say \bold{x_0}. Set  \bold{x'}  = \bold{x - x_0}. This is the displacement of the point x from \bold{x_0}.

    Then  \bold{y'} = A\bold{x'} = A(\bold{x-x_0}) is the rotated position vector, relative to the point \bold{x_0}, and therefore  \bold{y} = \bold{y'+x_0} = A\bold{(x-x_0)}+\bold{x_0} is the formula for the rotation about \bold{x_0}.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Apr 2005
    Posts
    12
    Hi can't it be made correct if I add the distance between the original point and the center of the figure ??

    /Fred

    Quote Originally Posted by hpe
    Let's say a rotation of the desired form abou the origin is of the form \bold{y} = A \bold{x} where A is a matrix and x and y are 2-vectors. You'll have to determine the matrix A - your solution isn't correct since it doesn't pay attention to the 90 degrees. Such a rotation is linear.

    Suppose now you want to rotate about a different point, say \bold{x_0}. Set  \bold{x'}  = \bold{x - x_0}. This is the displacement of the point x from \bold{x_0}.

    Then  \bold{y'} = A\bold{x'} = A(\bold{x-x_0}) is the rotated position vector, relative to the point \bold{x_0}, and therefore  \bold{y} = \bold{y'+x_0} = A\bold{(x-x_0)}+\bold{x_0} is the formula for the rotation about \bold{x_0}.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Lissajou Figures
    Posted in the Algebra Forum
    Replies: 2
    Last Post: January 16th 2010, 08:01 AM
  2. Three dimensional figures
    Posted in the Geometry Forum
    Replies: 1
    Last Post: July 17th 2009, 11:23 AM
  3. significant figures
    Posted in the Algebra Forum
    Replies: 1
    Last Post: January 11th 2009, 12:06 PM
  4. Replies: 1
    Last Post: October 7th 2008, 02:18 AM
  5. significant figures
    Posted in the Math Topics Forum
    Replies: 4
    Last Post: September 6th 2008, 04:11 PM

Search Tags


/mathhelpforum @mathhelpforum