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Math Help - Rotation Proof

  1. #1
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    Rotation Proof

    Prove that the two dimensional rotation matrix preserves dot products (e.g. show that  \bar{A}_{y} \bar{B}_{y} + \bar{A}_{z} \bar{B}_{z} = A_{y}B_{y} + A_{z}B_{z} ). So

     \begin{bmatrix} \bar{A}_{y} \\ \bar{A}_{z} \end{bmatrix} = \begin{bmatrix} \cos \phi & \sin \phi \\ - \sin \phi & \cos \phi \end{bmatrix} \begin{bmatrix} A_{y} \\ A_{z} \end{bmatrix}

     \begin{bmatrix} \bar{B}_{y} \\ \bar{B}_{z} \end{bmatrix} = \begin{bmatrix} \cos \phi & \sin \phi \\ - \sin \phi & \cos \phi \end{bmatrix} \begin{bmatrix} B_{y} \\ B_{z} \end{bmatrix}


    So  \bar{A}_{y} = A_{y} \cos \phi + A_{z} \sin \phi \ \ \ \ \ \bar{A}_{z} = -A_{y} \sin \phi + A_{z} \cos \phi

     \bar{B}_{y} = B_{y} \cos \phi + B_{z} \sin \phi \ \ \ \ \ \ \bar{B}_{z} = -B_{y} \sin \phi + B_{z} \cos \phi .


    And  \bar{A}_{y} \bar{B}_{y} + \bar{A}_{z} \bar{B}_{z} = \left(A_{y}B_{y} \cos^{2} \phi + A_{y}B_{z} \sin \phi \cos \phi + A_{z}B_{y} \sin \phi \cos \phi + A_{z}B_{z} \sin^{2} \phi \right) +  \left(A_{y}B_{y} \sin^{2} \phi + A_{y}B_{z} \sin \phi \cos \phi + A_{z}B_{y} \sin \phi \cos \phi + A_{z}B_{z} \sin^{2} \phi \right) .

    How is this equaled to  A_{y}B_{y} + A_{z}B_{z} ?
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  2. #2
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    Quote Originally Posted by heathrowjohnny View Post
    Prove that the two dimensional rotation matrix preserves dot products (e.g. show that  \bar{A}_{y} \bar{B}_{y} + \bar{A}_{z} \bar{B}_{z} = A_{y}B_{y} + A_{z}B_{z} ). So

     \begin{bmatrix} \bar{A}_{y} \\ \bar{A}_{z} \end{bmatrix} = \begin{bmatrix} \cos \phi & \sin \phi \\ - \sin \phi & \cos \phi \end{bmatrix} \begin{bmatrix} A_{y} \\ A_{z} \end{bmatrix}

     \begin{bmatrix} \bar{B}_{y} \\ \bar{B}_{z} \end{bmatrix} = \begin{bmatrix} \cos \phi & \sin \phi \\ - \sin \phi & \cos \phi \end{bmatrix} \begin{bmatrix} B_{y} \\ B_{z} \end{bmatrix}


    So  \bar{A}_{y} = A_{y} \cos \phi + A_{z} \sin \phi \ \ \ \ \ \bar{A}_{z} = -A_{y} \sin \phi + A_{z} \cos \phi

     \bar{B}_{y} = B_{y} \cos \phi + B_{z} \sin \phi \ \ \ \ \ \ \bar{B}_{z} = -B_{y} \sin \phi + B_{z} \cos \phi .


    And  \bar{A}_{y} \bar{B}_{y} + \bar{A}_{z} \bar{B}_{z} = \left(A_{y}B_{y} \cos^{2} \phi + A_{y}B_{z} \sin \phi \cos \phi + A_{z}B_{y} \sin \phi \cos \phi + A_{z}B_{z} \sin^{2} \phi \right) +  \left(A_{y}B_{y} \sin^{2} \phi + A_{y}B_{z} \sin \phi \cos \phi + A_{z}B_{y} \sin \phi \cos \phi + A_{z}B_{z} \sin^{2} \phi \right) .

    How is this equaled to  A_{y}B_{y} + A_{z}B_{z} ?
    First a couple of mistakes that need fixing up:

     \bar{A}_{y} \bar{B}_{y} + \bar{A}_{z} \bar{B}_{z} = \left(A_{y}B_{y} \cos^{2} \phi + A_{y}B_{z} \sin \phi \cos \phi + A_{z}B_{y} \sin \phi \cos \phi + A_{z}B_{z} \sin^{2} \phi \right) +  \left(A_{y}B_{y} \sin^{2} \phi - A_{y}B_{z} \sin \phi \cos \phi - A_{z}B_{y} \sin \phi \cos \phi + A_{z}B_{z} \cos^{2} \phi \right) .

    See how it'll work now?
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  3. #3
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    Hello, heathrowjohnny!

    You dropped a few minus-signs . . .


    Prove that the two dimensional rotation matrix preserves dot products
    (e.g. show that  \overline{A}_y \overline{B}_y + \overline{A}_z \overline{B}_z = A_yB_y + A_zB_z ).

     \begin{bmatrix} \,\overline{A}_y \\ \overline{A}_z \end{bmatrix} \;= \;\begin{bmatrix}\cos\phi & \sin\phi \\ \text{-}\sin\phi & \cos\phi \end{bmatrix} \begin{bmatrix}A_y \\ A_z \end{bmatrix}

     \begin{bmatrix} \,\overline{B}_{y} \\ \overline{B}_{z} \end{bmatrix} = \begin{bmatrix} \cos \phi & \sin \phi \\ \text{-}\sin\phi & \cos\phi \end{bmatrix} \begin{bmatrix} B_{y} \\ B_{z} \end{bmatrix}


    Then: . \begin{array}{ccccccc} \overline{A}_y &=&  A_y\cos\phi + A_z\sin\phi & & \overline{A}_z &=& \text{-}A_y\sin\phi + A_z\cos\phi \\<br />
\overline{B}_y &=& B_y\cos\phi + B_z\sin\phi & & \overline{B}_z &=& \text{-}B_y\sin\phi + B_z\cos\phi \end{array} .


    And: . \begin{array}{ccc}\overline{A}_y\overline{B}_y &=&A_yB_y\cos^2\!\phi + A_yB_z\sin\phi\cos\phi + A_zB_y\sin\phi\cos\phi + A_zB_z\sin^2\!\phi \\<br />
\overline{A}_z\overline{B}_z &=&A_yB_y\sin^2\!\phi - A_yB_z\sin\phi\cos\phi - A_zB_y\sin\phi\cos\phi + A_zB_z\cos^2\!\phi\end{array}


    \text{Add: }\;\overline{A}_y\overline{B}_y+\overline{A}_z\ove  rline{B}_z \;=\;A_yB_y\underbrace{(\sin^2\!\phi + \cos^2\!\phi)}_{\text{This is 1}} + A_zB_z\underbrace{(\sin^2\!\phi + \cos^2\!\phi)}_{\text{This is 1}}

    . . . . . . . . . . . . . = \;A_yB_y + A_zB_z

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