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December 20th, 2017, 02:12 AM   #1
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How rotation matrix work?

Hi guys and gals
Attached Images 2851c9dc2031127e6dacfb84b96446d8.jpg (12.8 KB, 0 views) December 20th, 2017, 02:33 AM #2 Senior Member   Joined: Feb 2016 From: Australia Posts: 1,826 Thanks: 646 Math Focus: Yet to find out. What exactly, do you need help with? December 20th, 2017, 02:35 AM #3 Banned Camp   Joined: Nov 2017 From: india Posts: 204 Thanks: 2 i want to know if we place 1 cosx =0.xx how screen work with it December 20th, 2017, 04:17 AM #4 Math Team   Joined: Jan 2015 From: Alabama Posts: 3,264 Thanks: 902 First can we assume that you know how to do matrix multiplication? $\begin{bmatrix}a & b & c \\ d & e & f \\ g & h & i \end{bmatrix}\begin{bmatrix} x \\ y \\ z \end{bmatrix}= \begin{bmatrix} ax+ by+ cz \\ dx+ ey+ fz \\ gx+ hy+ iz\end{bmatrix}$. So $\begin{bmatrix}1 & 0 & 0 \\ 0 & \cos(\theta) & -\sin(\theta) \\ 0 & \sin(\theta) & \cos(\theta) \end{bmatrix}\begin{bmatrix} x \\ y \\ z \end{bmatrix}= \begin{bmatrix} x \\ y\cos(\theta)- z\sin(\theta) \\ y\sin(\theta)+ z\cos(\theta)\end{bmatrix}$. Notice that the first row being $\begin{bmatrix}1 & 0 & 0 \end{bmatrix}$ ad the first column $\begin{bmatrix}1 \\ 0 \\ 0\end{bmatrix}$ means that the x-axis remains unchanged- it is the "axis of revolution". Any point in an x= constant plane remains in that plane. Further, the other coordinates depend only on y and z and are independent of x. Finally, in each such plane (y, z) is mapped to $(y\cos(\theta)- z \sin(\theta), y\sin(\theta)+ z\cos(\theta))$. The distance from the origin to that new point is $\sqrt{(y\cos(\theta)- z\sin(\theta))^2+ (y\sin(\theta)+ z\cos(\theta)^2}$ $= \sqrt{y^2 \cos^2(\theta)- 2yz\sin(\theta)\cos(\theta)+ z^2 \sin^2(\theta)+ y^2\sin^2(\theta)+ 3yz \sin(\theta)\cos(\theta)+ z^2 \cos^2(\theta)}$ The $-2yz\sin(\theta)\cos(\theta)$ and $2yz\sin(\theta)\cos(\theta)$ cancel, leaving $\sqrt{y^2(\cos^2(\theta)+ \sin^2(\theta)+ z^2(\sin^2(\theta)+ \cos^2(\theta))}= \sqrt{y^2+ z^2}$. The distance from the origin has not changed - the point moves in a circle about the x-axis. A little trigonometry shows it has moved around that circle through angle $\theta$. The second, with second row and column $\begin{bmatrix}0 & 1 & 0 \end{bmatrix}$ and $\begin{bmatrix}0 \\ 1 \\ 0 \end{bmatrix}$ is a rotation about the y-axis and the third, a rotation about the z-axis. Last edited by skipjack; December 20th, 2017 at 06:34 AM. Tags matrix, rotation, work Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post arsenal15 Linear Algebra 2 January 9th, 2015 05:09 AM daveollie Linear Algebra 1 November 15th, 2014 07:39 AM Poldek Linear Algebra 1 November 23rd, 2012 07:28 AM Vasily Abstract Algebra 2 September 16th, 2012 11:46 AM n757 Linear Algebra 2 April 16th, 2011 07:21 PM

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