My Math Forum How rotation matrix work?

 Linear Algebra Linear Algebra Math Forum

December 20th, 2017, 02:12 AM   #1
Banned Camp

Joined: Nov 2017
From: india

Posts: 204
Thanks: 2

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 Display Modes Linear 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

 Contact - Home - Forums - Cryptocurrency Forum - Top