![]() |
|
Linear Algebra Linear Algebra Math Forum |
![]() |
| LinkBack | Thread Tools | Display Modes |
December 20th, 2017, 03:12 AM | #1 |
Banned Camp Joined: Nov 2017 From: india Posts: 204 Thanks: 2 | How rotation matrix work?
Hi guys and gals please help me with rotation matrix work? |
![]() |
December 20th, 2017, 03:33 AM | #2 |
Senior Member Joined: Feb 2016 From: Australia Posts: 1,764 Thanks: 623 Math Focus: Yet to find out. |
What exactly, do you need help with?
|
![]() |
December 20th, 2017, 03:35 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, 05:17 AM | #4 |
Math Team Joined: Jan 2015 From: Alabama Posts: 3,261 Thanks: 896 |
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 07:34 AM. |
![]() |
![]() |
|
Tags |
matrix, rotation, work |
Thread Tools | |
Display Modes | |
|
![]() | ||||
Thread | Thread Starter | Forum | Replies | Last Post |
Eigenvalues of rotation matrix | arsenal15 | Linear Algebra | 2 | January 9th, 2015 06:09 AM |
Rotation Matrix | daveollie | Linear Algebra | 1 | November 15th, 2014 08:39 AM |
Rotation matrix | Poldek | Linear Algebra | 1 | November 23rd, 2012 08:28 AM |
Rotation matrix | Vasily | Abstract Algebra | 2 | September 16th, 2012 12:46 PM |
Rotation of a matrix | n757 | Linear Algebra | 2 | April 16th, 2011 08:21 PM |