integration

Hi guys and gals
please help me with rotation matrix work?

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Joppy

What exactly, do you need help with?

integration

i want to know if we place 1 cosx =0.xx
how screen work with it

Country Boy

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.