wolfe8
  • wolfe8
Find the combined transformation matrix for a rotation of an orthogonal coordinate system counterclockwise about the z-axis through an angle θ, followed by a counterclockwise rotation about the new y-axis through an angle φ. My problem is getting the transformation matrices. I drew the transformation and can see the angles I am looking for, but my problem is choosing whether to use cos, sin, or -sin.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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amistre64
  • amistre64
just trying to see if i understand the informations .... what did you come up with? |dw:1378730441438:dw|
amistre64
  • amistre64
im trying to see turning the faces of a rubiks cube :)
wolfe8
  • wolfe8
I think you have the angles switched. The first transformation has theta and the other phi(?) But other than that yeah that's it I think.

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amistre64
  • amistre64
doesnt the usual construct deal with cosines from each axis? |dw:1378730672194:dw|
wolfe8
  • wolfe8
That's what I'm having problems with... For the first trans., I have cos theta 0 0 sin theta cos theta 0 0 0 1 Is that right?
amistre64
  • amistre64
to turn about one of the axis, we leave that axis as is ... \[\begin{pmatrix} -&-&0\\ -&-&0\\ 0&0&1 \end{pmatrix}\] then sin cos the remaining parts \[\begin{pmatrix} cos(\theta)&-sin(\theta)&0\\ sin(\theta)&cos(\theta)&0\\ 0&0&1 \end{pmatrix}\] this is one rotation counter clockwise about the z axis
amistre64
  • amistre64
then we can apply the next trasformation leaving the y parts alone
wolfe8
  • wolfe8
Oh man I got it wrong AGAIN. So the next will have the second row and columns 0 except at position 22?
amistre64
  • amistre64
\[Zx = x'\]\[Yx'=b\] so we would want the composite function \[Y(Zx)=b\]
amistre64
  • amistre64
To define the Y matrix we procede the same way getting us \[\begin{pmatrix} cos(\phi)&0&-sin(\phi)\\ sin(\phi)&1&cos(\phi)\\ 0&0&0 \end{pmatrix}\] what does YZ look like?
wolfe8
  • wolfe8
Just the product of YZ?
amistre64
  • amistre64
of course, YZ is the composite transformation ... the "combined" effect of operating Z and then Y
wolfe8
  • wolfe8
Yup that's what I thought. Thanks a bunch! I really needed help with the trigonometry!
amistre64
  • amistre64
\[YZ=\begin{pmatrix} cos(\phi)&0&-sin(\phi)\\ sin(\phi)&1&cos(\phi)\\ 0&0&0 \end{pmatrix}\begin{pmatrix} cos(\theta)&-sin(\theta)&0\\ sin(\theta)&cos(\theta)&0\\ 0&0&1 \end{pmatrix} \] yep
wolfe8
  • wolfe8
That is one messy matrix
amistre64
  • amistre64
if i see it right :) i get \[\begin{pmatrix} cos(\theta)cos(\phi)&-sin(\theta)cos(\phi)&-sin(\phi)\\ cos(\theta)sin(\phi)+sin(\theta)&-sin(\theta)sin(\phi)+cos(\theta)&cos(\phi)\\ 0&0&0 \end{pmatrix}\]
amistre64
  • amistre64
bit that leads me to think its wrong, since the Z component would zero out ....
amistre64
  • amistre64
did I typo something ?
wolfe8
  • wolfe8
Oh dang it I got the order switched! I think I had ZY instead
amistre64
  • amistre64
ugh, my Y is off :) the cos sin parts should be in row 3 leaving row 2 with just a 1 in the middle
wolfe8
  • wolfe8
I did think of that. So let's do it again if you don't mind
amistre64
  • amistre64
\[YZ=\begin{pmatrix} cos(\phi)&0&-sin(\phi)\\ 0&1&0\\ sin(\phi)&0&cos(\phi) \end{pmatrix}\begin{pmatrix} cos(\theta)&-sin(\theta)&0\\ sin(\theta)&cos(\theta)&0\\ 0&0&1 \end{pmatrix}\]
wolfe8
  • wolfe8
Right. So because Z happened first, we put Y in front? Is that it?
amistre64
  • amistre64
Correct
amistre64
  • amistre64
cos(x) cos(y) | -cos(x) sin(y) | -sin(x) sin(y) | cos(y) | 0 cos(y) sin(x) | -sin(x) sin(y) | cos(x) the wolf spat out that for me, i used x for phi and y for theta
amistre64
  • amistre64
let me format that better cos(x) cos(y) | -cos(x) sin(y) | -sin(x) sin(y) | cos(y) | 0 cos(y) sin(x) | -sin(x) sin(y) | cos(x)
wolfe8
  • wolfe8
Yup I got that too. Thanks a bunch!
amistre64
  • amistre64
your welcome
amistre64
  • amistre64
be sure to dbl chk it with some points ....
wolfe8
  • wolfe8
I did and I think everything is okay. I will before submitting again
amistre64
  • amistre64
i think it just slightly off, but i dont have time to correct it. the Y I suggested does not base its rotation off of the NEW y position axis, but from the standard y position ... is what im thinking
wolfe8
  • wolfe8
Oh man
amistre64
  • amistre64
theres prolly an easy fix, or basing it off of the alpha, beta, gamma angles .... but ive got class starting soon. think about it and tell me what you get
wolfe8
  • wolfe8
Well figuring out how to translate it into trig. takes me awhile. I'll try to find out
amistre64
  • amistre64
the composition \(\phi(\theta)\) is equal to the composition of \(\theta(\phi)\). the hardest part is in determining the construction of \(\phi\). \(z'=z~cos(\phi)\) regardless of when we rotate for \(\phi\) |dw:1378741081193:dw|
amistre64
  • amistre64
if we do \(\phi\) to start with, the rotation into z' projects onto the original xy plne and produces a scaled version of the vector (x,y,0) |dw:1378741289481:dw|
amistre64
  • amistre64
its the (kx,ky) parts that we can then \(\theta\) into a new positions
amistre64
  • amistre64
|dw:1378743593408:dw|
amistre64
  • amistre64
\[cos~\alpha=\frac{x^2+y^2}{\sqrt{(x^2+y^2)(x^2+y^2+z^2)}}\] \[\alpha=cos^{-1}\left(\frac{x^2+y^2}{\sqrt{(x^2+y^2)(x^2+y^2+z^2)}}\right)\]
amistre64
  • amistre64
\[z'=\sqrt{x^2+y^2+z^2}~sin(\phi+\alpha)\] lol, not too sure how to pull that from a matrix tho
wolfe8
  • wolfe8
Holy! Haha alright. Thanks a bunch!

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