Mixing
3d matrix rotation
$begingroup$
I've been reading up on rotation transformation and following this PowerPoint.
The assignment I'm working on is asking to perform a rotation and find the image $Q$ of the point P = (1, 2, -1) after a $45$ degree $y$-roll. I was under the impression that a roll was on the $z$-axis; which is where I'm confused.
I'm trying to set up the problem like such (from WikiPedia):
But I'm lost as to translate the 3 points into a matrix. Is this correct? The way I'm trying to visualize it as below without the translation.
matrices rotations
edited Jan 13 at 10:09
Glorfindel
3,41981830
asked Apr 4 '13 at 20:04
KermitKermit
140116
$endgroup$
|
show 3 more comments
$begingroup$
I've been reading up on rotation transformation and following this PowerPoint.
The assignment I'm working on is asking to perform a rotation and find the image $Q$ of the point P = (1, 2, -1) after a $45$ degree $y$-roll. I was under the impression that a roll was on the $z$-axis; which is where I'm confused.
I'm trying to set up the problem like such (from WikiPedia):
But I'm lost as to translate the 3 points into a matrix. Is this correct? The way I'm trying to visualize it as below without the translation.
matrices rotations
edited Jan 13 at 10:09
Glorfindel
3,41981830
asked Apr 4 '13 at 20:04
KermitKermit
140116
$endgroup$
$begingroup$
I think you are reading too much meaning into the word 'roll'. The $y$ coordinate remains unchanged, the $x,z$ coordinates are rotated in the $x,z$ plane by $frac{pi}{4}$.
$endgroup$
– copper.hat
Apr 4 '13 at 20:16
$begingroup$
@copper.hat Would this be an accurate representation?
$endgroup$
– Kermit
Apr 4 '13 at 20:19
$begingroup$
Looks good... ${}{}{}$
$endgroup$
– copper.hat
Apr 4 '13 at 20:23
$begingroup$
I'm with copper.hat about the interpretation of "roll". It seems like it is just referring to "twisting" around a specified axis, and it doesn't prefer any special axis.
$endgroup$
– rschwieb
Apr 4 '13 at 20:23
1
$begingroup$
Not sure what you are talking about when you mean "three points into a matrix". We've got one point and one transformation matrix. The image of the point under the transformation is again a single point: the image of $P$ under the rotation.
$endgroup$
– rschwieb
Apr 4 '13 at 20:32
|
show 3 more comments
$begingroup$
I've been reading up on rotation transformation and following this PowerPoint.
The assignment I'm working on is asking to perform a rotation and find the image $Q$ of the point P = (1, 2, -1) after a $45$ degree $y$-roll. I was under the impression that a roll was on the $z$-axis; which is where I'm confused.
I'm trying to set up the problem like such (from WikiPedia):
But I'm lost as to translate the 3 points into a matrix. Is this correct? The way I'm trying to visualize it as below without the translation.
matrices rotations
edited Jan 13 at 10:09
Glorfindel
3,41981830
asked Apr 4 '13 at 20:04
KermitKermit
140116
$endgroup$
I've been reading up on rotation transformation and following this PowerPoint.
The assignment I'm working on is asking to perform a rotation and find the image $Q$ of the point P = (1, 2, -1) after a $45$ degree $y$-roll. I was under the impression that a roll was on the $z$-axis; which is where I'm confused.
I'm trying to set up the problem like such (from WikiPedia):
But I'm lost as to translate the 3 points into a matrix. Is this correct? The way I'm trying to visualize it as below without the translation.
matrices rotations
matrices rotations
edited Jan 13 at 10:09
Glorfindel
3,41981830
asked Apr 4 '13 at 20:04
KermitKermit
140116
edited Jan 13 at 10:09
Glorfindel
3,41981830
asked Apr 4 '13 at 20:04
KermitKermit
140116
edited Jan 13 at 10:09
Glorfindel
3,41981830
edited Jan 13 at 10:09
Glorfindel
3,41981830
edited Jan 13 at 10:09
Glorfindel
3,41981830
3,41981830
asked Apr 4 '13 at 20:04
KermitKermit
140116
asked Apr 4 '13 at 20:04
KermitKermit
140116
asked Apr 4 '13 at 20:04
KermitKermit
140116
140116
$begingroup$
I think you are reading too much meaning into the word 'roll'. The $y$ coordinate remains unchanged, the $x,z$ coordinates are rotated in the $x,z$ plane by $frac{pi}{4}$.
$endgroup$
– copper.hat
Apr 4 '13 at 20:16
$begingroup$
@copper.hat Would this be an accurate representation?
$endgroup$
– Kermit
Apr 4 '13 at 20:19
$begingroup$
Looks good... ${}{}{}$
$endgroup$
– copper.hat
Apr 4 '13 at 20:23
$begingroup$
I'm with copper.hat about the interpretation of "roll". It seems like it is just referring to "twisting" around a specified axis, and it doesn't prefer any special axis.
$endgroup$
– rschwieb
Apr 4 '13 at 20:23
1
$begingroup$
Not sure what you are talking about when you mean "three points into a matrix". We've got one point and one transformation matrix. The image of the point under the transformation is again a single point: the image of $P$ under the rotation.
$endgroup$
– rschwieb
Apr 4 '13 at 20:32
|
show 3 more comments
$begingroup$
I think you are reading too much meaning into the word 'roll'. The $y$ coordinate remains unchanged, the $x,z$ coordinates are rotated in the $x,z$ plane by $frac{pi}{4}$.
$endgroup$
– copper.hat
Apr 4 '13 at 20:16
$begingroup$
@copper.hat Would this be an accurate representation?
$endgroup$
– Kermit
Apr 4 '13 at 20:19
$begingroup$
Looks good... ${}{}{}$
$endgroup$
– copper.hat
Apr 4 '13 at 20:23
$begingroup$
I'm with copper.hat about the interpretation of "roll". It seems like it is just referring to "twisting" around a specified axis, and it doesn't prefer any special axis.
$endgroup$
– rschwieb
Apr 4 '13 at 20:23
1
$begingroup$
Not sure what you are talking about when you mean "three points into a matrix". We've got one point and one transformation matrix. The image of the point under the transformation is again a single point: the image of $P$ under the rotation.
$endgroup$
– rschwieb
Apr 4 '13 at 20:32
$begingroup$
I think you are reading too much meaning into the word 'roll'. The $y$ coordinate remains unchanged, the $x,z$ coordinates are rotated in the $x,z$ plane by $frac{pi}{4}$.
$endgroup$
– copper.hat
Apr 4 '13 at 20:16
$begingroup$
I think you are reading too much meaning into the word 'roll'. The $y$ coordinate remains unchanged, the $x,z$ coordinates are rotated in the $x,z$ plane by $frac{pi}{4}$.
$endgroup$
– copper.hat
Apr 4 '13 at 20:16
$begingroup$
@copper.hat Would this be an accurate representation?
$endgroup$
– Kermit
Apr 4 '13 at 20:19
$begingroup$
@copper.hat Would this be an accurate representation?
$endgroup$
– Kermit
Apr 4 '13 at 20:19
$begingroup$
Looks good... ${}{}{}$
$endgroup$
– copper.hat
Apr 4 '13 at 20:23
$begingroup$
Looks good... ${}{}{}$
$endgroup$
– copper.hat
Apr 4 '13 at 20:23
$begingroup$
I'm with copper.hat about the interpretation of "roll". It seems like it is just referring to "twisting" around a specified axis, and it doesn't prefer any special axis.
$endgroup$
– rschwieb
Apr 4 '13 at 20:23
$begingroup$
I'm with copper.hat about the interpretation of "roll". It seems like it is just referring to "twisting" around a specified axis, and it doesn't prefer any special axis.
$endgroup$
– rschwieb
Apr 4 '13 at 20:23
1
1
$begingroup$
Not sure what you are talking about when you mean "three points into a matrix". We've got one point and one transformation matrix. The image of the point under the transformation is again a single point: the image of $P$ under the rotation.
$endgroup$
– rschwieb
Apr 4 '13 at 20:32
$begingroup$
Not sure what you are talking about when you mean "three points into a matrix". We've got one point and one transformation matrix. The image of the point under the transformation is again a single point: the image of $P$ under the rotation.
$endgroup$
– rschwieb
Apr 4 '13 at 20:32
|
show 3 more comments
1 Answer
1
active
oldest
votes
$begingroup$
To roll $theta=pi/4$ around the $y$ axis, the matrix becomes:
$$begin{bmatrix}sqrt{2}/2&0&sqrt{2}/2\0&1&0\-sqrt{2}/2&0&sqrt{2}/2end{bmatrix}$$
Applying this to the point:
$$begin{bmatrix}sqrt{2}/2&0&sqrt{2}/2\0&1&0\-sqrt{2}/2&0&sqrt{2}/2end{bmatrix}begin{bmatrix}1\2\-1end{bmatrix}=begin{bmatrix}0\2\-sqrt{2}end{bmatrix}$$
So, $Q=(0,2,-sqrt{2})$
answered Apr 4 '13 at 20:31
rschwiebrschwieb
106k12102250
$endgroup$
$begingroup$
What I meant in my comment was just what you did in the "applying this point."
$endgroup$
– Kermit
Apr 4 '13 at 20:35
1
$begingroup$
@FreshPrinceOfSO OK :) Like the name, btw.
$endgroup$
– rschwieb
Apr 4 '13 at 20:39
add a comment |
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1 Answer
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$begingroup$
To roll $theta=pi/4$ around the $y$ axis, the matrix becomes:
$$begin{bmatrix}sqrt{2}/2&0&sqrt{2}/2\0&1&0\-sqrt{2}/2&0&sqrt{2}/2end{bmatrix}$$
Applying this to the point:
$$begin{bmatrix}sqrt{2}/2&0&sqrt{2}/2\0&1&0\-sqrt{2}/2&0&sqrt{2}/2end{bmatrix}begin{bmatrix}1\2\-1end{bmatrix}=begin{bmatrix}0\2\-sqrt{2}end{bmatrix}$$
So, $Q=(0,2,-sqrt{2})$
answered Apr 4 '13 at 20:31
rschwiebrschwieb
106k12102250
$endgroup$
$begingroup$
What I meant in my comment was just what you did in the "applying this point."
$endgroup$
– Kermit
Apr 4 '13 at 20:35
1
$begingroup$
@FreshPrinceOfSO OK :) Like the name, btw.
$endgroup$
– rschwieb
Apr 4 '13 at 20:39
add a comment |
$begingroup$
To roll $theta=pi/4$ around the $y$ axis, the matrix becomes:
$$begin{bmatrix}sqrt{2}/2&0&sqrt{2}/2\0&1&0\-sqrt{2}/2&0&sqrt{2}/2end{bmatrix}$$
Applying this to the point:
$$begin{bmatrix}sqrt{2}/2&0&sqrt{2}/2\0&1&0\-sqrt{2}/2&0&sqrt{2}/2end{bmatrix}begin{bmatrix}1\2\-1end{bmatrix}=begin{bmatrix}0\2\-sqrt{2}end{bmatrix}$$
So, $Q=(0,2,-sqrt{2})$
answered Apr 4 '13 at 20:31
rschwiebrschwieb
106k12102250
$endgroup$
$begingroup$
What I meant in my comment was just what you did in the "applying this point."
$endgroup$
– Kermit
Apr 4 '13 at 20:35
1
$begingroup$
@FreshPrinceOfSO OK :) Like the name, btw.
$endgroup$
– rschwieb
Apr 4 '13 at 20:39
add a comment |
$begingroup$
To roll $theta=pi/4$ around the $y$ axis, the matrix becomes:
$$begin{bmatrix}sqrt{2}/2&0&sqrt{2}/2\0&1&0\-sqrt{2}/2&0&sqrt{2}/2end{bmatrix}$$
Applying this to the point:
$$begin{bmatrix}sqrt{2}/2&0&sqrt{2}/2\0&1&0\-sqrt{2}/2&0&sqrt{2}/2end{bmatrix}begin{bmatrix}1\2\-1end{bmatrix}=begin{bmatrix}0\2\-sqrt{2}end{bmatrix}$$
So, $Q=(0,2,-sqrt{2})$
answered Apr 4 '13 at 20:31
rschwiebrschwieb
106k12102250
$endgroup$
To roll $theta=pi/4$ around the $y$ axis, the matrix becomes:
$$begin{bmatrix}sqrt{2}/2&0&sqrt{2}/2\0&1&0\-sqrt{2}/2&0&sqrt{2}/2end{bmatrix}$$
Applying this to the point:
$$begin{bmatrix}sqrt{2}/2&0&sqrt{2}/2\0&1&0\-sqrt{2}/2&0&sqrt{2}/2end{bmatrix}begin{bmatrix}1\2\-1end{bmatrix}=begin{bmatrix}0\2\-sqrt{2}end{bmatrix}$$
So, $Q=(0,2,-sqrt{2})$
answered Apr 4 '13 at 20:31
rschwiebrschwieb
106k12102250
answered Apr 4 '13 at 20:31
rschwiebrschwieb
106k12102250
answered Apr 4 '13 at 20:31
rschwiebrschwieb
106k12102250
answered Apr 4 '13 at 20:31
rschwiebrschwieb
106k12102250
106k12102250
$begingroup$
What I meant in my comment was just what you did in the "applying this point."
$endgroup$
– Kermit
Apr 4 '13 at 20:35
1
$begingroup$
@FreshPrinceOfSO OK :) Like the name, btw.
$endgroup$
– rschwieb
Apr 4 '13 at 20:39
add a comment |
$begingroup$
What I meant in my comment was just what you did in the "applying this point."
$endgroup$
– Kermit
Apr 4 '13 at 20:35
1
$begingroup$
@FreshPrinceOfSO OK :) Like the name, btw.
$endgroup$
– rschwieb
Apr 4 '13 at 20:39
$begingroup$
What I meant in my comment was just what you did in the "applying this point."
$endgroup$
– Kermit
Apr 4 '13 at 20:35
$begingroup$
What I meant in my comment was just what you did in the "applying this point."
$endgroup$
– Kermit
Apr 4 '13 at 20:35
1
1
$begingroup$
@FreshPrinceOfSO OK :) Like the name, btw.
$endgroup$
– rschwieb
Apr 4 '13 at 20:39
$begingroup$
@FreshPrinceOfSO OK :) Like the name, btw.
$endgroup$
– rschwieb
Apr 4 '13 at 20:39
add a comment |
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$begingroup$
I think you are reading too much meaning into the word 'roll'. The $y$ coordinate remains unchanged, the $x,z$ coordinates are rotated in the $x,z$ plane by $frac{pi}{4}$.
$endgroup$
– copper.hat
Apr 4 '13 at 20:16
$begingroup$
@copper.hat Would this be an accurate representation?
$endgroup$
– Kermit
Apr 4 '13 at 20:19
$begingroup$
Looks good... ${}{}{}$
$endgroup$
– copper.hat
Apr 4 '13 at 20:23
$begingroup$
I'm with copper.hat about the interpretation of "roll". It seems like it is just referring to "twisting" around a specified axis, and it doesn't prefer any special axis.
$endgroup$
– rschwieb
Apr 4 '13 at 20:23
1
$begingroup$
Not sure what you are talking about when you mean "three points into a matrix". We've got one point and one transformation matrix. The image of the point under the transformation is again a single point: the image of $P$ under the rotation.
$endgroup$
– rschwieb
Apr 4 '13 at 20:32