Mixing
Transformation matrix for rotation about arbitrary axis
$begingroup$
Specifically, I don't know what approach to take in answering Griffiths' question 1.9 in his introduction to electrodynamics:
Find the transformation matrix R that describes a rotation by 120
degrees about an axis from the origin through the point $(1,1,1)$. The
rotation is clockwise as you look down the axis towards the origin.
From Glen Murray's page on rotations, the supposed approach to take is successively rotate space so that the rotation axis sits along the z axis $T$, :
$$T =
begin{pmatrix} 1&0&0 \ 0& cos{alpha} & -sin{alpha} \ 0&sin{alpha} & cos{alpha}
end{pmatrix} cdot begin{pmatrix} cos{beta} & 0 & sin{beta} \ 0 &1&0 \ -sin{beta} &0 & cos{beta} end{pmatrix}
$$
perform the rotation $theta$,:
begin{pmatrix} cos{theta} & -sin{theta} &0 \ sin{theta} & cos{theta}&0 \ 0&0&1
end{pmatrix}
and rotate space back to its original orientation $T^{-1}$.
$T$ first rotates space such that the rotation axis sits in the xz plane. Secondly, it rotates space such that the rotation axis lies along the z axis.
This approach seems overly tedious, as this is an introductory question in an introductory chapter. Am I missing something here?
Where I to proceed in this approach, for the arbitrary axis from the origin through the point $(a,b,c)$, then I would need to derive the angles $alpha$ and $beta$ as follows.
$alpha = arctan{frac{b}{a}}$
$beta = arctan{frac{acos{alpha} +bsin{alpha}}{c}}$
linear-algebra linear-transformations rotations
asked Jan 28 at 9:16
Mussé RediMussé Redi
735826
$endgroup$
add a comment |
$begingroup$
Specifically, I don't know what approach to take in answering Griffiths' question 1.9 in his introduction to electrodynamics:
Find the transformation matrix R that describes a rotation by 120
degrees about an axis from the origin through the point $(1,1,1)$. The
rotation is clockwise as you look down the axis towards the origin.
From Glen Murray's page on rotations, the supposed approach to take is successively rotate space so that the rotation axis sits along the z axis $T$, :
$$T =
begin{pmatrix} 1&0&0 \ 0& cos{alpha} & -sin{alpha} \ 0&sin{alpha} & cos{alpha}
end{pmatrix} cdot begin{pmatrix} cos{beta} & 0 & sin{beta} \ 0 &1&0 \ -sin{beta} &0 & cos{beta} end{pmatrix}
$$
perform the rotation $theta$,:
begin{pmatrix} cos{theta} & -sin{theta} &0 \ sin{theta} & cos{theta}&0 \ 0&0&1
end{pmatrix}
and rotate space back to its original orientation $T^{-1}$.
$T$ first rotates space such that the rotation axis sits in the xz plane. Secondly, it rotates space such that the rotation axis lies along the z axis.
This approach seems overly tedious, as this is an introductory question in an introductory chapter. Am I missing something here?
Where I to proceed in this approach, for the arbitrary axis from the origin through the point $(a,b,c)$, then I would need to derive the angles $alpha$ and $beta$ as follows.
$alpha = arctan{frac{b}{a}}$
$beta = arctan{frac{acos{alpha} +bsin{alpha}}{c}}$
linear-algebra linear-transformations rotations
asked Jan 28 at 9:16
Mussé RediMussé Redi
735826
$endgroup$
add a comment |
$begingroup$
Specifically, I don't know what approach to take in answering Griffiths' question 1.9 in his introduction to electrodynamics:
Find the transformation matrix R that describes a rotation by 120
degrees about an axis from the origin through the point $(1,1,1)$. The
rotation is clockwise as you look down the axis towards the origin.
From Glen Murray's page on rotations, the supposed approach to take is successively rotate space so that the rotation axis sits along the z axis $T$, :
$$T =
begin{pmatrix} 1&0&0 \ 0& cos{alpha} & -sin{alpha} \ 0&sin{alpha} & cos{alpha}
end{pmatrix} cdot begin{pmatrix} cos{beta} & 0 & sin{beta} \ 0 &1&0 \ -sin{beta} &0 & cos{beta} end{pmatrix}
$$
perform the rotation $theta$,:
begin{pmatrix} cos{theta} & -sin{theta} &0 \ sin{theta} & cos{theta}&0 \ 0&0&1
end{pmatrix}
and rotate space back to its original orientation $T^{-1}$.
$T$ first rotates space such that the rotation axis sits in the xz plane. Secondly, it rotates space such that the rotation axis lies along the z axis.
This approach seems overly tedious, as this is an introductory question in an introductory chapter. Am I missing something here?
Where I to proceed in this approach, for the arbitrary axis from the origin through the point $(a,b,c)$, then I would need to derive the angles $alpha$ and $beta$ as follows.
$alpha = arctan{frac{b}{a}}$
$beta = arctan{frac{acos{alpha} +bsin{alpha}}{c}}$
linear-algebra linear-transformations rotations
asked Jan 28 at 9:16
Mussé RediMussé Redi
735826
$endgroup$
Specifically, I don't know what approach to take in answering Griffiths' question 1.9 in his introduction to electrodynamics:
Find the transformation matrix R that describes a rotation by 120
degrees about an axis from the origin through the point $(1,1,1)$. The
rotation is clockwise as you look down the axis towards the origin.
From Glen Murray's page on rotations, the supposed approach to take is successively rotate space so that the rotation axis sits along the z axis $T$, :
$$T =
begin{pmatrix} 1&0&0 \ 0& cos{alpha} & -sin{alpha} \ 0&sin{alpha} & cos{alpha}
end{pmatrix} cdot begin{pmatrix} cos{beta} & 0 & sin{beta} \ 0 &1&0 \ -sin{beta} &0 & cos{beta} end{pmatrix}
$$
perform the rotation $theta$,:
begin{pmatrix} cos{theta} & -sin{theta} &0 \ sin{theta} & cos{theta}&0 \ 0&0&1
end{pmatrix}
and rotate space back to its original orientation $T^{-1}$.
$T$ first rotates space such that the rotation axis sits in the xz plane. Secondly, it rotates space such that the rotation axis lies along the z axis.
This approach seems overly tedious, as this is an introductory question in an introductory chapter. Am I missing something here?
Where I to proceed in this approach, for the arbitrary axis from the origin through the point $(a,b,c)$, then I would need to derive the angles $alpha$ and $beta$ as follows.
$alpha = arctan{frac{b}{a}}$
$beta = arctan{frac{acos{alpha} +bsin{alpha}}{c}}$
linear-algebra linear-transformations rotations
linear-algebra linear-transformations rotations
asked Jan 28 at 9:16
Mussé RediMussé Redi
735826
asked Jan 28 at 9:16
Mussé RediMussé Redi
735826
asked Jan 28 at 9:16
Mussé RediMussé Redi
735826
asked Jan 28 at 9:16
Mussé RediMussé Redi
735826
asked Jan 28 at 9:16
Mussé RediMussé Redi
735826
735826
add a comment |
add a comment |
2 Answers
2
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oldest
votes
$begingroup$
Hint:
Looking down the axis toward the origin, you see the three axis forming angles of $120°$. This rotation is just a circular permutation of the axis.
answered Jan 28 at 9:23
Yves DaoustYves Daoust
131k676229
$endgroup$
$begingroup$
Could you elaborate? I see that you mean the x,y and z axes forming the angles, only how to derive this algebraically?
$endgroup$
– Mussé Redi
Jan 28 at 9:28
$begingroup$
@MusséRedi: form an orthogonal frame with $z'$ in the direction $(1,1,1)$ and find the matrix of the change of basis. Then project the axis on the plane $x'y'$.
$endgroup$
– Yves Daoust
Jan 28 at 9:38
$begingroup$
How do you do that?
$endgroup$
– Mussé Redi
Jan 28 at 9:41
$begingroup$
How do you see the three axes forming angles of $120^circ$, looking down the axis toward the origin?
$endgroup$
– Mussé Redi
Jan 28 at 9:50
add a comment |
$begingroup$
You could also just form an orthogonal (change of basis) matrix, $R$, with $frac1{sqrt{a^2+b^2+c^2}}(a,b,c)$ as the last row, to take the $z$-axis to $(a,b,c)$.
Then if $S$ is the rotation about the $z$-axis, our $T$ would be $T=R^tSR$.
See this example.
answered Jan 28 at 10:55
Chris CusterChris Custer
14.2k3827
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Hint:
Looking down the axis toward the origin, you see the three axis forming angles of $120°$. This rotation is just a circular permutation of the axis.
answered Jan 28 at 9:23
Yves DaoustYves Daoust
131k676229
$endgroup$
$begingroup$
Could you elaborate? I see that you mean the x,y and z axes forming the angles, only how to derive this algebraically?
$endgroup$
– Mussé Redi
Jan 28 at 9:28
$begingroup$
@MusséRedi: form an orthogonal frame with $z'$ in the direction $(1,1,1)$ and find the matrix of the change of basis. Then project the axis on the plane $x'y'$.
$endgroup$
– Yves Daoust
Jan 28 at 9:38
$begingroup$
How do you do that?
$endgroup$
– Mussé Redi
Jan 28 at 9:41
$begingroup$
How do you see the three axes forming angles of $120^circ$, looking down the axis toward the origin?
$endgroup$
– Mussé Redi
Jan 28 at 9:50
add a comment |
$begingroup$
Hint:
Looking down the axis toward the origin, you see the three axis forming angles of $120°$. This rotation is just a circular permutation of the axis.
answered Jan 28 at 9:23
Yves DaoustYves Daoust
131k676229
$endgroup$
$begingroup$
Could you elaborate? I see that you mean the x,y and z axes forming the angles, only how to derive this algebraically?
$endgroup$
– Mussé Redi
Jan 28 at 9:28
$begingroup$
@MusséRedi: form an orthogonal frame with $z'$ in the direction $(1,1,1)$ and find the matrix of the change of basis. Then project the axis on the plane $x'y'$.
$endgroup$
– Yves Daoust
Jan 28 at 9:38
$begingroup$
How do you do that?
$endgroup$
– Mussé Redi
Jan 28 at 9:41
$begingroup$
How do you see the three axes forming angles of $120^circ$, looking down the axis toward the origin?
$endgroup$
– Mussé Redi
Jan 28 at 9:50
add a comment |
$begingroup$
Hint:
Looking down the axis toward the origin, you see the three axis forming angles of $120°$. This rotation is just a circular permutation of the axis.
answered Jan 28 at 9:23
Yves DaoustYves Daoust
131k676229
$endgroup$
Hint:
Looking down the axis toward the origin, you see the three axis forming angles of $120°$. This rotation is just a circular permutation of the axis.
answered Jan 28 at 9:23
Yves DaoustYves Daoust
131k676229
answered Jan 28 at 9:23
Yves DaoustYves Daoust
131k676229
answered Jan 28 at 9:23
Yves DaoustYves Daoust
131k676229
answered Jan 28 at 9:23
Yves DaoustYves Daoust
131k676229
131k676229
$begingroup$
Could you elaborate? I see that you mean the x,y and z axes forming the angles, only how to derive this algebraically?
$endgroup$
– Mussé Redi
Jan 28 at 9:28
$begingroup$
@MusséRedi: form an orthogonal frame with $z'$ in the direction $(1,1,1)$ and find the matrix of the change of basis. Then project the axis on the plane $x'y'$.
$endgroup$
– Yves Daoust
Jan 28 at 9:38
$begingroup$
How do you do that?
$endgroup$
– Mussé Redi
Jan 28 at 9:41
$begingroup$
How do you see the three axes forming angles of $120^circ$, looking down the axis toward the origin?
$endgroup$
– Mussé Redi
Jan 28 at 9:50
add a comment |
$begingroup$
Could you elaborate? I see that you mean the x,y and z axes forming the angles, only how to derive this algebraically?
$endgroup$
– Mussé Redi
Jan 28 at 9:28
$begingroup$
@MusséRedi: form an orthogonal frame with $z'$ in the direction $(1,1,1)$ and find the matrix of the change of basis. Then project the axis on the plane $x'y'$.
$endgroup$
– Yves Daoust
Jan 28 at 9:38
$begingroup$
How do you do that?
$endgroup$
– Mussé Redi
Jan 28 at 9:41
$begingroup$
How do you see the three axes forming angles of $120^circ$, looking down the axis toward the origin?
$endgroup$
– Mussé Redi
Jan 28 at 9:50
$begingroup$
Could you elaborate? I see that you mean the x,y and z axes forming the angles, only how to derive this algebraically?
$endgroup$
– Mussé Redi
Jan 28 at 9:28
$begingroup$
Could you elaborate? I see that you mean the x,y and z axes forming the angles, only how to derive this algebraically?
$endgroup$
– Mussé Redi
Jan 28 at 9:28
$begingroup$
@MusséRedi: form an orthogonal frame with $z'$ in the direction $(1,1,1)$ and find the matrix of the change of basis. Then project the axis on the plane $x'y'$.
$endgroup$
– Yves Daoust
Jan 28 at 9:38
$begingroup$
@MusséRedi: form an orthogonal frame with $z'$ in the direction $(1,1,1)$ and find the matrix of the change of basis. Then project the axis on the plane $x'y'$.
$endgroup$
– Yves Daoust
Jan 28 at 9:38
$begingroup$
How do you do that?
$endgroup$
– Mussé Redi
Jan 28 at 9:41
$begingroup$
How do you do that?
$endgroup$
– Mussé Redi
Jan 28 at 9:41
$begingroup$
How do you see the three axes forming angles of $120^circ$, looking down the axis toward the origin?
$endgroup$
– Mussé Redi
Jan 28 at 9:50
$begingroup$
How do you see the three axes forming angles of $120^circ$, looking down the axis toward the origin?
$endgroup$
– Mussé Redi
Jan 28 at 9:50
add a comment |
$begingroup$
You could also just form an orthogonal (change of basis) matrix, $R$, with $frac1{sqrt{a^2+b^2+c^2}}(a,b,c)$ as the last row, to take the $z$-axis to $(a,b,c)$.
Then if $S$ is the rotation about the $z$-axis, our $T$ would be $T=R^tSR$.
See this example.
answered Jan 28 at 10:55
Chris CusterChris Custer
14.2k3827
$endgroup$
add a comment |
$begingroup$
You could also just form an orthogonal (change of basis) matrix, $R$, with $frac1{sqrt{a^2+b^2+c^2}}(a,b,c)$ as the last row, to take the $z$-axis to $(a,b,c)$.
Then if $S$ is the rotation about the $z$-axis, our $T$ would be $T=R^tSR$.
See this example.
answered Jan 28 at 10:55
Chris CusterChris Custer
14.2k3827
$endgroup$
add a comment |
$begingroup$
You could also just form an orthogonal (change of basis) matrix, $R$, with $frac1{sqrt{a^2+b^2+c^2}}(a,b,c)$ as the last row, to take the $z$-axis to $(a,b,c)$.
Then if $S$ is the rotation about the $z$-axis, our $T$ would be $T=R^tSR$.
See this example.
answered Jan 28 at 10:55
Chris CusterChris Custer
14.2k3827
$endgroup$
You could also just form an orthogonal (change of basis) matrix, $R$, with $frac1{sqrt{a^2+b^2+c^2}}(a,b,c)$ as the last row, to take the $z$-axis to $(a,b,c)$.
Then if $S$ is the rotation about the $z$-axis, our $T$ would be $T=R^tSR$.
See this example.
answered Jan 28 at 10:55
Chris CusterChris Custer
14.2k3827
edited Jan 28 at 11:07
answered Jan 28 at 10:55
Chris CusterChris Custer
14.2k3827
answered Jan 28 at 10:55
Chris CusterChris Custer
14.2k3827
answered Jan 28 at 10:55
Chris CusterChris Custer
14.2k3827
14.2k3827
add a comment |
add a comment |
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