- #1
Living_Dog
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Problem 1.9 of DJGriffiths asks for the rotation matrix about the (1,1,1) direction.
I thought I could rotate about z 45 degrees (R': x -> x'), then rotate about y' (R'': x' -> x''). How do I combine the two rotations to determine the final single rotation matrix... R = R''*R' or R = R'*R'' ?
thx in advance!
-LD
EDIT: after working on this some more, I realized that this is not the same rotation. The rotation is not from x->x'->x''. Rather it is a rotation around the (1,1,1) direction.
Having checked what few texts I have I still do not know how to make the rotation matrix.
After some thought ...wouldn't x go to the z position to satisfy this rotation? I'm thinking that if you look down the (1,1,1) line toward the origin, then the 3 axes make a 120o angle with each other. Thus the x-axis will spin into the original z-axis position. Also, z->y and y->x.
tia!
-LD
EDIT: after working on this some more, I found the solution. Perhaps it is a bit pragmatic since I used the above fact. So if anyone knows how to do this in general - for any direction - then I would like the answer. (I think I need a book on crystallographic rotations - yikes!)
100 -> 001
010 -> 100
001 -> 010
Now apply these conditions to the equation: A' = R*A. Then each equation will result in the immediate solution for 3 of the rotation matrix elements.
(0 0 1) = R (1 0 0) yields:
[tex]R_{11} = 0[/tex]
[tex]R_{21} = 0[/tex]
[tex]R_{31} = 1[/tex]
(1 0 0) = R (0 1 0) yields:
[tex]R_{12} = 1[/tex]
[tex]R_{22} = 0[/tex]
[tex]R_{32} = 0[/tex]
(0 1 0) = R (0 0 1) yields:
[tex]R_{13} = 0[/tex]
[tex]R_{23} = 0[/tex]
[tex]R_{33} = 1[/tex]
Thus, the R matrix is:
[tex]
\left( \begin{array}{ccc}
0 & 1 & 0 \\
0 & 0 & 1 \\
1 & 0 & 0
\end{array} \right)
[/tex]
...sorry I asked this and then found the solution... but as I said above, this is very pragmatic and I would really like to know how to do this for any angle of rotation about any direction in general.
Thanks!
-LD
I thought I could rotate about z 45 degrees (R': x -> x'), then rotate about y' (R'': x' -> x''). How do I combine the two rotations to determine the final single rotation matrix... R = R''*R' or R = R'*R'' ?
thx in advance!
-LD
EDIT: after working on this some more, I realized that this is not the same rotation. The rotation is not from x->x'->x''. Rather it is a rotation around the (1,1,1) direction.
Having checked what few texts I have I still do not know how to make the rotation matrix.
After some thought ...wouldn't x go to the z position to satisfy this rotation? I'm thinking that if you look down the (1,1,1) line toward the origin, then the 3 axes make a 120o angle with each other. Thus the x-axis will spin into the original z-axis position. Also, z->y and y->x.
tia!
-LD
EDIT: after working on this some more, I found the solution. Perhaps it is a bit pragmatic since I used the above fact. So if anyone knows how to do this in general - for any direction - then I would like the answer. (I think I need a book on crystallographic rotations - yikes!)
100 -> 001
010 -> 100
001 -> 010
Now apply these conditions to the equation: A' = R*A. Then each equation will result in the immediate solution for 3 of the rotation matrix elements.
(0 0 1) = R (1 0 0) yields:
[tex]R_{11} = 0[/tex]
[tex]R_{21} = 0[/tex]
[tex]R_{31} = 1[/tex]
(1 0 0) = R (0 1 0) yields:
[tex]R_{12} = 1[/tex]
[tex]R_{22} = 0[/tex]
[tex]R_{32} = 0[/tex]
(0 1 0) = R (0 0 1) yields:
[tex]R_{13} = 0[/tex]
[tex]R_{23} = 0[/tex]
[tex]R_{33} = 1[/tex]
Thus, the R matrix is:
[tex]
\left( \begin{array}{ccc}
0 & 1 & 0 \\
0 & 0 & 1 \\
1 & 0 & 0
\end{array} \right)
[/tex]
...sorry I asked this and then found the solution... but as I said above, this is very pragmatic and I would really like to know how to do this for any angle of rotation about any direction in general.
Thanks!
-LD
Last edited: