Problem 1.9 of DJGriffiths asks for the rotation matrix about the (1,1,1) direction.(adsbygoogle = window.adsbygoogle || []).push({});

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 isnotfrom x->x'->x''. Rather it is a rotationaroundthe (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 120^{o}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 thesolution. 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, theRmatrix 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

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# Homework Help: Rotation Matrix?

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