- #1
mjordan2nd
- 177
- 1
Homework Statement
Find the transformation matrix R that describes a rotation by 120 about an axis from the origin through the point (1,1,1). The rotation is clockwise as you look down the axis toward the origin.
Homework Equations
Rotations about the z-axis are given by
[tex]R_{z}(\alpha) = \left( \begin{array}{ccc}
cos(\alpha) & sin(\alpha) & 0 \\
-sin(\alpha) & cos(\alpha) & 0 \\
0 & 0 & 1
\end{array} \right)[/tex]
whereas rotations about the x-axis are given by
[tex]R_{x}(x) = \left( \begin{array}{ccc}
1 & 0 & 0 \\
0 & cos(x) & sin(x) \\
0 & -sin(x) & cos(x)
\end{array} \right)[/tex].
The Attempt at a Solution
My strategy in solving this problem was to rotate the coordinate system in such a way as to align the z-axis along the axis extending from the origin to (1,1,1). Once this was done, I was to rotate the system as a regular rotation in a two-dimensional x-y system.
The first rotation should be such that the x-axis is aligned perpendicular to the x-y projection of [tex]\hat{x} + \hat{y} + \hat{z}[/tex], or perpendicular to [tex]\hat{x} + \hat{y}[/tex]. This was done with a rotation about the z-axis, more specifically [tex]R_{z}(\frac{3 \pi}{4})[/tex].
I intended the second rotation to be about the x-axis to orient the z-axis as desired. Working now with primed coordinates after the previous rotation the desired axis lied in the y'-z plane. The coordinates of the original vector <1, 1, 1> in the primed system was [tex]\sqrt{2} \hat{y} + \hat{z}[/tex]. Therefore, I wanted to rotate the x-axis clockwise by [tex]Cos^{-1}(\frac{1}{\sqrt{3}})[/tex] degress. However, the way my matrices in section 2 were set up should have all rotations going counterclockwise, so I wanted my rotation matrix to be [tex]R_{x}(2 \pi - Cos^{-1}(\frac{1}{\sqrt{3}}))[/tex].
Now that the z-axis was properly aligned, I could rotate about it, so my final rotation matrix should be [tex]R_{z}(\frac{2 \pi}{3})[/tex].
If my logic is correct then the final rotation should be
[tex]R = R_{z}(\frac{2 \pi}{3}) * R_{x}(2 \pi - Cos^{-1}(\frac{1}{\sqrt{3}})) * R_{z}(\frac{3 \pi}{4})[/tex].
That said, I know my answer should be
[tex]R = \left( \begin{array}{ccc}
0 & 0 & 1 \\
1 & 0 & 0 \\
0 & 1 & 0
\end{array} \right)[/tex]
however this is not what I am getting. I am getting something very messy. Where have I gone wrong?