Why does the following image equal what it equals?
Why does x,y,z equal that? 1/sqrt(3),1/sqrt(3),1/sqrt(3)
Asked
Active
Viewed 6,165 times
1 Answers
0
You can solve this by a direct application of the Rodrigues' rotation formula.
In order to apply the formula, we need a unit vector along the rotation axis, $(1,1,1)$. In other words, we need a vector $\mathbf{k}$ which is parallel to $(1,1,1)$, but with the property that $|\mathbf{k}|=1$. This is the vector $(x,y,z)$ in the solution; note that \begin{align} \left|\left(\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}}\right)\right| = \sqrt{\left(\frac{1}{\sqrt{3}}\right)^2+\left(\frac{1}{\sqrt{3}}\right)^2+\left(\frac{1}{\sqrt{3}}\right)^2}=\sqrt{\frac{1}{3}+\frac{1}{3}+\frac{1}{3}}=\sqrt{1}=1. \end{align}
If we denote the aforementioned unit vector by $\mathbf{k}$, and the rotation matrix which we are trying to find by $\mathbf{R}$, we have by Rodrigues' formula that \begin{align} \mathbf{R} = \mathbf{I} + (\sin\theta) \mathbf{K} + (1-\cos\theta)\mathbf{K}^2, \end{align} where \begin{align} \mathbf{K} = \left(\matrix{ 0 & -z & y \\ z & 0 & -x \\ -y & x & 0}\right), \end{align} and $\mathbf{I}$ denotes the identity matrix as usual.
Morten
- 537