The question spurred in my mind when I was asked the following:

Find the transformation matrix T that describes a rotation by $120^\circ$ 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.

I imagined the rotation of the identity matrix $\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{bmatrix} $ and wrote the answer as $\begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0 \\ \end{bmatrix}$ through intuition. But, I would have had an issue with problem if the axis would have passed through the origin and an arbitrary point $(\alpha, \beta, \gamma)$.

I have a proposition: We look towards the origin through the given axis of rotation and imagine a plane passing through origin and perpendicular to this axis. Then, we take the horizontal components of the axis unit vectors on the plane, while their vertical components lie on the axis. Now that we have three components on the plane, we rotate them through the desired angle $\phi$. After rotation, we get the original vectors "back" by adding the planer components and their corresponding axial components vectorially.

Since the original problem had a lot of symmetry associated with the point $(1,1,1)$, I was able to solve the problem "in the head". But this scheme might turn out to be cumbersome for an arbitrary axis. Do you know any other simpler way to tackle the problem?

Thank you for your time. :-)

I believe my answer is not in correspondence with that of the formula here.

Could you tell what is wrong with my approach? Thanks.

  • $\begingroup$ Pick two vectors $u,v$ in the plane orthogonal to $(1,1,1)$ separated by $60^\circ$. Then the matrix you are looking for is $(I-2vv^T)(I-2uu^T)$. $\endgroup$ – Aweygan Jul 28 '16 at 1:09

We have the following formula for the rotation matrix of angle $\theta$ about a vector $u \in \Bbb R^3$.

$$R = (\cos \theta) I_3 + (\sin \theta) [u]_{\times} + (1 - \cos \theta) u \otimes u$$

Where $I_3$ is the identity matrix, $[\cdot]_{\times}$ is the cross-product matrix and $\otimes$ is the tensor product.

See the Wikipedia page.

  • $\begingroup$ Oh, this is extremely helpful! Thank you! $\endgroup$ – Vibhu Jul 28 '16 at 1:33
  • $\begingroup$ @Vibhu you're welcome. $\endgroup$ – user258700 Jul 28 '16 at 1:35

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