Determining the transformation matrix R

Question

Find the transformation matrix R that describes a rotation of $120$ degrees 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.

It matters not which axis about which I wish for the rotation to occur. Let's suppose the rotation of the coordinate system is about the z-axis.

This means only the x and y axis will be rotating clockwise.

Let the rotated system be the $\bar{x}$ and $\bar{y}$ axis. Let $A$ be the vector through $(1,1,1)$, $A_{x}=A\cos\theta$ and $A_{y}=A\sin\theta$

I've drawn diagrams but unsure how to proceed. Any help is appreciated.

Answer 1

Perhaps I’ve misunderstood the problem, but it seems pretty straightforward to me.

You’re being asked to express a $120$-degree clockwise rotation about the line through $(1,1,1)$ and the origin. If you sight back along this line towards the origin, the coordinate axes (i.e., their projections onto the plane through the origin and normal to the vector $\langle1,1,1\rangle^T$) are evenly spaced. So, a $120$-degree rotation will simply permute the coordinate axes. Remembering that the columns of a transformation matrix are the images of the basis vectors, we can immediately write down the matrix for this rotation:$$R=\pmatrix{0&1&0\\0&0&1\\1&0&0}.$$

As a check, the eigenvalues of $R$ are $1$ and $-\frac12\pm i\frac{\sqrt3}2$ (i.e., the cube roots of unity), which indeed corresponds to a $120$-degree rotation. $\langle1,1,1\rangle^T$ is an eigenvector of $1$, so we have the correct axis, too.

Answer 2

Rotation about any freely chosen axis in 3 dimensional space you can calculate using Rodrigues Formula

$R=I+\sin(\theta)S(v)+(1-\cos(\theta))S^2(v)$

where $S(v)$ is a skew-symmetric matrix assigned to the vector $v$

$S(v)=\begin{bmatrix} 0 & -v_z & v_y \\ v_z & 0 & -v_x \\ -v_y & v_x & 0 \end{bmatrix}$

The only prerequisite for using this formula is to have the axis represented by the unit vector - in your case it would be the vector $v= \left[ \dfrac {1} {\sqrt{3} } \ \dfrac{1}{\sqrt{3}} \ \ \dfrac {1} {\sqrt{3}} \right]^T$.

As you wish to have clockwise rotation about this axis you should assign $\theta \ = -120^{\circ}$, minus because according to the right hand rule the positive angle is assigned to anticlockwise rotation.