Specifically, I don't know what approach to take in answering Griffiths' question 1.9 in his introduction to electrodynamics:

Find the transformation matrix R that describes a rotation by 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.

From Glen Murray's page on rotations, the supposed approach to take is successively rotate space so that the rotation axis sits along the z axis $T$, :

$$T = \begin{pmatrix} 1&0&0 \\ 0& \cos{\alpha} & -\sin{\alpha} \\ 0&\sin{\alpha} & \cos{\alpha} \end{pmatrix} \cdot \begin{pmatrix} \cos{\beta} & 0 & \sin{\beta} \\ 0 &1&0 \\ -\sin{\beta} &0 & \cos{\beta} \end{pmatrix} $$

perform the rotation $\theta$,:

\begin{pmatrix} \cos{\theta} & -\sin{\theta} &0 \\ \sin{\theta} & \cos{\theta}&0 \\ 0&0&1 \end{pmatrix}

and rotate space back to its original orientation $T^{-1}$.

$T$ first rotates space such that the rotation axis sits in the xz plane. Secondly, it rotates space such that the rotation axis lies along the z axis.

This approach seems overly tedious, as this is an introductory question in an introductory chapter. Am I missing something here?

Where I to proceed in this approach, for the arbitrary axis from the origin through the point $(a,b,c)$, then I would need to derive the angles $\alpha$ and $\beta$ as follows.


$\alpha = \arctan{\frac{b}{a}}$


$\beta = \arctan{\frac{a\cos{\alpha} +b\sin{\alpha}}{c}}$



Looking down the axis toward the origin, you see the three axis forming angles of $120°$. This rotation is just a circular permutation of the axis.

  • $\begingroup$ Could you elaborate? I see that you mean the x,y and z axes forming the angles, only how to derive this algebraically? $\endgroup$ – Mussé Redi Jan 28 '19 at 9:28
  • $\begingroup$ @MusséRedi: form an orthogonal frame with $z'$ in the direction $(1,1,1)$ and find the matrix of the change of basis. Then project the axis on the plane $x'y'$. $\endgroup$ – Yves Daoust Jan 28 '19 at 9:38
  • $\begingroup$ How do you do that? $\endgroup$ – Mussé Redi Jan 28 '19 at 9:41
  • $\begingroup$ How do you see the three axes forming angles of $120^\circ$, looking down the axis toward the origin? $\endgroup$ – Mussé Redi Jan 28 '19 at 9:50

You could also just form an orthogonal (change of basis) matrix, $R$, with $\frac1{\sqrt{a^2+b^2+c^2}}(a,b,c)$ as the last row, to take the $z$-axis to $(a,b,c)$. Then if $S$ is the rotation about the $z$-axis, our $T$ would be $T=R^tSR$.

See this example.


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