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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.

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  • $\begingroup$ It does matter a bit which axis you are rotating about; if the axis is one of the coordinate axes already, then the problem is rather easy (it essentially reduces to the 2D case). But if it is not then you have to change coordinates first. With that in mind, can you find two vectors perpendicular to each other and to $(1,1,1)$? $\endgroup$ – Ian Jan 4 '16 at 13:27
  • $\begingroup$ Would that vector be the projection of the vector A (1,1,1) on the x-axis?@Ian $\endgroup$ – Mathematicing Jan 4 '16 at 13:31
  • $\begingroup$ No, because the x axis is not perpendicular to $(1,1,1)$. The vectors perpendicular to $(1,1,1)$ are the solutions to $x+y+z=0$. Can you find two of these which are perpendicular to each other? (Note that there is not a unique solution to this part of the question at all.) $\endgroup$ – Ian Jan 4 '16 at 13:34
  • $\begingroup$ A followup hint: written as $x+y+z=0$, the variable "x" is dependent while "y" and "z" are independent. So you can get one basis for the solution space by choosing linearly independent vectors $(y_1,z_1)$ and $(y_2,z_2)$, then set $x_1=-y_1-z_1$ and $x_2=-y_2-z_2$. Once you have one basis, do you know how to orthogonalize it? $\endgroup$ – Ian Jan 4 '16 at 18:01
  • $\begingroup$ Are you sure you’re interpreting the problem correctly? The way I understand your first paragraph is that you’re being asked to express as a matrix a rotation about the line through $(1,1,1)$ and the origin, not about one of the coordinate axes. If so, the fact that it’s a 120-degree rotation makes it very simple to construct this matrix. $\endgroup$ – amd Jan 4 '16 at 20:15
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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.

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  • $\begingroup$ Sir, Nice Answer But Now you have to shift that to original position. for that is there is any easy way? I know one way that find orthonormal basis containg 1 1 1 by GS process and then change of basis $\endgroup$ – SRJ Nov 21 '18 at 15:05
  • $\begingroup$ @Shubham What “shift ... to original position?” $R$ is exactly the transformation that you asked for in your question. $\endgroup$ – amd Nov 21 '18 at 20:39
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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.

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