Rotation matrix about an arbitrary axis

[Happiness]

Suppose a position vector v is rotated anticlockwise at an angle $\theta$ about an arbitrary axis pointing in the direction of a position vector p, what is the rotation matrix R such that Rv gives the position vector after the rotation?

Suppose p = $\begin{pmatrix}1\\1\\1\end{pmatrix}$ and $\theta$ = -120$^\circ$.

My approach is as follows.

First, perform transformation $T_1$: rotate the xy plane 45$^\circ$ anticlockwise about the z axis. (I treat this as keeping v fixed but expressing v in terms of the new coordinate system CS1.)

Next, perform transformation $T_2$: rotate the xz plane $\tan^{-1}\frac{1}{\sqrt2}$ clockwise about the y-axis (clockwise when the y-axis points into your eye) so that the x-axis is now pointing in the direction of p. (Again, I treat this as keeping v fixed but expressing v in terms of the new coordinate system CS2.)

Then, perform transformation $T_3$: rotate the yz plane 120$^\circ$ anticlockwise about the x axis, since rotating v by -120$^\circ$ has the same effect as rotating the yz plane 120$^\circ$ (turning the coordinate system about the rotation axis by 120$^\circ$ while keeping v fixed).

We have $T_1=\begin{pmatrix}\frac{1}{\sqrt2}&\frac{1}{\sqrt2}&0\\-\frac{1}{\sqrt2}&\frac{1}{\sqrt2}&0\\0&0&1\end{pmatrix}$ , $T_2=\begin{pmatrix}\frac{\sqrt2}{\sqrt3}&0&-\frac{1}{\sqrt3}\\0&1&0\\\frac{1}{\sqrt3}&0&\frac{\sqrt2}{\sqrt3}\end{pmatrix}$ , $T_3=\begin{pmatrix}1&0&0\\0&-\frac{1}{2}&\frac{\sqrt3}{2}\\0&-\frac{\sqrt3}{2}&-\frac{1}{2}\end{pmatrix}$

Next, perform the inverse of transformation $T_2$. (Changing the expression of v from coordinate system CS2 back to CS1.)

Lastly, perform the inverse of transformation $T_1$. (Changing the expression of v from coordinate system CS1 back to CS0, the original coordinate system.)

Thus, R = $T_1^{-1}T_2^{-1}T_3T_2T_1 = \begin{pmatrix}0&0&-1\\1&0&0\\0&-1&0\end{pmatrix}$

But R should be $\begin{pmatrix}0&1&0\\0&0&1\\1&0&0\end{pmatrix}$ since rotating v about p by -120$^\circ$ has the effect of turning the x-axis to the y axis, and the y-axis to the z axis, and the z axis to the x-axis (turning the coordinate system about the rotation axis p by 120$^\circ$ while keeping v fixed).

Why do I get different answers? What's wrong with my approach?

 

[Happiness]

Thanks for reading! The mistake has been found. $T_2$'s entries are wrong.

 

[HallsofIvy]

Here is how I would do this, in general, to find the matrix that rotates a vector through angle \theta about axis <a, b, c>. (I will assume that <a, b, c> has length 1: \sqrt{a^2+ b^2+ c^2}= 1.)

First, find the matrix that rotates < a, b, c> into < 0, 0, 1>. To do that, first find the matrix, A, that rotates, around the z-axis, mapping <a, b c> to <0, r, c> where r= \sqrt{a^2+ b^2}. I presume you know that, in two dimensions, the matrix rotating thorough angle \theta is given by
\begin{bmatrix}cos(\theta) &amp; - sin(\theta) \\ sin(\theta) &amp; cos(\theta) \end{bmatrix}.

So we can write a rotating about the z-axis, as
\begin{bmatrix}\alpha &amp; \ -beta &amp; 0 \\ \beta &amp; \alpha &amp; 0 \\ 0 &amp; 0 &amp; 1\end{bmatrix}
so that such a matrix mapping <a, b, c> to < 0, r, c> must give
\begin{bmatrix}\alpha &amp; \ -beta &amp; 0 \\ \beta &amp; \alpha &amp; 0 \\ 0 &amp; 0 &amp; 1\end{bmatrix}\begin{bmatrix}a \\ b \\ c\end{bmatrix}= \begin{bmatrix}a\alpha- b\beta \\ a\beta+ b\alpha \\ c\end{bmatrix}= \begin{bmatrix} 0 // r // z\end{bmatrix}
That gives the two equations a\alpha- b\beta= 0, a\beta+ b\alpha= r.

If we multiply the first of those equations by a, to get a^2\alpha- ab\beta= 0, multiply the second equation by b, to get b^2\alpha+ ab\beta= br, then add the two equations, we eliminate b getting (a^2+ b^2)\alpha= br so that \alpha= \frac{br}{a^2+ b^2}= \frac{b}{r} since r= \sqrt{a^2+ b^2}. Putting that into a^2\alpha- ab\beta= 0 we get ab\beta= a^2\alpha= \frac{a^2}{r} so that \beta= \frac{a}{r}.

That is, this matrix is
A= \begin{bmatrix}\frac{b}{r} &amp; -\frac{a}{r} &amp; 0 \\ \frac{a}{r} &amp; \frac{b}{r} &amp; 0 \\ 0 &amp; 0 &amp; 1\end{bmatrix}.

Now we want to find the matrix, B, that rotates, around the x-axis, mapping < 0, r, c> to < 0, 0 , 1>. We can write that as
\begin{bmatrix}1 &amp; 0 &amp; 0 \\ 0 &amp; \alpha &amp; -beta \\ 0 &amp; \beta &amp; \alpha \end{bmatrix}\begin{bmatrix} 0 &amp; r &amp; c \end{bmatrix}= \begin{bmatrix} 0 \\ r\alpha- c\beta \\ r\beta+ c\alpha \end{bmatrix}= \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix}.

So we have the two equations r\alpha- c \beta= 0 and c\alpha+ r\beta= 1. Multiply the first equation by r, to get r^2\alpha- rc\beta= 0, multiply the second equation by c to get c^2\alpha + rc\beta= c, then add, eliminating \beta and getting (r^2+ c^2)\alpha= c so that \alpha= \frac{c}{r^2+ c^2}= c since \sqrt{r ^2+ c^2}= \sqrt{a^2+ b^2+ c^2}= 1. Putting that into r\alpha- c\beta= 0 we have \beta= \frac{r}{c}(c)= r.

That is, this is the matrix
B= \begin{bmatrix}1 &amp; 0 &amp; 0 \\ 0 &amp; c &amp; -r \\ 0 &amp; r &amp; 1\end {bmatrix}

Now, to rotate a vector through angle \theta about the vector <a, b, c>, multiply by A and B to rotate <a, b, c> into <0, 0, 1> then rotate through angle \theta around the z-axis, then multiply by B^{-1} and A^{-1} to return to the original axis of rotation.