# How do you find a matrix that maps a rotation of $60º$ around a line $r(t) = 0 + t (1, 1, 1)$

Find a matrix that maps a rotation of $$60º$$ around a line $$r(t) = 0 + t (1, 1, 1)$$

I have some problems with these kind of rotations, because you aren't rotating around a $$x, y$$ or $$z$$ axis, so you can't directly use the rotational formula i.e.: $$\begin{pmatrix}\cos\theta & -\sin\theta\\ \sin\theta & \cos\theta\\ & & 1 \end{pmatrix}$$ I have tried to change basis, where the orthonormal basis is $$v_1 = \frac{\sqrt 3}{3} (1, 1, 1)$$, $$v_2 = \frac{\sqrt2}{2} (1, -1, 0)$$, $$v_3 = \frac{\sqrt 6}{6} (1, 1, -2)$$, but then got stuck. I know I should somehow make a matrix that maps from standard basis to my new basis.

Am I even on the right path or is this completely wrong?

Consider the linear map such that its matrix with respect to $$B=\{v_1,v_2,v_3\}$$ is$$M=\begin{bmatrix}1&0&0\\0&\frac12&-\frac{\sqrt3}2\\0&\frac{\sqrt3}2&\frac12\end{bmatrix}.$$So, if$$P=\begin{bmatrix}\frac1{\sqrt3}&\frac1{\sqrt2}&\frac1{\sqrt6}\\\frac1{\sqrt3}&-\frac1{\sqrt2}&\frac1{\sqrt6}\\\frac1{\sqrt3}&0&-\frac2{\sqrt6}\end{bmatrix}$$and if $$T$$ is the matrix of your rotation with respect to the standard basis, then $$M=P^{-1}TP$$ and therefore$$T=PMP^{-1}=\begin{bmatrix}\frac{2}{3} & -\frac{1}{3} & \frac{2}{3} \\ \frac{2}{3} & \frac{2}{3} & -\frac{1}{3} \\ -\frac{1}{3} & \frac{2}{3} & \frac{2}{3}\end{bmatrix}.$$
• But $\frac{1}{\sqrt 6} = \frac{\sqrt 6}{6}$, isn't it? Yeah that is clear to me, but that is not the whole answer, I have to also find some other matrices don't I?
• Concerning $v_3$, I misread what you wrote. I have edited my answer. May 17, 2020 at 8:54
• No, it's the other way around: it maps the standard basis to $B$. It is the change of bases matrix. May 17, 2020 at 9:00