$3$d rotation for easy Lorentz boost I need to apply a Lorentz boost to a couple of vectors ($a_1$ and $a_2$) in an arbitrary direction, let's say I want to switch them to a frame of reference with the following speed vector $v$ = $(0.3,0.8,0.2)$, to obtain $b_1$ and $b_2$.
I can also express this direction in terms of $\theta$ and $\phi$.
Now, I seem to understand that the best way to do this is to:


*

*Find a frame of reference such that $v$ lies along one of the axis, so that $v'$ = $(0.88,0,0).$

*Express the 2 vectors $a_1$ and $a_2$ in the rotated frame of reference to obtain $a'_1$ and $a'_2$.

*Apply easily Lorentz boost to $a'_1$ and $a'_2$ (which now applies to only one of their components), obtaining $b'_1$ and $b'_2$ .

*Re-rotate $b'_1$ and $b'_2$ back to the original frame of reference to obtain $b_1$ and $b_2$.
But I can't find the rotation matrix that would bring me from one frame to another.
 A: Usually the action of a boost of direction $\vec v$ on  a vector $\vec r$  is represented using the decomposition of $\vec r$ in two components: parallel and orthogonal to $\vec v$.  You can see how this works here.
If $\vec \beta=\vec v / c$ is the velocity vector  in units of the light velocity and $\gamma =1/\sqrt{1-\beta^2}$ , with $\beta=||\vec \beta||$, than the boost is represented by the matrix
$$
\begin{bmatrix}
\gamma&-\gamma \beta_x&-\gamma \beta_y&-\gamma \beta_z\\
-\gamma \beta_x&1+(\gamma-1)\dfrac{\beta_x^2}{\beta^2}&(\gamma-1)\dfrac{\beta_x\beta_y}{\beta^2}&(\gamma-1)\dfrac{\beta_x\beta_z}{\beta^2}\\
-\gamma \beta_y&(\gamma-1)\dfrac{\beta_y\beta_x}{\beta^2}&1+(\gamma-1)\dfrac{\beta_y^2}{\beta^2}&(\gamma-1)\dfrac{\beta_y\beta_z}{\beta^2}\\
-\gamma \beta_z&(\gamma-1)\dfrac{\beta_z\beta_x}{\beta^2}&(\gamma-1)\dfrac{\beta_z\beta_y}{\beta^2}&1+(\gamma-1)\dfrac{\beta_z^2}{\beta^2}
\end{bmatrix}
$$
As you can see here (or here for a more general and detailed exposition) .
It seems that the other method that you suggest can works, but I never seen  used. And this  method seems more simple.
