Special relativity - angles between fields

[Aleolomorfo]

Homework Statement

In an electromagnetic filed, the elctric field ##\vec{E}## forms an angle ##\theta## with the magnetic field ##\vec{B}##, and ##\theta## is invariant for all inertial observers. Finding the value of ##\theta##.

Homework Equations

Tranformations of fields perpendicular to the boost:
$$\vec{E'}=\gamma(\vec{E}+(\vec{v}\times\vec{B})_\perp)$$
$$\vec{B'}=\gamma(\vec{B}-(\vec{v}\times\vec{E})_\perp)$$

The Attempt at a Solution

I know that the only ##\theta## that is invariant in ##\frac{\pi}{2}## or ##\frac{3\pi}{2}##, so when the fields are perpendicular. I have always showed it in this way: I use the invariance of ##\vec{E}\cdot\vec{B}=\vec{E'}\cdot\vec{B'}##, so ##EB\cos{\theta}=E'B'\cos{\theta'}##. If ##\theta'=0##, ##\theta## must also be ##0## to hold the equality. Consequently if they are perpendicular in one frame, they are perpendicular in all the other, otherwise the invariance of ##\vec{E}\cdot\vec{B}## does not hold.
I know this is not the best proof, but I think it is in some way logically. However, the exercise wants a rigorous proof.
I have tried using ##\vec{E}\cdot\vec{B}## better but it is a vicious circle. I have aligned the x-axis along ##\vec{E}## and y-axis in order to have the two vectors in the ##xy## plane.
$$\vec{E}=E\vec{x}$$
$$\vec{B}=B(\cos{\theta}\vec{x}+\sin{\theta}\vec{y})$$
I have written ##E'## and ##B'## using the fields-transfromations for a boost along z:
$$E'_x=\gamma(E-vB\sin{\theta})$$
$$E'_y=\gamma v B\cos{\theta}$$
$$B'_x=\gamma B\cos{\theta}$$
$$B'_y=\gamma(B\sin{\theta}-vE)$$
But then everything I have done is inconclusive. I do not know how to conclude the demonstration.

 

[Orodruin]

I know this is not the best proof

Why? It uses invariants. As physicists we love invariants.