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.