My question is on problem 4.1 of Gravitation. In a generic case of electric field and magnetic field(i.e not $E=0$ or $B=0$ or $E$ and $B$ perpendicular), define the direction $\hat{n}$ unit vector ,
$$\hat{n}\tanh (2\alpha)=\frac{2\vec{E}\times\vec{B}}{\vec{E}^{2}+\vec{B}^2}$$
and $\vec{\beta}=\tanh(\alpha)\hat{n}$
is the velocity vector.
Show in the frame of the rocket with velocity $\vec{\beta}$, the Poynting vector vanishes.
I tried the following but I am stuck at the cancellation.
Let the $\bar{\vec{E}}$ and $\bar{\vec{B}}$ be the field in the rocket frame and the field without bars be the field of the rest frame. direction parallel along the velocity of rocket is denoted as subscript $\parallel$ and direction perpendicular to velocity of rocket direction is denoted by $\perp$ as the subscript.
By lorentz transformation. \begin{align} \bar{\vec{E_{\parallel}}}&=\vec{E_{\parallel}}, \\ \bar{\vec{E_{\perp}}}&=\frac{\vec{E_{\perp}}+\vec{\beta}\times\vec{B_\perp}}{\sqrt{1-\beta^{2}}},\\ \bar{\vec{B_{\parallel}}}&=\vec{B_{\parallel}}, \\ \bar{\vec{B_{\perp}}}&=\frac{\vec{B_{\perp}}-\vec{\beta}\times\vec{E_\perp}}{\sqrt{1-\beta^{2}}}. \end{align} Note $\beta\times X_{\perp}$=$\beta\times X$
In barred frame, \begin{align} \bar{\vec{E}}\times\bar{\vec{B}} &=(\bar{\vec{E_{\perp}}}+\bar{\vec{E_{\parallel}}})\times(\bar{\vec{B_{\perp}}}+\bar{\vec{B_{\parallel}}})\\ &=\bar{\vec{E_{\perp}}}\times\bar{\vec{B_{\parallel}}}+\bar{\vec{E_{\parallel}}}\times\bar{\vec{B_{\perp}}}. \end{align} Plug in the lorentz transformation. One gets
$$\frac{\vec{E}\times\vec{B}-\vec{E}_{\parallel}\times(\vec{\beta}\times\vec{E})-\vec{B}_{\parallel}\times(\vec{\beta}\times\vec{B})}{\sqrt{1-\beta^{2}}}.$$
Now the last two terms looks generically like $\vec{X_{\parallel}}\times(\vec{\beta}\times\vec{X}$) simplifies to $\vec{X_{\perp}}(\vec{X}\cdot\vec{\beta})$.
However, I could not see the cancellation at this stage. Did I do something wrong?