= Solution
Suppose $\varepsilon_{ij}e_j=\varepsilon_e e_i$. Taking the scalar product of the equation in part (d) with $\mathbf k$ gives $\mathbf k\cdot\mathbf e=0$. The <vector triple product> then reduces the equation to
$$
(-k^2+\omega^2\mu\varepsilon_e)\mathbf e=0,
$$
so $k^2=\omega^2\mu\varepsilon_e$. From part (d),
$$
\mathbf H=\frac{\mathbf B}{\mu}
=\frac{\mathbf k\times\mathbf e}{\mu\omega}
\sin(\mathbf k\cdot\mathbf x-\omega t).
$$
Consequently
$$
\mathbf N=\mathbf E\times\mathbf H
=\frac{\sin^2(\mathbf k\cdot\mathbf x-\omega t)}{\mu\omega}
\mathbf e\times(\mathbf k\times\mathbf e)
=\frac{|\mathbf e|^2\sin^2(\mathbf k\cdot\mathbf x-\omega t)}{\mu\omega}\mathbf k.
$$
\b[Thus the instantaneous flux, except at its zeros, and its time average both point in the wave-propagation direction $\mathbf k$.]
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