= Solution
Substitute the harmonic <plane wave>
$$
\mathbf u=\mathbf A e^{i(\mathbf K\cdot\mathbf x-\omega t)}
$$
into the <elastic wave> equation. The vector identity
$$
\mathbf K\times(\mathbf K\times\mathbf A)
=\mathbf K(\mathbf K\mathbin{\cdot}\mathbf A)-K^2\mathbf A
$$
gives
$$
\rho\omega^2\mathbf A
=(\lambda+\mu)\mathbf K(\mathbf K\mathbin{\cdot}\mathbf A)
+\mu K^2\mathbf A.
$$
If $\mathbf A$ is parallel to $\mathbf K$, this becomes
$$
\omega^2=c_P^2K^2,
\qquad
\boxed{c_P=\sqrt{\frac{\lambda+2\mu}{\rho}}},
$$
and gives a longitudinal <P wave>. If $\mathbf A\mathbin{\cdot}\mathbf K=0$, it becomes
$$
\omega^2=c_S^2K^2,
\qquad
\boxed{c_S=\sqrt{\frac{\mu}{\rho}}},
$$
and gives either polarization of an <S wave>.
For $\mathbf K=(k,0,m)$, convenient polarization vectors are
$$
\begin{array}{c|c}
\text{wave}&\mathbf A\\ \hline
\text{P}&(k,0,m)\\
\text{SV}&(m,0,-k)\\
\text{SH}&(0,1,0).
\end{array}
$$
The <SV-wave> lies in the $xz$ propagation plane, whereas the <SH-wave> is perpendicular to that plane.
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