= Solution
Let $\mathbf e$ be the downward vertical unit vector and let the rod's weight after buoyancy be $W\mathbf e$. Force balance gives
$$
\mathbf R\mathbf U=W\mathbf e.
$$
Because
$$
\mathbf R
=c_1\mathbf t\mathbf t
+c_2(\mathbf 1-\mathbf t\mathbf t),
$$
its <matrix inverse> is
$$
\mathbf R^{-1}
=\frac1{c_1}\mathbf t\mathbf t
+\frac1{c_2}(\mathbf 1-\mathbf t\mathbf t).
$$
Using $\mathbf e\cdot\mathbf t=\cos\theta$ gives
$$
\mathbf U
=W\left[
\frac{\cos\theta}{c_1}\mathbf t
+\frac1{c_2}\left(\mathbf e-\cos\theta\,\mathbf t\right)
\right].
$$
Therefore
$$
\mathbf e\cdot\mathbf U
=W\left(\frac{\cos^2\theta}{c_1}
+\frac{\sin^2\theta}{c_2}\right),
$$
and
$$
|\mathbf U|
=W\sqrt{
\frac{\cos^2\theta}{c_1^2}
+\frac{\sin^2\theta}{c_2^2}
}.
$$
The required angle is consequently
$$
\boxed{
\cos\alpha
=
\frac{\cos^2\theta/c_1+\sin^2\theta/c_2}
{\sqrt{\cos^2\theta/c_1^2+\sin^2\theta/c_2^2}}
}.
$$
If $c_1=c_2$, then $\cos\alpha=1$, so the rod falls vertically. For an anisotropic rod with $c_2>c_1$, the component parallel to its long axis encounters less drag, and the settling direction is generally deflected toward that axis.
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