Computing the unit vector for a generalised helix The space curve $$\mathbf x (t) = \begin{pmatrix} \cosh t \\ \sinh t \\ t \end{pmatrix}$$ is an example of a generalized helix, meaning that its tangent vector makes a constant angle $\theta$ with a fixed unit vector $\mathbf A$, so that $\mathbf T \cdot \mathbf A = \cos  \theta$.  Find the unit vector $\mathbf A$ in this case and the angle $\theta$.
Attempt:
I could write the unit vector $\mathbf A = \cos \theta \mathbf T + \sin \theta \mathbf B$ since then the magnitude is one and the condition $\mathbf T \cdot \mathbf A = \cos \theta$ is easily recovered. Then sub in expressions for $\mathbf T$ and $\mathbf B$, thereby expressing $\mathbf A$ in terms of the basis vectors in Euclidean space, but this doesn't help with finding the angle.
Thanks for any pointers.

Computing gives that (in the Euclidean basis)
$${\bf T} = \frac{{\bf x}'}{|{\bf x}'|} = \frac{1}{\sqrt{2}}(\tanh t, 1, \text{sech} t).$$
Also in the Euclidean basis, decompose $${\bf A} = (a, b, c).$$ Now, we know that $\cos \theta = {\bf T} \cdot {\bf A}$ is constant, so $$({\bf T} \cdot {\bf A})' = 0.$$ If we expand this quantity in the Euclidean basis using the above expressions, we get constraints on $a, b, c$, and these together with the fact that $|{\bf A}| = 1$ determines $\bf A$ up to sign.
