Let $(t, n, b)$ be the Frenet trihedron of some curve with curvature $k \neq 0$ and torsion $\tau \neq 0$, defined as usual with tangent $t$, normal $n$ and binormal $b$.
I would like to prove the following: if $t$ makes a constant angle with some fixed vector $a$, then $b$ also makes a constant angle with $a$. Here is what I came up with:
$t$ and $a$ make a constant angle, so
$$t \cdot a = c_t$$
for some constant $c_t$. Differentiating this yields $t' \cdot a + t \cdot a' = 0$, or since $a' = 0$ and $t' = kn$
$$n \cdot a = 0$$
I need to show that
$$b \cdot a = c_b$$
for some constant $c_b$. Writing the equations above in matrix form, with $T=[t, n, b]^T$ being the "thrihedron matrix" whose rows are $t$, $n$ and $b$, I get
$$T a = (c_t, 0, c_b)^T$$
Since $T$ is orthonormal, it does not change the length of $a$, so taking the squared norm on both sides yields
$$\|Ta\|^2 = \|a\|^2 = c_t^2 + c_b^2$$
$a$ and $c_t$ are constant, therefore $c_b$ is constant as well (continuity will prevent sign flips...) QED
I wonder if there is a simpler way to demonstrate this result - in particular, I don't like bringing in the matrix equation. I would prefer to show it more directly, using just "some dot products" and the Frenet formulas... is that possible?