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?

  • $\begingroup$ I'm more accustomed to "trihedral", but sure... anyway, you're aware that the binormal is the cross product of the tangent and normal vectors, yes? $\endgroup$ Commented Aug 7, 2011 at 10:32
  • $\begingroup$ Yes... by definition, $b = t \times n$, but how can this be put to use here? $\endgroup$ Commented Aug 7, 2011 at 11:42

1 Answer 1


That's rather easy... to show that $b \cdot a$ is constant, it suffices to show that its derivative is 0:

$$(b \cdot a)' = b' \cdot a + b \cdot a' = \tau n \cdot a = 0$$

(using that $b' = \tau n$, $a' = 0$ and $n \cdot a = 0$)


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