I am looking at the following exercise of the book of Andrew Pressley:
Let $P$ be an $n \times n$ orthogonal matrix and let $a \in \mathbb{R}^n$, so that $M(v) =Pv + a$ is an isometry of $\mathbb{R}^3$ (see Appendix 1).
Show that, if $\gamma$ is a unit-speed curve in $\mathbb{R}^n$, the curve $\Gamma = M(\gamma )$ is also unit-speed.
Show also that, if $t, n, b$ and $T, N, B$ are the tangent vector, principal normal and binormal of $\gamma$ and $\Gamma$, respectively, then $T = Pt$, $N = Pn$ and $B = Pb$.
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The only point I get stuck is at showing that $B = Pb$.
I did this:
We have $$B T \times N=Pt \times Pn=\det (P) P( t \times n)$$
Since $P$ is an orthogonal matrix, we have $\det (P)=\pm 1$.
How do we get $\det (P)=1$ so that $B=P(t \times n)=Pb$ ?
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Edit:
How do we conclude that in this case $P$ is orientation preserving?
Do we assume that the isometry is direct?
If $\gamma$ lies entirely in the $x$-$y$ plane, $a=0$ and $P=\mathrm{diag}(1,1,-1)$ is a reflection in the $z$ direction, then $\gamma = \Gamma$ but $P B = - B$. This shows the statement is not true: you need the extra assumption that $M$ is orientation-preserving, i.e. $\det P = 1$.
This shouldn't be too surprising: the binormal vector is defined via the cross product, which depends on a choice of orientation.