Find a rigid motion

Question

I am looking at the following curve: $$c(t)=(t + \sqrt3\sin t, 2\cos t, \sqrt3t-\sin t)$$ My goal is to find a rigid motion $L$ of $\Bbb R^3$ and a helix of the form: $$\gamma (t)=(a\cos(t),a\sin(t),bt)$$ such that $L \circ c = \gamma$. Now I found the curvature $\kappa=1/4$ and the torsion $\tau=-1/4$ of $c$. From there I could find $a=2$ and $b=-2$. But know I got stuck to find a rigid motion... How can I proceed? Many thanks for some help!

Answer 1

Since $\gamma(0) = c(0) = \vec{0}$, you can see that $L(\vec{0}) = \vec{0}$ so $L$ is a linear map which is represented by orthogonal matrix. Also, we have $L \circ c = \gamma \iff c = L^{-1} \circ \gamma$.

Now, note that each coordinate of $c(t)$ is a linear combination of the coordinates of $\gamma(t)$. This implies that it is easier to find $L^{-1}$ and then invert it. If we denote by $O$ the matrix representing $L^{-1}$ with respect to the standard basis, then the equation $c = L^{-1} \circ \gamma$ is equivalent to the equation $$ \begin{pmatrix} t + \sqrt{3}\sin t \\ 2 \cos t \\ \sqrt{3}t - \sin t \end{pmatrix} = O \begin{pmatrix} 2 \cos t \\ 2 \sin t \\ -2 t \end{pmatrix}. $$ By comparing coefficients, we can see that $$ O = \begin{pmatrix} 0 & \frac{\sqrt{3}}{2} & -\frac{1}{2} \\ 1 & 0 & 0 \\ 0 & -\frac{1}{2} & -\frac{\sqrt{3}}{2} \end{pmatrix}. $$ Then $L$ is represented by $O^{-1} = O^T$ and so $$ L(x,y,z) = \left( \frac{\sqrt{3}y}{2} - \frac{z}{2}, x, -\frac{y}{2} - \frac{\sqrt{3}z}{2} \right).$$