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I have the following system of simultaneous dot products in $\mathbb{R}^3$ which I am trying to solve for $x$:

\begin{eqnarray} x \cdot t & = & p \cdot t \\ x \cdot n & = & \frac{1}{k} + p \cdot n \\ x \cdot (k'n - k^2t - k\tau b) & = & p \cdot (k'n - k^2t - k\tau b) \end{eqnarray}

in which "$\cdot$" is the dot product, $(t, n, b)$ is an orthonormal basis of vectors, $p$ is a constant vector and $k, \tau$ are scalars (if it helps, this is a differential geometry problem taken from do Carmo, Exercise 3.3/10c with $(t, n, b)$ the moving trihedron and $k, \tau$ the curvature and torsion of a curve).

The solution is known to be

x = p + \frac{1}{k}n + \frac{k'}{k^2\tau}b

which I can verify but not derive... I tried a lot to coerce the above into the form $Ax = b$, but to no avail.

My question: how can the solution be derived from the given equations?

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    @rcollyer: you are right... I had some problems deciphering my own writing2011-08-27

1 Answers 1

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If we decompose $x$ and $p$ on the $(t,n,b)$ basis : $x = x_t t + x_n n + x_b b$ $p = p_t t + p_n n + p_b b$ The first two equations gives $x_t = p_t$ and $x_n = \frac{1}{k} + p_n$.

Replacing it in the last equation removes the terms in $x_t$ and $p_t$, leaving : k' x_n - k \tau x_b = k' p_n - k \tau p_b The second equation tells us that $x_n - p_n = \frac{1}{k}$ so it gives : x_b = \frac{k'}{k^2 \tau} + p_b