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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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Since $(t,n,b)$ is an orthonormal basis, let $x = \alpha t + \beta n + \gamma b$, and find $\alpha, \beta, \gamma$. –  Rahul Aug 27 '11 at 18:54
@Rahul: thanks, that does it... it's really obvious, I should have seen this. I even hesitate to suggest that you submit this as an answer (because it's so trite), but I would accept it if you did. –  koletenbert Aug 27 '11 at 19:04
I'll just point out that, in the last equation, if you expand the left-hand side and substitute values for $x\cdot t$ and $x\cdot n$ from the first two equations, you get a straightforward formula for $x\cdot b$. This makes @Rahul's hint go faster. –  Blue Aug 27 '11 at 19:05
@Day: yeah, I just did that... it's really easy now that I know how to do it :-) –  koletenbert Aug 27 '11 at 19:07
You have a $t$ in the denominator of the $b$ term in your known solution. Is that supposed to be a $\tau$? –  rcollyer Aug 27 '11 at 19:08

1 Answer 1

up vote 4 down vote accepted

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 : $$ x_b = \frac{k'}{k^2 \tau} + p_b $$

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