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Given Data in question

  1. I have following relations in vector space$\begin{eqnarray}n_0^{'}(s)=-\kappa(s) \times n_0(s)\\n_1^{'}(s)=-\kappa(s) \times n_1(s)\\n_2^{'}(s)=-\kappa(s) \times n_2(s)\end{eqnarray}\tag 1$
  2. $n_0,n_1,n_2$ are orthogonal unit vectors

  3. $\kappa(s)$ is not a unit vector

  4. Derivative of a unit vector is a vector that is normal to the given vector. We can normalize it to get normalized normal unit vector. Click here for Reference if needed for this data

Question

  1. Can we rewrite the equation (1) in the following form$\begin{eqnarray}n_0(s)=-\kappa(s) \times \psi_0(s)\\n_1(s)=-\kappa(s) \times \psi_1(s)\\n_2(s)=-\kappa(s) \times \psi_2(s)\end{eqnarray}\tag 2$?

If so what could be the vectors $\psi_0(s),\psi_1(s),\psi_2(s)$ ?

or can we rewrite equation (1) with a LHS in equation (2) irrespective of what is on R.H.S(just avoiding R.H.S condition for more relaxation in complexity of the problem )?

Nirvana
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  • You need the inverse of the cross product which I assume exists, but I don't know it off hand. You are looking for $\vec{\psi}\times\left(\vec{\kappa}\times\vec{n}\right) = \vec{n}$ (the switching is trivial since $\vec{a}\times\vec{b} = -\vec{b}\times\vec{a}$). – Jared Oct 04 '14 at 03:54
  • Yes switching is correct. But it is $\psi\times \kappa$ .That is another way of asking it – Nirvana Oct 04 '14 at 04:06
  • Thinking out loud here (and I'm pretty sure I convinced myself earlier this was wrong), but if you have $\vec{\kappa}\times\vec{n}$, then this produces a vector that is perpendicular to both $\vec{\kappa}$ and $\vec{n}$, so if you cross that vector (the cross product) with $\vec{\kappa}$, then shouldn't you get back a vector parallel with $\vec{n}$? – Jared Oct 04 '14 at 04:09
  • It is not Parallel I feel. You can simply imagine the geometry using right hand rule – Nirvana Oct 04 '14 at 04:20

1 Answers1

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Alright, let's look for an inverse here (it sounds like one does not exist)--which means your problem is either impossible or requires the constraints that the $\vec{n}$'s are orthogonal unit vectors.

The cross product is restricted to $\mathbb{R}^3$ (assuming reals--but that isn't really a constraint here, I don't think). So we have:

$$ \langle x_1, y_1, z_1\rangle\times\langle x_2, y_2, z_2\rangle = \langle y_1z_2 - z_1y_2, z_1x_2 - x_1z_2, x_1y_2 - y_1x_2\rangle $$

Then we have:

\begin{align} \langle x_3, y_3, z_3\rangle\times\left(\langle x_1, y_1, z_1\rangle\times\langle x_2, y_2, z_2\rangle\right) = & \langle y_3(x_1y_2 - y_1x_2) - z_3(z_1x_2 - x_1z_2),\\ &z_3(y_1z_2 - z_1y_2) - x_3(x_1y_2 - y_1x_2),\\ & x_3(z_1x_2 - x_1z_2) - y_3(y_1z_2 - z_1y_2)\rangle \end{align}

Now this looks very complicated (and it is--if you need to solve it "as is"). Instead we can imagine that we already know $\vec{n}' = \vec{v}_1\times\vec{v_2}$. Then this becomes:

\begin{align} \langle x_3, y_3, z_3\rangle\times\vec{n}' = & \langle y_3n'_z - z_3n'_y, z_3n'_x - x_3n'_z, x_3n'_y - y_3n'_x\rangle \end{align}

Now, setting this equal to the second argument of the original cross product, we have a set of linear equations for three unknowns ($x_3, y_3, z_3$):

$$ x_2 = y_3n'_z - z_3n'_y \\ y_2 = z_3n'_x - x_3n'_z \\ z_2 = x_3n'_y - y_3n'_x $$

The matrix gives:

$$ \begin{pmatrix} 0 & n_z' & -n_y' \\ -n_z' & 0 &n_x' \\ n_y' & -n_x' & 0 \end{pmatrix} \times \begin{pmatrix}x_3 \\ y_3 \\ z_3\end{pmatrix} = \begin{pmatrix}x_2 \\ y_2 \\ z_2\end{pmatrix} $$

There is no solution when the determinate of this matrix is zero (which could happen and if this occurs with $\langle x_2, y_2, z_2\rangle = \vec{0}$ then there are an infinite amount of solutions)--but otherwise you should be able to find an inverse.

Alright, so I just noticed that the determinate here (I'm assuming of all anti-symmetric matices) is zero: $n_z'x_x'n_y'- n_y'n_z'n_x' = 0$. This means that unless $\langle x_2, y_2, z_2\rangle$ is $\vec{0}$ there is no solution (since a matrix that has a determinant equal to zero only has a solution when the RHS is zero--in which case it has infinite solutions).

I'm going to leave this answer here, although I now realize it does not address your problem at all, rather is just a proof of the previous link that the inverse of the cross product does not exist.

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Jared
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