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For example, $a \times b = c$

If you only know $a$ and $c$, what method can you use to find $b$?

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It's a set of linear equations. So, you use linear algebra to solve. –  Raskolnikov Apr 12 '11 at 19:13
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$b$ is not uniquely determined by the knowledge of $a$ and $c$. Do you want to add some additional conditions on $b$ or are you interested in all possible solutions? –  Fabian Apr 12 '11 at 19:13

3 Answers 3

up vote 14 down vote accepted

As Fabian wrote, $b$ is not uniquely determined by $a$ and $c$. Moreover, there is no solution unless $a$ and $c$ are orthogonal. If $a$ and $c$ are orthogonal, then the solutions are $(c \times a)/(a\, . a) + t a$ for arbitrary scalars $t$.

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To spell this out in words I understand, $b$ lies in the plane orthogonal to $c$ (as must $a$). The component of $b$ orthogonal to $a$ is $|c|/|a|$ in the direction consistent with the right-hand rule. –  Henry Apr 13 '11 at 0:12

The name "product" for the cross product is unfortunate. It really should not be thought of as a product in the ordinary sense; for example, it is not even associative. Thus one should not expect it to have properties analogous to the properties of ordinary multiplication.

What the cross product really is is a Lie bracket.

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Anything called a "product" has to be associative? Then what should we call the Cartesian product of sets, seeing as $(A\times B)\times C\ne A\times(B\times C)$? To say nothing of the inner product of vectors, another misnomer. –  bof Oct 4 at 5:59

Alright, let's look for an inverse here:

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 we are trying to find $\langle x_3, y_3, z_3\rangle$ which is the inverse. So we set 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 is always the case: $n_z'x_x'n_y'- n_y'n_z'n_x' = 0$. 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).

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