What's the opposite of a cross product?

Question

For example, $a \times b = c$

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

Answer 1

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 $t$.

Answer 2

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.

Answer 3

Okay, so if we look at the geometric interpretation of the cross product it should be pretty intuitive why it is not invertible. All the answers already given are good, but I hope to give a simpler explanation.

Motivation of dot and cross products

As we know, there are two widely used operations that resemble multiplication with vectors.

There is the dot product (scalar) which is used when we only care about parallel components of the vectors — for example when calculating work done $W = \mathbf{\vec F} \cdot \mathbf{\vec s}$, we only care about how far an object is moved in the direction of the force.

Then there is the cross product (vector), which is used when we only care about perpendicular components of the vectors — for example when calculating torque on a door being opened $\vec{\boldsymbol \tau} = \mathbf{\vec F} \times \mathbf{\vec r}$, we only care about the component of the force applied perpendicular to the door. As a bonus, the cross product tells us the plane containing the two vectors in the form of its normal.

Why the cross product is non-invertible

Since we can interpret the cross product $\mathbf{\vec u} = \mathbf{\vec v} \times \mathbf{\vec w}$ as the product of the magnitudes of $\mathbf{\vec v}$ and the perpendicular component of $\mathbf{\vec w}$, the cross product only "cares" about the perpendicular component of $\mathbf{\vec w}$ and therefore disregards the component that is parallel to $\mathbf{\vec v}$. Therefore, we can change the parallel component of $\mathbf{\vec w}$ however we please without affecting the final cross-product $\mathbf{\vec u}$. Hence, in most cases, there exist infinite vectors $\mathbf{\vec w}$ along the line parallel to $\mathbf{\vec v}$ which will yield the same result $\mathbf{\vec u}$.

Note that a very similar argument can be made with the dot product disregarding the perpendicular component.