Solution vector equation involving cross product?

Question

I am starting to study linear algebra, and a problem appeared at the beginning of the textbook, the problem is the following: What 3-vector u satisfies $(1,1,0)\times u=(0,1,1)$. My immediate response was to try to find the inverse operation of the cross product but I have no idea if that operation even exist. Putting my question in more general terms, how could one find $b$ in the equation $\vec a\times \vec b=\vec c$, given that $\vec a,\vec b,\vec c$ are vectors in $\mathbb{R^3}$ space and orthogonal to each other ?

Answer 1

There is no vector $u$ such that $(1,1,0)\times u=(0,1,1)$, since $(1,1,0)$ and $(0,1,1)$ are not orthogonal.

If $a=(a_1,a_2,a_3)$, $c=(c_1,c_2,c_3)$ and $a$ and $c$ are orthogonal, in order to find a $b=(b_1,b_2,b_3)$ such that $a\times b=c$, you can solve the system$$\left\{\begin{array}{l}a_2b_3-a_3b_2=c_1\\a_3b_1-a_1b_3=c_2\\a_1b_2-a_2b_1=c_3.\end{array}\right.$$

Answer 2

Let $u=(a,b,c)$, computing leads to: $$(1,1,0)\times u=(c,-c,b-a).$$ Therefore, the given equation has no solution.

Besides, in this case we know from scratch that no solution can exist as $(1,1,0)\times u$ must be orthogonal to $(1,1,0)$ and $(0,1,1)$ is not.

Answer 3

Why not change it to a system of equations?

$$a \times b= (a_2b_3-a_3b_2,a_3b_1-a_1b_3,a_1b_2-a_2b_1)=(c_1,c_2,c_3)$$

Or on other words, we have the system of equations

$$\begin{pmatrix}0&-a_3&a_2\\a_3&0&-a_1\\-a_2&a_1&0 \end{pmatrix}\begin{pmatrix}b_1\\b_2\\b_3\end{pmatrix}=\begin{pmatrix}c_1\\c_2\\c_3\end{pmatrix}$$

It is worth noting that this matrix is skew symmetric, and has determinant $0$, so it will not always have a solution.

Its image is precisely the span of $(0,a_3,-a_2),(-a_3,0,a_1),(a_2,-a_1,0)$

the vector $c_1,c_2,c_3$ lays in its image (there exists a $b$ at all) if and only if it can be written as a linear combination of three vectors that span the matrix's image. In particular, if it can be written as a linear combination with scalars $\lambda_1,\lambda_2\lambda_3$, i.e: is of the form

$$(\lambda_1 a_2- \lambda_2a_3,\lambda_1 a_3-\lambda_3 a_1,\lambda_2 a_1-\lambda_1 a_2),$$

and after taking the dot product with $(a_1,a_2,a_3)$ is zero, so it is orthogonal to $(a_1,a_2,a_3)$.