Find a vector that is equal to the cross product of two given vectors

Consider the vectors

$A = 2 \mathbf{i} - \mathbf{j} + 2 \mathbf{k}, \qquad C = 3 \mathbf{i} + 4 \mathbf{j} - \mathbf{k}.$

Find $B$ satisfying $A \times B = C$ . How many such solutions are there?
Find $B$ such that $A \times B = C$ and $A \cdot B = 1$ . How many such solutions are there?

Let $B = (b_1, b_2, b_3)$ . For $A \times B = C$ we must have
$\begin{align*} && \begin{vmatrix*} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 2 & -1 & 2 \\ b_1 & b_2 & b_3 \end{vmatrix*} &= 3 \mathbf{i} + 4 \mathbf{j} - \mathbf{k} \\[9pt] \implies && \begin{vmatrix*} -1 & 2 \\ b_2 & b_3 \end{vmatrix*} \mathbf{i} - \begin{vmatrix*} 2 & 2 \\ b_1 & b_3 \end{vmatrix*} \mathbf{j} + \begin{vmatrix*} 2 & -1 \\ b_1 & b_2 \end{vmatrix*} \mathbf{k} &= 3 \mathbf{i} + 4 \mathbf{j} - \mathbf{k} \\[9pt] \implies && (-b_3 - 2b_2) \mathbf{i} + (2b_1 - 2b_3) \mathbf{j} + (2b_2 + b_1) \mathbf{k} &= 3 \mathbf{i} + 4 \mathbf{j} - \mathbf{k}. \end{align*}$

Therefore, we have the three equations,

$\begin{align*} -b_3 - 2b_2 &= 3 \\ 2b_1 - 2b_3 &= 4 \\ 2b_2 + b_1 &= -1. \end{align*}$

From the first equation we have $b_3 = -3-2b_2$ . From the second equation we then have $b_1 = -1 - 2b_2$ . Since any value of $b_2$ then satisfies the third equation we have that $b_2$ is arbitrary. Letting $b_2 = 0$ we then have $b_1 = -1$ and $b_3 = -3$ . Hence, a solution is

$B = (-1,0,-3) = -\mathbf{i} - 3\mathbf{k}.$

There are infinitely many solutions since we can take any value for $b_2$ to obtain another solution.
From part (a) we know that the vectors $B$ such that $A \times B = C$ are of the form
$B = (-1-2b_2, b_2, -3-2b_2)$

for any value of $b_2$ . Then,

$\begin{align*} && A \cdot B &= 1 \\ \implies && 2 \cdot (-1-2b_2) + (-1)(b_2) + 2(-3-2b_2) &= 1 \\ \implies && -2-4b_2 -b_2 - 6 - 4b_2 &=1 \\ \implies && -8-9b_2 &= 1 \\ \implies && b_2 &= -1. \end{align*}$

From part (a) we then have $b_1 = 1$ and $b_3 = -1$ . Hence, $B = \mathbf{i} - \mathbf{j} - \mathbf{k}$ is the only solution.