The cross product is defined by the formula
$$ \vec A_i \times \vec B_j = \|A\|\,\|B\| \sin\alpha = \vec C_k \tag{1}$$
Where $\alpha$ is angle between vectors $A$ and $B$. $C$ is a pseudovector perpendicular to $A$ and $B$. And i, j, k is standard basis direction mutually perpendicular to each other.
Assuming $\alpha$ it equals $90$ degrees then $\sin90=1$. This discussion is only for this case.
Reverse cross product we encounter for a problem
$$ \vec C_k \times \vec A_i = \vec B`_j \tag{2}$$
The $B`$ vector has the correct direction but its value usually not equal $B$, because use $(1)$
$$ \left(\|A\| \, \|B\|\right)_k \times \vec A_i =\|A\|^2 \, \|B\| = \vec B`_j \tag{3}$$
but using the proportion $(1)$ we can easily reverse them
$$AB=C \rightarrow B={C\over A} \tag{4}$$
If we find the inverse of the vector $A$, we can in this case when $\alpha=90$ degrees reverse cross product. How to get a vector $1/A$? The value of the vector is $$\|A\|= \sqrt{a_x^2 + a_y ^2 + a_z^2} = A $$ we must fulfill the condition $$\|A\|\;\left\|{1\over A}\right\| =1 \\[20 pt] \sqrt{a_x^2 + a_y ^2 + a_z^2}\frac{1}{\sqrt{ a_x^2 + a_y ^2 + a_z^2}} = 1 $$
this condition fulfill vector
$$\overrightarrow {1\over A} =\left( \frac{a_x}{\|A\|^2} ,\;\frac{a_y}{\|A\|^2},\; \frac{a_z}{\|A\|^2}\right) \tag{5} $$
This vector has the same direction as the vector A but inverse value.
Reverse cross product $(2)$ we have now $$ \vec C_k \times \overrightarrow {\frac {1} {A_i}} =\|A\|\;\|B\|\;\left\|{1\over A}\right\| = \|B\| = B`_j \tag{6}$$
Whether saving a fraction of a vector $1/A$ is correct?