Vtyjkyuk

The cross product is defined by the formula

$$ vec A_i times vec B_j = |A|,|B| sinalpha = vec C_k tag1$$

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

$$ vec C_k times vec A_i = vec B`_j tag2$$

The $B`$ vector has the correct direction but its value usually not equal $B$, because use $(1)$

$$ left(|A| , |B|right)_k times vec A_i =|A|^2 , |B| = vec B`_j tag3$$

but using the proportion $(1)$ we can easily reverse them

$$AB=C rightarrow B=Cover A tag4$$

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
$$|A|= sqrta_x^2 + a_y ^2 + a_z^2 = A $$
we must fulfill the condition
$$|A|;left|1over Aright| =1 \[20 pt]
sqrta_x^2 + a_y ^2 + a_z^2frac1sqrt a_x^2 + a_y ^2 + a_z^2 = 1 $$

this condition fulfill vector

$$overrightarrow 1over A =left( fraca_xA ,;fraca_yA,; fraca_zAright) tag5 $$

This vector has the same direction as the vector A but inverse value.

Reverse cross product $(2)$ we have now
$$ vec C_k times overrightarrow frac 1 A_i =|A|;|B|;left|1over Aright| = |B| = B`_j tag6$$

Whether saving a fraction of a vector $1/A$ is correct?