general vector solution given two others Reading this topic What's the opposite of a cross product?
I found that in order to get $a\times b=c$, b must be $ta+ (a\times c)/a^2$. Which makes sense, it obviously fit the equation, my question is, how can I arrive to that solution, and is that the only possibility?
 A: $c$ is perpendicular to $a$ and to $b$. Then, the first step is to find the plane of $a$ and $b$. For that, we need two vectors in the plane. One of them is obviously $a$, the other is $a\times c$. Why this choice? First, $a\times c$ is perpendicular to $c$, so it must be in the plane. Secondly, it is perpendicular to $a$. If you have two vectors in the plane, any other vector (in particular $b$) can be written as a linear combination of the two:$$b=\alpha a+\beta (a\times c)$$Here $\alpha$ and $\beta$ are some arbitrary numbers. However, you also know that $a\times b=c$, so $$a\times(\alpha a+\beta(a\times c))=c$$
By expanding the outer parenthesis, and using the properties of the cross product, we have $a\times a=0$ and $a\times(a\times c)=a^2c-(ac)a=a^2c$. I've used here that $a$ and $c$ are perpendicular, so $ac=0$. Then you get $\beta a^2c=c$ or $\beta=1/a^2$. So your final solution for $b$ is $$b=\alpha a+\frac 1{a^2}a\times c$$
A: You need $a\cdot c=0$ for $a\times b=c$ to be soluble for $b$.
Granted this, take the vector product with $a$ and use the formula
for the vector triple product:
$$a\times (a\times b)=a\times c$$
$$(a\cdot b)a-(a\cdot a)b=a\times c$$
$$ua-|a|^2 b=a\times c$$
$$b=\frac{ua-a\times c}{|a|^2}$$
where $u=(a\cdot b)$ is some scalar. Taking the vector product of this
with $a$, and using $a\cdot c=0$ shows this solution works.
