Hello
Originally Posted by
caelestis Do you know why this happens or what it means??
Given two vectors $\displaystyle \vec{a}$ and $\displaystyle \vec{c}$, if a vector $\displaystyle \vec{b}$ is a solution of $\displaystyle \vec{a}\times\vec{v}=\vec{c}\,\,\,\,\,\,(1)$ then for all $\displaystyle t\in\mathbb{R}$ one has $\displaystyle \vec{a}\times(\vec{b}+t\vec{a})=\vec{a}\times \vec{b}+\vec{0}=\vec{c}$ : every vector $\displaystyle \vec{b}+t\vec{a}$ is also a solution of (1). That's why if there is one solution then there is an infinite number of solutions.
The initial question can also be answered using another method : Given two vectors $\displaystyle \vec{a}$ and $\displaystyle \vec{c}$, (both different from $\displaystyle \vec{0}$) we want to solve $\displaystyle \vec{a}\times \vec{v}=\vec{c}\,\,\,\,\,(1)$ for $\displaystyle \vec{v}$.
_ If $\displaystyle \vec{a}$ and $\displaystyle \vec{c}$ aren't orthogonal, this equation has no solution.
_ If $\displaystyle \vec{a}\perp \vec{c}$ then $\displaystyle \vec{v}$ has to be orthogonal to $\displaystyle \vec{c}$ too. As $\displaystyle \vec{a}$ and $\displaystyle \vec{a}\times \vec{c}$ aren't collinear and are both orthogonal to $\displaystyle \vec{c}$, the solutions of (1) can be written $\displaystyle \vec{v}=\alpha \vec{a}+\beta \vec{a}\times \vec{c}$ for $\displaystyle \alpha,\beta \in\mathbb{R}$. If we put this in (1), we get
$\displaystyle \begin{aligned}\vec{a}\times(\alpha \vec{a}+\beta \vec{a}\times \vec{c})=\vec{c}& \Longleftrightarrow \beta \vec{a}\times (\vec{a}\times \vec{c})=\vec{c}\\
& \Longleftrightarrow -\beta \vec{a}^2\vec{c}=\vec{c}\,\,\,\,\,(\text{triple product expansion})\\
& \Longleftrightarrow \beta=-\frac{1}{\vec{a}^2}\end{aligned}$
There is no condition on $\displaystyle \alpha$ and the value of $\displaystyle \beta$ is $\displaystyle -\frac{1}{a^2}$. In your case, $\displaystyle \vec{a}=(-1,2,3)$ and $\displaystyle \vec{c}=1,5,-3)$ which gives $\displaystyle \vec{a}^2=1+2^2+3^2=14$ hence the solutions are
$\displaystyle
\vec{v}=\alpha\begin{pmatrix}-1\\2\\3\end{pmatrix}+\frac{1}{14}\begin{pmatrix}-1\\2\\3\end{pmatrix}\times\begin{pmatrix}1\\5\\-3\end{pmatrix}=\alpha\begin{pmatrix}-1\\2\\3\end{pmatrix}+\begin{pmatrix}3/2\\0\\1/2\end{pmatrix}$
which happens to be Mr. Fantastic's solution. $\displaystyle \left(\alpha=\frac{t}{3}-\frac{1}{6}\right)$