TITLE: Line for set of three-dimensional vectors
QUESTION [6 upvotes]: If there is a set for 3D vectors $v$ where
$ v \times \begin{pmatrix} -1 \\ 1 \\ 4 \end{pmatrix} = \begin{pmatrix} 5 \\ -27 \\ 8 \end{pmatrix}$
is a line, what is this line's equation?
I'm not sure how to solve this, except for setting $v = \begin{pmatrix} a \\ b \\ c \end{pmatrix}$ and maybe having a linear system of equations, but I don't see how I could use that to solve the problem.

REPLY [2 votes]: Let $\vec{v}=\langle a, b, c\rangle, \;\; \vec{u}=\langle-1, 1, 4\rangle,\;\;\vec{w}=\langle5, -27, 8\rangle$.
Then $\vec{v}\times\vec{u}=\langle b-c, -4a-c, a+b\rangle,$ 
so $\vec{v}\times\vec{u}=\vec{w}\iff 4b-c=5,\; -4a-c=-27,\; a+b=8\iff c=4b-5 \text{  and  } a=8-b$,
so $\vec{v}=\langle8-b, b, 4b-5\rangle=\langle8,0,-5\rangle+b\langle-1, 1, 4\rangle$.
Therefore the line has parametric equations $\color{red}{x=8-t, \; y=t, \;z=4t-5}$.

$\textbf{Alternate solution}$:
With the same notation as above, let $\vec{z}=\vec{u}\times\vec{w}=\langle116, 28, 22\rangle$.
Then $\vec{z}\times\vec{u}=(\vec{u}\times\vec{w})\times\vec{u}=(\vec{u}\cdot\vec{u})\vec{w}-(\vec{u}\cdot\vec{w})\vec{u}=18\vec{w}$ since $\vec{u}\cdot\vec{u}=18$ and $\vec{u}\cdot\vec{w}=0$.
Then if $\vec{s}=\frac{1}{18}\vec{z}=\langle\frac{58}{9}, \frac{14}{9}, \frac{11}{9}\rangle\;$ and $\;\color{red}{\vec{r}=\vec{s}+t\vec{u}=\langle\frac{58}{9}-t, \frac{14}{9}+t, \frac{11}{9}+4t\rangle}$, 
$\hspace{.3 in}\vec{r}\times\vec{u}=(\vec{s}+t\vec{u})\times\vec{u}=\vec{s}\times\vec{u}+t(\vec{u}\times\vec{u})=\vec{w}+t\vec{0}=\vec{w}.$