Line for set of three-dimensional vectors 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.
 A: 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 $\;\color{red}{\vec{r}=\vec{s}+t\vec{u}=\langle\frac{58}{9}-t, \frac{14}{9}+t, \frac{11}{9}+4t\rangle}$, 
$\hspace{.3 in}\vec{r}\times\vec{u}=(\vec{s}+t\vec{u})\times\vec{u}=\vec{s}\times\vec{u}+t(\vec{u}\times\vec{u})=\vec{w}+t\vec{0}=\vec{w}.$
A: It certainly comes down to solving a system of linear equations 
You need the cross product formula here, for which a determinant form exists.
Information about that can be found here:
http://tutorial.math.lamar.edu/Classes/CalcII/CrossProduct.aspx
So if you call your vector $v=<a,b,c>$ Then all you need to do is to set up the determinant with the first row $x,y,z$, the second row $a,b,c$ and the third row $-1,1,4$
Working out the determinant results in:
$x(4b-c)-y(4a+c)+z(a+b)$. Bring the negative in front of the $y$ inside to obtain the following equations:
$4b-c=5$ , $-4a-c=-27$ and $a+b=8$ Can you solve these equations? Now how about that line? Remember that the zero vector technically also work so a parametric equation of this "line" can thus be obtained.
A: $
\newcommand{\i}{\hat{\mathbf{i}}}
\newcommand{\j}{\hat{\mathbf{j}}}
\newcommand{\k}{\hat{\mathbf{k}}}
\newcommand{\v}{\vec{\mathbf{v}}}
$
Let $v =  \v$ and 
$$
\v = \begin{bmatrix} a \\ b \\ c \end{bmatrix},
\quad 
\i = \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix},
\quad 
\j = \begin{bmatrix} 0 \\ 1 \\ 0  \end{bmatrix},
\quad 
\k = \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix}
$$
Then
$$
v \times \begin{bmatrix} -1 \\ 1 \\ 4 \end{bmatrix} 
=
\begin{bmatrix} 5 \\ -27 \\ 8 \end{bmatrix}
$$
can be rewritten as 
$$
\v \times \begin{bmatrix} -1 \\ 1 \\ 4 \end{bmatrix} =
\begin{bmatrix} a \\ b \\ c \end{bmatrix} 
\times 
\begin{bmatrix} -1 \\ 1 \\ 4 \end{bmatrix}
=
\begin{bmatrix} 5 \\ -27 \\ 8 \end{bmatrix}
$$
Let us expand vector product:
$$
\begin{aligned}
\v \times \begin{bmatrix} -1 \\ 1 \\ 4 \end{bmatrix} 
=
\begin{vmatrix} 
\i & \j & \k \\
a & b & c \\
-1 & 1 & 4
\end{vmatrix}
=
\begin{vmatrix} b & c \\ 1 & 4 \end{vmatrix} \i-
\begin{vmatrix} a & c \\ -1 & 4 \end{vmatrix} \j+
\begin{vmatrix} a & b \\ -1 & 1 \end{vmatrix} \k
=
\begin{bmatrix} 5 \\ -27 \\ 8 \end{bmatrix}
\end{aligned}
$$
Thus, we conclude 
$$
\begin{cases}
4 b - c = 5 \\ 
4 a + c = 27 \\
a + b = 8
\end{cases}
\implies 
\begin{cases}
4 b - c = 5 \\ 
4 a + 4b = 32 \\
a  = 8 - b
\end{cases}
\implies 
\begin{cases}
4 b - c = 5 \\ 
32 + 8b = 32 \\
a  = 8 - b
\end{cases}
\implies 
\begin{cases}
c = -5 \\ 
b = 0 \\
a  = 8
\end{cases}
$$
Finally, we write 
$$
\bbox[9px, border:3px solid #FF0000]{v =   \begin{pmatrix} 8 \\ 0 \\ -5 \end{pmatrix}}
$$
