Find all vectors $\vec{v}=\begin{bmatrix}x\\y\\z\end{bmatrix}$ orthogonal to both $\vec{u_1}=\begin{bmatrix}2\\0\\-1\end{bmatrix}$ and $\vec{u_2}=\begin{bmatrix}-4\\0\\2\end{bmatrix}$
$\vec{u_1}$ and $\vec{u_2}$ are parallel, so the cross product will be $\vec{0}$. This won't help. The other thing I tried was based on the dot product, so:
$\vec{u_1}\cdot\vec{v}=0\\
\vec{u_2}\cdot\vec{v}=0$
leading to the system of equations:
$2x-z=0\\-4x+2z=0$
and then
$\vec{v}=s\begin{bmatrix}1\\0\\2\end{bmatrix}+t\begin{bmatrix}0\\1\\0\end{bmatrix}$
which makes sense to me because the orthogonal vectors would "rotate" around $\vec{u_1}$ and $\vec{u_2}$ in infinitely many directions.
Best Answer
Note that actually your two equations are the same equation. (Divide the second one by $-4$.) So you have $$ 2x-z=0 $$ as your constraint, so any vector parallel to $$ \begin{pmatrix} 1 \\ 0 \\ 2 \end{pmatrix} $$ will be orthogonal to both vectors (as will any with just a $y$ component, and by linearity, any linear combination thereof).