I got a question which says :
Given $$\vec{v} = <1,0,-1> $$ and line $$L_1 : (1-2t)\vec{i}+(4+3t)\vec{j}+(9-4t)\vec{k}$$ Find an equation of plane $P$ which is parallel to the vector $\vec{v}$ and containing the line $L_1$.
In solution section, it says ;
$\vec{n}$ is to be normal vector of $P$ , $\vec{v_1}$ is to be direction vector of $L_1$, $\vec{n}$ // ($\vec{v}$ x $\vec{v_1}$)
and I couldn't understand that step of solution.
$n \perp P$ and $v \parallel P$, so $n \perp v$.
Then $L \in P$ and $n \perp P$, so $n \perp L$ and $v_1 \parallel L$, so $n \perp v_1$. So $\DeclareMathOperator{span}{span}n \in \span(v, v_1)^\top$.
Further $v\times v_1 \perp v$ and $v \times v_1 \perp v_1$. This is a property of the vector product. So $v \times v_1 \in \span(v, v_1)^\top$.
$\span(v,v_1)$ is two-dimensional and in $\mathbb{R}^3$ then $\span(v,v_1)^\top$ is one-dimensional, this leaves only $n \,\|\, v \times v_1$.
Note: This assumes that $v$ and $v_1$ are not linear dependent, $v \nparallel v_1$.