$\vec{v} = \left[\begin{matrix}0\\2\\2\end{matrix}\right]$ and $\vec{w} = \left[\begin{matrix}2\\2\\3\end{matrix}\right]$
I know that the linear combination is $a\left[\begin{matrix}0\\2\\2\end{matrix}\right] + b\left[\begin{matrix}2\\2\\3\end{matrix}\right] = \left[\begin{matrix}2b\\2a+2b\\2a+3b\end{matrix}\right]$ where $a$ and $b$ are scalars.
But I'm a little confused when it comes to imagine what the geometry is like for this linear combination. To me, I think this linear combination fill a plane, because $\left[\begin{matrix}1\\0\\0\end{matrix}\right]$ is not a linear combination of $\vec{v}$ and $\vec{w}$.
I wonder if two vectors in 3D could only fill a line or a plane. Hope someone can help me clarify it. Thanks in advance!
It is clear that two vectors that are not both $0$ can span either a plane (if they are linearly independent), or a line (if they are linearly dependent). The problem, then, asks you in fact to check whether those two vectors are linearly independent or not. You can check this either in the quick way (notice that if $w = \alpha v$ then $2=0$, and if $v = \alpha w$ then $\alpha=0$ so $v=0$ - which are both contradictions), or in the long way (check that the rank of $\begin{pmatrix} 0 & 2 \\ 2 & 2 \\ 2 & 3 \end{pmatrix}$ is maximal, i.e. $2$). The second approach, even if longer, works for arbitrarily many vectors, whereas the first, even though quick, does not help for more than two vectors.