Basis of of subspaces of $R^3$ I'm having a little trouble with this question:
How do I find the basis of the set of vectors lying in the plane 2x − y − z = 0?
I'm stuck on how to start on this question
I tried to start by setting y= 2x-z
I'm not sure where to go from here 
 A: If we set $y=0$, then $x$ and $z$ satisfy the equation $2x-z=0$. Note that for every real number $\alpha$, the vector 
$$\begin{pmatrix} x  \\ y \\ z \end{pmatrix}=\alpha\begin{pmatrix} 1 \\ 0 \\ 2 \end{pmatrix}$$
satisfies this equation. Now set $z=0$, then for every real number $\beta$ the vector
$$\begin{pmatrix} x  \\ y \\ z \end{pmatrix}=\beta\begin{pmatrix} 1 \\ 2 \\ 0 \end{pmatrix}$$
satisfies $2x-y=0$. If we add these vectors together, they satisfy the equation $2x-y-z=0$. Indeed:
$$2x-y-z = 2(\alpha+\beta)-\alpha-\beta=0$$
So a basis of this plane is given by
$$\mathcal{B} =\left\{\begin{pmatrix} 1 \\ 0 \\ 2 \end{pmatrix},\begin{pmatrix} 1 \\ 2 \\ 0 \end{pmatrix}\right\}$$
A: Your start $y= 2x-z$ is good.  It means that every vector
$$\begin{pmatrix}x\\2x-z\\z\end{pmatrix}, \quad\text{$x,y$ being real numbers,}$$
belongs to the plane.  Hence every point in that plane may be expressed as
$$\begin{pmatrix}x\\2x-z\\z\end{pmatrix}=x\begin{pmatrix}1\\2\\0\end{pmatrix}+
z\begin{pmatrix}0\\-2\\1\end{pmatrix}.$$
