How to find $rk(x+2y-4z)$? I'm currently studying for a test on Friday. I'm doing the following exercise:

Find the dimension of $$P = \big\{(x,y,z) \in \mathbb{R}^{3} \mid x+2y-4z=0\big\}$$

We know that 
$$f\begin{pmatrix}x\\y\\z\\\end{pmatrix} = x+2y-4z$$
is linear and that $P = \ker(f)$. So, thanks to the rank–nullity theorem, we can say that
$$\dim(\mathbb{R}^{3}) = rk(f) + \dim(\ker(f))$$
In class we said that the rank has something to do with the surjectivity of $f$. However I still cannot figure it out. Thanks.

$P = \big\{(x,y,z) \in \mathbb{R}^{3}\ |\ x+2y-4z=0\big\}$
So we can notice that the set $P$ is a plane in $\mathbb R^3$ since it have the equation $x+2y-4z=0$, thus $dim(P)=2$
$P$ is a vector space also since $O(0,0,0) \in P$
