Proving a set of vectors are a basis Question  Let A $\subseteq {\mathbb R}^{3}$ be the plane $\{(x,y,z) \in \mathbb R^3 : 2x+3y-4z = 0\}$. Prove that $B = \{(-6,4,0),(5,-2,1)\}$ is a basis of the real vector space $A$.
So I have set $u = (a,b,c) \in A$
such that  $a = \frac{4c-2b}{2}$, then using this 
$u$ is now $(\frac{4c-2b}{2},b,c)$
$$ u = c (2,0,1)+ b(\frac{-3}{2},1,0)= c (2,0,1)+4b(-6,4,0)$$
I'm uncertain how to show the combination of both vectors span $A$.
 A: First, note that both vectors are in $A$. Second, note that $A$ has codimension 1, so dimension 2, so any two linearly independent vectors in $A$ form a basis of it (if that made no sense to you, simply note that it's not all of $\mathbb{R}^3$, so has dimension at most 2). It's then reasonably easy to check that the vectors in question are linearly independent, so the basis theorem gives that they span $A$ for free. 
To actually do that check: if $a(-6,4,0) + b(5,-2,1) = (0,0,0)$, then $(5b-6a,4a-2b,b)=(0,0,0)$. The last coordinate then gives us that $b = 0$, and either of the first two gives us that $a = 0$. Thus, $B$ is linearly independent. 
A: You mean $a = \frac{4c-3b}2$.
If you let $b=4, c=0$, then $a=\frac{4(0)-3(4)}2=-6$, hence you obtain the first vector.
If you let $b=-2, c=1$, then $a=\frac{4(1)-3(-2)}2=5$, that is how you get the second vector.
Since they are not multiple of each other and you know the dimension is $2$, it is a basis.
A: Both the vectors would span $A$ if they are perpendicular to its normal. The normal is $(2,3,-4)$ and the dot products $(-6,4,0).(2,3,-4)$ and $(5,−2,1).(2,3,-4)$ are $0$. Hence, they are two non-parallel vectors in the plane $A$ any combination of which in the form $a(-6,4,0)+b(5,−2,1)$ can get one to any point in the plane $A$.
