Find the basis for all vectors perpendicular to plane $x-2y+3z=0$ in $\mathbb{R}^3$ This can be rewritten as $A\vec{x}=0$, where $A=(1, -2, 3), \vec{x}=(x, y, z)^T$.
I understand that one can find vectors perpendicular to this place by finding the basis for the null space, as the null space is perpendicular to the row space.
This basis is $(2, 1, 0), (-3, 0, 1)$. I believe that the space describing all perpendicular vectors can be written as $a(2, 1, 0)^T+b (-3, 0, 1)^T$? In order to find the basis for this space, we solve for the null space of 
$$
  B=
  \left[ {\begin{array}{cc}
   2, 1, 0 \\
   3, 0, 1 \\
  \end{array} } \right]
$$
This gives you $(1, -2, 3)^T$ which makes sense as this described the original plane. But I don't understand why we combine the two solutions for the null space and transpose them to find the basis for all vectors $a(2, 1, 0)^T+b (-3, 0, 1)^T$. Could somebody explain this to me? Thank you.
Note: This question is about the same problem but has a different misunderstanding.
 A: Because we're working in a $3$-dimensional vector space and we have found $2$ linearly independent solutions $s_1$ ans $s_2$. Therefore, the space $S$ of all solutions must have dimension at least $2$. But it can't be $3$, since that would meant that $S=\mathbb{R}^3$, but it isn't ($(1\ \ -2\ \ 3)^T\notin S$). Therefore, $\dim S=2$ and $\{s_1,s_2\}$ is a basis of $S$.
A: This is because of linearity of the dot product. First, write your $A$ matrix as a vector $\vec y$, it will be simpler in a second, and note that what you mean by $A\vec x$ is nothing more than $\vec y\cdot \vec x$. I.e. your plane consists in all vectors perpendicular to the vector $y$. I guess you know this but I'm just settling notation.
Now, you have found two vectors $\vec u,\vec v$ such that $\vec y\cdot\vec u=0$ and $\vec y\cdot \vec v=0$. This means you can write a whole family of other vectors verifying the same. For instance, $2\vec u$ is also perpendicular to $y$. Now, go further and take any two real or complex values $a,b$. Then we have
$$\vec y \cdot (a\vec u+b\vec v)= a\vec y\cdot\vec u+b\vec y\cdot\vec v= 0.$$
Actually this family of vectors (one for each pair $(a,b)$) contains all vectors in the plane. To see why this happens, see @José Carlos Santos answer.
