Say I have a vector $\vec{b}$ that projects onto the column space of $\vec{a}$ to create a projected vector $\vec{p}$. This would also result in a perpendicular vector, which I call it $\vec{e}=\vec{b}-\vec{p}$. I used to understand that the vector $\vec{e}$ is in the nullspace of $N(\vec{a}^{T})$ because the dot product of the vector $\vec{e}\cdot\vec{p}=0$.
Recently, I thought that I could have misinterpreted the whole thing because I suddenly thought that $\vec{e}$ could have not been in the nullspace of $N(\vec{a}^{T})$ at all. I drew a sketch of a 2 dimension graph to visualise this thing but I think it applies to an extension of higher dimensions. In the picture below, it seems that $\vec{e}$ is only in the nullspace if it is $\vec{e^{'}}$ which I drew it in pink along the nullspace - that's $\vec{e}$ being moved all the way back.
So in this case, yes, any perpendicular vector along $\vec{a}$ would result in a dot product of zero scalar value. But how is $\vec{e}$ in the nullspace?
If $\vec{e}$ is not in the nullspace, then why does the projection formula $A^{T}\vec{e} =\vec{0} \Rightarrow A^{T}(\vec{b}-\vec{p})=\vec{0}$ works by really returning zero? It wouldn't have returned zero if $\vec{e}$ wasn't in the nullspace, would it?
Thanks for any help on this!