Let $T:R^3 \to R^3$ projection plane to line Let $T:R^3 \to R^3$ projection plane to line
If $T(a,b,c)=(a-c,b,0)$, show that T is  the projection on the $xy-$plane along the line $L=\{ (a,0,a): a \in \mathbb{R} \}$

The solution which i do not understand is

since $(a,b,c)=(a-c,b,0)+(c,0,c)$ we have $T(a,b,c)=(a-c,b,0)$


My intuition which is problaby wrong is that points in the xy plane are place along the line.
What is the right way to visualize this. Maybe it will help me understand the simple solution.


Hint:
look at the problem from the other side. Your transformation $T$ goes from a point $P=(x,y,z)$ to the point $T(P)=(x-z,y,0)$, so the vector from the two points is 
$$\vec v=\overline {PT(P)}=P-T(P)=(x-x+z,y-y,z-0)^T=(z,0,z)^T \quad \forall z$$
so the straight line that gives the transformation is a line parallel to:
$$
(x,y,z)^T=(1,0,1)^Tt=(t,0,t)^T
$$ 
