Intersection of two planes - vector notation So we have the following planes:


*

*$x+2y-z=4$

*$x=z$
And we want to find the intersection of the two planes. So what I would do is substitute $x=z$ into the above equation, which would yield $$x+2y-x = 2y = 4 $$ hence $y=2$, $x=z$ is the intersection. How do I write this down in vector notation?
I would suspect $(0,2,0) + t(1,0,1)$ where $t$ is a scalar, but geogebra gives me $(2,2,2) + t(2,0,2)$. Now I know the second term is equivalent, but why the $(2,2,2)$ instead of the $(0,2,0)$?
 A: The general solution to the system of $2$ equations above is: $y = 2, x = z$, thus a line of intersection can be defined  to consist of points $(x,y,z)$ with $x = z, y = 2$. This gives: $(x,y,z) = (t,2,t) = (t,0,t) + (0,2,0)=t(1,0,1)+ (0,2,0)$, $t$ is any real number. If you let $t = 2$, you get the point $(2,2,2)$ and this point is on this line as well. This means you can select any point on the line and add it with $t(1,0,1)$ to complete the expression.
A: The vector equation of a line has the form $$r(t) = vt + b$$ where $v$ is a vector parallel to the line and $b$ is a point on the line.
Knowing this, to see that these are just two different parametrizations of the same line we just need to $(1)$ check that the "direction vectors" -- $(1,0,1)$ and $(2,0,2)$ in your equations -- are parallel to each other and $(2)$ check that $(0,2,0)$ is a point on the line $r_2(t_2) = (2,2,2)+t_2(2,0,2)$ OR that $(2,2,2)$ is a point on the line $r_1(t_1) = (0,2,0)+t_1(1,0,1)$.
If both of those conditions hold then these two equations must describe the same line.
