Find Parametrizations For The Lines In Which The Planes Intersect: X + Y + Z = 1, X + Y = 2

Question

We need a point on the line of intersection. To get it, use the equations of the given planes as a system of linear equations. If we set z = 0

Can´t understand how x+y=2 and x+y= 1 when z = 0

Answer 1

Can´t understand how x+y=2 and x+y= 1 when z = 0

Because they don't. That just means that 0 is not in the intersection of the two planes.

The intersection of two planes is defined as the set of points that satisfies BOTH equations. You see that points $(x, y, 0)$ cannot satisfy the equation therefore they are not on the intersection of these planes.

The question here is to find the parametrization of the lines ($x(t), y(t), z(t)$) that satisfies the assertion $x+y+z = 1$ and $x+y=2$.

So we search for the set of points $(x, y, z)$ that satisfies the system:

$ \left \{ \begin{matrix} x+y+z = 1 \\ x+y = 2 \end{matrix} \right. \\ \left \{ \begin{matrix} x+y+z - (x + y) = 1 - 2\\ x+y = 2 \end{matrix} \right. \\ \left \{ \begin{matrix} z = -1\\ x+y = 2 \end{matrix} \right. \\ \left \{ \begin{matrix} z = -1\\ y = 2-x \end{matrix} \right. \\ $

Thus the points of form $(x, 2-x, -1)$ are the intersection of the planes. And the parametrization of this line is :

$x(t) = t \\ y(t) = 2-t \\ z(t) = -1$