I am having trouble finding the intersection line between two planes:
$\prod_1 :x - y - z = 1$
and
$\prod_2 :2x-y=3$
I have managed to find the vector of intersection between these two planes by calculating the cross product between planes normals which equals to: $(-1,-2,1)$
Could someone please give me a hint or a hand of how to calculate a point that lies in both these planes? Since $\prod_2$ doesn't have any z coordinate it makes me confused on what to do next.
First we need check that the planes $\Pi_1$ and $\Pi_2$ they are intersects. In order to see it, noticed that $\frac{2}{1}\not=\frac{-1}{-1}$ so $\Pi_{1}\cap\Pi_2=L$ with $L$ the line intersection. Here, we can use two way to find $L$.
Solving the linear system: Indeed, we have $$\begin{cases}x-y-z=1\\2x-y=3\end{cases},$$then $$\begin{pmatrix}1&-1&-1&1\\2&-1&0&3 \end{pmatrix}\sim \begin{pmatrix}1&-1&-1&1\\0&1&2&1 \end{pmatrix}$$ So we have $y+2z=1$ then $y=1-2z$ and $x-y-z=1$ then $x=1+y+z=1+(1-2z)+z=2-z$. Setting $z:=t\in{\bf R}$ we get $$L:\begin{cases}x=2-t\\y=1-2t\\z=t\end{cases},\quad t\in {\bf R}$$
Find the director vector $s$ and a point $P\in L$: The director vector $s$ can be find using the vector product between the normal vectors of the planes, that is,$$s=\begin{vmatrix}\vec{i}&\vec{j}&\vec{k}\\1&-1&-1\\2&-1&0 \end{vmatrix}=(-1,-2,1)$$ For the point $P\in L$ it is enough to fix one of the variables and solve the system simultaneously for the other variables. Setting $z=0$, we have $$\begin{cases}x-y=1\\2x-y=3\end{cases}$$ Then, $x=2$ and $y=1$ so $P(2,1,0)\in L$ and therefore $$L:\begin{cases}x=2-t\\y=1-2t\\z=0+t\end{cases},\quad t\in {\bf R}$$
Any point of intersection between the planes would be a solution for the linear system composed with the equations of the two planes:
$$P(x,y,z)\in \Pi_1\cap \Pi_2\Rightarrow x-y-z=1, 2x-y=3\Rightarrow (x,y,z)=(2-z,1-2z,z)$$
The vector equation of a line is $$\vec r-\vec r_0=t\cdot \vec a, t\in \mathbb R, \vec r_0\text{ point on the line}, \vec a \text{ direction}$$
The difference between any two vectors pointing on the line would give me the direction vector: $$\Delta \vec r=t\vec a\Leftrightarrow (-\Delta z, -2\Delta z, \Delta z)=t\vec a\Leftrightarrow \Delta z\cdot(-1,-2,1)=t\vec a, \Delta z\ne 0$$
Therefore $$\vec a=(a_x, a_y, a_z)=(-1,-2,1)$$
For $\vec r_0$ I would choose $$z=0: \vec r_0=(x_0, y_0, z_0)=(2,1,0)$$
The line equation would be $$\frac{x-x_0}{a_x}=\frac{y-y_0}{a_y}=\frac{z-z_0}{a_z}$$
