How to find the union&intersection of two lines by their equations?

Question

I will try to be as clear as possible concerning my confusion, and I will use some examples(several ones).

Case number 1.

Assume two equations(in cartesian form) of two planes.

$2x+2y-5z+2=0$ and $x-y+z=0$ Now,we need to find their vectors. For the first on, we get: {1,-1,0}, {0,5/2,1} and {1,0,2/5}. For the second equation, we get: {1,1,0}, {0,1,1},and {1,0,-1}. Now, I have a hard time understanding how I have to figure out where is their intersection and their union?

Case 2:

Assume the one of the previous planes $x-y+z=0$ and the line $x-y=0$. How do I find the intersection and union of these two?

Answer 1

Case 2 is simple:

$x-y+z=0$ and $x−y=0$

means that (substitute second into first)

$z=0$,

i.e. your solution lies on the plane $z=0$. This means that your solution is just the line $y=x$ since $x−y=0$.

You can see this because $x-y+z=0$ is a plane that intersects the $z=0$ plane exactly on that line.

Since the line defined by $x−y=0$ is contained in the plane $x-y+z=0$, then the union is just the plane $x-y+z=0$.

Answer 2

First of all: $Ax+By+Cz+D=0$ is plane equation.

Case 1: Intersection of to planes is line. To find equation of that line you have to solve system of equations: $$ 2x+2y-5z+2=0\\ x-y+z=0 \Rightarrow x=y-z \\ $$ If we substitute second equation into first we got $$ 2(y-z)+2y-5z+2=0 \Rightarrow 4y-7z+2=0 \Rightarrow y=\frac{7z-2}{4} $$ and for x $$ x=\frac{7z-2}{4}-z=\frac{3z-2}{4} $$ Now we have parametric equation of the intersection $$ x=-\frac{1}{2}+\frac{3}{4}z\\ y=-\frac{1}{2}+\frac{7}{4}z\\ z=0+1z $$ It's line equation which can be written in another form $$ \frac{x+\frac{1}{2}}{\frac{3}{4}}=\frac{y+\frac{1}{2}}{\frac{7}{4}}=\frac{z}{1} $$ If we multiply denominators by 4 we get shorter equation $$ \frac{x+\frac{1}{2}}{3}=\frac{y+\frac{1}{2}}{7}=\frac{z}{4} $$

Case 2: $x-y=0$ is not line equation - it's plane equation $x-y+0z=0$. So this case is also intersection of two planes - like case 1.

Solution: $$ x-y+z=0 \Rightarrow x=y-z\\ x-y=0 \Rightarrow y-z-y=0 \Rightarrow z=0 $$ Using $x=y-z$ we have $x=y$. So, intersection is a line whose equation is $$ x=0+1y\\ y=0+1y\\ z=0+0y $$ or in another form $$ \frac{x-0}{1}=\frac{y-0}{1}=\frac{z-0}{0} $$