Types of intersections

1. Planes are parallel

When two planes have the same normal, and do not share a common point, they are called parallel.

example

Find the points of intersection of the planes $x + 2y + z = 4$ and $x + 2y + z = 5$ if any exist.

answer

We can see if the planes are parallel by looking their normals:

$\vec {n_1} = \vec {n_2} = (1, \ 2, \ 1) \ \Rightarrow \$ Both planes are parallel to one another

Let's rearrange the equations:

$x + 2y + z = 4 \ \Rightarrow \ x + 2y + z - 4 = 0$

$x + 2y + z = 5 \ \Rightarrow \ x + 2y + z - 5 = 0$

Equating the equations, we get:

$(x + 2y + z - 4) = (x + 2y + z - 5)$

$\quad \quad \quad \quad \quad \ \ \, \, -4 = - 5$

$\quad \quad \quad \ \ \ \, \, - 4 + 5 = 0$

$\quad \quad \quad \quad \quad \quad \ \, \, 1 \ne 0$

Obviously $1 \ne 0$, so there is no line that the planes intersect.

$\therefore$ The two planes are parallel.

2. Planes intersect at a line

When two planes are not parallel, they will share a line of intersection that contains an infinite number of points.

example

Find the points of intersection of the planes $3x + 2y - z = 1$ and $-x + 2y + 2z = 2$ if any exist.

answer

We can tell the planes are not parallel since their normals are not equal:

$\vec {n_1} = (3, \ 2, \ -1) \ne (-1, \ 2, \ 2) = \vec {n_2}$

This means they must intersect at a line

If we rearrange the values for z, we get:

$3x + 2y - z = 1 \ \Rightarrow \ z = 3x + 2y - 1$

$-x + 2y + 2z = 2 \ \Rightarrow \ z = \frac {1} {2} x - y + 1$

If we equate them to each other, we get the equation of a line:

$3x + 2y - 1 = \frac {1} {2} x - y + 1$

$6x + 4y - 2 = x - 2y + 2$

$5x + 6y - 4 = 0$

$\quad \quad \quad \quad \ \, \, y = - \frac {5} {6} x + \frac {2} {3}$

This is in fact the line of intersection of the planes!

$\therefore$ The planes intersect at the line $y = - \frac {5} {6} x + \frac {2} {3}$.

3. Plane contains plane

If one plane has the same or a multiple of the normal vector of another plane while also sharing a point, they contain each other.

example

Find the points of intersection of the planes $3x + 3y + 3z = 12$ and $x + y + z = 4$ if any exist.

answer

As usual, we should first check if they are parallel:

$\vec {n_1} = (3, \ 3, \ 3)$ and $\vec {n_2} = (1, \ 1, \ 1) \ \Rightarrow \ \vec {n_1} = 3 \vec {n_2}$

So, they are parallel.

Now, let's see if they share a line:

$3x + 3y + 3z = 12 \ \Rightarrow \ z = 4 - x - y$ for the first plane

$x + y + z = 4 \ \Rightarrow \ z = 4 - x - y$ for the second plane

As you can see, both planes are the same, which means they coincide.

$\therefore$ The planes are coincident.