Types Of Consistent Intersections

Consistent systems are when all three planes intersect at either the same point, line, or are coincident.

The four types of consistent intersections are:

1. Planes intersect at a point

example

Find the intersection of the following planes, if any exist:

$x + 2y - z + 2 = 0 \quad (1) \newline 2x + y - z + 1 = 0 \quad (2) \newline x - 3y + z = 0 \quad (3)$

answer

First, let's create two new equations by subtracting the equations we already have:

$\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \ \ \, \, 2 \times (1) - (2) = 0$

$(2x + 4y - 2z + 4) - (2x + y - z + 1) = 0$

$\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \ \ \ \, 3y - z + 3 = 0 \quad (4)$

$\quad \quad \quad \quad \quad \quad \quad \quad \quad \ \ \, \, (1) - (3) = 0$

$(x + 2y - z + 2) - (x - 3y + z) = 0$

$\quad \quad \quad \quad \quad \quad \quad \quad \ \ \, 5y -2z + 2 = 0 \quad (5)$

Now, let's subtract the equations to get a value for $y$:

$\quad \quad \quad \quad \quad \quad \quad \, 2 \times (4) - (5) = 0$

$(6y - 2z + 6) - (5y - 2z + 2) = 0$

$\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \ \, \, y + 4 = 0$

$\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \, \, y = -4$

Let's find a value for $x$ and $z$ using equations $(4)$ and $(1)$:

$3y - z + 3 = 0 \ \Rightarrow \ 3(-4) - z + 3 = 0 \ \Rightarrow \ z = -9$

$x + 2y - z + 2 = 0 \ \Rightarrow \ x + 2(-4) - (-9) + 2 = 0 \ \Rightarrow \ x = -3$

So, all three planes intersect at the point $(-3, \ -4, \ -9)$

$\therefore$ The planes intersect at the point $(-3, \ -4, \ -9)$.

2a. Planes intersect at the same line, but with different angles

example

Find the intersection of the following planes, if any exist:

$2x - y + z - 1 = 0 \quad (1) \newline 3x - 5y + 4z - 3 = 0 \quad (2) \newline 3x + 2y - z = 0 \quad (3)$

answer

Let's find the intersection between planes 1 and 2:

$\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \ \, \, 3 \times (1) - 2 \times (2) = 0$

$(6x - 3y + 3z - 3) - (6x - 10y + 8z - 6) = 0$

$\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \ \ \ \, 7y - 5z + 3 = 0 \quad (4)$

Now let's do the same for planes 2 and 3:

$\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \, \, (2) - (3) = 0$

$(3x - 5y + 4z - 3) - (3x + 2y - z) = 0$

$\quad \quad \quad \quad \quad \quad \quad \quad \quad \ \, -7y + 5z - 3 = 0$

$\quad \quad \quad \quad \quad \quad \quad \quad \quad \ \ \ \ \, 7y - 5z + 3 = 0 \quad (4)$

So we actually end up getting the same line.

Let's just check to see if planes 1 and 3 also intersect at the same line:

$\quad \quad \quad \quad \quad \quad \quad \quad \ \, 3 \times (1) - 2 \times (3) = 0$

$(6x - 3y + 3z - 3) - (6x + 4y - 2z) = 0$

$\quad \quad \quad \quad \quad \quad \quad \quad \quad \ \ \ \, -7y + 5z - 3 = 0$

$\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \ \ \, 7y - 5z + 3 = 0 \quad (4)$

Again we get the same line.

This means that all the planes intersect at the same line $7y - 5z + 3 = 0$.

However, we can tell that they all intersect at different angles since their normals are different:

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

$\vec {n_2} = (3, \ -5, \ 4)$

$\vec {n_3} = (3, \ 2, \ -1)$

$\vec {n_1} \ne \vec {n_2} \ne \vec {n_3}$

$\therefore$ The planes intersect at the same line, but with different angles.

2b. Planes intersect at the same line, but two of them are coincident

example

$x + 3y - 2z + 1 = 0 \quad (1) \newline 2x + 6y - 4z + 2 = 0 \quad (2) \newline 2x - y - z = 0 \quad (3)$

answer

Before doing anything, we can see that equations $(1)$ and $(2)$ are parallel:

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

$\vec {n_2} = (2, \ 6, \ -4)$

$\vec {n_2} = 2 \vec {n_1}$

Let's check if they share a point by plugging in $x = 0$ and $y = 0$ in both equation:

$x + 3y - 2z + 1 = 0 \ \Rightarrow \ (0) + 3(0) - 2z + 1 = 0 \ \Rightarrow \ z = \frac {1} {2}$

$2x + 6y - 4z + 2 = 0 \ \Rightarrow \ 2(0) + 6(0) - 4z + 2 = 0 \ \Rightarrow \ z = \frac {1} {2}$

Both planes share the point $(0, \ 0, \ \frac {1} {2})$, and therefore they coincide

This means the three planes intersect at a line, and two of those planes coincide.

Now we must find the line at which they intersect:

$\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \ \ \, \, (2) - (3) = 0$

$(2x + 6y - 4z + 2) - (2x - y - z) = 0$

$\quad \quad \quad \quad \quad \quad \quad \quad \quad \ \ \, 7y - 3z + 2 = 0$

$\therefore$ The planes intersect at the line $7y - 3z + 2 = 0$, and two of them are coincident.

3. Planes are coincident

example

Find the intersection of the following planes, if any exist:

$x + y + z - 1 = 0 \quad (1) \newline 2x + 2y + 2z - 2 = 0 \quad (2) \newline 3x + 3y + 3z - 3 = 0 \quad (3)$

answer

Obviously, these planes appear to be parallel since all of their normals are scalar multiples of one another:

$\vec {n_1} = (1, \ 1, \ 1)$

$\vec {n_2} = (2, \ 2, \ 2)$

$\vec {n_3} = (3, \ 3, \ 3)$

$\vec {n_2} = 2 \vec {n_1}$

$\vec {n_3} = 3 \vec {n_1}$

Let's check if planes $(1)$, $(2)$ and $(3)$ all share a point by plugging in $x = 0$ and $y = 0$:

$x + y + z - 1 = 0 \ \Rightarrow \ (0) + (0) + z - 1 = 0 \ \Rightarrow \ z = 1$

$2x + 2y + 2z - 2 = 0 \ \Rightarrow \ 2(0) + 2(0) + 2z - 2 = 0 \ \Rightarrow \ z = 1$

$3x + 3y + 3z - 3 = 0 \ \Rightarrow \ 3(0) + 3(0) + 3z - 3 = 0 \ \Rightarrow \ z = 1$

$\therefore$ Since all planes are parallel and share the point $(0, \ 0, \ 1)$, they coincide each other.

Types Of Inconsistent Intersections

There are four inconsistent systems, at which not all of the planes intersect at the same point, line or plane.

Here is what they look like:

1. planes form a triangular prism

example

Find the solution to the following system:

$3x + 2y + z = 0 \quad (1) \newline x + 2y + 3z - 4 = 0 \quad (2) \newline x + y + z - 16 = 0 \quad (3)$

answer

Before starting, we know that these planes are not parallel since none of them have scalar multiples of each other's normals.

Let's find the intersection between planes $(1)$ and $(2)$:

$\quad \quad \quad \quad \quad \quad \quad \quad \quad \ \, \, 3 \times (1) - (2) = 0$

$(9x + 6y + 3z) - (x + 2y + 3z - 4) = 0$

$\quad \quad \quad \quad \quad \quad \quad \quad \quad \ \ \ \, \, 8x + 4y + 4 = 0 \quad (4)$

Now let's find the intersection between $(2)$ and $(3)$:

$\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \ \ \, \, (2) - 3 \times (3) = 0$

$(x + 2y + 3z - 4) - (3x + 3y + 3z - 48) = 0$

$\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \ \, \, -2x - y + 44 = 0 \quad (5)$

Finally, let's calculate the intersection between $(1)$ and $(3)$:

$\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \, \, (1) - (3) = 0$

$(3x + 2y + z) - (x + y + z - 16) = 0$

$\quad \quad \quad \quad \quad \quad \quad \quad \quad \, 2x + y + 16 = 0 \quad (6)$

So all of these lines are different, but we don't know if they will intersect or not.

Let's check if the lines $(4)$ and $(5)$ intersection:

$\quad \quad \quad \quad \quad \quad \quad \quad \quad \, (4) + 4 \times (5) = 0$

$(8x + 4y + 4) + (-8x - 4y + 176) = 0$

$\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \, 180 \ne 0$

Since we can't find a valid solution, lines $(4)$ and $(5)$ don't intersect.

Now let's check if the lines $(5)$ and $(6)$ intersect:

$\quad \quad \quad \quad \quad \quad \quad \quad \quad \ \ \, \, (5) + (6) = 0$

$(-2x - y + 44) + (2x + y + 16) = 0$

$\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \ \ \, 60 \ne 0$

So it looks like lines $(5)$ and $(6)$ don't intersect either.

Finally, let's check if lines $(4)$ and $(6)$ intersect:

$\quad \quad \quad \quad \quad \quad \quad \ \ \, \, (4) - 4 \times (6) = 0$

$(8x + 4y + 4) - (8x + 4y + 64) = 0$

$\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \ \ \, -60 \ne 0$

Again, we see that lines $(4)$ and $(6)$ don't intersect.

Since none of the lines of intersection intersect, then the solution to the three planes is a triangular prism.

$\therefore$ The planes intersect in a triangular prism.

2. Two parallel planes intersect a third plane

example

Find the solution to the following system:

$2x + 4y - 2z - 4 = 0 \quad (1) \newline x + 2y - z - 6 = 0 \quad (2) \newline x + y - 2z = 0 \quad (3)$

answer

Right off the bat, we can see that the planes $(1)$ and $(2)$ are parallel, since they have normals that are scalar multiples of one another:

$\vec {n_1} = (2, \ 4, \ -2)$

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

$\vec {n_1} = 2 \vec {n_2}$

Plane $(3)$ does not seem to be parallel, so we may be dealing with two parallel planes and another that intersects through both of them.

Let's find the intersection between planes $(1)$ and $(3)$:

$\quad \quad \quad \quad \quad \quad \quad \quad \ \ \ \, \, (1) - 2 \times (3) = 0$

$(2x + 4y - 2z - 4) - (x + y - 2z) = 0$

$\quad \quad \quad \quad \quad \quad \quad \quad \quad \ \ \ \, \, x + 3y - 4 = 0$

So planes $(1)$ and $(2)$ intersect in a line.

Now let's find the intersection for planes $(2)$ and $(3)$:

$\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \, \, 2 \times (2) - (3) = 0$

$2 \times (2x + 4y - 2z - 12) - (x + y - 2z) = 0$

$\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \ \ \, \, x + 3y - 12 = 0$

By simply looking at the normals of both of the lines, we can see that they are parallel:

$\vec {n_1} = (1, \ 3, \ 0)$

$\vec {n_2} = (1, \ 3, \ 0)$

However, they seem to be shifted from each other since they have a different value for their term not attached to a variable.

Since we have two planes that are paralled to each other, that also intersect another plane at parallel lines, we have solved the system.

$\therefore$ Two planes are parallel and they both intersect the third plane.

3. Three parallel planes

example

Find the solution to the following system:

$x + y + z - 1 = 0 \quad (1) \newline 2x + 2y + 2z - 5 = 0 \quad (2) \newline 4x + 4y + 4z - 1 = 0 \quad (3)$

answer

Starting off, we can see that all the planes are parallel, since their normals are scalar multiples of each other:

$\vec {n_1} = (1, \ 1, \ 1)$

$\vec {n_2} = (2, \ 2, \ 2)$

$\vec {n_3} = (4, \ 4, \ 4)$

$\vec {n_3} = 2 \vec {n_2} = 4 \vec {n_1}$

Let's check if they coincide, by testing if they share a point at $x = 0$ and $y = 0$:

$x + y + z - 1 = 0 \ \Rightarrow \ (0) + (0) + z - 1 = 0 \ \Rightarrow \ z = 1$

$2x + 2y + 2z - 5 = 0 \ \Rightarrow \ 2(0) + 2(0) + 2z - 5 = 0 \ \Rightarrow \ z = \frac {5} {2}$

$4x + 4y + 4z - 1 = 0 \ \Rightarrow \ 4(0) + 4(0) + 4z - 1 = 0 \ \Rightarrow \ z = \frac {1} {4}$

We can see that none of the planes share a point, so they are all parallel.

$\therefore$ The planes are parallel.

4. Three parallel planes with two coinciding each other

example

Find the solution to the following system:

$x + y + z - 1 = 0 \quad (1) \newline 2x + 2y + 2z - 2 = 0 \quad (2) \newline 4x + 4y + 4z - 1 = 0 \quad (3)$

answer

Again, we can see that all the planes are parallel by looking at their normals:

$\vec {n_1} = (1, \ 1, \ 1)$

$\vec {n_2} = (2, \ 2, \ 2)$

$\vec {n_3} = (4, \ 4, \ 4)$

$\vec {n_3} = 2 \vec {n_2} = 4 \vec {n_1}$

Now let's check whether or not the planes share a point at $x = 0$ and $y = 0$:

$x + y + z - 1 = 0 \ \Rightarrow \ (0) + (0) + z - 1 = 0 \ \Rightarrow \ z = 1$

$2x + 2y + 2z - 2 = 0 \ \Rightarrow \ 2(0) + 2(0) + 2z - 2 = 0 \ \Rightarrow \ z = 1$

$4x + 4y + 4z - 1 = 0 \ \Rightarrow \ 4(0) + 4(0) + 4z - 1 = 0 \ \Rightarrow \ z = \frac {1} {4}$

So it appears that two of them coincide.

$\therefore$ The planes are all paralle, and two of them coincide each other.