Intersection of 2 Planes
1 Introduction

Two distinct planes in can interact in one of two ways:

  • They intersect in a line.
  • They are parallel and distinct (no intersection).

A third possibility is that the equations represent the same plane (coincident), in which case they "intersect" everywhere on the plane.

Let the two planes be:

2 Analyzing Normals
  • If and are not parallel (i.e., for any scalar ), the planes must intersect in a line. The direction vector of this line of intersection will be orthogonal to both and , so .

  • If and are parallel (i.e., ), the planes are parallel.

    • To check if they are distinct or coincident:
      If and .
      This means !
      If , the planes are coincident.
      If , the planes are parallel and distinct.
3 Finding the Line of Intersection

To find the equation of the line of intersection, we solve the system of two equations with three variables. Since there are more variables than equations (when they intersect in a line), there will be infinitely many solutions. We express these solutions in terms of a parameter.

3.1 Example : Planes Intersecting in a Line

Find the equation of the line of intersection of the planes:

Solution

Normals: , . These are not parallel, so the planes intersect in a line.

We need to solve the system. Eliminate one variable, say .

Subtract (1) from (2):

Let (parameter). Then .

Substitute and into equation (1) to find :

So, the parametric equations of the line of intersection are:

This can be written as

To avoid fractions, we can choose (so ).

Then

Line:

This is an equivalent representation.

Alternative method for direction vector:

This is , so it's a valid direction vector. A point on the line can be found by setting one variable to 0 (e.g., ) in the original system:

Subtracting: . Then . Point .

So

This matches.

3.2 Example : Parallel and Distinct Planes

Show that the planes and are parallel and distinct.

Solution

Normals:

,

We see that

so the normals are parallel, the planes are therefore parallel.

Now check if they are distinct or coincident.

The equation for is .

  • Compare with

    If and were coincident, then would be a scalar multiple of .

    If we multiply by 2, we get

  • Comparing

    with

    The coefficients of match (after scaling ), but the constant terms and do not satisfy the same scaling ( , here ). Therefore, the planes are parallel and distinct.

Alternatively, try to solve the system:

Multiply (1) by 2:

Subtract (1') from (2):

This is a contradiction, so there is no solution.
The planes do not intersect. Since their normals are parallel, they must be parallel and distinct.

3.3 Example : Coincident Planes

Determine the nature of intersection of and

Solution

Normals:

so the normals are parallel, the planes are parallel.

Check for coincidence:

Multiply by 3:

This is identical to the equation for . Therefore, the planes are coincident. They intersect at every point on the plane.

The "solution" is any point satisfying .
We can parameterize this: Let . Then .

So

This describes all points on the plane.

4 Homework
4.1 Question 3.1 (Easy)

Consider the planes and .
Determine if these planes are parallel, coincident, or intersect in a line. If they intersect in a line, find a direction vector for the line of intersection.

Solution

Normal vectors: ,

Check if normal vectors are parallel:
Is ?

Since for all components, .
The normal vectors are parallel, so the planes are parallel.

Now, check if they are distinct or coincident.

  • Plane 1: .
  • Plane 2: .

Multiply Plane 1 by :

Comparing this with Plane 2 ( ), the coefficients of match, but the constant terms are different ( ).
Therefore, the planes are parallel and distinct.

Answer: The planes are parallel and distinct. They do not intersect.

4.2 Question 3.2 (Medium)

Find the parametric equations of the line of intersection of the planes and .

Solution

The normal vectors are and . They are not parallel, so the planes intersect in a line.

We need to solve the system of equations:

Eliminate one variable, say .
Add (1) and (2):

Now express one variable in terms of another, or introduce a parameter.
From (3), .
Let (parameter).
Then .

Substitute and into one of the original plane equations (e.g., (1)) to find :

The parametric equations of the line of intersection are:

This can be written in vector form as .

Alternative method for direction vector:

This matches the direction vector found above.
To find a point on the line, set in parametric equations: .
Check if is on both planes:

  • . (Correct)
  • . (Correct)

Answer: The parametric equations are .

4.3 Question 3.3 (Harder)

Find the Cartesian equation of a plane that passes through the point and is perpendicular to the line of intersection of the planes and .

Solution

The line of intersection, , of and will have a direction vector that is perpendicular to both and , where and

The plane is perpendicular to the line . This means the normal vector of , let's call it , must be parallel to the direction vector of , .
So, we can choose .

The Cartesian equation of a plane with normal is , so:

The plane passes through the point . Substitute these coordinates into the equation to find :

So, the Cartesian equation of plane is .

Answer: The Cartesian equation of the plane is .

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