Intersection of Lines & Planes
1 Introduction

In this section, we explore how lines and planes can interact in three-dimensional space ( ), and how two lines can intersect in both and . Understanding these intersections is fundamental to solving geometric problems.

2 Intersection of a Line with a Plane

A line and a plane in can have three possible types of intersections:

  1. No Points of Intersection: The line is parallel to the plane and distinct from it.
  2. One Point of Intersection: The line passes through the plane.
  3. Infinite Points of Intersection: The line lies entirely within the plane.

To find the intersection, we typically use the parametric equation of the line and the Cartesian equation of the plane.

  • Line

  • Plane : .

Substitute the parametric equations of the line into the Cartesian equation of the plane and solve for the parameter .

2.1 Example : Line Intersecting a Plane at a Point

Find the point of intersection of the plane and the line

Solution

Substitute the parametric equations of into :

Since we found a unique value for , there is one point of intersection. Substitute back into the line's equations:

The point of intersection is .

2.2 Example : Line Parallel to a Plane

Determine if the line intersects the plane

Solution

Parametric equations for

Substitute into :

This was a mistake in setting up the example for "parallel". Let's adjust the plane or line. Consider the line

and plane

  • Direction vector of line: .
  • Normal vector of plane: .
  • If , the line is parallel to the plane.

So, the line is parallel to the plane. Now we check if it lies on the plane or is distinct.

Substitute

into

:

This is a contradiction. Thus, there is no solution for .
The line is parallel to the plane and does not intersect it.

2.3 Example : Line Lying in a Plane

Determine if the line intersects the plane

Solution
  • Direction vector of line: .
  • Normal vector of plane: .

Thus, the line is parallel to the plane.

Parametric equations for

, substitute into :

This statement is true for all values of .

Therefore, the line lies entirely within the plane.

3 Intersection of Two Lines

Two lines in or can:

  1. Intersect at a Single Point.
  2. Be Parallel and Distinct: No intersection.
  3. Be Coincident: Infinite intersection points (they are the same line).
  4. Be Skew (in only): Not parallel and do not intersect.

To find the intersection, we use parametric equations for both lines and equate the corresponding coordinates.

  • Line :
  • Line :

Note the use of different parameters ( and ).

Equating components:

Solve the system for and .

  • In : Solve the two equations for and . If a unique solution exists, they intersect.

  • In : Use two equations (e.g., (1) and (2)) to find and . Substitute these values into the third equation (3).

    • If (3) is satisfied: Single point of intersection.
    • If (3) is not satisfied: Lines are skew (if not parallel).

    • If the system for from (1) and (2) yields no solution (e.g. ): check if direction vectors are parallel.

      • If yes, parallel lines.
      • If no, this case shouldn't happen if handled correctly (usually leads to skew if (3) then fails).
    • If the system for yields infinite solutions (e.g. ): lines are coincident (if direction vectors are parallel).
3.1 Example : Intersecting Lines in

Find the intersection of and

Solution

Parametric equations:

Equate components:

Solve (1) and (2) for and :
Add (1) and (2):

Substitute into (1):

Check these values in (3): . The left side is . The right side of (3) is . Since , the values are consistent.

Substitute into (or into ):

The point of intersection is .

3.2 Example : Skew Lines in

Determine if and intersect.

Solution

Direction vectors: , . These are not scalar multiples, so the lines are not parallel or coincident.

Parametric equations:

Equate components:

Solve (1) and (2):
Add (1) and (2):

Substitute into (1):

Check these values in (3): Left side: . Right side: . Since , the values are not consistent.
The lines do not intersect. Since they are not parallel, they are skew lines.

3.3 Example : Parallel and Distinct Lines in

Consider

Determine their intersection.

Solution

Direction vectors: , . Since , the lines are parallel.

Equate components: (1) (2)

From (2), . Substitute into (1):

This is a contradiction. There is no solution for (and ).
Thus, the lines are parallel and distinct.

4 Homework
4.1 Question 1.1 (Easy)

Find the point of intersection, if any, between the line and the plane .

Solution

Substitute the parametric equations of the line into the equation of the plane :

Combine terms with and constant terms:

Since we found a unique value for , there is a single point of intersection.
Substitute back into the parametric equations of the line :

Answer: The point of intersection is .

4.2 Question 1.2 (Medium)

Determine whether the lines and intersect, are parallel, or are skew. If they intersect, find the point of intersection.

Solution

Direction vectors: ,

Check if direction vectors are parallel:
Is for some scalar ?

Since for all components, .
The lines are parallel.

Now, check if they are coincident or distinct.
To be coincident, a point from must lie on .
Let be a point on .
If is on , then for some :

Equate components:

Since the same value of satisfies all three equations, the point from is also on .
Since the lines are parallel and share at least one point, they are coincident.

Answer: The lines are coincident (and therefore intersect at infinitely many points, they are the same line).

4.3 Question 1.3 (Harder)

Consider the line and the plane .
For what value of does the line lie entirely within the plane ?

Solution

For the line to lie entirely within the plane , two conditions must be met:

  1. The direction vector of the line, , must be perpendicular to the normal vector of the plane, . This ensures the line is parallel to the plane.

  2. A point from the line must lie on the plane .
    Let's use the point on when : .
    Substitute into the plane equation:

    This is a contradiction. This means that with the current plane equation , the line can never lie in the plane if its direction vector is and passes through .

Let's re-read the question. It asks for what value of the line lies in the plane. The method is generally to substitute the parametric equations of the line into the plane equation. If the resulting equation in becomes , then the line lies in the plane.

Substitute into :

Collect terms with and constant terms:

  • For the line to lie entirely within the plane, this equation must be true for all values of . This can only happen if the form is .
  • For our equation , if the coefficient of is zero, we get , which is a contradiction.

This implies that for this specific plane and a line of the form , the line cannot lie within the plane, regardless of . It will either intersect at one point or be parallel and distinct.

Let's check my interpretation.
Perhaps the question implies that we first satisfy the parallel condition, then check a point.
If (from ), the line is .
Direction vector is . Normal vector is

So the line is parallel to the plane.
A point on the line (for ) is . Substitute into plane:

Since , the point is not on the plane. So the line is parallel and distinct.

There might be an issue with how the question is posed relative to the expected outcome, or I'm missing a nuance.
If the question meant for "what value of (and possibly adjusting the constant term of the plane or a point on the line) does it lie in the plane?", then is necessary for parallelism.
However, as strictly stated, for this line and this plane: For this to be true for all (line in plane), we need AND .
The part is impossible.
This means there is NO value of for which this specific line lies in this specific plane.

If the question intended to ask for the line to be parallel to the plane (but not necessarily in it), then makes , leading to , or , which means no solution for , hence parallel and distinct.

Let's assume the question has a solvable setup, and the equation should become .
The constant terms must cancel.
From , we got .
For to occur, we need:

  1. Coefficient of to be zero: .
  2. The constant part of the equation to be zero: . This is impossible.

Conclusion based on strict interpretation: There is no value of for which the given line lies entirely within the given plane. The line will always be parallel and distinct if , or intersect at a single point if .

Self-correction

(Self-correction: Perhaps the problem intended a different constant for the plane, or a point that forces the line to be in the plane. Given the phrasing, the above is the correct deduction. If a student got this, they should state "no such k exists".)

Revised approach if a solutionmustexist: If the question had been, for instance, , and the line must lie in it:

For this to be , we need: and .
So if the plane was , and , then the line would lie in it.

Final Answer (as the question is written): For the line to lie in the plane , the equation derived from substituting the line into the plane must be true for all . This requires (so ) AND . Since is false, there is no such value of .

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