Two pencils of planes have a common plane

AI Thread Summary
To determine the parameter α for which the pencil of planes through line AB shares a common plane with the pencil of planes through line CD, it is essential to find a plane that contains points A, B, C, and D. The process involves deriving the equation of the plane formed by points A, B, and C, then ensuring that point D also satisfies this plane's equation. The discussion highlights that a common plane exists if the two pencils are either parallel or intersecting; if the lines are skew, no common plane can exist. Ultimately, solving for α requires ensuring all four points lie within the same plane.
beglor
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Homework Statement


Find the value of the parameter α for which the pencil of planes through the straight line AB has a common plane with the pencil of planes through the straight line CD, where A(1, 2α, α), B(3, 2, 1), C(−α, 0, α) and D(−1, 3, −3).

Homework Equations


Let Δ be a line given by two equations:
A1x+B1y+C1z+D1=0
A2x+B2y+C2z+D2=0
The collection of all planes containing a given straight line Δ is called the pencil of planes through Δ.
The plane π belongs to the pencil of planes through the line Δ if and only if there exists λ,μ∈ℝ such that the equation of the plane π is:
λ(A1x+B1y+C1z+D1)+μ(A2x+B2y+C2z+D2)=0

The Attempt at a Solution


I wrote the equations of the lines AB and CD. But I don't know the condition for a plane to be common to two pencil of planes in the same time.
 
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I took a course in projective geometry in 1959 taught by C. R. Wylie Jr. at the University of Utah. Never used it since and about the only thing I remember about it is the use of pencils of lines and pencils of planes. I always figured the terms came from projective geometry.
 
beglor said:

Homework Statement


Find the value of the parameter α for which the pencil of planes through the straight line AB has a common plane with the pencil of planes through the straight line CD, where A(1, 2α, α), B(3, 2, 1), C(−α, 0, α) and D(−1, 3, −3).

Homework Equations


Let Δ be a line given by two equations:
A1x+B1y+C1z+D1=0
A2x+B2y+C2z+D2=0
The collection of all planes containing a given straight line Δ is called the pencil of planes through Δ.
The plane π belongs to the pencil of planes through the line Δ if and only if there exists λ,μ∈ℝ such that the equation of the plane π is:
λ(A1x+B1y+C1z+D1)+μ(A2x+B2y+C2z+D2)=0

The Attempt at a Solution


I wrote the equations of the lines AB and CD. But I don't know the condition for a plane to be common to two pencil of planes in the same time.
There will be a common plane if either
  1. the pencil of planes through the line AB is parallel to the pencil of planes through CD, or
  2. the pencil of planes through the line AB intersects the pencil of planes through CD
In case 1, the vector ##\overrightarrow{AB}## will be a scalar multiple of the vector ##\overrightarrow{CD}##.
In case 2, the equations of the two lines have to have a common solution.

If the lines are skew, there can't be a common plane.

I think I've covered all the possibilities...
 
beglor said:

Homework Statement


Find the value of the parameter α for which the pencil of planes through the straight line AB has a common plane with the pencil of planes through the straight line CD, where A(1, 2α, α), B(3, 2, 1), C(−α, 0, α) and D(−1, 3, −3).

Homework Equations


Let Δ be a line given by two equations:
A1x+B1y+C1z+D1=0
A2x+B2y+C2z+D2=0
The collection of all planes containing a given straight line Δ is called the pencil of planes through Δ.
The plane π belongs to the pencil of planes through the line Δ if and only if there exists λ,μ∈ℝ such that the equation of the plane π is:
λ(A1x+B1y+C1z+D1)+μ(A2x+B2y+C2z+D2)=0

The Attempt at a Solution


I wrote the equations of the lines AB and CD. But I don't know the condition for a plane to be common to two pencil of planes in the same time.

You want to find a plane that contains both lines AB and CD, so contains the four points A, B, C and D. It is a standard exercise to find the plane containing the three points A, B and C; then you can fix ##\alpha## by requiring that the fourth point, D, must also lie in the same plane.
 
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