Friday, May 22, 2015
Lines and Planes - Problems 4: UNIZOR.COM - Geometry3D
Unizor - Creative Minds through Art of Mathematics - Math4Teens
Construct a common perpendicular h to two given skew lines a and b.
Assume a common perpendicular h to skew lines a and b is constructed. Let points of its intersection with these lines be A and B correspondingly. Then h must be an intersection of two planes: plane σ that is perpendicular to line a at point A and plane τ that is perpendicular to line b at point B:
h = σ∩τ
We don't know positions of points A and B. Consider, instead of point B, any other point B' on line b and construct plane τ' perpendicular to line b at this point B':
B'∈b; τ'⊥b; B'∈τ'
Since planes τ and τ' are parallel their intersections with plane σ are parallel. So, if we know plane σ, using any plane τ' perpendicular to line b allows us to construct a line h'=σ∩τ' that is parallel to a common perpendicular h we need.
Let's make another step analogous to the previous one. We don't know the position of point A. Consider, instead of it, we choose any point A' on line a and draw a plane σ' perpendicular to line a at this point:
A'∈a; σ'⊥a; A'∈σ'
It is parallel to plane σ and, therefore, lines of intersection of plane τ' with these two planes, σ and σ', are parallel. Let new line of intersection between τ' and σ' be h":
h" = σ'∩τ'
Since h ∥ h' and h' ∥ h", lines h and h" are parallel.
We came to a situation, when the construction of any two planes σ' and τ' correspondingly perpendicular to lines a and b allows us to construct a line h" which is parallel to the one we need to construct.
Next step is to shift in space our line h" parallel to itself to a place where it intersects both lines a and b.
We can do it in two steps:
Step 1 involves drawing a line parallel to a given line h" through a given point A' within plane σ'. We know how to do it from plane geometry. Let this new line that is perpendicular to a, parallel to h" and going through A' be line p.
Step 2 can be accomplished by drawing a plane ρ through lines a and p and finding a point of intersection of this plane with line b. This point must be point B on line b - the base of a common perpendicular h we are constructing. Dropping a perpendicular from point B to line a within plane ρ produces a common perpendicular h.
1.[A',σ': A'∈a; σ'⊥a; A'∈σ']
2.[B',τ': B'∈b; τ'⊥b; B'∈τ']
4.[p: p∈σ'; A'∈p; p ∥ h"]
5.[ρ: a∈ρ; p∈ρ]
7.[h: B∈h; h∈ρ; h⊥a]
More details are presented in the notes to this lecture at UNIZOR.COM