## Tuesday, May 19, 2015

### Unizor - Geometry3D - Lines and Planes - Problems 2

Unizor - Creative Minds through Art of Mathematics - Math4Teens

We would like to address this construction problem in a very detailed way to exemplify the rigorousness required from mathematics, if we want to approach this subject on an advanced level.

This rigorousness requires to

(1) analyze the construction problem to properly plan the construction on a step by step level;

(2) present the construction steps and

(3) prove that this construction leads to required results.

Problem

Given a straight line a in the three-dimensional space and a point M outside of it.

Construct a plane σ that passes through a given point M and is perpendicular to a given line a.

Analysis

Assume that such plane σ has been constructed (a⊥σ, M∈σ) and its intersection with line a, a base of a perpendicular, is point P (P=a∩σ).

Construct a new plane τ that is defined by line a and point M (a∈τ, M∈τ). Plane τ contains both points P (since P∈a and a∈τ) and M (by construction). Hence, plane τ contains line a and segment MP.

Segment MP is also contained in plane σ because point P is a base of the perpendicular to plane σ.

So, segment MP belongs to the intersection of planes σ and τ (MP∈σ∩τ).

In plane τ segment MP is perpendicular to line a (MP⊥a) since line a is perpendicular to an entire plane σ and, therefore, to any line on this plane that passes through a base point P.

Therefore, we can construct segment MP even without plane σ but using only plane τ (defined by given line a and point M) and drawing a perpendicular from point M to line a within that plane.

To determine the position of plane σ, it is sufficient to have two intersecting lines going through a base point P. The point P and one such line, segment MP, we have just constructed by dropping a perpendicular from point M to line a within plane τ.

To construct another line also perpendicular to a at the base point P, construct another plane ρ through line a that is different from plane τ. In that plane draw a perpendicular NP to line a at point P.

Now we have two lines defined by segments MP and NP perpendicular to line a. These two lines define a plane perpendicular to line a and containing both of these lines, in particular segment MP and point M in it.

Therefore, this plane is the one we need - plane σ.

Construction

Step 1. Construct a plane τ by given line a and point M.

Step 2. Drop a perpendicular MP from point M to line a, point P is the base of this perpendicular.

Step 3. Construct any other plane ρ through a, different from plane τ.

Step 4. Draw a perpendicular NP in plane ρ to line a at point P.

Step 5. Construct a plane σ by lines defined by segments MP and NP.

Proof of construction

(a) σ∋MP

⇒ σ∋M

(b) a⊥MP; a⊥NP;

σ∋MP; σ∋NP

⇒ a⊥σ

Proof of uniqueness

Assume there exist two planes σ and σ', both containing point M and both perpendicular to line a. Let points P and P' be intersections of line a with these planes, correspondingly.

Assume, P≠P' (case P=P' is considered below).

Draw a plane τ through line a and point M. It intersects plane σ along segment MP, which is perpendicular to line a. It also intersects plane σ' along segment MP', which is also perpendicular to line a.

So, within plane τ we have two perpendiculars MP and MP' from the same point M to line a, which is impossible in plane Euclidean geometry.

Therefore, points P and P' of intersection of line a with planes σ and σ' are one and the same.

Now we drop the initial assimption that P≠P' as invalid and consider only the case when P=P'.

Our planes σ and σ' have now two common points - given point M and the intersection with line a - point P=P'.

To say that planes σ and σ' are identical, we need the third common point.

Draw a plane ρ through line a that is different from plane τ. It intersects plane σ along a line that goes through point P and it intersects plane σ' also along a line that goes through point P. If we assume that these lines of intersection are different, we would have two different perpendiculars to line a at its point P within one plane ρ, which is impossible in plane geometry.

Therefore, these two lines perpendicular to line a at point P within plane ρ are one and the same. Now we have plenty of new points along this line that are common between planes σ and σ', which completes the proof that these planes are one and the same.

The uniqueness of a plane perpendicular to a line and passing through a point outside of this line is proven.

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