## Monday, May 18, 2015

### Unizor - Geometry3D - Lines and Planes - 2 Planes Perpendicular to a Line

Unizor - Creative Minds through Art of Mathematics - Math4Teens

Theorem 1

If one of two parallel planes is perpendicular to some line, the other one is perpendicular as well.

Proof

Assume that planes γ and δ are parallel and line a is perpendicular to plane γ:

γ ∥ δ;

a⊥γ.

We will prove that a⊥δ.

Let points P=a∩γ and Q=a∩δ be intersection points of line a with two given planes.

Construct a new plane σ that contains line a. It also contains points P and Q of intersection of line a with planes γ and δ and, therefore, intersects γ along some line PP' and intersects δ along some line QQ'. Notice that a⊥PP' since a⊥γ.

If a plane σ intersects parallel planes γ and δ along lines PP' and QQ' correspondingly, these lines are parallel (see Theorem 2 of "Plane ∥ Plane" lecture), that is PP' ∥ QQ'.

Similarly, construct a different new plane τ that contains line a. It also contains points P and Q of intersection of line a with planes γ and δ and, therefore, intersects γ along some line PP" and intersects δ along some line QQ". Notice that a⊥PP" since a⊥γ.

If a plane τ intersects parallel planes γ and δ along lines PP" and QQ" correspondingly, these lines are parallel (the same Theorem 2 of "Plane ∥ Plane" lecture as above), that is PP" ∥ QQ".

We have now the following relationships between the lines.

In plane σ:

a⊥PP' and PP' ∥ QQ', from which follows that a⊥QQ'.

In plane τ:

a⊥PP" and PP" ∥ QQ", from which follows that a⊥QQ".

Since a⊥QQ' and a⊥QQ", we conclude that a⊥δ.

End of proof.

Theorem 2

If two planes, γ and δ are perpendicular to the same line a, they are parallel to each other.

Proof

Given: a⊥γ; a⊥δ.

To prove: γ∥δ.

Let point P=a∩γ and point Q=a∩δ.

Construct a new plane α that contains line a. It also contains points P and Q of intersection of line a with planes γ and δ and, therefore, intersects γ along some line PP' and intersects δ along some line QQ'.

Within plane α both lines, PP' and QQ', are perpendicular to line a and, consequently are parallel to each other.

Similarly, construct a different new plane β that contains line a. It also contains points P and Q of intersection of line a with planes γ and δ and, therefore, intersects γ along some line PP" and intersects δ along some line QQ".

Within plane β both lines, PP" and QQ", are perpendicular to line a and, consequently are parallel to each other.

Now we have a pair of lines PP' and PP" on plane γ correspondingly parallel to a pair of lines QQ' and QQ" on plane δ. According to one of the theorems proven earlier (see Theorem 1 in "Plane ∥ Plane" lecture), this is a sufficient condition for these planes to be parallel to each other.

End of proof.

Theorem 3

If from a point M lying outside line a two different planes, γ and δ, are constructed in such a way that a⊥γ and a∩δ≠∅, then plane δ is NOT parallel to plane γ and is NOT perpendicular to line a.

Proof

Obviously, γ and δ are NOT parallel because they intersect at point M.

Let point A=a∩γ and point B=a∩δ. Assume, these are two different points on line a.

If a⊥δ, we have two perpendiculars from point M to line a, segments MA and MB, which is impossible in plane Euclidean geometry. So, a is NOT perpendicular to δ. Actually, it is slant or oblique line relative to this plane.

End of proof.

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