Prove that for every pair of skew lines $a$ and $b$, there is a unique pair of planes: one passing through $a$ and parallel to $b$, the other passing through $b$ and parallel to $a$, and that these planes are parallel.
Useful Knowledge (Axioms, Theories, and Corollaries, and previous Exercises)
- Through a line and a point outside it, one can draw a plane, and such a plane is unique.
- Through two intersecting lines, one can draw a plane, and such a plane is unique.
- Two angles whose respective sides are parallel and have the same direction, are congruent and lie either in parallel planes or in the same plane.
- Two skew lines $a$ and $b$.
Draw an arbitrary plane $M$ through $a$ and an arbitrary point $B$ on $b$. (Useful Knowledge 1)
Draw an arbitrary plane $N$ through $b$ and an arbitrary point $A$ on $a$. (Useful Knowledge 1)
Draw a line $a'$ parallel to $a$ on $M$ through $B$.
Draw a line $b'$ parallel to $b$ on $N$ through $A$.
Draw a plane $P$ through $a$ and $b'$. (Useful Knowledge 2)
Draw a plane $Q$ through $b$ and $a'$. (Useful Knowledge 2)
$P$ and $Q$ are parallel. (Useful Knowledge 3)
I'm confident that I've proof the significant piece. However, note the word "unique" in the exercise. I did not prove the uniqueness. How would I go about proving it's uniquness? Also, in general, I'm not particularly good at proving uniqueness. Is there some generic method of proving things unique?
- I suggest drawing the model described throughout the proof.
- This exercise is from Kiselev's Geometry; Book II: Stereometry (English Adaptation). It is Exercise 14, found in: Chapter 1 (Lines and planes), Section 2 (Parallel lines and planes).