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)

  1. Through a line and a point outside it, one can draw a plane, and such a plane is unique.
  2. Through two intersecting lines, one can draw a plane, and such a plane is unique.
  3. 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).

1 Answer 1


Here are some hints to hopefully point you in the right direction.

How would you define a pair of skew lines? That property is essential for this to work. Depending on what terms you'd use for such a definition, I'd suggest using the same terms to formulate the proof of uniqueness. (Likely terms include notions of “vectors” or “directions”.) If these terms are not sufficient, add the terms you'd use to define parallel planes.

Which step of your algorithm fails if the two lines are not skew but parallel? What implicit assumption does your proof make which would be unsatisfied in that situation? Make sure to make the assumption explicit, and show that it holds for the case of skew lines. Do the same analysis and justification for the case of intersecting lines.

Edit: Based on your comment I'm assuming that you define skew lines as lines with no common plane, parallel lines as lines in a common plane but with no common point, and intersecting lines at lines that have exactly one point in common. Similarly, parallel planes would be planes with no points in common, while non-parallel planes intersect in a line. Since such a incidence-based formalism doesn't include many notions of directions, I'll have to think of something else.

One practical approach to show uniqueness would be a proof by contradiction. Suppose the pair $(P,Q)$ was not unique. There are essentially two possible cases to consider. One case is where you have one choice for one of the planes but at least two for the other. So assume $(P,Q_1)$ and $(P,Q_2)$ to be solutions to your problem. Then $Q_1\parallel P$ and $P\parallel Q_2$, so $Q_1\parallel Q_2$ by transitivity of parallelism (which you may have to show first). Therefore (and since $Q_1$ and $Q_2$ are distinct) $Q_1$ and $Q_2$ have no points in common (which may be your definition of parallelism or something to show as a lemma), but since both must contain $b$ you have a contradiction.

The other case is two pairs of pairwise distinct parallel planes, namely $(P_1,Q_1)$ and $(P_2,Q_2)$. Then $P_1$ and $P_2$ must intersect in $a$ while $Q_1$ and $Q_2$ must intersect in $b$. Intuitively it's clear that this can only happen if $a\parallel b$, but how does one show that? Well, consider your plane $M$ contaning all of $a$ and one point $B$ of $b$. If $M$ were to intersect $Q_1$ and $Q_2$ in the same line, that line would be $b$, so $M$ would contain all of $b$, so $a$ and $b$ would be parallel not skew lines. So $M$ has to intersect $Q_1$ and $Q_2$ in distinct lines intersecting at $B$. Let's call these lines $q_1$ and $q_2$. On the other hand, $M$ intersects $P_1$ and $P_2$ both in $a$. Now since a pair of parallel planes intersects any other non-parallel plane in a pair of parallel lines, you have $a\parallel q_1$ (due to $P_1\parallel Q_1$) and $a\parallel q_2$ (from $P_2\parallel Q_2$) which by transitivity (this time in the plane) implies $q_1\parallel q_2$. This is a contradiction to the fact that $q_1$ and $q_2$ need to intersect in $B$, thus the whole assumption of two distinct pairs of solutions has been disproven.

  • $\begingroup$ How would you define a pair of skew lines? Skew lines are two lines positioned in space in such a way that no plane can be drawn through them. Which step of your algorithm fails if the two lines are not skew but parallel? The first step fails (and for the same reason, the second) because $a$ and $b'$ would be the same line. What implicit assumption does your proof make which you would be unsatisfied in that situation? I'm not sure; given your previous question, I guess the answer ought to be that the lines are indeed skew. However, the exercise already says that this is the case. $\endgroup$
    – Fine Man
    Sep 1, 2016 at 6:26
  • $\begingroup$ @SirJony: Yes. The implicit assumption would be that $a$ and $b'$ are distinct lines. Which is the case for skew lines, but that's a property you might have to show unless you already have that established. I'll try to make my uniqueneness hint more explicit based on your definition of skew lines. $\endgroup$
    – MvG
    Sep 1, 2016 at 6:34
  • $\begingroup$ The book that presents this exercise makes the assumption that, when talking about multiple objects (points, lines, planes, etc.), such objects are distinct, unless stated otherwise. Should I include this information in my question (as well as future questions)? $\endgroup$
    – Fine Man
    Sep 1, 2016 at 6:38
  • $\begingroup$ @SirJony: While that may be true for objects given in the problem statement, I would be very careful about extending that to objects you derive from these yourself. Sure, you could adopt the book's convention and say that objects are distinct unless stated otherwise. But then it would be your task to state that the step of drawing $b'$ might well lead to a line that is identical to $a$, only to show later on that thanks to the skew property this cannot happen after all. I wouldn't suggest adopting such a convention for your own proofs since it makes it easy to miss relevant corner cases. $\endgroup$
    – MvG
    Sep 1, 2016 at 10:19
  • $\begingroup$ Oy vey! If only intuition was proof enough. :) I've use proof-by-contradiction a couple times in elementary proofs. You mention "practical approach" regarding proof-by-contradiction, so would you recommend for most proofs of uniqueness that I use this method? $\endgroup$
    – Fine Man
    Sep 1, 2016 at 19:28

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