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Let $p_{1}, p_{2} \text{ and } p_{3}$ be three planes which intersect in a straight line (and not a point, which is generally the case).

Let a fourth plane $p_{4}$ cut these planes (not at the line of intersection of the three planes). Evidently, the intersections of the planes with $p_{4}$ would form three distinct straight lines.

Introduction to Higher Geometry by Graustein says that the three lines are concurrent. Any help regarding the proof would be great.

Thanks in advance.

2 Answers2

1

Let $l$ be the line where $p_1,p_2,p_3$ intersect. The following cases cases are possible:

  1. $p_4$ intersects $l$ in a point. Then this point is on all three intersection lines $p_i\cap p_4$.
  2. $p_4$ contains $l$. This case was explicitly disallowed.
  3. $p_4$ is parallel to $l$. Then the intersection linse $p_i\cap p_4$ are all parallel to $l$, hence are parallel. (In case of projective geometry, this case does not occur; or to put it differently: If $p_4$ and $l$ intersect at a point at infinity, then so do the $p_i\cap p_4$, i.e. we have just case 1 with a point at infinity (which of course doesn't matter as points at infinity are not distinguished))
Hagen von Eitzen
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  • A plane is parallel to line $l$ if the two do not intersect. The intersection lines $p_{i}\bigcap p_{4}$ do not have to be parallel to each other to lie on a plane parallel to $l$. –  May 24 '13 at 16:00
  • @AyushKhaitan If $l$ is parallel to $p_4$, the line $p_i\cap p_4$ is in $p_4$, hence does not intersect $l$. Thus the lines $p_i\cap p_4$ and $l$ are two lines in $p_i$ without intersection - that makes them parallel. – Hagen von Eitzen May 24 '13 at 16:21
  • It would be nice if it turned out that the question is really about projective geometry, so that case 3 is subsumed by case 1. – Andreas Blass May 24 '13 at 20:47
  • @AndreasBlass Oh, you're right- I read only the question, not the tag. – Hagen von Eitzen May 24 '13 at 21:22
  • @HagenvonEitzen- As far as I understand this, that makes the lines skew, and not necessarily parallel. In 3-D geometry, non-intersecting lines do not have to be parallel. They can be skew too. –  May 25 '13 at 03:40
  • It just struck me that you're right. any point of intersection between any two lines $p_{i}\bigcap p_{4}$ has to lie on the intersection of the three planes $p_{1}, p_{2} \text{and} p_{3}$, which is a line, and the plane $p_{4}$. If the line of intersection of the three planes does not lie on $p_{4}$, then $p_{4}$ can intersect said line only at one point. Hence, the three lines are concurrent. Thanks :) –  May 26 '13 at 02:04
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As you mentioned, generally three planes intersect in a point. You supposed that the common line $l$ of $p_1$, $p_2$, $p_3$ is not in $p_4$ . So, for example, $p_1$, $p_2$ and $p_4$ meet in a point $P$. This point must be on the lines $p_1\cap p_4$ and $p_2\cap p_4$. $P$ must also be on $l$, because it is a common point of $p_1$ and $p_2$. Thus $P=l\cap p_4$. Similarly, consider the planes $p_1$, $p_3$ and $p_4$, and let their common point be $Q$. $Q$ is on $p_1\cap p_4$ and $p_3\cap p_4$, and $Q$ is also on $l$. So $Q=l\cap p_4$, thus $P=Q$. Summarizing: $P$ is on $p_1\cap p_4$, $p_2\cap p_4$ and $p_3\cap p_4$ as we had to prove.

Mr. VV
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  • Oh yes, thank you. In fact I had logged in just to tell @Hagen Von Eitzen that I'd finally understood his proof. –  May 26 '13 at 02:01