Proving the concurrence of three lines.

Question

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.

Answer 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))

Answer 2

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.