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Is it true that two planes may intersect in a point ?


If they intersect then, they always make a straight line ?

I have some doubt; please explain.

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In $\Bbb R^3$ two distinct planes either intersect in a line or are parallel, in which case they have empty intersection; they cannot intersect in a single point. In $\Bbb R^n$ for $n>3$, however, two planes can intersect in a point. In $\Bbb R^4$, for instance, let $$P_1=\big\{\langle x,y,0,0\rangle:x,y\in\Bbb R\big\}$$ and $$P_2=\big\{\langle 0,0,x,y\rangle:x,y\in\Bbb R\big\}\;;$$ $P_1$ and $P_2$ are $2$-dimensional subspaces of $\Bbb R^4$, so they are planes, and their intersection $$P_1\cap P_2=\big\{\langle 0,0,0,0\rangle\big\}$$ consists of a single point, the origin in $\Bbb R^4$. Similar examples can easily be constructed in any $\Bbb R^n$ with $n>3$.

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Seems correct. Still, how do we demonstrate that two planes in $\mathbb{R}^3$ cannot intersect in a single point. – Teddy Oct 23 '12 at 8:40
@Teddy WLOG remove one zero from P1 and P2. Now there's no way to not get $x=y$, unless they describe the same plane. – Mark Hurd Oct 23 '12 at 8:45
@MarkHurd I think I understand your intention. However, this reasoning only demonstrates that planes of this particular type intersect in a line. What about other planes. What about planes which are not subspaces of $\mathbb{R}^3$ (but rather affine spaces) ? – Teddy Oct 23 '12 at 9:38
@Teddy: Simultaneously translate them by the additive inverse of a point of intersection, thereby bringing that point to the origin. Now you’re dealing with subspaces, and their intersection can be translated back to give the intersection of the original planes. – Brian M. Scott Oct 23 '12 at 9:42
I might be wrong (have no rigorous proof of what follows), but a curious example that comes to mind would be the (fourth-dimensional) "graph" of the complex identity function w = f(z) = z. It seems plausible that the "graph" should be a plane (it has to be a 2D surface immersed in 4D, and I see no reason why it would have any curvature). It is also evident that it should meet the "x axis" (i.e., z-plane, w = 0) of this function at z = 0 only. – NicolasMiari Dec 21 '15 at 2:02

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