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To check if two planes have a line in common is quite simple, but how do I determine if two planes have a point in common?

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Do you mean at least a point or exactly a point? In 3D the second case is not possible. For the first, they have to be non parallel. – enzotib Oct 14 '12 at 18:29
Well the question is: "Do the following planes have a point in common?" So I'm wondering if two planes can have just a point in common and not a line. – Nima Oct 14 '12 at 18:31
up vote 4 down vote accepted

Write the equation of the planes as $ax+by+cz=d$ where $a^2+b^2+c^2=1$ so that $d $ is the distance of the plane from the origin in $\mathbb R^3$. The only case when they don't intersect is when they are skew planes, parallel but at different distance from the origin. Otherwise, they are either the same plane or their intersection is a line.

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So two planes can't have a point in common? – Nima Oct 14 '12 at 18:36
@Nima Do you mean "at least one point" or do you mean "exactly one point"? – Rudy the Reindeer Oct 14 '12 at 18:49
@MattN.Thats what I'm not sure of, the question is exactly this: Do the following planes have a point in common? a) x+y+z = 4 and 2x+3y-z = 6. b) x+y+z = 4, 2x+3y-z=6, -x-y+2z=-4. – Nima Oct 14 '12 at 18:59
a. has a line in common, b. has a point in common. but if the question is True/False, would I write a) True or a) False? – Nima Oct 14 '12 at 19:01
@Nima Typically "have a point in common" is interpreted to mean "does there exist at least one point which is on all of the planes in the collection". With this interpetation, the answers would be True/True. – KReiser Oct 14 '12 at 19:09

Two planes have just a point in common in spaces with dimension 4 or higher. Let's name the planes V2 and V'2, dimension "dim". Sign "_" will be conjunction of spaces (linear span of their two basis), sign "^" will be their intersection (which is also a space). Then according to theorem about "dimension of conjunction and intersection of spaces" we can write:

dim (V2_V'2)+dim(V2^V'2)=dim(V2)+dim(V'2)=2+2=4 .

If dim(V2_V'2)=4 then din(V2^V'2)=0 . That means that the dimension of the intersection of the spaces is 0 --> the intersection is just a point. The problem is that it is a little bit hard to imagine it. Have a good fun!

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