# Plane of evidence

1. Mar 27, 2015

### point

in classical mathematics plane is taught as an axiom, I question whether the plane can be a proof, let's start from the assumption that there are $$n ( n>1)$$ line that intersect at a single point in space , from them can be proved that the plane can be used as evidence, ..

2. Mar 27, 2015

### phinds

You can have multiple lines that all intersect at the same point but which are not in the same plane unless of course you want to first posit the existance of a plane and then say that all the lines have to be in that plane and thus they prove that the plane exists but first you would have to posit that that plane exists and then ...

3. Mar 27, 2015

### Staff: Mentor

There are definitions for planes, but maybe that's what you mean.
???
What does this mean? One definition (from Wikipedia) is that a plane "is a flat, two-dimensional surface."
The plane can be used as evidence for what?
As @phinds has already said, you can have multiple (i.e., more than two) lines that intersect at a common point without the lines determining a plane.

Where are you trying to go with this?

4. Mar 28, 2015

### point

2-plane

1. two line intersect at point A , space
3. straight lines DB , BE , CE , CD
The first process - point B slides on the line to point A , straight lines DB (BE , CE , CD ) intersecting space , formed a final surface BDCE
Second process - point B slides on the line moving away from point A , straight lines DB (BE , CE , CD ) intersecting space , formed infinitive surface

Merging the surface, we get what we now know plane

how to get out of this function plane ? , because the Cartesian coordinate system is constant

5. Mar 28, 2015

### phinds

I have no idea what you are talking about. Yes, your figure makes sense. You have some lines in a plane. So what? What is your point? What is your question?

6. Mar 28, 2015

### HallsofIvy

It is true that two intersecting lines define a unique plane. Take the cross product of vectors in the directions of the lines, then use that as normal vector to the plane.

For example, the two lines x= 3t+ 2, y= 2t- 4, z= t+ 1 and x= t, y= -t- 2, z= t- 1 intersect at (2, -4, 1) (with t= 0 for the first line, t= 2 for the second). A vector in the direction of the first line is <3, 2, 1> and a vector in the direction of the second line is <1, -1, 1>. The cross product of those two vectors is <3, -2, -5>. The plane having that normal vector and containing the point (2, -4, 1) is 3(x- 2)- 2(y+ 4)- 5(z- 1)= 0.