Test for coplanarity of four points

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Homework Help Overview

The discussion revolves around the conditions for coplanarity of four points defined by their position vectors. The original poster questions whether a specific test involving scalar coefficients is necessary to determine if four points are coplanar, and they express uncertainty about their assumptions regarding coplanarity in general.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants explore the necessity of a mathematical test for coplanarity, with some suggesting that any four points might be coplanar. Others question this assumption and discuss the implications of specific geometric configurations, such as a regular tetrahedron.

Discussion Status

The conversation is ongoing, with participants providing insights into the nature of coplanarity and discussing the validity of the original poster's assumptions. Some guidance has been offered regarding the conditions under which four points can be considered coplanar.

Contextual Notes

There is a mention of a regular tetrahedron as a specific case, which raises questions about the generality of the coplanarity test. The discussion also reflects on the requirement of three non-collinear points to define a plane.

Nero26
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Hi all,
If a,b,c,d are position vectors of four points A,B,C,D.The points will be coplanar if xa+yb+zc+td=0,x+y+z+t=0,provided x,y,z,t are not all 0,and they are scalars.Is this test needed to show 4 points are coplanar?
If we consider two lines joining A,B and C,D then this will give us two vectors which are always coplanar.So points A,B,C,D are also coplanar.So I assumed that any 4 points are coplanar and no test is needed for it.
Or is this the test to verify coplanarity of D with the plane containing A,B,C ?
I'm wondering if my assumption is true?Please help me clarifying it.
I'm new here ,Please treat my mistakes with forgiveness. :smile:
Thanks.
 
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Nero26 said:
Hi all,
If a,b,c,d are position vectors of four points A,B,C,D.The points will be coplanar if xa+yb+zc+td=0,x+y+z+t=0,provided x,y,z,t are not all 0,and they are scalars.Is this test needed to show 4 points are coplanar?
If we consider two lines joining A,B and C,D then this will give us two vectors which are always coplanar.So points A,B,C,D are also coplanar.So I assumed that any 4 points are coplanar and no test is needed for it.

What if A,B,C,D are the vertices of a regular tetrahedron?
 
LCKurtz said:
What if A,B,C,D are the vertices of a regular tetrahedron?
:smile:Thanks a lot for your clue.I think I'm getting near to it.Can you please take a look on the attachment...
And please mention if I need some more things to do.
 

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Nero26 said:
:smile:Thanks a lot for your clue.I think I'm getting near to it.Can you please take a look on the attachment...
And please mention if I need some more things to do.

My point was that your statement that any 4 points are coplanar is false. Remember that it takes three non-collinear points to determine a plane (their triangle is part of the plane). Four points are coplanar only if the 4th point lies in the plane determined by the first three.

The test I would use for co-plane-ness of points A,B,C,D would be to make vectors of the sides like this: u = AB, v = AC, w = AD and calculate the triple scalar product or "box" product ##u\cdot v \times w##. If that is non-zero they aren't coplanar and if it is zero they are.
 
Thanks a lot for your help.I think I got your point " Four points are coplanar only if the 4th point lies in the plane determined by the first three."
Have a nice day!:smile:
 

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