Test for coplanarity of four points

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SUMMARY

The coplanarity of four points A, B, C, and D can be determined using the scalar equations xa + yb + zc + td = 0 and x + y + z + t = 0, where x, y, z, and t are non-zero scalars. However, it is incorrect to assume that any four points are coplanar; the fourth point must lie in the plane defined by the first three points. A reliable method to verify coplanarity is to calculate the triple scalar product of vectors formed by the points, specifically using the formula u · (v × w), where u, v, and w are vectors from the points A, B, C, and D.

PREREQUISITES
  • Understanding of vector representation and operations
  • Knowledge of scalar equations and their implications in geometry
  • Familiarity with the concept of coplanarity in three-dimensional space
  • Basic understanding of the triple scalar product
NEXT STEPS
  • Study vector operations, particularly the cross product and dot product
  • Learn about the geometric interpretation of the triple scalar product
  • Explore the properties of coplanarity in higher dimensions
  • Investigate applications of coplanarity tests in computer graphics and computational geometry
USEFUL FOR

Students and professionals in mathematics, physics, computer graphics, and engineering who need to understand the geometric relationships between points in three-dimensional space.

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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