# Finding 4 points in a plane, when random points given

1. Oct 21, 2009

### soopo

1. The problem statement, all variables and given/known data
Points (-3, -1, 4), (0, -1, -2), (2, 5, 1), (3, 2, 7) and (5, 1, -2) are the vertexes of an 3D quadrilateral.
Find four points which are on a plane in the 3D quadrilateral.

3. The attempt at a solution

I know that you can find the points by counting the normal vector of two given vectors and then multiplying this with the given vector, not one in the cross product. If the dot product is zero for each vector, then you have found the plane.

However, this takes many steps to count.

The correct answer to the question is apparently ABDC.

How can you find efficiently the plane?

2. Oct 21, 2009

### lanedance

is that exactly how the question is written? it seems a little ambiguous...

coudl it be wihch 4 vertices fall in a plane?

and what are ABDC?

3. Oct 22, 2009

### soopo

4 vertices fall in a plane.
A refers to the first point in the exercise, B to the second, ... D to the fourth and so on.

4. Oct 22, 2009

### lanedance

i still don't get it, doesn't a quadrilateral only have 4 sides & vertices?

but you could do it the way you suggest, though a little tedious

5. Oct 22, 2009

### lanedance

i think i get it now, its a quadrilateral base with another point, find which 4 points make the base, all falling in a plane?

6. Oct 22, 2009

### soopo

Exactly. This is what I mean and trying to find a faster way than the one which I proposed in my question to find the four points in the plane.