Finding 4 points in a plane, when random points given

[soopo]

Homework Statement

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.

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?

 

[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?

 

[soopo]

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?

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.

 

[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

 

[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?

 

[soopo]

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?

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.