Find the center of the curve in 3D made by 4-points

In summary: Az) = D[(By-Cy)(Az) - (Bz-Cz)(Ay)](Ax) + [(Cx-Ax)(Bz-Cz) - (Bx-Cx)(Cz-Az)](Ay) + [(Bx-Ax)(Cy-Ay) - (By-Ay)(Cx-Bx)](Az) = D[ByAz - AyCz - BzAy + AzCy](Ax) + [CxAy - AxCz - BxCy + CzAx](Ay) + [BxAz - AxBz - ByCx + AxCy](Az) = D[ByAz - AyCz - BzAy + Az
  • #1
aminjam
1
0
A(Ax,Ay,Az)
B(Bx,By,Bz)
C(Cx,Cy,Cz)
D(Dx,Dy,Dz)

Example:
Points X Y Z
a -2.00 0.00 -1.00
b -2.00 0.00 -1.59
c -1.59 0.00 -2.00
d -1.00 0.00 -2.00
The Answer is supposed to be ? Center: X:-1.0 Y:0.0 Z:-1.0

What I am trying to solve is to find the equation of the plane based on two vectors AB , AC; then find the normal vector, and then find the plane equation, which look like this:

mX+nY+oZ = m(Ax) + n(Ay) + o(Az) where
m = (By-Ay)(Cz-Az) - (Bz-Az)(Cy-Ay)
n = (Cx-Ax)(Bz-Az) - (Bx-Ax)(Cz-Az)
o = (Bx-Ax)(Cy-Ay) - (By-Ay)(Cx-Ax)

Now that I have the equation of the plane I am trying to find the the center of curve by using the equation of the plane.
I believe i need to find a perpendicular vector to AB vector in my plane, and another perpendicular vector to CD vector in my plane, and find where the two vector meets, and that point would be the center of curve drawn. But I don't know how to do that??
 

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  • #2


First, let's clarify some of the terminology being used. In your example, A, B, C, and D are points, not vectors. A vector is a quantity that has both magnitude and direction. In this case, the vectors would be AB, AC, and CD. It's important to distinguish between points and vectors because they are not the same thing and have different properties.

To find the equation of a plane, you need three points that lie on the plane. In this case, you have four points (A, B, C, and D), so you can choose any three of them to find the equation. Let's choose points A, B, and C. The equation of a plane can be written in the form Ax + By + Cz = D, where A, B, and C are the coefficients and D is a constant. To find these coefficients, you can use the method you described in your post:

m = (By-Ay)(Cz-Az) - (Bz-Az)(Cy-Ay)
n = (Cx-Ax)(Bz-Az) - (Bx-Ax)(Cz-Az)
o = (Bx-Ax)(Cy-Ay) - (By-Ay)(Cx-Ax)

In this case, A, B, and C would be equal to m, n, and o, respectively. To find D, you can substitute any of the three points (A, B, or C) into the equation and solve for D. Let's use point A:

m(Ax) + n(Ay) + o(Az) = D
Substituting in the values for m, n, and o:
[(By-Ay)(Cz-Az) - (Bz-Az)(Cy-Ay)](Ax) + [(Cx-Ax)(Bz-Az) - (Bx-Ax)(Cz-Az)](Ay) + [(Bx-Ax)(Cy-Ay) - (By-Ay)(Cx-Ax)](Az) = D
Simplifying:
[(By-Ay)(Cz-Az) - (Bz-Az)(Cy-Ay)](Ax) + [(Cx-Ax)(Bz-Az) - (Bx-Ax)(Cz-Az)](Ay) + [(Bx-Ax)(Cy-Ay) - (By-Ay)(Cx-Ax)]
 

1. How do I determine the center of a 3D curve made by 4 points?

The center of a 3D curve made by 4 points can be determined by finding the intersection of the perpendicular bisectors of the lines connecting each pair of points. This point will be the center of the curve.

2. Is there a formula for finding the center of a 3D curve made by 4 points?

Yes, the formula for finding the center of a 3D curve made by 4 points is (x,y,z) = ((x1+x2)/2, (y1+y2)/2, (z1+z2)/2), where (x1,y1,z1) and (x2,y2,z2) are two of the four points on the curve.

3. Can I use the same method to find the center of any 3D curve?

No, this method specifically applies to 3D curves made by 4 points. For other types of curves, different methods may need to be used to find the center.

4. How accurate is the center determined by this method?

The accuracy of the center determined by this method depends on the precision of the coordinates of the 4 points used. If the points are precise, the center will be accurate. However, if the points are rounded or approximate, the center may not be completely accurate.

5. Can I use this method for curves with more than 4 points?

No, this method only works for 3D curves made by 4 points. For curves with more points, a different method that takes into account additional points will need to be used to find the center.

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