如何调整一个3D点距另一个3D点的距离给定距离

Question

I am working with point3D and vector3D classes and I need some help adjusting a point by a given distance.

• Point A - point residing at coordinate 0,0,0.
• Point B - point residing at coordinate 1,1,1.
• Vector AB - vector AB which tells me the length between the two points A and B is distance = 1.73205078.

Code:

Point3D A = new Point3D { X = 0, Y = 0, Z = 0 };
Point3D B = new Point3D { X = 1, Y = 1, Z = 1 };
Vector3D AtoB = A - B;
Double distanceBetweenAandB = AtoB.Length; // the distance will be 1.73205078 here.

I would like to adjust point B. I would like to reduce the distance between point A and point B to 0.5 instead of 1 (adjusting to position C as shown in the diagram). I am trying to work out how to do this.

Point A (0,0,0) is known, point B (1,1,1) is known and the distance to adjust by is known (0.5). How do I calculate?

Pseudo code:

Point3D A = new Point3D { X = 0, Y = 0, Z = 0 };
Point3D B = new Point3D { X = 1, Y = 1, Z = 1 };
Double distanceToAdjust = 0.5;

Point3D newCoordinate = B - distanceToAdjust; // this doesnt work!

Adjusted point B shown in diagram below:

I am using my own defined Point3D class and Vector3D class.

Answer 1

Let's assume your given parameters for your points, and create a 3rd, which we'll call newCoordinate , and that point A will be your reference:

Point3D A = new Point3D { X = 0, Y = 0, Z = 0 };
Point3D B = new Point3D { X = 1, Y = 1, Z = 1 };

Double distanceToAdjust = 0.5;

Point3D newCoordinate = new Point3D { 
                                        A.X + ((B.X - A.X) * distanceToAdjust),
                                        A.Y + ((B.Y - A.Y) * distanceToAdjust),
                                        A.Z + ((B.Z - A.Z) * distanceToAdjust)
                                    }

Here we see the original points:

Assuming this values, newCoordinate would sit at X=0.5, Y=0.5, Z=0.5. Nice graph follows:

There it is, sitting right in between the two original points.

As a simulation, if you change A and B and assume this values instead:

Point3D A = new Point3D { X = -8, Y = 4, Z = 3 };
Point3D B = new Point3D { X = 3, Y = 2, Z = 1 };

Then newCoordinate position would be X=-2.5, Y=3, Z=2.

Now, same points, but using distanceToAdjust = 1.2 :

Keep this two things in mind:

• Changes in distance always need a reference point. In my sample, I assumed A; that's why it appears as the first portion of each newCoordinate parameter initialization.
• distanceToAdjust was taken as a multiplier factor .

Addendum: The nifty tool I used to help visualization can be found here .

Answer 2

Assuming you implemented vector operations:

if point A is always [0,0,0]

Point3D new = B.Normalize() * distance;

for any two points

Point3D newCoord = A + ((B - A).Normalize() * distance); //move to origin, normalize, scale and move back

not fast solution though.

Answer 3

"the length between the two points A and B is distance = 1"

No, the distance is the square root of three, about 1.732.

The distance from (0,0,0) to (0,0,1) is 1. The distance from (0,0,0) to (0,1,1) is the square root of two. (Think a triangle in two dimensions, and Pythagoas theorem.) The distance from (0,0,0) to (1,1,1) is the square root of three. (Think a triangle in two dimensions, where that dimension is on a plane along the hypothenuse of the previous triangle. AB = √(1² + (√2)²).)

---

I assume that you don't want to subtract 0.5 from anything, but actually multiply the distance by 0.5, ie getting halfways from A to B. You can calculate the point C by taking that part of the distance between point A and point B in each dimension:

Point3D C = new Point3D {
  A.X + (B.X - A.X) * distanceToAdjust,
  A.Y + (B.Y - A.Y) * distanceToAdjust,
  A.Z + (B.Z - A.Z) * distanceToAdjust
};

Answer 4

In pseudo code, here's how I ended up implementing

pointA = …
pointB = …
vectorAB = B-A
desiredDistance = 0.5;  // where 0.5 is vectorAB.Length/desiredDistance
vectorAC = vectorAB * desiredDistance ; 

pointC = A+vectorAC;

Actual code:

Vector3D pointC = (Vector3D)(A + (float)desiredDistance  * (B - A)); 

Answer 5

I'm unsure if this is what you would need but is it possible to create a method within your Point3D class to allow subtraction/addition?

(Just guessing the Point3D class as simply as it could be)Something like

   public class Point3D
   {

       public double X,Y,Z

       public void ChangeCord(Point3D point)
       {
           X =- point.X;
           Y =- point.Y;
           Z =- point.Z;
       }
   }

So it could just be:

    Point3D A = new Point3D { X = 0, Y = 0, Z = 0 };
    Point3D B = new Point3D { X = 1, Y = 1, Z = 1 };
    Double distanceToAdjust = 0.5;

    Point3D newCoordinate = B.ChangeCord(new Point3d{ X = 0.5, Y = 0.5, Z = 0.5 });