CODEWITHSUNDEEP
c++math3drotationtrigonometry
user985611
user985611

Reputation: 139

How to place a point in 3D so that it creates a certain angle?

I have a small question about 3D.

What follows is an example of my problem.

I have two points:

A: 12 4 5 B: 6 8 -10

I have another point: C: 5 6 7

I need to transform(?) point C so that the angle ABC is 48 degrees.

How do I do this? I would appreciate if someone can help me with the formulas or maybe even make the above example into a working one.

Another way to put it: How do I transform C.x, C.y, and C.z so that the angle ABC is 48 degrees?

I would really appreciate some help on this as I am stuck at the moment.

Side note: I already implemented a method for finding the angle:

float Angle( float x1, float y1, float z1,
             float x2, float y2, float z2 )
{
  float x, y, z;
  CrossProduct( x1, y1, z1, x2, y2, z2, &x, &y, &z );

  float result = atan2 ( L2Norm( x, y, z ),
                         DotProduct( x1, y1, z1, x2, y2, z2 ) );

  return result;
}

You use it: Angle( B.x - A.x, B.y - A.y, B.z - A.z, C.x - B.x, C.y - B.y, C.z - B.z );

Upvotes: 3

Views: 238

Answers (2)

Aki Suihkonen
Aki Suihkonen

Reputation: 20027

       A------C    
       |     
 c''   |    c'
       B

As three point in 3D define a plane, there are only 2 possible candidates for a transform C-->c' or C-->c'' at that plane.

c' would be then c' = A+t*(B-A) + u*(C-A) with constraint Normalize(c'-A) dot Normalize(B-A) == cos (48 / 180 * pi).

I'd first suggest normalizing D=(B-A), after that:

D dot D+u*(C-A) = 1 * |D+u(C-A)| * cos (48 degrees)

Dx*(Dx+u*(Cx-Ax))+ Dy*(Dy+u*(Cy-Ay))+Dz*(Dz+u*(Cz-Az)) ==
    0.669 * sqrt ((Dx+u*(Cx-Ax))^2+(Dy+u*(Cy-Ay))^2+(Dz+u*(Cz-Az))^2)

This is of form a+u*b == 0.669*sqrt(c+du+e*u^2), which will be simplified to a second degree polynomial in u by squaring both sides.

Upvotes: 2

forthefreedom
forthefreedom

Reputation: 11

The track of point C is actually a cone, you can imagine, B is the vertex and line AB is the central line of the cone, means the 3D cone is symmetric on AB.

Upvotes: 1

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