How to crete 3D modelCurves to avoid exception 'curve must be in the plane' ?

[Anonymous]

Hi!

I want to create new group of lines (3d). My code looks like:

List<ElementId> setOfLines = new List<ElementId>();
for (var i = 0; i < Points.Length - 1; i++)
{
  Line line = Line.CreateBound(Points[i], Points[i + 1]);
  XYZ normal = line.Direction.Normalize();
  Plane plane = Plane.CreateByNormalAndOrigin(normal, Points[i]);
  SketchPlane sketchPlane = SketchPlane.Create(CurrentRevitDocument, plane);
  ModelCurve modelCurve = CurrentRevitDocument.Create.NewModelCurve(line, sketchPlane);
  setOfLines.Add(modelCurve. Id);
}
CurrentRevitDocument.Create.NewGroup(setOfLines);

But i have an exception "curve must be in the plane, pCurveCopy".

I think this is because of the coordinates of the normal, they are wrong. How can i create my 3D modelCurves right? Should I use another methods for this?

Thanks for your help.

[BardiaJahan · Accepted solution]

It seems the normal of your sketch plane is parallel to your line. This means the line is definitely not in that plane. For the normal, you should use a vector that is perpendicular to the line so that the line lies in the plane.

Basically if XYZ n is the normal then n.DotProduct(line.Direction) must be zero.

Now I propose these two options:

1- You could use any arbitrary vector XYZ v that is not parallel to your line and use XYZ u = v.CrossProduct(line.Direction) as the normal of your sketch plane. For simplicity, v could be one of the basis vectors - e.g. XYZ.BasisZ as long as it is not parallel to your line. (a simple if-else) => XYZ.BasisZ.CrossProduct(line.Direction)

2- Or as I prefer construct vector n as following and use it as the normal:

XYZ dir = line.Direction;
double x = dir.X, y = dir.Y, z = dir.Z;
XYZ n = new XYZ (z - y, x - z, y - x);

[Anonymous]

Thank you very much for your help!