Gets a value indicating whether or not this is a valid plane. A plane is considered to be valid when all fields contain reasonable information and the equation jibes with point and zaxis.

Thank you, so that means in this case that the second plane does not pass the test and therefor qualified as not valid by rhino? What difference does it make if the plane itselft works for my case(getting transform information)? Will there be a case where I absolutly have to make sure my plane is valid?
Kinda confused as in how to handle such tests in the future, if necessary at all.

The question is, how is it possible to create an InValid Plane from a point and two Vectors?

rgr code (probably) uses “Transforms” (Plane to Plane) and the functions below to create Planes.

public static Plane PlaneFromNormal(Point3d origin, Vector3d normal)
{
Plane plane = new Plane(origin, normal);
return plane;
}
public static Plane PlaneFromNormal(Point3d origin, Vector3d normal, Vector3d xAxis)
{
xAxis = new Vector3d(xAxis);
xAxis.Unitize();
Vector3d yAxis = Vector3d.CrossProduct(normal, xAxis);
Plane plane = new Plane(origin, xAxis, yAxis);
return plane;
}
/// <summary>
/// Construct a plane from a point, and two vectors in the plane.
/// </summary>
/// <param name="origin">A 3D point identifying the origin of the plane.</param>
/// <param name="xAxis">A non-zero 3D vector in the plane that determines the X axis direction.</param>
/// <param name="yAxis"> A non-zero 3D vector not parallel to x_axis that is usedto determine the Y axis direction. Note, y_axis does not have to be perpendicular to x_axis.</param>
/// <returns>The plane if successful</returns>
public static Plane PlaneFromFrame(Point3d origin, Vector3d xAxis, Vector3d yAxis)
{
return new Rhino.Geometry.Plane(origin, xAxis, yAxis);
}

Hi @dale, thank you for that method, I’ll put it to use and run some tests probably next week and report back.

@arendvw: it could well be, but both planes are approx.at the same place/distance from origin in the document. It would be interesting to know if, then why.