So what exactly is an osculating plane? r'(t) X r"(t)? Am I right? From what I know it is a plane that kisses a point on the curve at point t of curve f(t) and it is a plane normal to r'(t) X r"(t)? Am I right? So to look for this osculating plane I simply have to find the cross product of r'(t) and r"(t)? What does it look like on a xyz graph? ==================== EDIT: By the way per definition of the equation of a plane: a(x-x0) + b(y-y0 + c(z-z0) I need to find a) a point on the plane (x0,y0,z0) b) a vector normal to the plane <a,b,c> Hence if I am looking for the osculating plane I would need: a) a point on the osculating plane (x0,y0,z0) b) B(t) which happens to be the vector normal to my osculating plane B(t) = T(t) X N(t), from this I find my desired <a,b,c> I put it all together and I get my osculating plane of form: a(x-x0) + b(y-y0 + c(z-z0). (Right or wrong?) ============================== So what exactly is the cross product r'(t) X r"(t) for? Cutting down time so I dont have to go through the tedious process of finding T(t) and N(t)?