给定3D点处的3D曲线的切线

Question

I'm trying to compute tangent line (or tangent vector) at 3D point of a 3D curve. The problem is how to compute the slope according to x,y and z at point ? I recall that for a 2D curve, the equation of the tangent line is:

tang=(x-x_k)*slope_k+y_k

Answer 1

The slope in 3D can be written as simple different between two 3D points. Here is why: https://math.stackexchange.com/questions/799783/slope-of-a-line-in-3d-coordinate-system