Yes. You are correct. In fact, the precise definition of the velocity of a particle whose position is given by the vector equation r(t) is another vector r'(t). Speed is defined as the magnitude of velocity, |r'(t)|.
For the second part, the vector r'(1) represents the trajectory of the particle at the instant of its position r(1). Since k is the normal of the plane of interest, the "triangle" of which you speak is really just the shape formed by the two vectors r'(1) and k crossing at an angle.