Vectors question: Is r'(t) technically a tangent (non-unit) vector to r(t)

I know T(t) = r'(t) / |r'(t)|, but here's my reasoning:

Since T(t) = r'(t)/|r'(t)| describes a vector at any given point t which is tangent to r(t), with a unique scalar |r'(t)| at that point, couldn't one simply remove the scalar and obtain a vector which is still tangent to r(t)?

My second question, which stems from the first: Is it then possible to calculate a non-unit normal vector N(t) by simply taking the second derivative of r(t), being r''(t)? And therefore, also a non-unit binomial vector B(t)?

Or is this bad math? If so, please explain where my reasoning went wrong.

Q1: Yes, but remember that r'(t) must be nonzero for this to work. If you parameterise the curve r in terms of the arc length s, this case is avoided. It also has the advantage that |r'(s)| = 1 (or, to put it another way, s'(t) = |r'(t)|), so the normalisation is taken care of.

Q2: Yes, but once again be careful of the case where the second derivative is zero. Interestingly, if you parameterise on arc length, then r''(s) has a geometric interpretation. The magnitude is the curvature of the curve at s, usually denoted κ. 1/κ is the radius of a circle which optimally fits the local shape of the curve, and the direction of r''(s) points to the centre of that circle. This is called the principal normal vector.

Q3: And yes! If T(s) is the tangent vector and N(s) is the principal normal vector, then B(s)=T(s)×N(s) is called the binormal vector, and together they form a local orthogonal coordinate system known as the Frenet frame.

Now let me answer the questions which you didn't ask:

If you think of three points on the curve very close to each other, those three points define a plane called the osculating plane. The tangent vector and the principal normal both lie in this plane, and the binormal B is tangent to it. The quantity