Suppose $\mathbf{T}(s)$ is the unit tangent vector to a curve parametrized by arclength, and $\mathbf{N}(s) = \mathbf{T}'(s)/||\mathbf{T}'(s)||$ assuming $\mathbf{T}(s) \ne \mathbf{0}$.
Define the binormal vector $\mathbf{B}(s)=\mathbf{T}(s)\times \mathbf{N}(s)$. Then $\left\{\mathbf{T}(s),\mathbf{N}(s),\mathbf{B}(s)\right\}$ forms a right handed orthonormal basis for $\mathbb{R}^3$.
I see they are orthogonal, but why are they basis? I feel like I'm missing a basic fact about cross products probably.
Best Answer
Whenever we have $3$ orthogonal vectors in $\mathbb{R}^3$, they form a basis if and only if each of them is non-null. But you are assuming that $\mathbf{T}(s)\neq0$, $\bigl\lVert\mathbf{N}(s)\bigr\rVert=1$ and the cross-product of two non-null orthogonal vectors is never $0$.