Given a twice-differentiable vector-valued function $r\left(t\right)$, what is the definition of the binormal vector $B\left(t\right)$?

Consider the given question,

A twice-differentiable vector-valued function is$r\left(t\right)$.

If $r\left(t\right)$is a differentiable vector function on the interval $I\subseteq R$, then unit tangent function to $r\left(t\right)$is given below,

$T\left(t\right)=\frac{r\text{'}\left(t\right)}{||r\text{'}\left(t\right)||}$

Where, $r\text{'}\left(t\right)\ne 0$.

From the principal unit normal vector,

If $r\left(t\right)$is a differential vector function on $I\subseteq R$, then the principal unit normal vector at $r\left(t\right)$is denoted as,

$N\left(t\right)=\frac{T\text{'}\left(t\right)}{||T\text{'}\left(t\right)||}$

Where, $T\text{'}\left(t\right)\ne 0$the binomial vector $B\left(t\right)$is the cross product product of the above$2$vectors.

Consider the definition of binomial vector,

Assume $r\left(t\right)$be a differential vector function on the interval $I\subseteq R$such that $T\text{'}\left(t_0\right)\ne 0$where $t_0\in I$.

The binomial vector B at $r\left(t_0\right)$is defined as,

$B\left(t_0\right)=T\left(t_0\right)\times N\left(t_0\right)$

As $T\left(t_0\right),n\left(t_0\right)$are the two orthogonal unit vectors, their cross product $B\left(t_0\right)$is another unit vector and is orthogonal to both of them.