3.3 Arc length and curvature  (Page 5/18)

The normal and binormal vectors

We have seen that the derivative $r^{\prime }\left(t\right)$ of a vector-valued function is a tangent vector to the curve defined by $\text{r}\left(t\right),$ and the unit tangent vector $\text{T}\left(t\right)$ can be calculated by dividing $r^{\prime }\left(t\right)$ by its magnitude. When studying motion in three dimensions, two other vectors are useful in describing the motion of a particle along a path in space: the principal unit normal vector    and the binormal vector    .

Note that, by definition, the binormal vector is orthogonal to both the unit tangent vector and the normal vector. Furthermore, $\text{B}\left(t\right)$ is always a unit vector. This can be shown using the formula for the magnitude of a cross product

where $\theta$ is the angle between $\text{T}\left(t\right)$ and $\text{N}\left(t\right).$ Since $\text{N}\left(t\right)$ is the derivative of a unit vector, property (vii) of the derivative of a vector-valued function tells us that $\text{T}\left(t\right)$ and $\text{N}\left(t\right)$ are orthogonal to each other, so $\theta =\pi \text{/}2.$ Furthermore, they are both unit vectors, so their magnitude is 1. Therefore, $\Vert \text{T}\left(t\right)\Vert \Vert \text{N}\left(t\right)\Vert \text{sin}\theta =\left(1\right)\left(1\right)\text{sin}\left(\pi \text{/}2\right)=1$ and $\text{B}\left(t\right)$ is a unit vector.

The principal unit normal vector can be challenging to calculate because the unit tangent vector involves a quotient, and this quotient often has a square root in the denominator. In the three-dimensional case, finding the cross product of the unit tangent vector and the unit normal vector can be even more cumbersome. Fortunately, we have alternative formulas for finding these two vectors, and they are presented in Motion in Space .