8. Properties of Curves

e. Tangent, Normal, and Binormal Vectors

1. Definitions

When dealing with vectors in $\mathbb{R}^{3}$, we use the standard basis vectors $\hat{\imath}$, $\hat{\jmath}$ and $\hat{k}$. These are three mutually perpendicular unit vectors which form a right handed system.

When dealing with vectors along a curve, it is nice to have three basis vectors (i.e. three mutually perpendicular unit vectors which form a right handed system) which are adapted to the curve. These are called $\hat{T}$, $\hat{N}$ and $\hat{B}$, and are collectively called a Frenet frame. To understand the definitions of the Frenet frame, first notice that the velocity, $\vec{v}$ determines the instantaneous direction of the curve while $\vec{v}$ and $\vec{a}$ determine the instantaneous plane of the curve.

Frenet Frame
The unit tangent vector $\hat{T}$ is the unit vector tangent to the curve, i.e. the direction of $\vec{v}$.
The unit normal vector $\hat{N}$ is the unit vector perpendicular to $\vec{v}$ in the plane of $\vec{v}$ and $\vec{a}$ on the same side of $\vec{v}$ as $\vec{a}$, i.e. $\hat{N}$ is the unit vector in the direction of $\text{proj}_{\bot \vec{v}}\vec{a}$.
The unit binormal vector $\hat{B}$ is the unit vector perpendicular to $\hat{T}$ and $\hat{N}$ related by the right hand rule, i.e. $\hat{B}=\hat{T}\times\hat{N}$

We first need to justify that $\hat{B}$ defined as $\hat{T}\times\hat{N}$ is in fact a unit vector. However, since $\hat{T}$ and $\hat{N}$ are perpendicular unit vectors $|\hat{T}|=1 \qquad |\hat{N}|=1 \qquad \text{and} \qquad \theta=90^\circ$ Hence $|\hat{B}| =|\hat{T}\times\hat{N}| =|\hat{T}|\,|\hat{N}|\sin\theta =1\cdot1\cdot1=1$

Recall that $\hat{\imath}$, $\hat{\jmath}$ and $\hat{k}$ form a right handed triplet of mutually perpendicular unit vectors in that they satisfy: $\hat{\imath}\times\hat{\jmath}=\hat{k} \qquad \hat{\jmath}\times\hat{k}=\hat{\imath} \qquad \hat{k}\times\hat{\imath}=\hat{\jmath}$ where we cyclically rotate $\hat{\imath}$, $\hat{\jmath}$ and $\hat{k}$ in these formulas. Similarly:

$\hat{T}$, $\hat{N}$ and $\hat{B}$ form a right handed triplet of mutually perpendicular unit vectors in that they satisfy: $\hat{T}\times\hat{N}=\hat{B} \qquad \hat{N}\times\hat{B}=\hat{T} \qquad \hat{B}\times\hat{T}=\hat{N}$ where we cyclically rotate $\hat{T}$, $\hat{N}$ and $\hat{B}$ in these formulas.   Justification

Find the unit tangent vector, $\hat{T}$, the unit normal vector $\hat{N}$, and the unit binormal vector $\hat{B}$ of the circle of radius $R$ in the $xy$-plane, parametrized as $\vec{r}(\theta)=(R\cos\theta,R\sin\theta,0)$ for general $\theta$ and at $\theta=\dfrac{\pi}{4}$.

First we find the velocity and its length: $\begin{aligned} \vec{v} &=\langle -R\sin\theta,R\cos\theta,0\rangle \\ |\vec{v}| &=\sqrt{(-R\sin\theta)^2+(R\cos\theta)^2+0^2}=R \end{aligned}$ From these, we calculate the unit tangent vector: $\begin{aligned} \hat{T} &=\dfrac{\vec{v}}{|\vec{v}|} =\dfrac{1}{R}\langle -R\sin\theta,R\cos\theta,0\rangle \\ &=\langle -\sin\theta,\cos\theta,0\rangle \end{aligned}$ Next, we compute the acceleration and notice that it is perpendicular to the velocity: $\begin{aligned} \vec{a} &=\langle -R\cos\theta,-R\sin\theta,0\rangle \\ \vec{v}\cdot\vec{a} &=R^2\sin\theta\cos\theta-R^2\cos\theta\sin\theta=0 \end{aligned}$ So the unit normal is the unit vector in the direction of the acceleration. We compute the length of the acceleration and the unit normal vector: $\begin{aligned} |\hat{a}| &=\sqrt{(-R\cos\theta)^2+(-R\sin\theta)^2+0^2}=R \\ \hat{N} &=\dfrac{\vec{a}}{|\vec{a}|} =\dfrac{1}{R}\langle -R\cos\theta,-R\sin\theta,0\rangle \\ &=\langle -\cos\theta,-\sin\theta,0\rangle \\ \end{aligned}$ Finally, the unit binormal vector is: $\begin{aligned} \hat{B} &=\hat{T}\times\hat{N} =\begin{vmatrix} \hat{\imath} & \hat{\jmath} & \hat{k} \\ -\sin\theta & \cos\theta & 0 \\ -\cos\theta & -\sin\theta & 0 \end{vmatrix} \\ &=\hat\imath(0)-\hat\jmath(0)+\hat k(\sin^2\theta+\cos^2\theta) =\langle 0,0,1\rangle \end{aligned}$

We check the three vectors are perpendicular by computing their dot products: $\begin{aligned} \hat{T}\cdot\hat{N}&=\sin\theta\cos\theta-\cos\theta\sin\theta=0 \\ \hat{T}\cdot\hat{B}&=0+0+0=0 \\ \hat{B}\cdot\hat{N}&=0+0+0=0 \end{aligned}$

In the examples of a circle and a helix, the acceleration is perpendicular to the velocity. So it is easy to find the normal vector. On the next page, we will find formulas which help us compute $\hat{T}$, $\hat{N}$ and $\hat{B}$ even if the acceleration is not perpendicular to the velocity.