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.

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.