# 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.
\[
\hat{B}=\hat{T}\times\hat{N}
\]

The figures at the right show:

a curve in **blue**,

the unit tangent vector, \(\hat{T}\), in
**red**,

the unit normal vector, \(\hat{N}\), in
**cyan** and

the unit binormal vector, \(\hat{B}\) in
**magenta**.

In the first figure, use the mouse to rotate the plot until \(\hat{N}\)
is pointing straight at you. Notice that the curve instantaneously lies in
the plane of \(\hat{T}\) and \(\hat{N}\).

Now rotate the plot until \(\hat{B}\) is pointing straight at you. Notice
that the curve instantaneously bends in the direction of \(\hat{N}\).

The second figure shows how \(\hat{T}\), \(\hat{N}\) and \(\hat{B}\) change as the point moves along the curve. Notice, \(\hat{T}\) is in the direction of the curve, \(\hat{N}\) is in the direction it is turning and \(\hat{B}\) is perpendicular to the plane of the motion.

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.

To see this, use your right hand to see that \(\hat{T}\times\hat{N}\) points along \(\hat{B}\). Then use the mouse to rotate the figure and use your right hand to see where \(\hat{N}\times\hat{B}\) and \(\hat{B}\times\hat{T}\) point.

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: \[\begin{aligned} \hat{B} &=\hat{T}\times\hat{N} =\begin{vmatrix} \hat{\imath} & \hat{\jmath} & \hat{k} \\ -\sin\theta & \cos\theta & 0 \\ -\cos\theta & -\sin\theta & 0 \end{vmatrix} \\ &=\hat\imath(0)-\hat\jmath(0)+\hat k(\sin^2\theta+\cos^2\theta) =\langle 0,0,1\rangle \end{aligned}\]

The figure shows \(\hat{T}\), \(\hat{N}\) and \(\hat{B}\) moving around the circle. In particular, for \(\theta=\dfrac{\pi}{4}\): \[\begin{aligned} \hat{T} &=\left\langle \dfrac{-1}{\sqrt{2}},\dfrac{1}{\sqrt{2}},0\right\rangle \\ \hat{N} &=\left\langle \dfrac{-1}{\sqrt{2}},\dfrac{-1}{\sqrt{2}},0\right\rangle \\ \hat{B} &=\langle 0,0,1\rangle \end{aligned}\]

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

Find the unit tangent vector, \(\hat{T}\), the unit normal vector \(\hat{N}\), and the unit binormal vector \(\hat{B}\) of the helix parametrized as \(\vec{r}(\theta)=(4\cos\theta,4\sin\theta,3\theta)\) for general \(\theta\) and at \(\theta=\dfrac{\pi}{4}\)

Like the circle, the acceleration is again perpendicular to the velocity which makes it easy to calculate the unit normal vector.

Generally: \[\begin{aligned} \hat{T} &=\left\langle \dfrac{-4}{5}\sin\theta,\dfrac{4}{5}\cos\theta,\dfrac{3}{5}\right\rangle \\ \hat{N} &=\langle -\cos\theta,-\sin\theta,0\rangle \\ \hat{B} &=\left\langle \dfrac{3}{5}\sin\theta,\dfrac{-3}{5}\cos\theta,\dfrac{4}{5}\right\rangle \end{aligned}\] For \(\theta=\dfrac{\pi}{4}\): \[\begin{aligned} \hat{T} &=\left\langle \dfrac{-2\sqrt{2}}{5},\dfrac{2\sqrt{2}}{5},\dfrac{3}{5}\right\rangle \\ \hat{N} &=\left\langle \dfrac{-1}{\sqrt{2}},\dfrac{-1}{\sqrt{2}},0\right\rangle \\ \hat{B} &=\left\langle \dfrac{3}{5\sqrt{2}},\dfrac{-3}{5\sqrt{2}},\dfrac{4}{5}\right\rangle \end{aligned}\]

First we find the velocity and its length: \[\begin{aligned} \vec{v} &=\langle -4\sin\theta,4\cos\theta,3\rangle \\ |\vec{v}| &=\sqrt{16\sin^2\theta+16\cos^2\theta+9}=5 \end{aligned}\] From these, we calculate the unit tangent vector: \[ \hat{T} =\dfrac{\vec{v}}{|\vec{v}|} =\left\langle \dfrac{-4}{5}\sin\theta,\dfrac{4}{5}\cos\theta,\dfrac{3}{5}\right\rangle \] Next, we compute the acceleration and check that it is perpendicular to the velocity: \[\begin{aligned} \vec{a} &=\langle -4\cos\theta,-4\sin\theta,0\rangle \\ \vec{v}\cdot\vec{a} &=16\sin\theta\cos\theta-16\cos\theta\sin\theta=0 \end{aligned}\] 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: \[\begin{aligned} |\hat{a}| &=\sqrt{(-4\cos\theta)^2+(-4\sin\theta)^2+0^2}=4 \\ \hat{N} &=\dfrac{\vec{a}}{|\vec{a}|} =\langle -\cos\theta,-\sin\theta,0\rangle \\ \end{aligned}\] Finally, the unit binormal vector is: \[\begin{aligned} \hat{B} &=\hat{T}\times\hat{N} =\begin{vmatrix} \hat{\imath} & \hat{\jmath} & \hat{k} \\[3 pt] \dfrac{-4}{5}\sin\theta & \dfrac{4}{5}\cos\theta & \dfrac{3}{5} \\[3 pt] -\cos\theta & -\sin\theta & 0 \end{vmatrix} \\ &=\hat\imath\left(\dfrac{3}{5}\sin\theta\right) -\hat\jmath\left(\dfrac{3}{5}\cos\theta\right) +\hat k\left(\dfrac{4}{5}\sin^2\theta+\dfrac{4}{5}\cos^2\theta\right) \\ &=\left\langle\dfrac{3}{5}\sin\theta,\dfrac{-3}{5}\cos\theta,\dfrac{4}{5}\right\rangle \end{aligned}\]

The figure shows \(\hat{T}\), \(\hat{N}\) and \(\hat{B}\) moving around the helix. In particular, for \(\theta=\dfrac{\pi}{4}\): \[\begin{aligned} \hat{T} &=\left\langle \dfrac{-2\sqrt{2}}{5},\dfrac{2\sqrt{2}}{5},\dfrac{3}{5}\right\rangle \\ \hat{N} &=\left\langle \dfrac{-1}{\sqrt{2}},\dfrac{-1}{\sqrt{2}},0\right\rangle \\ \hat{B} &=\left\langle \dfrac{3}{5\sqrt{2}},\dfrac{-3}{5\sqrt{2}},\dfrac{4}{5}\right\rangle \end{aligned}\] It is easy to check these are \(3\) mutually perpendicular unit vectors.

We check the three vectors \[\begin{aligned} \hat{T} &=\left\langle \dfrac{-4}{5}\sin\theta,\dfrac{4}{5}\cos\theta,\dfrac{3}{5}\right\rangle \\ \hat{N} &=\langle -\cos\theta,-\sin\theta,0\rangle \\ \hat{B} &=\left\langle\dfrac{3}{5}\sin\theta,\dfrac{-3}{5}\cos\theta,\dfrac{4}{5}\right\rangle \end{aligned}\] are perpendicular by computing their dot products: \[\begin{aligned} \hat{T}\cdot\hat{N}&=\dfrac{4}{5}\sin\theta\cos\theta-\dfrac{4}{5}\cos\theta\sin\theta=0 \\ \hat{T}\cdot\hat{B}&=-\dfrac{12}{25}\sin^2\theta-\dfrac{12}{25}\cos^2\theta+\dfrac{12}{25}=0 \\ \hat{B}\cdot\hat{N}&=-\dfrac{3}{5}\sin\theta\cos\theta+\dfrac{3}{5}\cos\theta\sin\theta=0 \end{aligned}\]

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.

## Heading

Placeholder text: Lorem ipsum Lorem ipsum Lorem ipsum Lorem ipsum Lorem ipsum Lorem ipsum Lorem ipsum Lorem ipsum Lorem ipsum Lorem ipsum Lorem ipsum Lorem ipsum Lorem ipsum Lorem ipsum Lorem ipsum Lorem ipsum Lorem ipsum Lorem ipsum Lorem ipsum Lorem ipsum Lorem ipsum Lorem ipsum Lorem ipsum Lorem ipsum Lorem ipsum