# Calculating Unit Normal and Unit Binormal Vectors Examples 1

We have just looked at a way to calculate the unit binormal $\hat{B}(t)$ and unit normal $\hat{N}(t)$ vectors of a vector-valued function $\vec{r}(t) = (x(t), y(t), z(t))$ from the Method for Calculating Unit Normal and Unit Binormal Vectors page with the following formulae:

**Unit Binormal Vector:**$\hat{B}(t) = \frac{\vec{r'}(t) \times \vec{r''}(t)}{\| \vec{r'}(t) \times \vec{r''}(t) \|}$.

**Unit Normal vector:**$\hat{N}(t) = \hat{B}(t) \times \hat{T}(t)$.

We will now look at some examples of finding the unit binormal and unit normal vectors given a vector-valued function.

## Example 1

**Find the unit normal and unit binormal vectors for the vector equation $\vec{r}(t) = (6, t^2, -t)$.**

We will first differentiate $\vec{r}(t)$ twice to get $\vec{r'}(t) = (0, 2t, -1)$ and $\vec{r''}(t) = (0, 2, 0)$. Now we will compute $\vec{r'}(t) \times \vec{r''}(t)$ as follows:

(1)The magnitude of $\vec{r'}(t) \times \vec{r''}(t)$ is $\| \vec{r'}(t) \times \vec{r''}(t) \| = \sqrt{2^2} = 2$. Therefore, we have that:

(2)We should note that this makes sense as our curve solely lies on the plane $x = 6$.

Now to calculate $\hat{N}(t)$ we must first find $\hat{T}(t) = \frac{\vec{r'}(t)}{\| \vec{r'}(t) \|}$. The magnitude of $\vec{r'}(t)$ is $\| \vec{r'}(t) \| = \sqrt{4t^2 + 1}$ and so we have that $\hat{T}(t) = \frac{1}{\sqrt{4t^2 + 1}} (0, -2t, 1) = \left ( 0, \frac{-2t}{\sqrt{4t^2 + 1}}, \frac{1}{ \sqrt{4t^2 + 1}} \right )$.

Therefore we have that:

(3)