Unit Vectors

# Unit Vectors

 Definition: A vector $\vec{u}$ is considered to be a Unit Vector if the norm of $\vec{u}$ is equal to $1$, that is $\| \vec{u} \| = 1$.

For example, consider the vector $\vec{u} = (\frac{1}{2}, \frac{\sqrt{3}}{2})$. This vector is a unit vector since $\| \vec{u} \| = \sqrt{ \left( \frac{1}{2} \right)^2 + \left( \frac{\sqrt{3}}{2} \right)^2} = \sqrt{\frac{1}{4} + \frac{3}{4}} = \sqrt{1} = 1$.

Furthermore, sometimes it may be necessary to find a unit vector given some other vector as reference. If we have some vector $\vec{u}$, we can find a unit vector $\vec{u}_{unit}$ that goes in the same direction as $\vec{u}$ with the following formula:

(1)
\begin{align} \vec{u}_{unit} = \vec{u} \frac{1}{\| \vec{u} \|} \end{align}

For example, consider the vector $\vec{u} = (3, 4)$. We note that $\| \vec{u} \| = \sqrt{3^2 + 4^2} = \sqrt{25} = 5$. If we wanted to find a unit vector that went in the same direction as $\vec{u}$, all we would do is apply our formula, that is $\vec{u}_{unit} = \vec{u}\frac{1}{\| \vec{u} \|} = (3, 4) \cdot \frac{1}{5} = (\frac{3}{5}, \frac{4}{5})$.

## Example 1

Find a unit vector that goes in the same direction as vector $\vec{u} = (1, 2, 3)$ and then verify this new vector has a magnitude of $1$ and goes in the same direction as $\vec{u}$.

To solve this equation, we first need to calculate $\| \vec{u} \|$:

(2)
\begin{align} \| \vec{u} \| = \sqrt{1^2 + 2^2 + 3^2} \\ \| \vec{u} \| = \sqrt{14} \end{align}

We can now apply the formula from earlier to find the unit vector that goes in the same direction as vector $\vec{u}$ and has a magnitude of 1.

(3)
\begin{align} \vec{u}_{unit} = \vec{u} \frac{1}{\| \vec{u} \|} \\ \vec{u}_{unit} = (1, 2, 3) \cdot \frac{1}{\sqrt{14}} \\ \vec{u}_{unit} = (\frac{1}{\sqrt{14}}, \frac{2}{\sqrt{14}}, \frac{3}{\sqrt{14}}) \end{align}

Verifying the magnitude of this vector is rather easy. All we have to do is evaluate the norm of this vector and show that it is 1 as follows:

(4)
\begin{align} \| \vec{u}_{unit} \| = \sqrt{(\frac{1}{\sqrt{14}})^2 + (\frac{2}{\sqrt{14}})^2 + (\frac{3}{\sqrt{14}})^2} \\ \| \vec{u}_{unit} \| = \sqrt{\frac{1}{14} + \frac{4}{14} + \frac{9}{14}} \\ \| \vec{u}_{unit} \| = \sqrt{1} \\ \| \vec{u}_{unit} \| = 1 \end{align}

Now we know that if $k$ is a scalar, then the vector $k\vec{u}$ will go in the same direction as $\vec{u}$. In this case, our scalar is $\frac{1}{\| \vec{u} \|}$, so our result will go in the same direction as our original vector.

## Example 2

Given the vector $\vec{a} = (a_1, \frac{1}{2}, \frac{1}{3})$, give all values for the component $a_1$ such that $\| \vec{a} \| = 1$.

Since $\vec{u} \in \mathbb{R}^3$, we substitute into the formula for the norm of a vector to obtain:

(5)
\begin{align} \| \vec{a} \| = \sqrt{a_1^2 + a_2^2 + a_3^2} \\ \| \vec{a} \| = \sqrt{a_1^2 + \left ( \frac{1}{2}\right )^2 + \left ( \frac{1}{3} \right)^2} \\ \| \vec{a} \| = \sqrt{a_1^2 + \frac{1}{4} + \frac{1}{9}} \\ \| \vec{a} \| = \sqrt{a_1^2 + \frac{13}{36}} 1 = \sqrt{a_1^2 + \frac{13}{36}} \\ 1 = a_1^2 + \frac{13}{36} \\ a_1^2 = \frac{23}{36} \\ a_1 = \pm \sqrt{\frac{23}{36}} \end{align}

Therefore if $a_1 = \pm \sqrt{\frac{23}{36}}$, then $\| \vec{u} \| = 1$.