# Unit Vectors

Vector quantities have a direction and a magnitude. However, sometimes one is interested in only the direction of the vector and not the magnitude. In such cases, for convenience, vectors are often "normalized" to be of unit length. These **unit vectors** are commonly used to indicate direction, with a scalar coefficient providing the magnitude. A vector decomposition can then be written as a sum of unit vectors and scalar coefficients.

Given a vector $\vec{V}$, one might consider the problem of finding the vector parallel to $\vec{V}$ with unit length. In other words, one might want to find some "scale factor" $a$ such that $\left\|a \vec{V} \right\| = 1$. Since $\left\|a \vec{V}\right\| = a \left\|\vec{V}\right\|$, it follows trivially that $a = \frac1{\left\|\vec{V}\right\|}.$

Thus, given a vector $\vec{V}$, the **unit vector** $\hat{V}$ defined by

$\hat{V} = \frac{\vec{V}}{\left\|\vec{V}\right\|}$

has unit length and is parallel to $\vec{V}$. A unit vector is frequently (though not always) written with "hat" symbol to indicate that it is of unit length.

Find the unit vector $\hat{D}$ in the direction of $\vec{D} = \langle 4, 3 \rangle$.

We have (using the Pythagorean theorem)

$|\vec{D}| = 5 \implies \hat{D} = \frac{\vec{D}}{5} = \left \langle \frac{4}{5}, \frac{3}{5} \right \rangle.\ _\square$

Find the unit vector $\hat{A}$ in the direction of $\vec{A} = \langle -12, 5 \rangle$.

We have

$|\vec{A}| = 13 \implies \hat{A} = \frac{\vec{A}}{13} =\left \langle -\frac{12}{13}, \frac{5}{13} \right \rangle.\ _\square$

Unit vectors are often used in the decomposition of a vector into orthogonal components. For instance, one can define the unit vectors that point in each of the positive coordinate axes as follows:

$\begin{aligned} \hat{x} &= \left \langle 1, 0, 0 \right \rangle \\ \hat{y} &= \left \langle 0, 1, 0 \right \rangle \\ \hat{z} &= \left \langle 0, 0, 1 \right \rangle. \end{aligned}$

It follows that for any vector $\vec{V} = \left \langle v_x, v_y, v_z \right \rangle$, it holds that

$\vec{V} = v_x \hat{x} + v_y \hat{y} + v_z \hat{z}.$

In this way, one frequently encounters vectors written in terms of the unit vectors $\hat{x}$, $\hat{y}$, and $\hat{z}$. An alternate notation uses $\hat{i} = \hat{x}$ , $\hat{j} = \hat{y}$, and $\hat{k} = \hat{z}$.

Note that the coordinate unit vectors are orthonormal. That is to say, $\hat{x} \cdot \hat{x} = \hat{y} \cdot \hat{y} = \hat{z} \cdot \hat{z} = 1$ and $\hat{x} \cdot \hat{y} = \hat{y} \cdot \hat{z} = \hat{x} \cdot \hat{z} = 0$.

When two vectors $\vec{V}= (v_x, v_y, v_z)$ and $\vec{W} = (w_x, w_y, w_z)$ add, we simply have

$\vec{V} +\vec{W}= (v_x \hat{x} + v_y \hat{y} + v_z \hat{z}) + (w_x \hat{x} + w_y \hat{y} + w_z \hat{z}) = (v_x + w_x) \hat{x} + (v_y + w_y) \hat{y} + (v_z + w_z) \hat{z},$

which is the same thing as $(v_x + w_x, v_y + w_y, v_z + w_z)$. Similarly, when one takes the dot product $\vec{V} \cdot \vec{W}$, one obtains

$\begin{aligned} \vec{V} \cdot \vec{W} &= (v_x \hat{x} + v_y \hat{y} + v_z \hat{z}) \cdot (w_x \hat{x} + w_y \hat{y} + w_z \hat{z}) \\ &= v_x w_x \hat{x} \cdot \hat{x} + v_y w_y \hat{y} \cdot \hat{y} + v_z w_z \hat{z} \cdot \hat{z} \\ &= v_x w_x + v_y w_y + v_z w_z, \end{aligned}$

as expected.