# 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:

\[ \begin{align} \hat{x} &= \left \langle 1, 0, 0 \right \rangle \\ \hat{y} &= \left \langle 0, 1, 0 \right \rangle \\ \hat{z} &= \left \langle 0, 0, 1 \right \rangle. \end{align} \]

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

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

as expected.