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 \( \vec{D} = \langle 4, 3 \rangle \).

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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 \( \vec{A} = \langle -12, 5 \rangle \).

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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.