3.5. Vector magnitude#

The magnitude of a vector \(\mathbf{a}\) is the distance between the head and tail of \(\mathbf{a}\) which we can calculate using an extension of Pythagoras’ theorem.

Definition 3.4 (Vector magnitude)

The magnitude of a vector \(\mathbf{a} = (a_1, a_2, \ldots, a_n)\) denoted by \(\|\mathbf{a}\|\) is calculated using

(3.1)#\[ \|\mathbf{a}\| = \sqrt{\sum_{i=1}^n a_i^2} = \sqrt{a_1^2 + a_2^2 + \cdots + a_n^2}. \]
../_images/3_magnitude.svg

Vector magnitude is also known as the Euclidean norm.

Example 3.1

Calculate the magnitudes of the following vectors

\[ \begin{align*} \mathbf{u} &= (1, 2, 3), & \mathbf{v} &= (5, -12, 0) & \mathbf{w} &= (1, 0, 1). \end{align*} \]

Solution

\[\begin{split} \begin{align*} \|\mathbf{u}\| &= \sqrt{1^2 + 2^2 + 3^2} = \sqrt{1+4+9} = \sqrt{14}, \\ \|\mathbf{v}\| &= \sqrt{5^2 + (-12)^2 + 0^2} = \sqrt{25+144+0} = \sqrt{169} = 13, \\ \|\mathbf{w}\| &= \sqrt{1^2 + 0^2 + 1^2} = \sqrt{1+0+1} = \sqrt{2}. \end{align*} \end{split}\]

Theorem 3.2 (Properties of vector magnitude)

For any two vectors \(\mathbf{a}, \mathbf{b} \in \mathbb{R}^n\) and \(k \in \mathbb{R}\) the following properties are satisfied

  • \(\| \mathbf{a} \| \geq 0\)   (the magnitude of a vector \(\mathbf{v}\) is always non-negative)

  • \(\|\mathbf{a}\| = 0 \iff \mathbf{a} = \mathbf{0}\)   (the magnitude of a vector \(\mathbf{v}\) is zero if and only if the vector is the zero vector)

  • \(\|k\mathbf{a}\| = |k| \| \mathbf{a} \|\)

  • \(\| \mathbf{a} + \mathbf{b} \| \leq \|\mathbf{a}\| + \| \mathbf{b} \|\)   (triangle inequality)

3.5.1. Unit vectors#

For every non-zero vector \(\mathbf{a}\) there exist a unique unit vector which is a vector in the same direction as \(\mathbf{a}\) and whose magnitude is 1.

Definition 3.5 (Unit vectors)

A unit vector is a vector with a magnitude of 1.

Theorem 3.3 (Normalising a vector)

Any non-zero vector can be scaled to transform it into a unit vector by dividing all its coordinates by its magnitude

(3.2)#\[ \hat{\mathbf{a}} = \frac{\mathbf{a}}{\|\mathbf{a}\|}. \]

This process is called normalising a vector. Unit vectors are denoted with a caret above the vector name, i.e., \(\hat{\mathbf{a}}\) which is read as ‘a hat’.

Proof. Let \(\mathbf{a}\) be a non-zero vector

\[\begin{split} \begin{align*} \left\|\frac{1}{\|\mathbf{a}\|}\mathbf{a}\right\| &= \left\|\frac{1}{\|\mathbf{a}\|}\right\| \|\mathbf{a}\| \\ &= \frac{1}{\|\mathbf{a}\|} \|\mathbf{a}\| \qquad \text{(since $\|\mathbf{a}\|>0$)}\\ &= 1. \end{align*} \end{split}\]

Example 3.2

Find the unit vectors parallel to the following:

(i)   \(\mathbf{u} = (1, 2, 3)\);   (ii)   \((5, -12, 0)\);   (iii)   \(\mathbf{w} = (1, 0, 1)\)

Solution

(i)

\[ \hat{\mathbf{u}} = \dfrac{\mathbf{u}}{\|\mathbf{u}\|} = \dfrac{(1, 2, 3)}{\sqrt{14}} = \left( \frac{1}{\sqrt{14}}, \frac{2}{\sqrt{14}}, \frac{3}{\sqrt{14}}\right) = \left( \frac{\sqrt{14}}{14}, \frac{\sqrt{14}}{7}, \frac{3\sqrt{14}}{14} \right)\]

Check magnitude of \(\hat{\mathbf{u}}\)

\[ \begin{align*} \|\hat{\mathbf{u}}\| &= \sqrt{ \left( \frac{\sqrt{14}}{14} \right)^2 + \left( \frac{\sqrt{14}}{7} \right)^2 + \left(\frac{3\sqrt{14}}{14} \right)^2} = \sqrt{ \frac{14}{196} +\frac{14}{49} + \frac{126}{196}} = \sqrt{1} = 1 \qquad \checkmark \end{align*} \]

(ii)

\[ \hat{\mathbf{v}} = \dfrac{\mathbf{v}}{\|\mathbf{v}\|} = \dfrac{(5, -12, 0)}{13} = \left( \frac{5}{13}, -\frac{12}{13}, 0 \right) \]

Check magnitude of \(\hat{\mathbf{v}}\)

\[ \begin{align*} \|\hat{\mathbf{v}}\| &= \sqrt{ \left(\frac{5}{13} \right)^2 + \left( -\frac{12}{13} \right)^2 + 0^2} = \sqrt{ \frac{25}{169} + \frac{144}{169} + 0} = \sqrt{1} = 1 \qquad \checkmark \end{align*} \]

(iii)

\[ \hat{\mathbf{w}} = \dfrac{\mathbf{w}}{\|\mathbf{w}\|} = \dfrac{(1, 0, 1)}{\sqrt{2}} = \left( \frac{1}{\sqrt{2}}, 0, \frac{1}{\sqrt{2}} \right) = \left( \frac{\sqrt{2}}{2}, 0, \frac{\sqrt{2}}{2} \right)\]

Check magnitude of \(\hat{\mathbf{w}}\)

\[ \begin{align*} \|\hat{\mathbf{w}}| &= \sqrt{ \left( \frac{\sqrt{2}}{2} \right)^2 + 0^2 + \left( \frac{\sqrt{2}}{2} \right)^2 } = \sqrt{ \frac{2}{4} + 0 + \frac{2}{4} } = \sqrt{1} = 1 \qquad \checkmark \end{align*} \]