Vectors

A vector is an object that has length and direction. In mathematical notation vectors are denoted in print using a boldface character or as an arrow over a character and underlined when handwritten

\[ \begin{align*} \mathbf{a}, \qquad \vec{a}, \qquad \underline{a}. \end{align*} \]

A vector in \(\mathbb{R}^n\) is defined by the signed distance along each axis by an \(n\)-tuple. For example, let \(\mathbf{a}\) be a vector in \(\mathbb{R}^3\) defined by the 3-tuple \(\mathbf{a} = (a_x, a_y, a_z)\) where \(a_x,a_y,a_z \in \mathbb{R}\), then \(\mathbf{a}\) can be represented geometrically as shown in Fig. 5.

Fig. 5 A vector in \(\mathbb{R}^3\).

The tuple representing a vector can be written as either a matrix consisting of a single row or a single column. So a vector in \(\mathbb{R}^3\) can be represented as

\[\begin{split} \begin{align*} \mathbf{a} = (a_x, a_y, a_z) = \begin{pmatrix} a_x \\ a_y \\ a_z \end{pmatrix}. \end{align*} \end{split}\]

These representations are called row vector and column vector respectively.

Vector magnitude

The length of a vector \(\mathbf{a}\) is known as the magnitude and is denoted using \(|\mathbf{a}|\).

Definition 1 (Vector magnitude)

The magnitude of a vector in \(\mathbb{R}^n\), \(\mathbf{a} = (a_1, a_2, \ldots, a_n)\), is calculated using

(1)\[|\mathbf{a}| = \sqrt{ \sum_{i=1}^n a_i^2 } = \sqrt{a_1^2 + a_2^2 + \cdots + a_n^2}.\]

Note \(|\mathbf{a}| > 0\).

Example 1

Calculate the magnitude of the vector \(\mathbf{a} = (3, 4, 0)\).

Solution

\[ \begin{align*} |\mathbf{a}| = \sqrt{3^2 + 4^2 + 0} = \sqrt{25} = 5. \end{align*} \]

Unit vectors

A unit vector is denoted by \(\hat{\mathbf{a}}\) (referred to as ‘\(\mathbf{a}\) hat’) is a vector parallel to \(\mathbf{a}\) with a magnitude of 1.

Definition 2 (Normalising a vector)

The unit vector \(\hat{\mathbf{a}}\) that is parallel to the vector \(\mathbf{a}\) can be calculated using

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

This is known as normalising a vector.

Example 2

Calculate a unit vector that is parallel to \(\mathbf{a} = (3, 4, 0)\).

Solution

\[ \begin{align*} \hat{\mathbf{a}} = \frac{(3, 4, 0)}{5} = \left( \frac{3}{5}, \frac{4}{5}, 0 \right). \end{align*} \]

Checking that \(|\hat{\mathbf{a}}|=1\)

\[\begin{split} \begin{align*} |\hat{\mathbf{a}}| &= \sqrt{\left(\frac{3}{5}\right)^2 + \left(\frac{4}{5}\right)^2 + 0^2} \\ &= \sqrt{\frac{9}{25} + \frac{16}{25}} = \sqrt{\frac{25}{25}} = 1. \qquad \checkmark \end{align*} \end{split}\]

Scalar multiplication of a vector

The scalar multiple of a vector \(\mathbf{a}=(a_1, a_2, a_3)\) by the scalar \(k\) is defined by

\[\begin{split} \begin{align*} k\mathbf{a} = \begin{pmatrix} k a_1 \\ k a_2 \\ k a_3 \end{pmatrix}. \end{align*} \end{split}\]

The effect of multiplying a vector by a scalar is that the magnitude of the vector is scaled by the value of the scalar. If \(k>0\) then the direction of the vector \(k\mathbf{a}\) is the same as \(\mathbf{a}\) whereas if \(k<0\) then the vector \(k\mathbf{a}\) points in the opposite direction to \(\mathbf{a}\) (Fig. 6).

Fig. 6 Scalar multiples of the vector \(\mathbf{a}\).

Vector addition and subtraction

The addition of two vectors is achieved by adding the corresponding elements in the tuples. Let \(\mathbf{a}=(a_1,a_2,a_3)\) and \(\mathbf{b}=(b_1,b_2,b_3)\) then the sum \(\mathbf{a}+\mathbf{b}\) is calculated by

\[\begin{split} \begin{align*} \mathbf{a} + \mathbf{b} &= \begin{pmatrix} a_1 \\ a_2 \\ a_3 \end{pmatrix} + \begin{pmatrix} b_1 \\ b_2 \\ b_3 \end{pmatrix} = \begin{pmatrix} a_1 + b_1 \\ a_2 + b_2 \\ a_3 + b_3 \end{pmatrix}. \end{align*} \end{split}\]

Similarly the subtraction of two vectors is achieved by subtracting the corresponding element in the tuples, i.e.,

\[\begin{split} \begin{align*} \mathbf{a} - \mathbf{b} &= \begin{pmatrix} a_1 \\ a_2 \\ a_3 \end{pmatrix} - \begin{pmatrix} b_1 \\ b_2 \\ b_3 \end{pmatrix} = \begin{pmatrix} a_1 - b_1 \\ a_2 - b_2 \\ a_3 - b_3 \end{pmatrix}. \end{align*} \end{split}\]

Thinking about this in a geometrical sence, the vector addition \(\mathbf{a} + \mathbf{b}\) is achieved by placing the tail of \(\mathbf{b}\) at the head of \(\mathbf{a}\). The resulting vector points from the tail of \(\mathbf{a}\) to the head of \(\mathbf{b}\). The vector subtraction \(\mathbf{a} - \mathbf{b}\) is achieved by reversing the direction of \(\mathbf{b}\) and placing the tail at the head of \(\mathbf{a}\).

Fig. 7 Vector addition and subtraction.

The dot product

The product of two vectors can be calculated in two ways: the dot product and the cross product. The dot product of two vectors \(\mathbf{a}\) and \(\mathbf{b}\) is denoted by \(\mathbf{a}\cdot \mathbf{b}\) and returns a scalar quantity and the dot product is often referred to as the scalar product.

Definition 3 (Geometric definition of the dot product)

The geometric definition of the dot product of two vectors, \(\mathbf{a}, \mathbf{b} \in \mathbb{R}^n\), is

(3)\[\mathbf{a} \cdot \mathbf{b} = |\mathbf{a}| |\mathbf{b}| \cos(\theta),\]

where \(\theta\) is the angle between the two vectors (Fig. 8).

Fig. 8 The two vectors \(\mathbf{a}\) and \(\mathbf{b}\) and the angle between them \(\theta\).

The value of a dot product can be computed using the algebraic definition of the dot product.

Definition 4 (Algebraic definition of the dot product)

The dot product of two vectors \(\mathbf{a}=(a_1, a_2, \ldots, a_n)\) and \(\mathbf{b}= (b_1, b_2, \ldots, b_n)\) in \(\mathbb{R}^n\) can be calculated using

(4)\[\mathbf{a} \cdot \mathbf{b} = \sum_{i=1}^n a_i b_i = a_1b_1 + a_2b_2 + \cdots + a_n b_n.\]

For vectors \(\mathbf{a}\), \(\mathbf{b}\) and \(\mathbf{c} \in \mathbb{R}^n\) the dot product has the following properties

• commutative: \(\mathbf{a} \cdot \mathbf{b} = \mathbf{b} \cdot \mathbf{a}\);

• distributive: \(\mathbf{a} \cdot (\mathbf{b} + \mathbf{c}) = \mathbf{a} \cdot \mathbf{b} + \mathbf{a} \cdot \mathbf{c}\);

• orthogonal: the non-zero vectors \(\mathbf{a}\) and \(\mathbf{b}\) are orthogonal (perpendicular) if \(\mathbf{a} \cdot \mathbf{b} = 0\).

Example 3

(i)   Calculate the dot product of the two vectors \(\mathbf{a}=(3,4,0)\) and \(\mathbf{b}=(5, 12, 0)\).

Solution

\[\begin{split} \begin{align*} \mathbf{a}\cdot \mathbf{b} = \begin{pmatrix} 3 \\ 4 \\ 0 \end{pmatrix} \cdot \begin{pmatrix} 5 \\ 12 \\ 0 \end{pmatrix} = 3 \times 5 + 4 \times 12 + 0 \times 0 = 15 + 48 = 63. \end{align*} \end{split}\]

(ii)   Calculate the angle between the two vectors \(\mathbf{a}=(3,4,0)\) and \(\mathbf{b}=(5, 12, 0)\).

Solution

\[\begin{split} \begin{align*} \mathbf{a}\cdot \mathbf{b} &= |\mathbf{a}||\mathbf{b}|\cos(\theta)\\ \therefore \theta &= \cos^{-1}\left( \frac{\mathbf{a} \cdot \mathbf{b}}{|\mathbf{a}||\mathbf{b}|}\right) = \cos^{-1}\left(\frac{63}{5 \cdot 13}\right) = 0.2487. \end{align*} \end{split}\]

(iii)   Two orthogonal vectors are \(\mathbf{a}=(1,2,3)\) and \(\mathbf{b}=(4, x, 6)\). Determine the value of \(x\).

Solution

\[\begin{split} \begin{align*} 0 &= \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix} \cdot \begin{pmatrix} 4 \\ x \\ 6 \end{pmatrix} = 4 + 2x + 18, \\ \therefore x &= -11. \end{align*} \end{split}\]

The cross product

The cross product of two vectors \(\mathbf{a}\) and \(\mathbf{b}\) is denoted by \(\mathbf{a} \times \mathbf{b}\) and returns a vector that is perpendicular to both \(\mathbf{a}\) and \(\mathbf{b}\) (Fig. 9).

Fig. 9 The cross product \(\mathbf{a} \times \mathbf{b}\) is a vector perpendicular to both \(\mathbf{a}\) and \(\mathbf{b}\).

Definition 5 (Geometric definition of the cross product)

The cross product of two vectors, \(\mathbf{a}, \mathbf{b} \in \mathbb{R}^3\), is defined by

(5)\[\mathbf{a} \times \mathbf{b} = |\mathbf{a}| |\mathbf{b}| \sin(\theta) \hat{\mathbf{n}},\]

where \(\theta\) is the angle between \(\mathbf{a}\) and \(\mathbf{b}\) and \(\hat{\mathbf{n}}\) is a unit vector perpendicular to both \(\mathbf{a}\) and \(\mathbf{b}\).

The cross product of two vectors in \(\mathbb{R}^3\) can be computed using the determinant formula.

Definition 6 (Determinant formula for computing a cross product)

The cross product of two vectors, \(\mathbf{a}=(a_x, a_y, a_z)\) and \(\mathbf{b}= (b_x, b_y, b_z)\), is computed using

(6)\[\begin{split}\mathbf{a} \times \mathbf{b} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ a_x & a_y & a_z \\ b_x & b_y & b_z \end{vmatrix} = \begin{pmatrix} a_y b_z - a_z b_y \\ a_z b_x - a_x b_z \\ a_x b_y - a_y b_z \end{pmatrix}.\end{split}\]

For vectors \(\mathbf{a}\), \(\mathbf{b}\), \(\mathbf{c} \in \mathbb{R}^3\) and \(k \in \mathbb{R}\) the cross product has the following properties:

• \(\mathbf{a} \times \mathbf{a} = \mathbf{0}\);

• \(\mathbf{a} \times \mathbf{b} = -(\mathbf{b} \times \mathbf{a})\);

• not commutative: \(\mathbf{a} \times \mathbf{b} \neq \mathbf{b} \times \mathbf{a}\);

• distributive: \(\mathbf{a} \times (\mathbf{b} + \mathbf{c}) = \mathbf{a} \times \mathbf{b} + \mathbf{a} \times \mathbf{c}\);

• scalar multiplication: \(k\mathbf{a} \times \mathbf{b} = \mathbf{a} \times k\mathbf{b} = k(\mathbf{a} \times \mathbf{b})\).

Example 4

Calculate the cross product of the two vectors \(\mathbf{a}=(3,4,0)\) and \(\mathbf{b}=(1, 2, 3)\).

Solution

\[\begin{split} \begin{align*} \mathbf{a} \times \mathbf{b} &= \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 3 & 4 & 0 \\ 1 & 2 & 3 \end{vmatrix} = \mathbf{i} \begin{vmatrix} 4 & 0 \\ 2 & 3 \end{vmatrix} - \mathbf{j} \begin{vmatrix} 3 & 0 \\ 1 & 3 \end{vmatrix} + \mathbf{k} \begin{vmatrix} 3 & 4 \\ 1 & 2 \end{vmatrix} \\ &= (4\cdot 3 - 0 \cdot 2)\mathbf{i} - (3 \cdot 3 - 0 \cdot 1)\mathbf{j} + (3 \cdot 2 - 4 \cdot 1)\mathbf{k} \\ &= \begin{pmatrix} 12 \\ -9 \\ 2 \end{pmatrix}. \end{align*} \end{split}\]

We can check that this vector is perpendicular to \(\mathbf{a}\) and \(\mathbf{b}\) using the dot product, e.g.,

\[\begin{split} \begin{align*} (\mathbf{a} \times \mathbf{b}) \cdot \mathbf{a} &= \begin{pmatrix} 12 \\ -9 \\ 2 \end{pmatrix} \cdot \begin{pmatrix} 3 \\ 4 \\ 0 \end{pmatrix} = 36 - 36 + 0 = 0, \\ (\mathbf{a} \times \mathbf{b}) \cdot \mathbf{b} &= \begin{pmatrix} 12 \\ -9 \\ 2 \end{pmatrix} \cdot \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix} = 12 - 18 + 6 = 0. \end{align*} \end{split}\]