Title: Default for 9.1 Basic concepts of vectors
Course: HELM Workbooks 2023-2024

Introduction

It is useful to separate physical quantities into two types: the first are called scalars; the second are known as vectors. A scalar is a quantity that can be described by a single number which can be positive, negative or zero. An example of a scalar quantity is the mass of an object, so we might state that ‘the mass of the stone is 3 kg’. It is important to give the units in which the quantity is measured. Obvious examples of scalars are temperature and length, but there are many other engineering applications in which scalars play an important role. For example, speed, work, voltage and energy are all scalars.

On the other hand, vectors are quantities which require the specification of a magnitude and a direction. An example of a vector quantity is the force applied to an object to make it move. When the object shown in the figure below is moved by applying a force to it we achieve different effects depending on the direction of the force.

[Image: Two blocks with arrows denoting forces acting on them]
A force is a vector quantity

In order to specify the force completely we must state not only its magnitude (its ‘strength’) but also the direction in which the force acts. For example we might state that ‘a force of 5 newtons is applied vertically from above’. Clearly this force would have a different effect from one applied horizontally. The direction in which the force acts is crucial.

There are many engineering applications where vectors are important. Force, acceleration, velocity, electric and magnetic fields are all described by vectors. Furthermore, when computer software is written to control the position of a robot, the position is described by vectors.

Sometimes confusion can arise because words used in general conversation have specific technical meanings when used in engineering calculations. An example is the use of the words ‘speed’ and ‘velocity’. In everyday conversation these words have the same meaning and are used interchangeably. However in more precise language they are not the same. Speed is a scalar quantity described by giving a single number in appropriate units; for example ‘the speed of the car is 40 kilometres per hour’. On the other hand velocity is a vector quantity and must be specified by giving a direction as well. For example ‘the velocity of the aircraft is 20 metres per second due north’.

In engineering calculations, the words speed and velocity cannot be used interchangeably. Similar problems arise from use of the words ‘mass’ and ‘weight’. Mass is a scalar which describes the amount of substance in an object. The unit of mass is the kilogramme. Weight is a vector, the direction of which is vertically downwards because weight arises from the action of gravity. The unit of weight is the newton. Displacement and distance are related quantities which can also cause confusion. Whereas distance is a scalar, displacement is ‘directed distance’, that is, distance together with a specified direction. So, referring to the figure below, if an object is moved from point A to point B, we can state that the distance moved is 10 metres, but the displacement is 10 metres in the direction from A to B.

[Image: A block has been moved left to right from A to B. An arrow from its original position to its new one denotes the displac]
Displacement means directed distance

You will meet many other quantities in the course of your studies and it will be helpful to know which are vectors and which are scalars. Some common quantities and their type are listed in the table below.  The S.I. units in which these are measured are also shown.

Some common scalar and vector quantities
quantity | type | S.I. unit
distance | scalar | metre, \(\mathrm{m}\)
mass | scalar | kilogramme, \(\mathrm{kg}\)
temperature | scalar | kelvin, \(\mathrm{K}\)
pressure | scalar | pascal, \(\mathrm{Pa}\)
work | scalar | joule, \(\mathrm{J}\)
energy | scalar | joule, \(\mathrm{J}\)
displacement | vector | metre, \(\mathrm{m}\)
force | vector | newton, \(\mathrm{N}\)
velocity | vector | metres per second, \(\mathrm{m}\, \mathrm{s}^{-1}\)
acceleration | vector | metres per second per second, \(\mathrm{m}\, \mathrm{s}^{-2}\)

The mathematical description of vector quantities

Because a vector has a direction as well as a magnitude we can represent a vector by drawing a line. The length of the line represents the magnitude of the vector given some appropriate scale, and the direction of the line represents the direction of the vector. We call this representation a directed line segment. For example, the figure below shows a vector which represents a velocity of \(3\ \mathrm{m}\,\mathrm{s}^{-1}\) north-west. Note that the arrow on the vector indicates the direction required.

[Image]
A vector quantity can be represented by drawing a line

More generally, the below figure shows an arbitrary vector quantity.

[Image: Three identical arrows are labelled with the different ways of denoting vectors.]
Representation of vectors

It is important when writing vectors to distinguish them from scalars. Various notations are used. In the figure we emphasise that we are dealing with the vector from \(A\) to \(B\) by using an arrow and writing \(\overrightarrow{AB}\). Often, in textbooks, vectors are indicated by using a bold typeface such as \(\boldsymbol{a}\). It is difficult when handwriting to reproduce the bold face and so it is conventional to underline vector quantities and write \(\underline{a}\) instead. So \(\overrightarrow{AB}\) and \(\underline{a}\) represent the same vector in the figure. We can also use the notation \(\underline{AB}\). In general in this quiz we will use underlining but we will also use the arrow notation where it is particularly helpful. Note that in these quizzes all math is in boldface so this won't distinguish vectors here but you should be aware of the convention for when you encounter vectors elsewhere.

Example

The figure below shows an object pulled by a force of \(10\ \mathrm{N}\) at an angle of \(60^{\circ}\) to the horizontal.

[Image: A an arrow sticks out from a block at an angle of 60 degrees to show the force acting on the block]

Show how this force can be represented by a vector.

Solution

The force can be represented by drawing a line of length 10 units at an angle of \(60^\circ\) to the horizontal, as shown below.

[Image: Arrow drawn at 60 degrees to the horizontal with a scale showing that the arrow is 10 units long.]

We have labelled the force \(\underline{F}\). When several forces are involved they can be labelled \(\underline{F}_1\), \(\underline{F}_2\) and so on.

When we wish to refer simply to the magnitude (or length) of a vector we write this using the modulus sign as \(|\overrightarrow{AB}|\), or \(|\underline{a}|\), or simply \(a\) (without the underline.)

In general two vectors are said to be equal vectors if they have the same magnitude and same direction. So, in the figure below the vectors \(\overrightarrow{CD}\) and \(\overrightarrow{AB}\) are equal even though their locations differ.

[Image: A pair of vectors that are parallel and of equal length.]
Equal vectors

This is a useful and important property of vectors: a vector is defined only by its direction and magnitude, not by its location in space. These vectors are often called free vectors.

The vector \(-\underline{a}\) is a vector in the opposite direction to \(\underline{a}\), but has the same magnitude as \(\underline{a}\), as shown below.

[Image: Two vectors which are parallel and equal in length but the arrows point in opposite directions.]

Geometrically, if \(\underline{a}=\overrightarrow{AB}\) then \(-\underline{a}=\overrightarrow{BA}\).

Addition of vectors

Vectors are added in a particular way known as the triangle law. To see why this law is appropriate to add them this way consider the following example:

An unmanned vehicle moves on tracks around a factory floor carrying components from the store at \(A\) to workers at \(C\) as shown in the figure below.

[Image: A triangle of vectors show the two paths the robot can take from the store (A) to the workers (C) either directly on via]

The vehicle may arrive at \(C\) either directly or via an intermediate point \(B\). The movement from \(A\) to \(B\) can be represented by a displacement vector \(\overrightarrow{AB}\). Similarly, movement from \(B\) to \(C\) can be represented by the displacement vector \(\overrightarrow{BC}\), and movement from \(A\) to \(C\) can be represented by \(\overrightarrow{AC}\). Since travelling from \(A\) to \(B\) and then \(B\) to \(C\) is equivalent to travelling directly from \(A\) to \(C\) we write \[\overrightarrow{AB} + \overrightarrow{BC} =\overrightarrow{AC}\] This is an example of the triangle law for adding vectors. We add vectors \(\overrightarrow{AB}\) and \(\overrightarrow{BC}\) by placing the tail of \(\overrightarrow{BC}\) at the head of \(\overrightarrow{AB}\) and completing the third side of the triangle, forming \(\overrightarrow{AC}\).

Consider the more general situation in the figure below:

[Image: Two vectors marked a and b]

Suppose we wish to add \(\underline{b}\) to \(\underline{a}\). To do this \(\underline{b}\) is translated, keeping its direction and length unchanged, until its tail coincides with the head of \(\underline{a}\). Then the sum \(\underline{a}+\underline{b}\) is defined by the vector represented by the third side of the completed triangle, that is \(\underline{c}\) in the figure below.

[Image: A triangle composed of three vectors, a,b and c]

Note, from this figure, that we can write \(\underline{c}=\underline{a}+\underline{b}\) since going along \(\underline{a}\) and then along \(\underline{b}\) is equivalent to going along \(\underline{c}\).

Key Point

\(\underline{a}+ \underline{b} = \underline{b} + \underline{a}\) vector addition is commutative.

\(\underline{a}+ ( \underline{b}+ \underline{c}) = ( \underline{a} + \underline{b} ) + \underline{c}\) vector addition is associative.

Resultant of two forces acting upon a body

A force \(\underline{F}_1\) of \(2\ \mathrm{N}\) acts vertically downwards, and a force \(\underline{F}_2\) of \(3\ \mathrm{N}\) acts horizontally to the right, upon the body shown below.

[Image: A rectangular object is acted on by two forces, shown as arrows. One acts vertically down and the other acts horizontall]

We can use vector addition to find the combined effect or resultant of the two concurrent forces. (Concurrent means that the forces act through the same point.) Translating \(\underline{F}_2\) until its tail touches the head of \(\underline{F}_1\), we complete the triangle ABC as shown. The vector represented by the third side is the resultant, \(\underline{R}\). We write \[\underline{R} = \underline{F_1} + \underline{F_2}\] and say that \(\underline{R}\) is the vector sum of \(\underline{F_2}\) and \(\underline{F_1}\). The resultant force acts at an angle of \(\theta\) below the horizontal where \(\tan\theta  = 2/3\), so that \(\theta = 33.7^\circ\), and has magnitude (given by Pythagoras’ theorem) \(\sqrt{13}\ \mathrm{N}\).

Resolving a force into two perpendicular directions

In the previous section we saw that two forces acting upon a body can be replaced by a single force which has the same effect. It is sometimes useful to reverse this process and consider a single force as equivalent to two forces acting at right-angles to each other. Consider the force \(\underline{F}\) inclined at an angle \(\theta\) to the horizontal as shown below.

[Image: An arrow represents a force acting at an angle theta to the horizontal plane. Vertical and horizontal arrows show how it]

\(\underline{F}\) can be replaced by two forces, one of magnitude \(F\cos\theta\) and one of magnitude \(F\sin\theta\) as shown. \(\underline{F}\) has been resolved into two perpendicular components. This is sensible because if we re-combine the two perpendicular forces of magnitudes \(F\cos\theta\) and \(F\sin\theta\) using the triangle law we find \(\underline{F}\) to be the resultant force.

Example

[Image: Arrow whcih represents a force of 5N angled at 30 degrees to the horizontal. A set of axes indicates that the x axis is]

The figure shows a force of \(5\ \mathrm{N}\) acting at an angle of \(30^\circ\) to the \(x\) axis. Resolve this force into its components parallel and perpendicular to the \(x\) axis.

Solution

[Image: The above image but now also showing horizontal and vertical arrows indicating the components of the force in the x and]

The force can be resolved into two components, one directed along the \(x\) axis with magnitude \(5\cos30^\circ\) and one perpendicular to this of magnitude \(5\sin30^\circ\). Together, these two components have the same effect as the original force as shown in the diagram.

The need to resolve a vector along a given direction occurs in other areas. For example, as a police car or ambulance with siren operating passes by, the pitch of the siren appears to increase as the vehicle approaches and decrease as it goes away. This change in pitch is known as the Doppler effect. This effect occurs in any situation where waves are reflected from a moving object.

A radar gun produces a signal which is bounced off the target moving vehicle so that when it returns to the gun, which also acts as a receiver, it has changed pitch. The speed of the vehicle can be calculated from the change in pitch. The speed indicated on the radar gun is the speed directly towards or away from the gun. However it is not usual to place oneself directly in front of moving vehicle when using the radar gun (panel (a) of the diagram below). Consequently the gun is used at an angle to the line of traffic (panel (b)).

[Image: Two panels. The first shows a radar gun positioned in line with an oncoming vehicle. The second shows it offset by an an]

This means that it registers only the component of the velocity towards the gun. Suppose that the true speed along the road is \(v\). Then the component measured by the gun (\(v\cos\theta\)) is less than \(v\).

Example

A safety inspector wishes to check the speed of a train along a straight piece of track. She stands \(10\ \mathrm{m}\) to the side of the track and uses a very accurate radar speed gun. If the reading on the gun is to be within 5% of the true speed of the train, how far away from the approaching train should the reading be taken?

Solution

[Image: A radar gun is located a distance, d, diagonally from a train moving horizontally to the left at speed,v. The radar gun]

For an error of 5%, the gun should read \(0.95|{\underline v}|\). So \[|{\underline v}|\cos\theta = 0.95|{\underline v}|\] or \[\cos\theta = 0.95\] If the distance to the side of the track at which the gun is used is \(s\) m and the distance between the radar gun and the train is \(d\) m, then \[\sin\theta = \frac{s}{d}\] Here \(s=10\) so \[\sqrt{1-\cos^2\theta}=\frac{10}{d}\] So if \(\cos\theta = 0.95\) \[d=\frac{10}{\sqrt{1-(0.95)^2}}=32.03\] This means that the reading should be taken when the train is over \(32\ \mathrm{m}\) from the radar gun to ensure an error of less than 5%.

The force vectors on an aeroplane in steady flight

The forces acting on an aeroplane are shown in the figure below.

[Image: Aeroplane flying at an angle beta to the horizon and alpha to its own horizontal. The forces of Thrust and drag are mark]

The magnitude (strength) of the forces are indicated by

\(T\): the thrust provided by the engines,

\(W\): the weight,

\(D\): the drag (acting against the direction of flight path) and

\(L\): the lift (taken perpendicular to the path.)

In a more realistic situation force vectors in three dimensions would need to be considered. As the plane is in steady flight the sum of the forces in any direction is zero. (If this were not the case, then, by Newton’s second law, the non-zero resultant force would cause the aeroplane to accelerate.)

So, resolving forces in the direction of the path: \[T\cos\alpha - D − W\sin\beta = 0\] Then, resolving forces perpendicular to the path: \[T\sin\alpha + L − W\cos\beta = 0\].

Example

If the plane has mass \(72\, 000\) tonnes, the drag is \(130\ \mathrm{kN}\), the lift is \(690\ \mathrm{kN}\) and \(\beta = 6^\circ\) find the magnitude of the thrust and the value of \(\alpha\) to maintain steady flight.

Solution

From these two equations we see:\[T\cos\alpha = D + W\sin\beta = 130000 + (72000)(9.81)\sin6^\circ = 203830.54\] and \[T\sin\alpha = W\cos\beta − L = (72000)(9.81)\cos 6^\circ − 690000 = 12450.71\] hence \[\tan\alpha=\frac{T\sin\alpha}{T\cos\alpha}=\frac{12450.71}{203830.54}=0.061084\quad\rightarrow\quad \alpha=3.50^\circ\] and consequently, for the thrust:\[T = 204210\ \mathrm{N}\].

Subtraction of vectors

Subtraction of one vector from another is performed by adding the corresponding negative vector. That is, if we seek \(\underline{a}-\underline{b}\) we form \(\underline{a}+(-\underline{b})\). This is shown geometrically in the figure below. Note that in the right-hand diagram the arrow on \(\underline{b}\) has been reversed to give \(-\underline{b}\).

[Image: On the left, two free vectors a and b are shown. On the right a and b have been composed into a triangle and the arrow o]
Subtraction of a vector is performed by adding a negative vector

Multiplying a vector by a scalar

If \(k\) is any positive scalar and \(\underline{a}\) is a vector then \(k\underline{a}\) is a vector in the same direction as \(\underline{a}\) but \(k\) times as long. If \(k\) is negative, \(k\underline{a}\) is a vector in the opposite direction to \(\underline{a}\) and \(k\) times as long (shown in the figure below). The vector \(k\underline{a}\) is said to be a scalar multiple of \(\underline{a}\).

[Image: Right side shows two parallel vectors, a and ka, with ka being longer. Left side shows two parallel vectors, a and -ka,]
Multiplying a vector by a scalar

The vector \({3\underline{a}}\) is three times as long as \(\underline{a}\) and has the same direction. The vector \(\frac{1}{2}\underline{r}\) is in the same direction as \(\underline{r}\) but is half as long. The vector \(-4\underline{b}\) is in the opposite direction to \(\underline{b}\) and four times as long.

For any scalars \(k\) and \(l\), and any vectors \(\underline{a}\) and \(\underline{b}\), the following rules hold:

Key Point

\[k(\underline{a}+\underline{b}) = k\underline{a} + k\underline{b}\] \[(k + l)\underline{a} = k\underline{a} + l\underline{a}\] \[k(l\underline{a}) = (kl)\underline{a}\]

Unit Vectors

A vector which has a magnitude of \(1\) is called a unit vector. If \(\underline{a}\) has magnitude \(3\), then a unit vector in the direction of \(\underline{a}\) is \(\frac{1}{3}\underline{a}\), as shown below.

[Image]
A unit vector has length one unit

A unit vector in the direction of a given vector is found by dividing the given vector by its magnitude.

A unit vector in the direction of \(\underline{a}\) is given the ‘hat’ symbol \(\hat{\underline{a}}\).

Key Point

A unit vector can be found by dividing a vector by its modulus: \[\hat{\underline{a}} = \frac{\underline{a}}{|\underline{a}|}\]

Where \(\underline{a}\) is a non-zero vector.

Exercise

[Image: An arrow representing a force points at an angle theta to the horizontal x axis]

A force acts on an object as follows: \( \vec{F} = [value]\ \mathrm{N}\) at \( \theta = {\displaystyle[value]}\) angle with the \(x\) axis. Find:

1. \( \vec{F_x} = \) \(\textrm{N}\)
2. \( \vec{F_y} = \)  \(\textrm{N}\)

Enter your answer precisely, do not enter an approximate floating point number.
Feedback: We find the force's \(x\) and \(y\) components: 1. \(\vec{F_x} = F\cos\theta = |[value]|\cos[value] = [value] \, \textrm{N}. \) 2. \(\vec{F_y} = F\sin\theta = |[value]|\sin[value] = [value] \, \textrm{N}. \)

Exercise

Two forces act on a point object as follows: \( \vec{F_1} = [value]\ \mathrm{N}\) in the direction of the positive \(x\) axis and \( \vec{F_2} = [value]\ \mathrm{N}\) in the direction of the positive \(y\) axis. Find:

1. The angle the resultant force makes with the positive \(x\)-axis, in degrees, to 1 d.p, \(\theta_{r} = \)
2. The magnitude of the force to 1 d.p, \(|F_R| = \) \( \textrm{N}\)
Feedback: 1. We calculate the angle of the resultant force using \(\tan\theta = [value]/[value]\). Therefore, \[\theta = \tan^{-1}\left( \frac{[value]}{[value]} \right) = [value]^\circ. \] We use Pythagoras' theorem to find the magnitude of the resultant force: \[\lvert\overrightarrow{F_R}\rvert = \sqrt{F_1^2 + F_2^2} = \sqrt{[value]^2 + [value]^2} = [value]\ \mathrm{N}.\]

Exercise

In the rectangle \(ABCD\) the side \(AB\) is represented by the vector \(\underline{p}\) and the side \(BC\) is represented by \(\underline{q}\). Find an expression in terms of \(\underline{p}\) and \(\underline{q}\) for:

1. The diagonal \(\overrightarrow{AC}\):  
2. The diagonal \(\overrightarrow{BD}\):  

Note: you should enter your answer in terms of the letters p and q without underline or boldface.
Feedback: 1. The diagonal \(\overrightarrow{AC}\) is the sum of the sides \(\overrightarrow{AB}\) and \(\overrightarrow{BC} \). Therefore \[\overrightarrow{AC} =\overrightarrow{AB} +\overrightarrow{BC} = \underline{p} + \underline{q}\] 2. The diagonal \(\overrightarrow{BD}\) is the sum of the sides \(\overrightarrow{BC}\) and \(\overrightarrow{CD}\). Therefore \[\overrightarrow{BD} = \overrightarrow{BC} + \overrightarrow{CD} = \underline{q} - \underline{p}\]

Exercise

Using the rules given in Key Point above, simplify the following:

1. \([value]\underline{a} + [value]\underline{a} =\)
2. \([value]([value]\underline{b}) =\)
3. Expand \([value](\underline{q} + \underline{r}) =\)

Write \(\underline{a}\) and \(\underline{b}\) as a and b, respectively.
Feedback: 1. The second rule \([value]\underline{a} + [value]\underline{a}\) can be simplified to \([value]\underline{a}.\) 2. Using the third rule, \([value]([value]\underline{b}) = [value]\underline{b}.\) 3. Using the first rule, \([value](\underline{q} + \underline{r}) = [value]\underline{q} + [value]\underline{r}.\)

Exercise

Which of the following are scalars and which are vectors?

1. [value]
2. [value]
3. [value]
4. [value]
Feedback: 1. [value] 2. [value] 3. [value] 4. [value]

Exercise

In triangle \(OAB\) the point \(P\) divides \(AB\) internally in the ratio \(m:n\).

If \(\overrightarrow{OA} = \underline{a}\) and \(\overrightarrow{OB} = \underline{b}\), then find an expression for \(\overrightarrow{OP}\) in terms of \(\underline{a}\) and \(\underline{b}\).

You might find it helpful to draw a diagram showing this triangle.

\(\overrightarrow{OP} = \)
Feedback: The figure below shows the triangle with the vector \(\overrightarrow{OP}\) marked with a dotted line within. [Image: A triangle OAB with vectors a and b marked on the edges OA and OB respectively. The point P is marked partway along the] From the diagram we can see \(\overrightarrow{OP} = \overrightarrow{OA} + \overrightarrow{AP}\). We are told \(\overrightarrow{OA} = \underline{a}\) and so we need to find \(\overrightarrow{AP}\). We know that \(P\) divides \(\overrightarrow{AB}\) in the ratio \(m:n\) so \[\overrightarrow{AP} = \frac{m}{m+n}\left(\overrightarrow{AB}\right),\] as \[\overrightarrow{AB} = \underline{b} - \underline{a}\] we find \[\overrightarrow{AP} = \frac{m}{m+n}\left(\underline{b} - \underline{a}\right)\] and therefore, \[\overrightarrow{OP} = \underline{a} + \frac{m}{m+n}\left(\underline{b} - \underline{a}\right).\] Note: You can arrive at an equivalent expression by starting from \(\overrightarrow{OP} = \overrightarrow{OB} + \overrightarrow{BP}\).

Exercise

Given a regular hexagon \(A\), \(B\), \(C\), \(D\), \(E\), \(F\), with \(O\) at the centre.

Let \(\underline{a} = \vec{OA}\) and \(\underline{b}=\vec{OB}\).

[Image: Regular hexagon centred at the origin with vertices starting at A, B, C, D, E and F.]

Express each of the following in terms of \(\underline{a}\) and \(\underline{b}\).

1. \(\vec{BC} = \)
2. \(\vec{BC}+\vec{DE}+\vec{FA} = \)
3. \(\vec{CF} = \)

Write \(\underline{a}\) and \(\underline{b}\) as a and b, respectively.
Feedback: 1. From the properties of the regular hexagon we know that \( \vec{CB} \) is parallel to \( \underline{a} \) and they have the same length. So, \( \vec{BC} = -\underline{a} \). 2. We notice that \( \vec{CO} \) is a translation of \( \vec{DE} \) and \( \vec{OB} \) is a translation of \( \vec{FA} \). So, \( \vec{BC} + \vec{DE} + \vec{FA} = \vec{BC} + \vec{CO} + \vec{OB} = 0 \). 3. From the properties of the hexagon we know that the diagonal \( \vec{CF} \) is parallel to the side \( \vec{BA} \) and has twice it's length. So \( \vec{CF} = 2 \vec{BA} = 2(\underline{a} - \underline{b}) \).