In Section 08 we described force as a vector quantity, because it has direction, and energy as a scalar quantity, because it does not. It might be worth reviewing that section if you find this one tricky. Here, in this section, we are going to introduce some general properties of all vectors, not just force, but using force to exemplify our points. It’s a fairly lengthy page: you might want to return several times after digesting small chunks each time.
Text books often use displacement as their example from which they explain how vectors work. Weirdly, though, this can introduce a very hidden, and quite deep, misconception. We discuss this in our blog post ‘Why don’t the microwaves get through the holes’.
This section will not tell you everything about vectors you might need to know; instead we will concentrate on two absolutely key points about vectors:
- How we add two vectors together to form a single one and, just as important, the opposite – how we decompose or ‘resolve’ a single vector into two others that we call its ‘components’
- How we multiply vectors together, why there are two ways of doing so, and when each of them is relevant
But we will need to take some detours to get to these key ideas, and we will start with a reminder (if you have been to Section 8, and a briefing if you haven’t) on notation…
Vector notation
There are various mathematical notations to express the fact that a quantity is a vector. Three of the most common are the use of:
- Bold text, such as [latex]\mathbf{F}[/latex] for a force vector (this is the notation we are going to use)
- An arrow above the symbol, as in [latex]\overrightarrow{F}[/latex]
- A line below the symbol (commonly used in handwritten script, where bold text is hard to achieve) like this: [latex]\underline{F}[/latex]
Then the symbol [latex]F[/latex] without any of these embellishments means the ‘magnitude’ or size of the vector. In the case of a force, [latex]\mathbf{F}[/latex], that would be how many Newtons of force, but with no directional information. The magnitude of a vector can also be written with the modulus sign, as in [latex]|\mathbf{F}|[/latex].
A vector is usually drawn as an arrow, showing its direction. The length of the vector represents its magnitude.
Adding vectors
Let’s start our first key point: adding vectors is a really common thing to do with them. That’s true for scalars, too – you wouldn’t question why it’s useful to be able to add: $3.50 + $1.75 = $5.25 is a useful thing to be able to do when shopping. So, just for now, don’t question why we are learning to add vectors, let’s just do it… Watch the video to see two equivalent methods.
Resolving vectors, and why we would want to…
Now we have seen that two vectors can be added to find a ‘resultant’, we can imagine doing the opposite: taking a single vector and splitting it into two vectors that add together to form the vector. By far the most sensible choice of vectors (to a scientist anyway, perhaps not for a mathematician) are vectors at right angles to each other – you may hear them described as ‘perpendicular components’ or ‘orthogonal components’ or simply ‘components’ – as shown in the following video. Strangely, this operation tends to feel less intuitive than adding vectors, even though it is so closely related.
The video concentrates on the ‘how’. ‘Why’ we would want to do all this anyway will follow shortly after.
So a vector can be split into two perpendicular components; here is a diagram to summarise the video.
It’s a really useful skill to be able to ‘see’ immediately which component is the cosine, and which is the sine, in any given situation, without having to ‘do’ the trigonometry explicitly. A quick way to proceed, is that the component ‘along’ the angle is the cosine component. Have a look at the following vectors and see if you can determine the components, [latex]A[/latex] and [latex]B[/latex] in terms of [latex]V[/latex] and [latex]\theta[/latex] in each case.
Answer (don’t read on if you want to spend more time trying the task in the diagram):
- All the green A arrows are the cosines: [latex]A=V \cos{\theta}[/latex]
- All the blue B arrows are the sines: [latex]B=V \sin{\theta}[/latex]
Although the choice of which directions to orientate those components is in theory arbitrary, in practice in scientific contexts, there is often a choice of direction that makes life far easier, and enables us to work things out, as shown in the examples in the video below.
A bit more about notation – Cartesian coordinates and unit vectors
Before we continue we need to describe how positions in space can be described by Cartesian coordinates, and how vectors can be used for that purpose. Because we are following on from our adding and resolving videos, which described 2-dimensional situations, the following video will begin with a 2-dimensional situation and then extend to the 3-dimensional space that we all inhabit.
So to summarise the video, unit vectors [latex]\mathbf{i}[/latex], [latex]\mathbf{j}[/latex], [latex]\mathbf{k}[/latex], are vectors of unit length in the [latex]x[/latex], [latex]y[/latex] and [latex]z[/latex] directions respectively. And then any vector [latex]\mathbf{A}[/latex] in 3-dimensional space can be expressed in terms of its [latex]x[/latex]-, [latex]y[/latex]- and [latex]z[/latex]-components:
[latex]\displaystyle \mathbf{A}={A_x}\mathbf{i}+{A_y}\mathbf{j}+{A_z}\mathbf{k}[/latex]
We can think of this expression as scaling up each unit vector [latex]\mathbf{i}[/latex], [latex]\mathbf{j}[/latex], [latex]\mathbf{k}[/latex] by an amount equal to the component in that direction. The vector [latex]\mathbf{A}[/latex] is sometimes written in a shorthand form, in which you are supposed to know that you are dealing in terms of the unit vectors [latex]\mathbf{i}[/latex], [latex]\mathbf{j}[/latex], [latex]\mathbf{k}[/latex]:
[latex]\displaystyle \mathbf{A}=\begin{pmatrix} {A_x}\\{A_y}\\{A_z} \end{pmatrix}[/latex]
This is called a column vector (for obvious reasons). It is a simple example of a matrix (it is a 3 x 1 matrix). And yes, we said we’d use forces rather than displacements to exemplify vectors, and instead have proceeded to talk about the position of hotels, but we needed to cover Cartesian coordinates and vector notation. We’ll get back to forces soon…
In Section 26 we’ll take a closer look at matrices, and their cousins – determinants.
All this can be useful because the components can be treated separately, and then combined at the end if necessary. For example, gravity affects motion in the vertical ([latex]z[/latex]) direction, but not the other two. In this next video, we will look at a problem involving projectiles. We will do it three different ways, to show an example of vectors being useful, and hopefully end with the same answer each time…
Cartesian co-ordinates are not the only kind of Cartesian co-ordinate system. We will say more about this in Sections 26 and 27.
Multiplying vectors – why more than one kind?
Now that we are familiar with vectors and their components, we can address our second key point from the top of this (by now, very long) page… multiplying vectors together, and the fact that there are two ways to do it. Not two techniques to get to the same result, but two completely different processes, both of which you could think of as multiplying.
You are already familiar with multiplying scalars. 2 x 3 = 6. There you go, that was multiplying scalars. So why the complication with vectors…? Well, it must be something to do with the fact that they have a direction, because that’s the only difference between vectors and scalars. The two different processes for ‘multiplying’ vectors are:
- The dot product, which takes two vectors as the inputs and provides a scalar as the output, and
- The cross product which takes two vectors as the inputs and provides a vector as the output
You might wonder what happens when you multiply a scalar by a vector (that’s the only combination not covered by the preceding paragraph). Well, the vector is ‘scaled up’ by a factor equal to the magnitude of the scalar, so in the diagram 6 x A (scalar x vector) = 6A (vector, with length multiplied by the value of the scalar). If you like, you can think of that as why a scalar is called a scalar – the scalar ‘scales’ the vector…
We’ll describe when each kind of multiplying is applicable as we describe how to do them.
The first kind of multiplying – the dot product
We’ll look at this through the example of ‘work done’, often defined as ‘force multiplied by the distance travelled in the direction of the force.’ You can return to Section 08 for a recap if you wish.
The equation for work is [latex]W=Fx \cos{\theta}[/latex], and you might have a hunch that the [latex]\cos {\theta}[/latex] has something to do with vector components. Read the definition of work again, and try to match it up to the equation. It corresponds to interpreting the equation [latex]W=Fx \cos{\theta}[/latex] as [latex]W=F\times(x \cos{\theta})[/latex], with [latex]x \cos{\theta}[/latex] being the “distance travelled in the direction of the force.”
In the lawn roller example below, the angle between the force and the direction of travel is 60°, and so the force, 10 N is multiplied by “the distance travelled in the direction of the force”, i.e. the component of [latex]x[/latex] in the direction of [latex]F[/latex], which in this case is [latex]x \cos{60^{\circ}}[/latex] or 4 m.
However, you might find it conceptually simpler to think of work as “that component of the force moving the object, multiplied by the distance moved”. This corresponds to interpreting the equation [latex]W=Fx \cos{\theta}[/latex] as [latex]W=(F \cos{\theta})\times x[/latex].
Both forms are mathematically equivalent, but perhaps the second makes more conceptual sense. The vertical portion of the force, which is not moving the roller, along with the weight, is balanced by the reaction force from the ground. By the way, there’s nothing special about the angles we have drawn and the isometric grid used – it’s just that [latex]\cos{60^{\circ}}=\frac{1}{2}[/latex] (rather than some irrational decimal), so it makes the numbers nice and easy to visualise.
In case that was tricky to follow with static diagrams, you can see a more animated version of the same argument in the video below.
To summarise the main points of the video:
The dot product of two vectors [latex]\mathbf{A}[/latex] and [latex]\mathbf{B}[/latex] is a scalar quantity and can be defined as:
[latex]\displaystyle \mathbf{A} \cdot \mathbf{B} = AB \cos{\theta}[/latex]
The dot product is therefore a number (scalar) that increases with increasing magnitudes of [latex]\mathbf{A}[/latex] and [latex]\mathbf{B}[/latex], but also increases the more ‘aligned’ the two vectors are. As we saw by looking at the lawn roller in two different ways, the dot products finds the ‘projection’ of one vector onto another, in other words the component of a first vector in the direction of a second (and then multiplies it by the magnitude of the second). So we tend to see dot products in situations where:
- ‘two things produce more effect the more aligned they are’, and
- ‘those two things produce more effect, the greater the value of each of them’.
We have already seen this in the context of work, force and energy (it might be a good time to revisit Section 08?). Another example of such a situation is magnetic flux, which is a useful quantity because the rate of change of it (see Section 1) governs the generation of electricity in power stations. Magnetic flux can be imagined as ‘the number of lines of magnetic force’, as visualised by looking at iron filings aligning around a compass (if it was good enough for Michael Faraday, it’s probably good enough for us). The magnetic flux [latex]\Phi[/latex] through an area depends on the field strength [latex]\mathbf{B}[/latex] and the area. It might not seem at first that area is a vector. But it is! Areas can be orientated differently, and really do point in certain directions. So if we label an area with the vector [latex]\mathbf{S}[/latex], which is perpendicular to the surface of the area, then [latex]\Phi[/latex] will be maximized when [latex]\mathbf{B}[/latex] and [latex]\mathbf{S}[/latex] are in the same direction, and
[latex]\displaystyle \Phi = \mathbf{B} \cdot \mathbf{S}[/latex]
Or, where [latex]\mathbf{B}[/latex] varies over the area [latex]\mathbf{S}[/latex]:
[latex]\displaystyle \Phi = \int \mathbf{B} \cdot d\mathbf{S}[/latex]
This idea is explored more visually in the diagram below. The area [latex]\mathbf{A}[/latex] rotates through 90° from case P to case R, and the effect is explained.
Read the following three points in conjunction with the diagram above:
- In case P, the vectors [latex]\mathbf{B}[/latex] and [latex]\mathbf{S}[/latex] are completely aligned – in other words, the angle [latex]\theta[/latex] between them is 0. Then [latex]\cos {\theta}=1[/latex], and the dot product [latex]\mathbf{B} \cdot \mathbf{S}[/latex] is equal to [latex]BS[/latex], which is the greatest value it can ever have. Notice that either making the area, [latex]\mathbf{S}[/latex], larger or increasing the value of field strength, [latex]\mathbf{B}[/latex], (which will produce more lines of flux closer together) will also increase the dot product (the flux through the area) by increasing the value of [latex]BS[/latex], rather than by ‘improving’ the alignment.
- In case Q, the ‘imperfect alignment’ reduces the number of lines of flux through the area, as compared to case P. This is catered for mathematically through the [latex]\cos{\theta}[/latex] term, which takes a value between 0 and 1, causing the dot product to have a value less than [latex]BS[/latex].
- In case R, the magnetic field is at right angles to the area, and so no lines of flux pass through it. Mathematically, [latex]\cos{\theta}=\cos{90^{\circ}}=0[/latex] and [latex]\mathbf{B} \cdot \mathbf{S}=0[/latex]
The other thing you might need to know about the dot product is that it can be defined a different way.
If [latex]\mathbf{A}[/latex] and [latex]\mathbf{B}[/latex] are two vectors expressed in component form in Cartesian coordinates, so that:
[latex]\mathbf{A}={A_x}\mathbf{i}+{A_y}\mathbf{j}+{A_z}\mathbf{k}[/latex]
[latex]\mathbf{B}={B_x}\mathbf{i}+{B_y}\mathbf{j}+{B_z}\mathbf{k}[/latex]
then
[latex]\mathbf{A} \cdot \mathbf{B}={A_x}{B_x}+{A_y}{B_y}+{A_z}{B_z}[/latex]
In other words, multiply the two [latex]x[/latex]-components, the two [latex]y[/latex]-components, and the two [latex]z[/latex]-components, and add those three products together.
- We won’t prove here that these two definitions of the dot product are equivalent, but they are. Those who are interested can find a fairly easy proof here. You will need to know about the cosine rule to follow it, but not much else.
- There’s a more satisfying proof in Boas [1]; all you have to believe is that the distributive property (look it up if unsure) holds for dot products (it does, and she proves that, too).
- You can also find a different explanation of the meaning of the dot product at Better Explained.
We might decide that our first definition suits us better conceptually for now, because:
- It encodes our point about ‘quantities having more effect when more aligned’
- It doesn’t assume you are working in a Cartesian coordinate system (there are other coordinate systems, such as spherical and cylindrical polar co-ordinates)
What happens if you take the dot product of a vector, [latex]\mathbf{A}[/latex], with itself? Well, there is no angle between a vector and itself, so [latex]\theta=0[/latex], [latex]\cos{\theta}=1[/latex], and the dot product [latex]\mathbf{A} \cdot \mathbf{A} =A^2[/latex]. So you can think of the dot product as similar to ‘squaring’. If someone ever asks you to ‘square a vector’, they mean take the dot product with itself.
If you have visited the ‘Complex numbers’ page, you might see the similarity between [latex]\mathbf{A} \cdot \mathbf{A}=A^2[/latex] and [latex]zz^*=|z|^2[/latex]. Apologies if you read this and it turns out we haven’t written the complex number page yet…
And what about the unit vectors [latex]\mathbf{i}[/latex], [latex]\mathbf{j}[/latex] and [latex]\mathbf{k}[/latex]? Well, by definition they have length 1, and 1 multiplied by 1 is 1. In addition, any of the Cartesian unit vectors is in the same direction as itself (!), and perpendicular ([latex]\theta=90^{\circ}[/latex]) to any of the others. These two facts mean that
[latex]\mathbf{i} \cdot \mathbf{i} = \mathbf{j} \cdot \mathbf{j} = \mathbf{k} \cdot \mathbf{k} = 1[/latex],
and
[latex]\mathbf{i} \cdot \mathbf{j} = \mathbf{i} \cdot \mathbf{k} = \mathbf{j} \cdot \mathbf{i} = \mathbf{j} \cdot \mathbf{k} = \mathbf{k} \cdot \mathbf{i} = \mathbf{k} \cdot \mathbf{j} = 0[/latex].
Think about these relationships and make sure you can see that they are true. You might not see why they are useful to know, and you may not need to manipulate Cartesian unit vectors in the near future, but if this paragraph makes sense to you, then your understanding of everything on this page so far is probably quite secure.
The second kind of multiplying – the cross product
We have seen that the dot product tends to crop up in situations in which something is maximised when two quantities are aligned. The more parallel (the less perpendicular) the two quantities, the greater is some effect. In contrast, there are some quantities in physics in which an effect is greater, the more perpendicular (rather than parallel) two quantities are, and this is where the cross product comes into play. Two examples are:
- The force on a charged particle moving through a magnetic field
- Torque
We’ll look at the second of these here.
And we’ll look at the first – the ‘Lorentz Force’ – in Section 20.
In the diagram below, four cases are shown of using a spanner (wrench). Forces of the same magnitude are applied to the spanner in all four cases, and are shown as the vector [latex]\mathbf{F}[/latex]. The position vector [latex]\mathbf{r}[/latex] represents the position of the point where the force is applied, relative to the centre of rotation (the pivot, or fulcrum). Hopefully, it will be intuitively obvious that case Z is ‘best’ for turning the nut, because the force is in the ‘right’ direction:
- In physics terms, this means that [latex]\mathbf{F}[/latex] is perpendicular to [latex]\mathbf{r}[/latex], so that the torque is greatest
- In mathematical terms, the cross product of [latex]\mathbf{F}[/latex] and [latex]\mathbf{r}[/latex] is maximised when they are perpendicular (we know we haven’t explained how to calculate a cross product yet, but we are coming to that – for now we are explaining what we want it to do for us)
Also, you might see that in case W, there will be no turning at all, since the force is aligned with the position vector. Cases X and Y are intermediate between W and Z, with Y corresponding to a greater torque than X, and both of them representing less torque than Z. Whereas with work done it is the component of the force parallel to the direction of the displacement that ‘has an effect’, for ‘turning objects’ it is the perpendicular component of the force that creates the torque.
The magnitude of the torque, [latex]\tau[/latex], is given by [latex]\tau=rF \sin{\theta}[/latex].
Note the following:
- Torque is maximised at [latex]\theta = 90 ^{\circ}[/latex], so that [latex]\sin{90 ^{\circ}} = 1 [/latex], and [latex]\tau = rF[/latex]
- Torque is minimised at [latex]\theta = 0[/latex], so that [latex]\sin{0}=0[/latex], and [latex]\tau =rF[/latex]
These bullet points describe mathematically what we had explained in words when looking at cases W – Z for the spanner.
We previously described:
- ‘Parallel components’ of a vector A as [latex]A\cos{\theta}[/latex] and ‘perpendicular components’ as [latex]A\sin{\theta}[/latex] (see the diagram of 5 vectors V that you analysed further up the page)
- the dot product of [latex]\mathbf{A}[/latex] and [latex]\mathbf{B}[/latex], as a way to ‘multiply’ two vectors, given by [latex]\mathbf{A} \cdot \mathbf{B} = AB\cos{\theta}[/latex]
Now we have seen that torque is given by [latex]\tau=rF\sin{\theta}[/latex], and we have described it as using the perpendicular component, [latex]F\sin{\theta}[/latex], of the force .
Surely then, if we describe the position and force as vectors [latex]\mathbf{r}[/latex] and [latex]\mathbf{F}[/latex] respectively, there must be a way to ‘multiply’ the vectors that is ‘similar to’ the dot product, but different in that it uses the perpendicular (sine) rather than parallel (cosine) component of one of the vectors…
There is! It is called the cross product. This probably doesn’t surprise you, since we have been leading up to it for a while now. The cross product of two vectors, A and B, is denoted by
[latex]\mathbf{A} \times \mathbf{B}[/latex]
It is pronounced “A cross B,” and its magnitude (shown with the modulus sign ‘| |’) is given by
[latex]|\mathbf{A} \times \mathbf{B}|=AB\sin{\theta}[/latex]
Notice that (and you can refer back to the discussion on torque to contextualise these points if you wish):
- [latex]\mathbf{A} \times \mathbf{B} = AB[/latex] (and is at its maximum value) when [latex]\theta = 90 ^{\circ}[/latex]
- [latex]\mathbf{A} \times \mathbf{B} =0[/latex] when [latex]\theta = 0[/latex]
- [latex]\mathbf{A} \times \mathbf{A} =0[/latex] (can you see why this must always be true? – it’s related to the reason why [latex]\mathbf{A} \cdot \mathbf{A} = A^2[/latex])
An important difference between the dot and cross product
Let’s return to the question of work done by the lawn roller. Consider the situation we have seen before, and then let’s create a new situation by reversing the direction of the force, as in the right hand portion of the diagram below.
What changes from one situation to another? Well, the roller obviously moves in the opposite direction, but in terms of work done, nothing changes. In both cases 40 J of work is done. Work is a scalar, and is blind to the direction of the movement relative to our Cartesian, or other, coordinate system (although, as we know, it does depend upon the direction of the movement relative to the direction of the force).
Now consider the spanner/wrench, and reverse the direction of the force on it (see diagram below).
This time, something really is different! In one case the nut will tighten onto the (invisible) bolt, and in the other it will untighten. So the direction of the force matters; unlike work done, torque must have a direction. Mathematically, this means that we need to define the cross product of two vectors so that it (unlike the dot product) is also a vector. So far we have only defined the magnitude of a cross product, so let’s complete its definition now, with an extra part to make sure that we always get the direction correct.
The magnitude of [latex]\mathbf{A} \times \mathbf{B}[/latex] is [latex]|\mathbf{A} \times \mathbf{B}|=AB \sin {\theta}[/latex], where [latex]\theta[/latex] is the positive angle (from zero to 180°) between [latex]\mathbf{A}[/latex] and [latex]\mathbf{B}[/latex]. The direction of [latex]\mathbf{A} \times \mathbf{B}[/latex] is perpendicular to the plane of [latex]\mathbf{A}[/latex] and [latex]\mathbf{B}[/latex] and in the direction of advance of a right-handed (normal) screw rotated from [latex]\mathbf{A}[/latex] to [latex]\mathbf{B}[/latex] as shown as [latex]\mathbf{C}[/latex] in the diagram below. Imagine [latex]\mathbf{C}[/latex] as a screw, and turning a screwdriver from [latex]\mathbf{A}[/latex] to [latex]\mathbf{B}[/latex]. The screw [latex]\mathbf{C}[/latex] would advance (‘screw itself in’) in the direction of the arrow [latex]\mathbf{C}[/latex].
An interesting implication is that whereas [latex]\mathbf{A} \cdot \mathbf{B} = \mathbf{B} \cdot \mathbf{A}[/latex], the same is not true for the cross product: [latex]\mathbf{A} \times \mathbf{B}[/latex] is not equal to [latex]\mathbf{B} \times \mathbf{A}[/latex]. In fact,
[latex]\mathbf{B} \times \mathbf{A} = -\mathbf{A} \times \mathbf{B}[/latex]
so that in the diagram above, [latex]\mathbf{B} \times \mathbf{A}[/latex] would be represented by an arrow the same length as [latex]\mathbf{C}[/latex], but pointing vertically downwards.
If we make in the diagram above, we have a Cartesian coordinate system! The [latex]x[/latex], [latex]y[/latex] and [latex]z[/latex] axes are usually very carefully labelled to form what is known as a right-handed co-ordinate system.
A right-handed coordinate system is one in which moving [latex]x[/latex] into [latex]y[/latex] ‘screws in [latex]z[/latex]’ (this is not the wording you will find in text books!).
Mathematically the relationship between the axes can be expressed in terms of the unit vectors of a right-handed Cartesian coordinate system as:
[latex]\mathbf{i} \times \mathbf{j}=\mathbf{k}[/latex]
And it follows (you can check this by looking at a 3D coordinate system and trying to ‘screw in’ each axis in turn) that
[latex]\mathbf{j} \times \mathbf{k} = \mathbf{i}[/latex]
[latex]\mathbf{k} \times \mathbf{i} = \mathbf{j}[/latex]
Scientists and mathematicians actually care which axis is which in a 3D Cartesian coordinate system (after all, you care whether your plumber turns the spanner clockwise or anticlockwise). If you return to the videos of the hotel position, you will see that we have been careful to make the coordinate system right-handed in this sense.
Returning to torque, we can now express torque as a vector quantity (in a more concise notation than previously) as [latex]\mathbf{\tau}=\mathbf{r} \times \mathbf{F}[/latex]; in the diagram of the oppositely directed forces on the spanners, the torque would be directed into the screen in one case, and out of the screen in the other case.
There are many properties of vectors, and of dot and cross products, that we have not covered in this Section. Our goal has been to give you some insight into vectors, so that if you ever have to manipulate them mathematically, the techniques you learn will feel physically meaningful.
Link to Section 8 – Work done and energy
Where we looked at work done as a link between forces and energies, and thus between a vector quantity and a scalar.
Link to Home Page
And you can always return here.
Reference [1]: Mathematical Methods in the Physical Sciences, Boas ML, Wiley 2nd edition 1983, Chapter 3