## 34. Displacements along a line and in a plane.

The ‘real number’ $$x$$, with which we have been concerned in the two preceding chapters, may be regarded from many different points of view. It may be regarded as a pure number, destitute of geometrical significance, or a geometrical significance may be attached to it in at least three different ways. It may be regarded as the measure of a length, viz. the length $$A_{0}P$$ along the line $$\Lambda$$ of Chap. I. It may be regarded as the mark of a point, viz. the point $$P$$ whose distance from $$A_{0}$$ is $$x$$. Or it may be regarded as the measure of a displacement or change of position on the line $$\Lambda$$. It is on this last point of view that we shall now concentrate our attention.

Imagine a small particle placed at $$P$$ on the line $$\Lambda$$ and then displaced to $$Q$$. We shall call the displacement or change of position which is needed to transfer the particle from $$P$$ to $$Q$$ the displacement $$\overline{PQ}$$. To specify a displacement completely three things are needed, its magnitude, its sense forwards or backwards along the line, and what may be called its point of application,  the original position $$P$$ of the particle. But, when we are thinking merely of the change of position produced by the displacement, it is natural to disregard the point of application and to consider all displacements as equivalent whose lengths and senses are the same. Then the displacement is completely specified by the length $$PQ = x$$, the sense of the displacement being fixed by the sign of $$x$$. We may therefore, without ambiguity, speak of the displacement $$[x]$$,1 and we may write $$\overline{PQ} = [x]$$.

We use the square bracket to distinguish the displacement $$[x]$$ from the length or number $$x$$.2 If the coordinate of $$P$$ is $$a$$, that of $$Q$$ will be $$a + x$$; the displacement $$[x]$$ therefore transfers a particle from the point $$a$$ to the point $$a + x$$.

We come now to consider displacements in a plane. We may define the displacement $$\overline{PQ}$$ as before. But now more data are required in order to specify it completely. We require to know: (i) the magnitude of the displacement, i.e. the length of the straight line $$PQ$$; (ii) the direction of the displacement, which is determined by the angle which $$PQ$$ makes with some fixed line in the plane; (iii) the sense of the displacement; and (iv) its point of application. Of these requirements we may disregard the fourth, if we consider two displacements as equivalent if they are the same in magnitude, direction, and sense. In other words, if $$PQ$$ and $$RS$$ are equal and parallel, and the sense of motion from $$P$$ to $$Q$$ is the same as that of motion from $$R$$ to $$S$$, we regard the displacements $$\overline{PQ}$$ and $$\overline{RS}$$ as equivalent, and write $\overline{PQ} = \overline{RS}.$

Now let us take any pair of coordinate axes in the plane (such as $$OX$$$$OY$$ in Fig. 19). Draw a line $$OA$$ equal and parallel to $$PQ$$, the sense of motion from $$O$$ to $$A$$ being the same as that from $$P$$ to $$Q$$. Then $$\overline{PQ}$$ and $$\overline{OA}$$ are equivalent displacements. Let $$x$$ and $$y$$ be the coordinates of $$A$$. Then it is evident that $$\overline{OA}$$ is completely specified if $$x$$ and $$y$$ are given. We call $$\overline{OA}$$ the displacement $$[x, y]$$ and write $\overline{OA} = \overline{PQ} = \overline{RS} = [x, y].$

## 35. Equivalence of displacements. Multiplication of displacements by numbers.

If $$\xi$$ and $$\eta$$ are the coordinates of $$P$$, and $$\xi’$$ and $$\eta’$$ those of $$Q$$, it is evident that $x = \xi’ – \xi,\quad y = \eta’ – \eta.$ The displacement from $$(\xi, \eta)$$ to $$(\xi’, \eta’)$$ is therefore $[\xi’ – \xi, \eta’ – \eta].$

It is clear that two displacements $$[x, y]$$, $$[x’, y’]$$ are equivalent if, and only if, $$x = x’$$, $$y = y’$$. Thus $$[x, y] = [x’, y’]$$ if and only if $\begin{equation*} x = x’,\quad y = y’. \tag{1}\end{equation*}$

The reverse displacement $$\overline{QP}$$ would be $$[\xi – \xi’, \eta – \eta’]$$, and it is natural to agree that \begin{aligned} &= -[\xi’ – \xi, \eta’ – \eta],\\ \overline{QP} &= -\overline{PQ},\end{aligned} these equations being really definitions of the meaning of the symbols $$-[\xi’ – \xi, \eta’ – \eta]$$, $$-\overline{PQ}$$. Having thus agreed that $-[x, y] = [-x, -y],$ it is natural to agree further that $\begin{equation*} \alpha[x, y] = [\alpha x, \alpha y], \tag{2}\end{equation*}$ where $$\alpha$$ is any real number, positive or negative. Thus (Fig. 19) if $$OB = -\frac{1}{2}OA$$ then $\overline{OB} = -\tfrac{1}{2}\overline{OA} = -\tfrac{1}{2}[x, y] = [-\tfrac{1}{2}x, -\tfrac{1}{2}y].$

The equations (1) and (2) define the first two important ideas connected with displacements, viz. equivalence of displacements, and multiplication of displacements by numbers.

We have not yet given any definition which enables us to attach any meaning to the expressions $\overline{PQ} + \overline{P’Q’},\quad [x, y] + [x’, y’].$ Common sense at once suggests that we should define the sum of two displacements as the displacement which is the result of the successive application of the two given displacements. In other words, it suggests that if $$QQ_{1}$$ be drawn equal and parallel to $$P’Q’$$, so that the result of successive displacements $$\overline{PQ}$$$$\overline{P’Q’}$$ on a particle at $$P$$ is to transfer it first to $$Q$$ and then to $$Q_{1}$$ then we should define the sum of $$\overline{PQ}$$ and $$\overline{P’Q’}$$ as being $$\overline{PQ_{1}}$$. If then we draw $$OA$$ equal and parallel to $$PQ$$, and $$OB$$ equal and parallel to $$P’Q’$$, and complete the parallelogram $$OACB$$, we have $\overline{PQ} + \overline{P’Q’} = \overline{PQ_{1}} = \overline{OA} + \overline{OB} = \overline{OC}.$

Let us consider the consequences of adopting this definition. If the coordinates of $$B$$ are $$x’$$$$y’$$, then those of the middle point of $$AB$$ are $$\frac{1}{2}(x + x’)$$, $$\frac{1}{2} (y + y’)$$, and those of $$C$$ are $$x + x’$$, $$y + y’$$. Hence $\begin{equation*} [x, y] + [x’, y’] = [x + x’, y + y’], \tag{3}\end{equation*}$ which may be regarded as the symbolic definition of addition of displacements. We observe that \begin{aligned} + [x, y] &= [x’ + x, y’ + y]\\ &= [x + x’, y + y’] = [x, y] + [x’, y’]\end{aligned} In other words, addition of displacements obeys the commutative law expressed in ordinary algebra by the equation $$a + b = b + a$$. This law expresses the obvious geometrical fact that if we move from $$P$$ first through a distance $$PQ_{2}$$ equal and parallel to $$P’Q’$$, and then through a distance equal and parallel to $$PQ$$, we shall arrive at the same point $$Q_{1}$$ as before.

In particular $\begin{equation*}[x, y] = [x, 0] + [0, y]. \tag{4}\end{equation*}$ Here $$[x, 0]$$ denotes a displacement through a distance $$x$$ in a direction parallel to $$OX$$. It is in fact what we previously denoted by $$[x]$$, when we were considering only displacements along a line. We call $$[x, 0]$$ and $$[0, y]$$ the components of $$[x, y]$$, and $$[x, y]$$ their resultant.

When we have once defined addition of two displacements, there is no further difficulty in the way of defining addition of any number. Thus, by definition, $\begin{gathered} + [x’, y’] + [x”, y”] = ([x, y] + [x’, y’]) + [x”, y”]\\ = [x + x’, y + y’] + [x”, y”] = [x + x’ + x”, y + y’ + y”].\end{gathered}$

We define subtraction of displacements by the equation $\begin{equation*}[x, y] – [x’, y’] = [x, y] + (-[x’, y’]), \tag{5}\end{equation*}$ which is the same thing as $$[x, y] + [-x’, -y’]$$ or as $$[x – x’, y – y’]$$. In particular $[x, y] – [x, y] = [0, 0].$

The displacement $$[0, 0]$$ leaves the particle where it was; it is the zero displacement, and we agree to write $$[0, 0] = 0$$.

Example XX

1. Prove that

(i) $$\alpha [\beta x, \beta y] = \beta [\alpha x, \alpha y] = [\alpha \beta x, \alpha \beta y]$$,

(ii) $$([x, y] + [x’, y’]) + [x”, y”] = [x, y] + ([x’, y’] + [x”, y”])$$,

(iii) $$[x, y] + [x’, y’] = [x’, y’] + [x, y]$$,

(iv) $$(\alpha + \beta) [x, y] = \alpha [x, y] + \beta [x, y]$$,

(v) $$\alpha \{[x, y] + [x’, y’]\} = \alpha [x, y] + \alpha [x’, y’]$$.

[We have already proved (iii). The remaining equations follow with equal ease from the definitions. The reader should in each case consider the geometrical significance of the equation, as we did above in the case of (iii).]

2. If $$M$$ is the middle point of $$PQ$$, then $$\overline{OM} = \frac{1}{2}(\overline{OP} + \overline{OQ})$$. More generally, if $$M$$ divides $$PQ$$ in the ratio $$\mu : \lambda$$, then $\overline{OM} = \frac{\lambda}{\lambda + \mu}\, \overline{OP} + \frac{\mu}{\lambda + \mu}\, \overline{OQ}.$

3. If $$G$$ is the centre of mass of equal particles at $$P_{1}$$, $$P_{2}$$, …, $$P_{n}$$, then $\overline{OG} = (\overline{OP_{1}} + \overline{OP_{2}} + \dots + \overline{OP_{n}})/n.$

4. If $$P$$$$Q$$$$R$$ are collinear points in the plane, then it is possible to find real numbers $$\alpha$$$$\beta$$$$\gamma$$, not all zero, and such that $\alpha \cdot \overline{OP} + \beta \cdot \overline{OQ} + \gamma \cdot \overline{OR} = 0;$ and conversely. [This is really only another way of stating Ex. 2.]

5. If $$\overline{AB}$$ and $$\overline{AC}$$ are two displacements not in the same straight line, and $\alpha \cdot \overline{AB} + \beta \cdot \overline{AC} = \gamma \cdot \overline{AB} + \delta \cdot \overline{AC},$ then $$\alpha = \gamma$$ and $$\beta = \delta$$.

[Take $$AB_{1} = \alpha \cdot AB$$, $$AC_{1} = \beta \cdot AC$$. Complete the parallelogram $$AB_{1}P_{1}C_{1}$$. Then $$\overline{AP_{1}} = \alpha \cdot \overline{AB} + \beta \cdot \overline{AC}$$. It is evident that $$\overline{AP_{1}}$$ can only be expressed in this form in one way, whence the theorem follows.]

6. $$ABCD$$ is a parallelogram. Through $$Q$$, a point inside the parallelogram, $$RQS$$ and $$TQU$$ are drawn parallel to the sides. Show that $$RU$$$$TS$$ intersect on $$AC$$.

[Let the ratios $$AT:AB$$, $$AR:AD$$ be denoted by $$\alpha$$$$\beta$$. Then $\begin{gathered} \overline{AT} = \alpha \cdot \overline{AB},\quad \overline{AR} = \beta \cdot \overline{AD}, \\ \overline{AU} = \alpha \cdot \overline{AB} + \overline{AD},\quad \overline{AS} = \overline{AB} + \beta \cdot \overline{AD}.\end{gathered}$

Let $$RU$$ meet $$AC$$ in $$P$$. Then, since $$R$$$$U$$$$P$$ are collinear, $\overline{AP} = \frac{\lambda}{\lambda + \mu}\, \overline{AR} + \frac{\mu}{\lambda + \mu}\, \overline{AU},$ where $$\mu/\lambda$$ is the ratio in which $$P$$ divides $$RU$$. That is to say $\overline{AP} = \frac{\alpha\mu}{\lambda + \mu}\, \overline{AB} + \frac{\beta\lambda + \mu}{\lambda + \mu}\, \overline{AD}.$

But since $$P$$ lies on $$AC$$, $$\overline{AP}$$ is a numerical multiple of $$\overline{AC}$$; say $\overline{AP} = k \cdot \overline{AC} = k \cdot \overline{AB} + k \cdot \overline{AD}.$ Hence (Ex. 5) $$\alpha\mu = \beta\lambda + \mu = (\lambda + \mu)k$$, from which we deduce $k = \frac{\alpha\beta}{\alpha + \beta – 1}.$ The symmetry of this result shows that a similar argument would also give $\overline{AP’} = \frac{\alpha\beta}{\alpha + \beta – 1}\, \overline{AC},$ if $$P’$$ is the point where $$TS$$ meets $$AC$$. Hence $$P$$ and $$P’$$ are the same point.]

7. $$ABCD$$ is a parallelogram, and $$M$$ the middle point of $$AB$$. Show that $$DM$$ trisects and is trisected by $$AC$$.3

## 37. Multiplication of displacements.

So far we have made no attempt to attach any meaning whatever to the notion of the product of two displacements. The only kind of multiplication which we have considered is that in which a displacement is multiplied by a number. The expression $[x, y] \times [x’, y’]$ so far means nothing, and we are at liberty to define it to mean anything we like. It is, however, fairly clear that if any definition of such a product is to be of any use, the product of two displacements must itself be a displacement.

We might, for example, define it as being equal to $[x + x’, y + y’];$ in other words, we might agree that the product of two displacements was to be always equal to their sum. But there would be two serious objections to such a definition. In the first place our definition would be futile. We should only be introducing a new method of expressing something which we can perfectly well express without it. In the second place our definition would be inconvenient and misleading for the following reasons. If $$\alpha$$ is a real number, we have already defined $$\alpha [x, y]$$ as $$[\alpha x, \alpha y]$$. Now, as we saw in § 34, the real number $$\alpha$$ may itself from one point of view be regarded as a displacement, viz. the displacement $$[\alpha]$$ along the axis $$OX$$, or, in our later notation, the displacement $$[\alpha, 0]$$. It is therefore, if not absolutely necessary, at any rate most desirable, that our definition should be such that $[\alpha, 0] [x, y] = [\alpha x, \alpha y],$ and the suggested definition does not give this result.

A more reasonable definition might appear to be $[x, y] [x’, y’] = [xx’, yy’].$ But this would give $[\alpha, 0] [x, y] = [\alpha x, 0];$ and so this definition also would be open to the second objection.

In fact, it is by no means obvious what is the best meaning to attach to the product $$[x, y] [x’, y’]$$. All that is clear is (1) that, if our definition is to be of any use, this product must itself be a displacement whose coordinates depend on $$x$$ and $$y$$, or in other words that we must have $[x, y] [x’, y’] = [X, Y],$ where $$X$$ and $$Y$$ are functions of $$x$$$$y$$$$x’$$, and $$y’$$; (2) that the definition must be such as to agree with the equation $[x, 0] [x’, y’] = [xx’, xy’];$ and (3) that the definition must obey the ordinary commutative, distributive, and associative laws of multiplication, so that \begin{aligned} &= [x’, y’] [x, y],\\ ([x, y] + [x’, y’]) [x”, y”] &= [x, y] [x”, y”] + [x’, y’] [x”, y”],\\ [x, y] ([x’, y’] + [x”, y”]) &= [x, y] [x’, y’] + [x, y] [x”, y”],\end{aligned} and\begin{aligned} [x, y] ([x’, y’] [x”, y”]) &= ([x, y] [x’, y’]) [x”, y”].\end{aligned}

## 38.

The right definition to take is suggested as follows. We know that, if $$OAB$$$$OCD$$ are two similar triangles, the angles corresponding in the order in which they are written, then $OB/OA = OD/OC,$ or $$OB \cdot OC = OA \cdot OD$$. This suggests that we should try to define multiplication and division of displacements in such a way that $\overline{OB}/\overline{OA} = \overline{OD}/\overline{OC},\quad \overline{OB} \cdot \overline{OC} = \overline{OA} \cdot \overline{OD}.$

Now let $\overline{OB} = [x, y],\quad \overline{OC} = [x’, y’],\quad \overline{OD} = [X, Y],$ and suppose that $$A$$ is the point $$(1, 0)$$, so that $$\overline{OA} = [1, 0]$$. Then $\overline{OA} \cdot \overline{OD} = [1, 0] [X, Y] = [X, Y],$ and so $[x, y] [x’, y’] = [X, Y].$ The product $$\overline{OB} \cdot \overline{OC}$$ is therefore to be defined as $$\overline{OD}$$, $$D$$ being obtained by constructing on $$OC$$ a triangle similar to $$OAB$$. In order to free this definition from ambiguity, it should be observed that on $$OC$$ we can describe two such triangles, $$OCD$$ and $$OCD’$$. We choose that for which the angle $$COD$$ is equal to $$AOB$$ in sign as well as in magnitude. We say that the two triangles are then similar in the same sense.

If the polar coordinates of $$B$$ and $$C$$ are $$(\rho, \theta)$$ and $$(\sigma, \phi)$$, so that $x = \rho\cos\theta,\quad y = \rho\sin\theta,\quad x’ = \sigma\cos\phi,\quad y’ = \sigma\sin\phi,$ then the polar coordinates of $$D$$ are evidently $$\rho\sigma$$ and $$\theta + \phi$$. Hence \begin{aligned} {2} X &= \rho\sigma\cos(\theta + \phi) &&= xx’ – yy’,\\ Y &= \rho\sigma\sin(\theta + \phi) &&= xy’ + yx’.\end{aligned} The required definition is therefore $[x, y] [x’, y’] = [xx’ – yy’, xy’ + yx’]. \quad{(6)}$

We observe (1) that if $$y = 0$$, then $$X = xx’$$, $$Y = xy’$$, as we desired; (2) that the right-hand side is not altered if we interchange $$x$$ and $$x’$$, and $$y$$ and $$y’$$, so that $[x, y] [x’, y’] = [x’, y’] [x, y];$ and (3) that \begin{gathered} \{[x, y] + [x’, y’]\} [x”, y”] = [x + x’, y + y’] [x”, y”]\\ \begin{aligned}[t] &= [(x + x’) x” – (y + y’) y”, (x + x’) y” + (y + y’) x”]\\ &= [xx” – yy”, xy” + yx”] + [x’x” – y’y”, x’y” + y’x”]\\ &= [x, y] [x”, y”] + [x’, y’] [x”, y”]. \end{aligned}\end{gathered}

Similarly we can verify that all the equations at the end of § 37 are satisfied. Thus the definition  fulfils all the requirements which we made of it in § 37.

Show directly from the geometrical definition given above that multiplication of displacements obeys the commutative and distributive laws. [Take the commutative law for example. The product $$\overline{OB} \cdot \overline{OC}$$ is $$\overline{OD}$$ (Fig. 22), $$COD$$ being similar to $$AOB$$. To construct the product $$\overline{OC} \cdot \overline{OB}$$ we should have to construct on $$OB$$ a triangle $$BOD_{1}$$ similar to $$AOC$$; and so what we want to prove is that $$D$$ and $$D_{1}$$ coincide, or that $$BOD$$ is similar to $$AOC$$. This is an easy piece of elementary geometry.]

1. It is hardly necessary to caution the reader against confusing this use of the symbol $$[x]$$ and that of Chap.II ( and ).↩︎
2. Strictly speaking we ought, by some similar difference of notation, to distinguish the actual length $$x$$ from the number $$x$$ which measures it. The reader will perhaps be inclined to consider such distinctions futile and pedantic. But increasing experience of mathematics will reveal to him the great importance of distinguishing clearly between things which, however intimately connected, are not the same. If cricket were a mathematical science, it would be very important to distinguish between the motion of the batsman between the wickets, the run which he scores, and the mark which is put down in the score-book.↩︎
3. The two preceding examples are taken from Willard Gibbs’ Vector Analysis.↩︎