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Quaternions.

Chapter I.
Vectors, and their Composition.

1. For at least two centuries the geometrical representation of the negative and imaginary algebraic quantities, −1 and −1 has been a favourite subject of speculation with mathematicians. The essence of almost all of the proposed processes consists in employing such expressions to indicate the Direction, not the length, of lines.

2. Thus it was long ago seen that if positive quantities were measured off in one direction along a fixed line, a useful and lawful convention enabled us to express negative quantities of the same kind by simply laying them off on the same line in the opposite direction. This convention is an essential part of the Cartesian method, and is constantly employed in Analytical Geometry and Applied Mathematics.

3. Wallis, towards the end of the seventeenth century, proposed to represent the impossible roots of a quadratic equation by going out of the line on which, if real, they would have been laid off. This construction is equivalent to the consideration of −1 as a directed unit-line perpendicular to that on which real quantities are measured.

4. In the usual notation of Analytical Geometry of two dimensions, when rectangular axes are employed, this amounts to reckoning each unit of length along Oy as +−1, and on Oy′ as −1, and on ; while on Ox each unit is +1, and on Ox′ it is −1. T. Q. I. 1 2 If we look at these four lines in circular order, i.e. in the order of positive rotation (that of the northern hemisphere of the earth about its axis, or opposite to that of the hands of a watch), they give

1, −1, −1, −−1.
In this series each expression is derived from that which precedes it by multiplication by the factor −1. Hence we may consider −1. as an operator, analogous to a handle perpendicular to the plane of xy, whose effect on any line in that plane is to make it rotate (positively) about the origin through an angle of 90°.

5. In such a system, (which seems to have been first developed, in 1805, by Buée) a point in the plane of reference is defined by a single imaginary expression. Thus a + b−1 may be considered as a single quantity, denoting the point, P, whose cöordinates are a and b. Or, it may be used as an expression for the line OP joining that point with the origin. In the latter sense, the expression a + b−1 implicitly contains the direction, as well as the length, of this line ; since, as we see at once, the direction is inclined at an angle tan−1b/a to the axis of x, and the length is a2 + b2. Thus, say we have

OP = a + b−1;
the line OP considered as that by which we pass from one extremity, O, to the other, P. In this sense it is called a Vector. Considering, in the plane, any other vector,
OQ = a′ + b′−1;
the addition of these two lines obviously gives
OR = a + a′ + ( b + b′ )−1;
and we see that the sum is the diagonal of the parallelogram on OP, OQ. This is the law of the composition of simultaneous velocities ; and it contains, of course, the law of subtraction of one directed line from another.

6. Operating on the first of these symbols by the factor −1, it becomes b + a−1 ; and now, of course, denotes the point whose x and y cöordinates are b and a ; or the line joining this point with the origin. The length is still a2 + b2, but the angle the line makes with the axis of x is tan−1(−ab) ; which is evidently greater by π/2 than before the operation.

3 7. De Moivre’s Theorem tends to lead us still further in the same direction. In fact, it is easy to see that if we use, instead of −1, the more general factor cos α + −1 sin α , its effect on any line is to turn it through the (positive) angle α. in the plane of x, y. [Of course the former factor, −1, is merely the particular case of this, when α = π/2.]

Thus (cos α + −1 sin α) (a + b −1 ) = a cos α b sin α + −1 (a sin αb cos α), by direct multiplication. The reader will at once see that the new form indicates that a rotation through an angle α has taken place, if he compares it with the common formulae for turning the cöordinate axes through a given angle. Or, in a less simple manner, thus—

Length = (a cos αb sin α)2 + (a sin α + b cos α)2
= a2 + b2 as before.

Inclination to axis of x

= tan−1 a sin α + b cos α ____________ a cos αb sin α = tan−1 tan α + (b/a) ____________ 1 − (b/a) tan α
= α + tan−1 b/a.

8. We see now, as it were, why it happens that

(cos α + −1sin α)m = cos mα + −1sin mα.
In fact, the first operator produces m successive rotations in the same direction, each through the angle α ; the second, a single rotation through the angle .

9. It may be interesting, at this stage, to anticipate so far as to remark that in the theory of Quaternions the analogue of

cos θ + −1 sin θ
is cos θ + ϖ sin θ,
where ϖ2 = −1.
Here, however, ϖ is not the algebraic −1, but is any directed unit-line whatever in space.

10. In the present century Argand, Warren, Mourey, and others, extended the results of Wallis and Buée. They attempted 1-2 4 to express as a line the product of two lines each represented by a symbol such a + b−1. To a certain extent they succeeded, but all their results remained confined to two dimensions.

The product, Π, of two such lines was defined as the fourth proportional to unity and the two lines, thus

1 :a + b−1::a′ + b′−1: Π,
or Π = (aa′bb′) + (a′bb′a)−1

The length of Π is obviously the product of the lengths of the factor lines ; and its direction makes an angle with the axis of x which is the sum of those made by the factor lines. From this result the quotient of two such lines follows immediately.

11. A very curious speculation, due to Servois and published in 1813 in Gergonne’s Annales, is one of the very few, so far as has been discovered, in which a well-founded guess at a possible mode of extension to three dimensions is contained. Endeavouring to extend to space the form a + b−1. for the plane, he is guided by analogy to write for a directed unit-line in space the form

p cos α + q cos β + r cos γ,
where α, β, γ are its inclinations to the three axes. He perceives easily that p, q, r must be non-reals : but, he asks, “seraient-elles imaginaires réductibles à la forme générale A + B−1?” The i, j, j of the Quaternion Calculus furnish an answer to this question. (See Chap. II.) But it may be remarked that, in applying the idea to lines in a plane, a vector OP will no longer be represented (as in § 5) by
OP = a + b−1,
but by OP = pa + qb.
And if, similarly, OQ = pa′ + qb′,
the addition of these two lines gives for OR (which retains its previous signification)
OP = p(a + a′) + q(b + b′).

12. Beyond this, few attempts were made, or at least recorded, in earlier times, to extend the principle to space of three dimensions; and, though many such had been made before 1843, none, with the single exception of Hamilton’s, have resulted in simple, practical methods ; all, however ingenious, seeming to lead almost at once to processes and results of fearful complexity.

5 For a lucid, complete, and most impartial statement of the claims of his predecessors in this field we refer to the Preface to Hamilton’s Lectures on Quaternions. He there shews how his long protracted investigations of Sets culminated in this unique system of tridimensional-space geometry.

13. It was reserved for Hamilton to discover the use and properties of a class of symbols which, though all in a certain sense square roots of −1, may be considered as real unit lines, tied down to no particular direction in space ; the expression for a vector is, or may be taken to be,

p = ix + jy + kz;
but such vector is considered in connection with an extraspatial magnitude w, and we have thus the notion of a Quaternion
w + p.
This is the fundamental notion in the singularly elegant, and enormously powerful, Calculus of Quaternions.

While the schemes for using the algebraic −1 to indicate direction make one direction in space expressible by real numbers, the remainder being imaginaries of some kind, and thus lead to expressions which are heterogeneous ; Hamilton’s system makes all directions in space equally imaginary, or rather equally real, thereby ensuring to his Calculus the power of dealing with space indifferently in all directions.

In fact, as we shall see, the Quaternion method is independent of axes or any supposed directions in space, and takes its reference lines solely from the problem it is applied to.

14. But, for the purpose of elementary exposition, it is best to begin by assimilating it as closely as we can to the ordinary Cartesian methods of Geometry of Three Dimensions, with which the student is supposed to be, to some extent at least, acquainted. Such assistance, it will be found, can (as a rule) soon be dispensed with; and Hamilton regarded any apparent necessity for an occasional recurrence to it, in higher applications, as an indication of imperfect development in the proper methods of the new Calculus.

We commence, therefore, with some very elementary geometrical ideas, relating to the theory of vectors in space. It will subsequently appear how we are thus led to the notion of a Quaternion.

6 15. Suppose we have two points A and B in space, and suppose A given, on how many numbers does B’s relative position depend ?

If we refer to Cartesian cöordinates (rectangular or not) we find that the data required are the excesses of B’s three cöordinates over those of A. Hence three numbers are required.

Or we may take polar cöordinates. To define the moon’s position with respect to the earth we must have its Geocentric Latitude and Longitude, or its Right Ascension and Declination, and, in addition, its distance or radius-vector. Three again.

16. Here it is to be carefully noticed that nothing has been said of the actual cöordinates of either A or B, or of the earth and moon, in space ; it is only the relative cöordinates that are contemplated.

Hence any expression, as AB, denoting a line considered with reference to direction and currency as well as length, (whatever may be its actual position in space) contains implicitly three numbers, and all lines parallel and equal to AB, and concurrent with it, depend in the same way upon the same three. Hence, all lines which are equal, parallel, and concurrent, may be represented by a common symbol, and that symbol contains three distinct numbers. In this sense a line is called a vector, since by it we pass from the one extremity, A, to the other, B, and it may thus be considered as an instrument which carries A to B: so that a vector may be employed to indicate a definite translation in space.

[The term “currency” has been suggested by Cayley for use instead of the somewhat vague suggestion sometimes taken to be involved in the word “direction.” Thus parallel lines have the same direction, though they may have similar or opposite currencies. The definition of a vector essentially includes its currency.]

17. We may here remark, once for all, that in establishing a new Calculus, we are at liberty to give any definitions whatever of our symbols, provided that no two of these interfere with, or contradict, each other, and in doing so in Quaternions simplicity and (so to speak) naturalness were the inventor’s aim.

18. Let AB be represented by α, we know that α involves three separate numbers, and that these depend solely upon the position of B relatively to A. Now if CD be equal in length to AB 7 and if these lines be parallel, and have the same currency, we may evidently write

CD = AB = α,
where it will be seen that the sign of equality between vectors contains implicitly equality in length, parallelism in direction, and concurrency. So far we have extended the meaning of an algebraical symbol. And it is to be noticed that an equation between vectors, as
α = β,
contains three distinct equations between mere numbers.

19. We must now define + (and the meaning of will follow) in the new Calculus. Let A, B, C be any three points, and (with the above meaning of = ) let

AB = α,  BC = β,  AC = γ,

If we define + (in accordance with the idea (§ 16) that a vector represents a translation) by the equation

α + β = γ,
or AB + BC = AC,

we contradict nothing that precedes, but we at once introduce the idea that vectors are to be compounded, in direction and magnitude, like simultaneous velocities. A reason for this may be seen in another way if we remember that by adding the (algebraic) differences of the Cartesian cöordinates of B and A, to those of the cöordinates of C and B, we get those of the cöordinates of C and A. Hence these cöordinates enter linearly into the expression for a vector. (See, again, § 5.)

20. But we also see that if C and A coincide (and C may be any point)

AC = 0,
for no vector is then required to carry A to C. Hence the above relation may be written, in this case,
AB + BA = 0,

or, introducing, and by the same act defining, the symbol ,

AB = −BA.

Hence, the symbol , applied to a vector, simply shews that its currency is to be reversed.

8 And this is consistent with all that precedes ; for instance,

AB + BC = AC,
and AB = ACBC,
or = AC + CB,
are evidently but different expressions of the same truth.

21. In any triangle, ABC, we have, of course,

AB + BC + CA = 0 ;
and, in any closed polygon, whether plane or gauche,
AB + BC + ..... + YZ + ZA = 0.

In the case of the polygon we have also

AB + BC + ..... + YZ = ZA.

These are the well-known propositions regarding composition of velocities, which, by Newton’s second law of motion, give us the geometrical laws of composition of forces acting at one point.

22. If we compound any number of parallel vectors, the result is obviously a numerical multiple of any one of them.

Thus, if A, B, C are in one straight line,

BC = x AB ;

where x is a number, positive when B lies between A and C, otherwise negative : but such that its numerical value, independent of sign, is the ratio of the length of BC to that of AB. This is at once evident if AB and BC be commensurable ; and is easily extended to incommensurables by the usual reductio ad absurdum.

23. An important, but almost obvious, proposition is that any vector may be resolved, and in one way only, into three components parallel respectively to any three given vectors, no two of which are parallel, and which are not parallel to one plane.

Let OA, OB, OC be the three fixed vectors, OP any other vector. From P draw PQ parallel to CO, meeting the plane BOA in Q. [There must be a definite point Q, else PQ, and therefore CO, would be parallel to BOA, a case specially excepted.] From Q draw QR parallel to BO, meeting OA in R. Then we have OP = OR + RQ + QP (§ 21), and these components are respectively parallel to the three given 9 vectors. By § 22 we may express OR as a numerical multiple of OA, RQ of OB, and OP of OC. Hence we have, generally, for any vector in terms of three fixed non-coplanar vectors, α, β, γ,

QP = ρ = + + ,

which exhibits, in one form, the three numbers on which a vector depends (§ 16). Here x, y, z are perfectly definite, and can have but single values.

24. Similarly any vector, as OQ, in the same plane with OA and OB can be resolved (in one way only) into components OR, RQ, parallel respectively to OA and OB ; so long, at least, as these two vectors are not parallel to each other.

25. There is particular advantage, in certain cases, in employing a series of three mutually perpendicular unit-vectors as lines of reference. This system Hamilton denotes by i, j, k.

Any other vector is then expressible as

ρ = xi + yj + zk.
Since i, j, k are unit-vectors, x, y, z are here the lengths of conterminous edges of a rectangular parallelepiped of which ρ is the vector-diagonal ; so that the length of ρ is, in this case,
x2 + y2 + z2 .
Let ϖ = ξi + ηj + ζk
be any other vector, then (by the proposition of § 23) the vector equation
ρ = ϖ
obviously involves the following three equations among numbers,
x = ξ, y = η, z = ζ.

Suppose i to be drawn eastwards, j northwards, and k upwards, this is equivalent merely to saying that if two points coincide, they are equally to the east (or west) of any third point, equally to the north (or south) of it, and equally elevated above (or depressed below) its level.

26. It is to be carefully noticed that it is only when α, β, γ are not coplanar that a vector equation such as

ρ = ϖ
or + + = ξα + ηβ + ζγ
necessitates the three numerical equations
x = ξ, y = η, z = ζ.
10 For, if α, β, γ be coplanar (§ 24), a condition of the following form must hold
γ = + ,
Hence ρ = (x + za)α + (y + zb)β,
ϖ = (ξ + ζa)α + (η + ζb)β,
and the equation ρ = ϖ
now requires only the two numerical conditions
x + za = ξ + ζa,   y + zb = η + ζb.

27. The Commutative and Associative Laws hold in the combination of vectors by the signs + and −. It is obvious that, if we prove this for the sign + , it will be equally proved for −, because − before a vector (§ 20) merely indicates that it is to be reversed before being considered positive.

Let A, B, C, D be, in order, the corners of a parallelogram ; we have, obviously,

AB = DC , AD = BC .
And AB + BC = AC = AD + DC = BC + AB .

Hence the commutative law is true for the addition of any two vectors, and is therefore generally true.

Again, whatever four points are represented by A, B, C, D, we have

AD = AB + BD = AC + CD ,

or substituting their values for AD, BD, AC respectively, in these three expressions,

AB + BC + CD = AB + (BC + CD) = (AB + BC) + CD .

And thus the truth of the associative law is evident.

28. The equation

ρ = α β ,

where ρ is the vector connecting a variable point with the origin, β a definite vector, and x an indefinite number, represents the straight line drawn from the origin parallel to β (§ 22).

The straight line drawn from A, where OA = α, and parallel to β , has the equation

  ρ = α + x β
 
 (1).

In words, we may pass directly from O to P by the vector OP or ρ ; or we may pass first to A, by means of OA or a, and then to P along a vector parallel to β (§ 16).

11 Equation (1) is one of the many useful forms into which Quaternions enable us to throw the general equation of a straight line in space. As we have seen (§ 25) it is equivalent to three numerical equations ; but, as these involve the indefinite quantity x, they are virtually equivalent to but two, as in ordinary Geometry of Three Dimensions.

29. A good illustration of this remark is furnished by the fact that the equation

ρ = y α + x β ,

which contains two indefinite quantities, is virtually equivalent to: only one numerical equation. And it is easy to see that it represents the plane in which the lines α and β lie ; or the surface which is formed by drawing, through every point of OA, a line parallel to OB. In fact, the equation, as written, is simply § 24 in symbols.

And it is evident that the equation

ρ = γ + y α + x β

is the equation of the plane passing through the extremity of γ, and parallel to α and β.

It will now be obvious to the reader that the equation

ρ = p1α1 + p2α2 + …… = Σ p α ,

where α1, α2, &c. are given vectors, and p1, p2, &c. numerical quantities, represents a straight line if p1, p2, &c. be linear functions of one indeterminate number ; and a plane, if they be linear expressions containing two indeterminate numbers. Later (§ 31(l)), this theorem will be much extended.

Again, the equation

ρ = x α + y β + zγ

refers to any point whatever in space, provided α, β, γ are not coplanar. (Ante, § 23.)

30. The equation of the line joining any two points A and B, where OA = α and OB = β, is obviously

ρ = α + x(βα) ,
or ρ = β + y(αβ) .

These equations are of course identical, as may be seen by putting 1 − y for x.

12 The first may be written

ρ + (x − 1)αxβ = 0;
or pρ + qα + rβ = 0 ,

subject to the condition p + q + r = 0 identically. That is — A homogeneous linear function of three vectors, equated to zero, expresses that the extremities of these vectors are in one straight line, if the sum of the coefficients be identically zero.

Similarly, the equation of the plane containing the extremities A, B, C of the three non-coplanar vectors α, β, γ is

ρ = x (βα + y (γβ) ,

where x and y are each indeterminate.

This may be written

pρ + qα + rβ + sγ = 0 ,

with the identical relation

p + q + r + s = 0 ,

which is one form of the condition that four points may lie in one plane.

section 31. not completed

31. We have already the means of proving, in a very simple manner, numerous classes of propositions in plane and solid geometry. A very few examples, however, must suffice at this stage ; since we have hardly, as yet, crossed the threshold of the subject, and are dealing with mere linear equations connecting two or more vectors, and even with them we are restricted as yet to operations of mere addition. We will give these examples with a painful minuteness of detail, which the reader will soon find to be necessary only for a short time, if at all.

(a) The diagonals of a parallelogram bisect each other.

Let ABCD be the parallelogram, the point of intersection of its diagonals. Then

AO + OB = AB = DC = DO + OO,

which gives AOOC = DOOB . The two vectors here equated are parallel to the diagonals respectively. Such an equation is, of course, absurd unless

(1) The diagonals are parallel, in which case the figure is not a parallelogram ;

(2) AO = OC, and DO = OB, the proposition. 13

(b) To shew that a triangle can be constructed, whose sides are parallel, and equal, to the bisectors of the sides of any triangle.

Let ABC be any triangle, Aa, Bb, Cc the bisectors of the sides.

Then Aa = AB + Ba = AB + ½ BC,
Bb - - - = BC + ½ CA,
Cc - - - = CA + ½ AB,
Hence Aa + Bb + Cc = 3/2 ( AB + BC + CA ) = 0 ;

which (§21) proves the proposition.

Also Aa = AB + ½BC
= AB − ½(CA + AB)
= ½(AB − CA) = ½(AB + AC)

results which are sometimes useful. They may be easily verified by producing Aa to twice its length and joining the extremity with B.

(b′) The bisectors of the sides of a triangle meet in a point, which trisects each of them.

Taking A as origin, and putting α, β, γ for vectors parallel, and equal, to the sides taken in order BC, CA, AB; the equation of Bb is (§28 (1)) That of Cc is, in the same way, p = -(l H At the point 0, where Bb and Cc intersect, -* Since 7 and /S are not parallel, this equation gives From these x = y = f.

Hence ZO = J (7 - 0) = | Aa. (See Ex. (6).)

This equation shews, being a vector one, that Aa passes through 0, and that AO : Oa :: 2:1. 14

(c) If 04 = 0, OC = la + mβ, be three given co-planar vectors, c the intersection of AB, OC, and if the lines indicated in the figure be drawn, the points av bv c l lie in a straight line.

We see at once, by the process indicated in 30, that c= l + m Hence we easily find mβ I-m mβ Oa = -i-r m lj l-l-2m l-21-m These give - (1 - I - 2m) ~0al + (l-2l- m) ~Obl - (m - I) Oc^ = 0. But - (1 - I - 2m) + (1 - 21 - m) - (m - I) = identically.

This, by 30, proves the proposition.

(d) Let OA = a, OB = /S, be any two vectors. If MP be a given line parallel to OB ; and OQ, BQ, be drawn parallel to AP, OP respectively ; the locus of Q is a straight line parallel to OA.

Let

Then OM=eoi. 15 Hence the equation of OQ is and that of BQ is p = ft + z (ea. + acj3). At Q we have, therefore, These give xy = e, and the equation of the locus of Q is p = eβ + 2/X i.e. a straight line parallel to 0.4, drawn through N in OjB pro duced, so that

Cor. If BQ meet MP in q, Pq = ; and if AP meet NQ in p, Qp = a.

Also, for the point R we have pR = AP, QR = Bq.

Further, the locus of R is a hyperbola, of which MP and NQ are the asymptotes. See, in this connection, 31 (k) below.

Hence, if from any two points, A and B, lines be drawn intercepting a given length Pq on a given line Mq ; and if, from R their point of intersection, Rp be laid off= PA, and RQ = qB ; Q and p lie on a fixed straight line, and the length of Qp is constant.

(e) To find the centre of inertia of any system of masses.

If OA = a, OB = a, l , be the vector sides of any triangle, the vector from the vertex dividing the base AB in C so that BC : CA :: m : ml ma + m,GL is

For AB is a x a, and therefore AC is

Hence 00 = OA + AC _ ma. + m1 1 m + ra t This expression shews how to find the centre of inertia of two masses ; m at the extremity of a, ml at that of ar Introduce mg 16 at the extremity of 2, then the vector of the centre of inertia of the three is, by a second application of the formula, (m -f m From this it is clear that, for any number of masses, expressed generally by m at the extremity of the vector a, the vector of the centre of inertia is 2 (ma) This may be written 2m (a - ft) = 0. Now al - ft is the vector of mx with respect to the centre of inertia. Hence the theorem, If the vector of each element of a mass, drawn from the centre of inertia, be increased in length in proportion to the mass of the element, the sum of all these vectors is zero.

(f)

We see at once that the equation where t is an indeterminate number, and a, ft given vectors, represents a parabola. The origin, 0, is a point on the curve, ft is parallel to the axis, i.e. is the diameter OB drawn from the origin, and a is OA the tangent at the origin. In the figure QP = at, OQ = ^2-.

The secant joining the points where t has the values t and t is represented by the equation *-*? (30) V } Write x for x (f - 1) [which may have any value], then put tf = t, and the equation of the tangent at the point (t) is 17 In this put x = t, and we have fit* p ~ "T or the intercept of the tangent on the diameter is equal in length to the abscissa of the point of contact, but has the opposite currency.

Otherwise : the tangent is parallel to the vector a + fit or at + βf or f + at + ^f or OQ + OP. But TP=TO + OP, L - hence TO = OQ.

(g) Since the equation of any tangent to the parabola is * + fit), let us find the tangents which can be drawn from a given point. Let the vector of the point be p=p + qβ (24). Since the tangent is to pass through this point, we have, as conditions to determine t and #, f 2 by equating respectively the coefficients of a and fi.

Hence t=pJp* 2q.

Thus, in general, m tangents can be drawn from a given point. These coincide if p2 = 2q ; that is, if the vector of the point from which they are to be drawn is p=pa + qP = pa + ^0, i.e. if the point lies on the parabola. They are imaginary if Zq pz , that is, if the point be r being positive. Such a point is evidently within the curve, as at R, where OQ = ^β, QP=pa, PR = rfi. T. Q. I. 2

18 (h) Calling the values of t for the two tangents found in (g) ^ and t 2 respectively, it is obvious that the vector joining the points of contact is which is parallel to 4-β -- s or, by the values of ^ and 2 in (g), a + pβ. Its direction, therefore, does not depend on q. In words, If pairs of tangents be drawn to a parabola from points of a diameter produced, the chords of contact are parallel to the tangent at the vertex of the diameter. This is also proved by a former result, for we must have OT for each tangent equal to QO.

(i) The equation of the chord of contact, for the point whose vector is Rt 2 is thus p = 0-^ + -- + y (a + pft). Suppose this to pass always through the point whose vector is p = act + 6β. Then we must have ^ + y = a, \ or t^p + Jp*- 2pa + 26. Comparing this with the expression in (g), we have q = pa - b ; that is, the point from which the tangents are drawn has the vector p=pa + (pa-b) fi = - bβ +p (a + aft), a straight line ( 28 (1)). The mere form of this expression contains the proof of the usual properties of the pole and polar in the parabola ; but, for the sake of the beginner, we adopt a simpler, though equally general, process.

Suppose a = 0. This merely restricts the pole to the particular 19 diameter to which we have referred the parabola. Then the pole is Q, where p = bβ; and the polar is the line TU, for which p=-bβ+pot. Hence the polar of any point is parallel to the tangent at the extremity of the diameter on which the point lies, and its intersection with that diameter is as far beyond the vertex as the pole is within, and vice versa.

(j) As another example let us prove the following theorem. If a triangle be inscribed in a parabola, the three points in which the sides are met by tangents at the angles lie in a straight line.

Since is any point of the curve, we may take it as one corner of the triangle. Let t and t t determine the others. Then, if w,, -sr 2, OT 3 represent the vectors of the points of intersection of the tangents with the sides, we easily find t," tt l f -\- 1 These values give j. 2 /2 -.- J ir-.=- v 2 t 2 Also ^- - - "* - " 1 - = identically. t tt.

Hence, by 30, the proposition is proved.

(k) Other interesting examples of this method of treating curves will, of course, suggest themselves to the student. Thus p = a. cos t + @ sin t represents an ellipse, of which the given vectors a and ft are semiconjugate diameters. If t represent time, the radius-vector of this ellipse traces out equal areas in equal times. [We may anticipate so far as to write the following : 2 Area = TJVpdp = TVap . fdt ; which will be easily understood later.]

2-2 20 Again, p = at + or p = a tan x + (3 cot x t evidently represents a hyperbola referred to its asymptotes. [If t represent time, the sectorial area traced out is proportional to log tt taken between proper limits.]

Thus, also, the equation p = a. (t + sin t) + β cos t in which a and ft are of equal lengths, and at right angles to one another, represents a cycloid. The origin is at the middle point of the axis (2/9) of the curve. [It may be added that, if t represent time, this equation shews the motion of the tracing point, provided the generating circle rolls uniformly, revolving at the rate of a radian per second.]

When the lengths of a, ft are not equal, this equation gives the cycloid distorted by elongation of its ordinates or abscissae : not a trochoid. The equation of a trochoid may be written p = a (et + sin f) + ft cos t, e being greater or less than 1 as the curve is prolate or curtate. The lengths of a and ft are still taken as equal.

But, so far as we have yet gone with the explanation of the calculus, as we are not prepared to determine the lengths or in clinations of vectors, we can investigate only a very limited class of the properties of curves, represented by such equations as those above written.

(l) We may now, in extension of the statement in 29, make the obvious remark that (where, as in 23, the number of vectors, a, can always be reduced to three, at most) is the equation of a curve in space, if the numbers pv pz, &c. are functions of one indeterminate. In such a case the equation is sometimes written P =t (t). But, if pv p2, &c. be functions of two indeterminates, the locus of the extremity of p is a surface; whose equation is sometimes written

[It may not be superfluous to call the reader’s attention to the fact that, in these equations, / (t) or c/ (t, u) is necessarily a vector expression, since it is equated to a vector, p,]

21 (m) Thus the equation p = a cos t + β sin t 4- yt ..................... (1) belongs to a helix, while p = a cos t + @ sin t + 7^ .................. (2) represents a cylinder whose generating lines are parallel to 7, and whose base is the ellipse p = a cos + β sin t. The helix above lies wholly on this cylinder.

Contrast with (2) the equation p = u(a cos t + β sin t + 7) ..................... (3) which represents a cone of the second degree : made up, in fact, of all lines drawn from the origin to the ellipse p = a cos t + sin t + 7.

If, however, we write p = u (a cos t + @ sin 4- yt), we form the equation of the transcendental cone whose vertex is at the origin, and on which lies the helix (1).

In general p = uf (t) is the cone whose vertex is the origin, and on which lies the curve while p = (f(t) + UOL is a cylinder, with generating lines parallel to a, standing on the same curve as base.

Again, p = pa + qβ + ry with a condition of the form op 2 + bf + cr 2 = I belongs to a central surface of the second order, of which a, (3, 7 are the directions of conjugate diameters. If a, b, c be all positive, the surface is an ellipsoid.

32. In Example (f) above we performed an operation equivalent to the differentiation of a vector with reference to a single numerical variable of which it was given as an explicit function. As this process is of very great use, especially in quaternion investigations connected with the motion of a particle or point ; and as it will afford us an opportunity of making a preliminary step towards 22 overcoming the novel difficulties which arise in quaternion differentiation ; we will devote a few sections to a more careful, though very elementary, exposition of it.

33. It is a striking circumstance, when we consider the way in which Newton’s original methods in the Differential Calculus have been decried, to find that Hamilton was obliged to employ them, and not the more modern forms, in order to overcome the characteristic difficulties of quaternion differentiation. Such a thing as a differential coefficient has absolutely no meaning in quaternions, except in those special cases in which we are dealing with degraded quaternions, such as numbers, Cartesian cöordinates, &c. But a quaternion expression has always a differential, which is, simply, what Newton called a fluxion.

As with the Laws of Motion, the basis of Dynamics, so with the foundations of the Differential Calculus ; we are gradually coming to the conclusion that Newton’s system is the best after all.

34. Suppose ρ to be the vector of a curve in space. Then, generally, ρ may be expressed as the sum of a number of terms, each of which is a multiple of a constant vector by a function of some one indeterminate ; or, as in §31 (l), if P be a point on the curve,

OP = ρ = φ(t).

And, similarly, if Q be any other point on the curve,

OQ = ρ1 = ρ + δρ = φ(t1) = φ(t + δt),

where δt is any number whatever.

The vector-chord PQ is therefore, rigorously,

δρ = ρ1 − ρ = φ(t + δt),
sections 35-etc. not completed

35. It is obvious that, in the present case, because the vectors involved in c/ are constant, and their numerical multipliers alone vary, the expression * (t + $t) is, by Taylor’s Theorem, equivalent to *, . *. . Hence, Sp = ^- ( + And we are thus entitled to write, when Bt has been made inde finitely small, LTi. mi.t /p\ = d- p = -d^f) (t) = , ,A / (*) 23 In such a case as this, then, we are permitted to differentiate, or to form the differential coefficient of, a vector, according to the ordinary rules of the Differential Calculus. But great additional insight into the process is gained by applying Newton’s method.

36. Let OP be and OQ1 Pl = (f (t + dt), where dt is any number whatever.

The number t may here be taken as representing time, i.e. we may suppose a point to move along the curve in such a way that the value of t for the vector of the point P of the curve denotes the interval which has elapsed (since a fixed epoch) when the moving point has reached the extremity of that vector. If, then, dt represent any interval, finite or not, we see that will be the vector of the point after the additional interval dt.

But this, in general, gives us little or no information as to the velocity of the point at P. We shall get a better approximation by halving the interval dt, and finding Q2 , where OQ2 = $ (t + |cft), as the position of the moving point at that time. Here the vector virtually described in \dt is PQ2 . To find,_on this supposition, the vector described in dt, we must double PQ2 , and we find, as a second approximation to the vector which the moving point would have described in time dt, if it had moved for that period in the direction and with the velocity it had at P, = 2 The next approximation gives - 8 And so on, each step evidently leading us nearer the sought truth. Hence, to find the vector which would have been described in time dt had the circumstances of the motion at P remained undisturbed, we must find the value of dp = Tq = (t 4 i dt} - 4> (t)\ \ X /

24 We have seen that in this particular case we may use Taylor’s Theorem. We have, therefore, dp =^= 00 X { (0 \ dt + f (t) A ^+ C.J = (#> (t) dt. And, if we choose, we may now write s -*

37. But it is to be most particularly remarked that in the whole of this investigation no regard whatever has been paid to the magnitude of dt. The question which we have now answered may be put in the form—A point describes a given curve in a given manner. At any point of its path its motion suddenly ceases to be accelerated. What space will it describe in a definite interval ? As Hamilton well observes, this is, for a planet or comet, the case of a ‘celestial Atwood’s machine.’

38. If we suppose the variable, in terms of which ρ is expressed, to be the arc, s, of the curve measured from some fixed point, we find as before

= φ′(s) ds .

From the very nature of the question it is obvious that the length of must in this case be ds, so that φ′(s) is necessarily a unit-vector. This remark is of importance, as we shall see later ; and it may therefore be useful to obtain afresh the above result without any reference to time or velocity.

39. Following strictly the process of Newton’s VIIth Lemma, let us describe on Pq2 an arc similar to PQ2, and so on. Then obviously, as the subdivision of ds is carried farther, the new arc (whose length is always ds) more and more nearly (and without limit) coincides with the line which expresses the corresponding approximation to .

sections 40-etc. not completed

40. As additional examples let us take some well-known plane curves; and first the hyperbola (31 Here db-1 -) 25 This shews that the tangent is parallel to the vector -f In words, if the vector (from the centre) of a point in a hyperbola be one diagonal of a parallelogram, two of whose sides coincide with the asymptotes, the other diagonal is parallel to the tangent at the point, and cuts off a constant area from the space between the asymptotes. (For the sides of this triangular area are t times the length of a, and I/t times the length of ft, respectively ; the angle between them being constant.) Next, take the cycloid, as in 31 (), p = a (t + sin t)+ft cos t. We have dp = {(!+ cos t) ft sin t} dt. At the vertex t = 0, cos t = l, sin t = 0, and dp = 2adt. At a cusp t = TT, cos = 1, sin = 0, and dp = 0.

This indicates that, at the cusp, the tracing point is (instantaneously) at rest. To find the direction of the tangent, and the form of the curve in the vicinity of the cusp, put t = TT + T, where powers of T above the second are omitted. We have dp = ftrdt + -J"- dt, 2* so that, at the cusp, the tangent is parallel to ft. By making the same substitution in the expression for p, we find that the part of the curve near the cusp is a semicubical parabola, or, if the origin be shifted to the cusp (p = TTOL - ft\

41. Let us reverse the first of these questions, and seek the envelop of a line which cuts off from two fixed axes a triangle of constant area.

If the axes be in the directions of a and ft, the intercepts may evidently be written at and ^-. Hence the equation of the line is (30) //? \ p = a.t 4- x ( j at) . 26 The condition of envelopment is, obviously, (see Chap. IX.) This gives = la - ( x f + a dt + - ctf * Hence (1 as) dt tdx = 0, x , dx and - - dt + = 0. 6 C From these, at once, x = J, since cfcc and cfa are indeterminate. Thus the equation of the envelop is the hyperbola as before ; a, β being portions of its asymptotes.

42. It may assist the student to a thorough comprehension of the above process, if we put it in a slightly different form. Thus the equation of the enveloping line may be written which gives dp = = CLd {t(l - as)} + @d . \*/ Hence, as a is not parallel to β, we must have d and these are, when expanded, the equations we obtained in the preceding section.

43. For farther illustration we give a solution not directly employing the differential calculus. The equations of any two of the enveloping lines are

* Here we have opportunity for a remark (very simple indeed, but) of the utmost importance. We are not to equate separately to zero the coefficients of dt and dx; for we must remember that this equation is of the form where p and q are numbers ; and that, so long as a and β are actual and non-parallel vectors, the existence of such an equation requires ( 24)
27 (R \ p = at + x ( - at ) , \i / t and t t being given, while x and xl are indeterminate.

At the point of intersection of these lines we have ( 20), These give, by eliminating xv ti n\L _ x).\ or x . Hence the vector of the point of intersection is _ aflt + ff P= tt + t and thus, for the ultimate intersections, where - = 1, t + ^- ) as before.

Cor. If, instead of the ultimate intersections, we consider the intersections of pairs of these lines related by some law, we obtain useful results. Thus let OL + β ; or the intersection lies in the diagonal of the parallelogram on a, 0.

If ^ = mt, where m is constant, 8 mtot + P = i But we have also x = m 1 m-f 1

Hence the locus of a point which divides in a given ratio a line cutting off a given area from two fixed axes, is a hyperbola of which these axes are the asymptotes.

28 If we take either W tt^t + tfj) = constant, or *- = constant, r + c, the locus is a parabola ; and so on.

It will be excellent practice for the student, at this stage, to work out in detail a number of similar questions relating to the envelop of, or the locus of the intersection of selected pairs from, a series of lines drawn according to a given law. And the process may easily be extended to planes. Thus, for instance, we may form the general equation of planes which cut off constant tetrahedra from the axes of cöordinates. Their envelop is a surface of the third degree whose equation may be written p = CCOL + yβ + 7 ; where ocyz = a3 .

Again, find the locus of the point of intersection of three of this group of planes, such that the first intercepts on β and 7, the second on 7 and a, the third on a and β, lengths all equal to one another, &c. But we must not loiter with such simple matters as these.

44. The reader who is fond of Anharmonic Ratios and Transversals will find in the early chapters of Hamilton’s Elements of Quaternions an admirable application of the composition of vectors to these subjects. The Theory of Geometrical Nets, in a plane, and in space, is there very fully developed; and the method is shewn to include, as particular cases, the corresponding processes of Grassmann’s Ausdehnungslehre and Möbius’ Barycentrische Calcul. Some very curious investigations connected with curves and surfaces of the second and third degrees are also there founded upon the composition of vectors.

Examples to Chapter I.

1. The lines which join, towards the same parts, the extremities of two equal and parallel lines are themselves equal and parallel. (Euclid, I. xxxiii.)

2. Find the vector of the middle point of the line which joins the middle points of the diagonals of any quadrilateral, plane or 29 gauche, the vectors of the corners being given ; and so prove that this point is the mean point of the quadrilateral.

If two opposite sides be divided proportionally, and two new quadrilaterals be formed by joining the points of division, the mean points of the three quadrilaterals lie in a straight line.

Shew that the mean point may also be found by bisecting the line joining the middle points of a pair of opposite sides.

3. Verify that the property of the coefficients of three vectors whose extremities are in a line (§ 30) is not interfered with by altering the origin.
4. If two triangles ABC, abc, be so situated in space that Aa, Bb, Cc meet in a point, the intersections of AB, ab, of BC, be, and of CA, ca, lie in a straight line.
5. Prove the converse of 4, i.e. if lines be drawn, one in each of two planes, from any three points in the straight line in which these planes meet, the two triangles thus formed are sections of a common pyramid.
6. If five quadrilaterals be formed by omitting in succession each of the sides of any pentagon, the lines bisecting the diagonals of these quadrilaterals meet in a point. (H. Fox Talbot.)
7. Assuming, as in §7, that the operator
cos θ + −1 sin θ

turns any radius of a given circle through an angle θ in the positive direction of rotation, without altering its length, deduce the ordinary formulae for cos (A + B), cos (AB), sin (A + B), and sin (AB), in terms of sines and cosines of A and B.

8. If two tangents be drawn to a hyperbola, the line joining the centre with their point of intersection bisects the lines joining the points where the tangents meet the asymptotes : and the secant through the points of contact bisects the intercepts on the asymptotes.
9. Any two tangents, limited by the asymptotes, divide each other proportionally.
10. If a chord of a hyperbola be one diagonal of a parallelogram whose sides are parallel to the asymptotes, the other diagonal passes through the centre.
30 11. Given two points A and B, and a plane, C. Find the locus of P, such that if AP cut C in Q, and BP cut C in R, QR may be a given vector.
12. Shew that ρ = x2α + y2β + (x + y)2γ

is the equation of a cone of the second degree, and that its section by the plane

ρ = pα + qβ + rγ ____________ p + q + r

is an ellipse which touches, at their middle points, the sides of the triangle of whose corners α, β, γ are the vectors. (Hamilton, Elements, p. 96.)

13. The lines which divide, proportionally, the pairs of opposite sides of a gauche quadrilateral, are the generating lines of a hyperbolic paraboloid. (Ibid. p. 97.)
14. Shew that ρ = x2α + y2β + z2γ where x + y + z = 0, represents a cone of the third order, and that its section by the plane
ρ = pα + qβ + rγ ____________ p + q + r

is a cubic curve, of which the lines

ρ = pα + qβ ____________ p + q , & c.

are the asymptotes and the three (real) tangents of inflection. Also that the mean point of the triangle formed by these lines is a conjugate point of the curve. Hence that the vector α + β + γ is a conjugate ray of the cone. (Ibid. p. 96.)