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a

The straight line drawn from A, where OA = a, and parallel to B, has the equation p=a+x2.

(1) In words, we may pass directly from 0 to P by the vector OP or p; or we may pass first to A, by means of OA or a, and then to P along a vector parallel to B (§ 16).

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

p=ya + , 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 a and ß 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

p=7+ya+ is the equation of a plane passing through the extremity of y, and parallel to a and B. It will now be obvious to the reader that the equation

P = P1 ay + P2 Q2 + ...... = Epa, where az, az, &c. are given vectors, and P1, P2, &c. numerical quan

&. tities, 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 (1)), this theorem will be much extended. Again, the equation

p=xa+y+zy refers to any point whatever in space, provided a, ß, y are not coplanar. (Ante, $ 23).

30.] The equation of the line joining any two points A and B, where A = a and OB OA

= B, is obviously
p= a + (-a),

p=B+ya-B). These equations are of course identical, as may be seen by putting 1- y for x.

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The first may be written

p+(x - 1)a-B = 0;

Pp+qa trß = 0, subject to the condition p +9+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 a, ß, y is

γ p= a + x (B-a) + y(y-B), where x and y are each indeterminate. This may be written

PP+qa+r3+ 8y = 0, with the identical relation

p+2+r+8 = 0. which is the condition that four points may lie in one plane.

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, 0 the point of intersection of its diagonals. Then

ÃO+OB = AB = DC = DO+OC, which gives

ÃO-OC = DO-OB. 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 ;

OC, and DO = OB, the proposition.

(2) AO

(6.) To show 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

A = AB + Ba = AB+ * BC,
BE

= BC+CA,
Cc

= CA + } AB. Hence

Aa + Bb +Cc=i (AB+BC+CA)= 0; which (§ 21) proves the proposition. Also

Aa = ĀB++ BC

= AB-} (CA+ AB)

- }(AB-CA) = } (AB+ĀC), results which are sometimes useful. They may be easily verified by producing Aa to twice its length and joining the extremity with B.

p=y+v(y + ) = (1+x)x+ R.

2

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(6'.) The bisectors of the sides of a triangle meet in a point, which

trisects each of them. Taking A as origin, and putting a, b, y for vectors parallel, and equal, to the sides taken in order BC, CA, AB; the equation of Bb is (§ 28 (1) =

1) y 5B That of Cc is, in the same way,

p=-(1+y)

BY

2

. At the point 0, where Bb and Cc intersect, p= (1+x)y+B=-(1+y)

BY. 1=- –

2
Since y and ß are not parallel, this equation gives

Y
1+=- %, and =-(1+y).

2

2 From these

X = y = - -3. Hence ÃO = $(y–B) = $ Aa. (See Ex. (6).) This equation shows, being a vector one, that Aa passes through 0, and that A0:0a :: 2:1. - (C) If OA=a,

OB=B,

OC=aa+, be three given co-planar

6 vectors, and the lines indicated in the figure be drawn, the points aq, b1, ci lie in a straight line.

C

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B

a

a a

Oc =

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1-a

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a a

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OB =d,

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M

OQ is

We see at once, by the process indicated in § 30, that a a+bB,

bB Ob

1-6 Hence we easily find Οα, 06

aa + bB

Oc, =
1-a-27'
1—2a-b'

b-a These give

-(1-a-26)0a,+(1 – 2a-6)00,-(6-a)Oc, = 0. But

-(1-a20)+(1 — 20–6)-(6-a) = 0 identically. This, by $ 30, proves the proposition. (d.) Let OA

B, be any two vectors. If MP be
parallel to OB; and OQ, BQ, be drawn parallel to AP,
OP respectively; the locus of Q is a straight line parallel
to 0A.

Let OM = ea.
R

Then

AP = e-la+aB.
Р.

Hence the equation of
А.

p= y(e 1a +2B); and that of BQ is

p= ß+2 (ea + xB). At we have, therefore,

xy = 1+ 22,

y (e-1) = ze. These give xy = e, and the equation of the locus of Qis

PE = + y'a, i. e. a straight line parallel to 04, drawn through N in OB produced, so that ON:OB::OM:04.

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

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

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. If OA = a, OB = a, be the vector sides of any triangle, the vector from the vertex dividing the base AB in C so that

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B

=}

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=.

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BC:CA::m:m, is
ma+m7Q1.

m + mi
For AB is az-a, and therefore AC is

mi

(az-a).

m + mi Hence

OC = OA+AC

mi
= a + (a,-a)

m+mi
ma+mQ.

m+mi This expression shows how to find the centre of inertia of two masses; m at the extremity of a, m, at that of a . Introduce m, at the extremity of an, then the vector of the centre of inertia of the three is, by a second application of the formula, (m + m) m+mi

ma+m, az + m2 Q2. (m+m2)+ m2

m+ m2 + m2 For any number of masses, expressed generally by m at the extremity of the vector a, we have the vector of the centre of inertia

Σ (ma)
ß

Σ (mm)
This may be written Im(a-B) = 0.

Now

aj is the vector of m, 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

R

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=

.

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p= att

Bt2

T

B

а

2 where t is an indeterminate number, and a, ß given vec

Q tors, represents a parabola. The origin, 0, is a point on the curve, B 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

312 QP = at, OQ

Q

2

P,

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