which is a known quaternion; and thus our condition becomes

This divides itself into two cases, according as n is an even or an odd number.

This gives one determinate direction, Va, for p1; and shews that there are two, and only two, solutions.

and therefore p1 may be drawn to any point in the great circle of the unit-sphere whose poles are on the vector a.

= 3.

235.] To illustrate these results, let us take first the case of n= Here we must have

S.aẞy = 0,

or the three given vectors must (as is obvious on other grounds) be parallel to one plane. Here aẞy, which lies in this plane, is (§ 106) the vector-tangent at the first corner of each of the inscribed triangles; and is obviously perpendicular to the vector drawn from the centre to that corner.

as might have been at once seen from § 106.

236.] Hamilton has given (Lectures, p. 674) an ingenious and simple process by which the above investigation is rendered applicable to the more difficult problem in which each side of the inscribed polygon is to pass through a given point instead of being parallel to a given line. His process depends upon the integration of a linear equation in finite differences. By an immediate application of the linear and vector function of Chapter V, the above solutions may be at once extended to any central surface of the second order.

237.] To find the equation of a cone of revolution, whose vertex is the origin.

Suppose a, where Ta 1, to be its axis, and e the cosine of its semi-vertical angle; then, if p be the vector of any point in the

238.] Change the origin to the point in the axis whose vector is xa, and the equation becomes

Let the radius of the section of the cone made by
Sα = 0

retain a constant value b, while a changes; this necessitates.

so that when a is infinite, e is unity. In this case the equation becomes S2aw+w2+b2 = 0,

which must therefore be the equation of a circular cylinder of radius b, whose axis is the vector a. To verify this we have only to notice that if be the vector of a point of such a cylinder we must (§ 201) have TVa=b,

which is the same equation as that above.

239.] To find, generally, the equation of a cone which has a circular section :

Take the origin as vertex, and let the circular section be the intersection of the plane Sap = 1

with the sphere (passing through the origin)

Hence, eliminating Tp, we find the following equation which Up must satisfy

or

Sa UpSB Up=-1, p2-Sap Sẞp = 0,

which is therefore the required equation of the cone.

As a and ẞ are similarly involved, the mere form of this equation proves the existence of the subcontrary section discovered by Apollonius.

240.] The equation just obtained may be written

or, since a and ẞ are perpendicular to the cyclic arcs (§ 59*),

where p and p' are arcs drawn from any point of a spherical conic perpendicular to the cyclic arcs. This is a well-known property of such curves.

241.] If we cut the cyclic cone by any plane passing through the origin, as

Syp = 0,

then Vay and Vẞy are the traces on the cyclic planes, so that

p = xUV ay+yUVßy (§ 29).

Substitute in the equation of the cone, and we get -x2-y2+Pxy = 0,

where P is a known scalar. Hence the values of x and y are the same pair of numbers. This is a very elementary proof of the proposition in § 59*, that PL = MQ (in the last figure of that section).

242.] When x and y are equal, the transversal arc becomes a tangent to the spherical conic, and is evidently bisected at the point of contact. Here we have

This is the equation of the cone whose sides are perpendiculars (through the origin) to the planes which touch the cyclic cone, and from this property the same equation may readily be deduced.

243.] It may be well to observe that the property of the Stereographic projection of the sphere, viz. that the projection of a circle is a circle, is an immediate consequence of the above form of the equation of a cyclic cone.

244] That § 239 gives the most general form of the equation of a cone of the second order, when the vertex is taken as origin, follows from the early results of next Chapter. For it is shewn in § 249 that the equation of a cone of the second order can always be put in the form

This may be written

22.Sap SBp+Ap2

ρφρ

= 0. Spopp = 0,

where is the self-conjugate linear and vector function

φρ = Σ. αρβ + (4+ Σβαβ) ρ.

By § 168 this may be transformed to

φρ = γρ + Γ. λρμ,

and the general equation of the cone becomes

(p −Sλμ) p2 +28λpSμp = 0,

which is the form obtained in § 239.

Hence op is perpendicular to the tangent-plane at the extremity of p. The equation of this plane is therefore ( being the vector of any point in it)

Spp (―p) = 0,

or, by the equation of the cone,

Supp=0.

246.] The equation of the cone of normals to the tangent-planes of a given cone can be easily formed from that of the cone itself. For we may write it in the form

S(p-1pp) pp = 0,

and if we put op=σ, a vector of the new cone, the equation becomes Sop-100.

Numerous curious properties of these connected cones, and of the corresponding spherical conics, follow at once from these equations. But we must leave them to the reader.

247.] As a final example, let us find the equation of a cyclic cone when five of its vector-sides are given―i. e. find the cone of the second order whose vertex is the origin, and on whose surface lie the vectors a, ß, y, d, e.

If we write

0 = S.V (Vaß Vde)V (VßyVep)V (VyòV pa),

(1)

we have the equation of a cone whose vertex is the origin-for the equation is not altered by putting xp for p. Also it is the equation of a cone of the second degree, since p occurs only twice. Moreover the vectors a, ẞ, y, d, e are sides of the cone, because if any one of them be put for p the equation is satisfied. Thus if we put ẞ for p the equation becomes

0 = S.V (VaßVde)V (VߥVeß)V (VyòV ßa)

= S.V (VaßVde) { VßaS.VydVߥVeß — VyòS.VßаVßуVeß}. The first term vanishes because

S.V (VaẞVòe) Vẞa = 0,

and the second because

S.VBaVBуVeẞ = 0,

since the three vectors Vßa, Vẞy, Veß, being each at right angles to ẞ, must be in one plane.

As is remarked by Hamilton, this is a very simple proof of Pascal's

Theorem for (1) is the condition that the intersections of the planes of a, ẞ and d, €; ß, y and e, p; y, d and p, a; shall lie in one plane; or, making the statement for any plane section of the cone, that the points of intersection of the three pairs of opposite sides, of a hexagon inscribed in a curve, may always lie in one straight line, the curve must be a conic section.

EXAMPLES TO CHAPTER VII.

1. On the vector of a point P in the plane

Sap = 1

a point Q is taken, such that QO.OP is constant; find the equation of the locus of Q.

2. What spheres cut the loci of P and Q in (1) so that both lines of intersection lie on a cone whose vertex is 0?

3. A sphere touches a fixed plane, and cuts a fixed sphere. If the point of contact with the plane be given, the plane of the intersection of the spheres contains a fixed line.

Find the locus of the centre of the variable sphere, if the plane of its intersection with the fixed sphere passes through a given point. 4. Find the radii of the spheres which touch, simultaneously, the four given planes

Sap=0,

SBp = 0,

Syp = 0, Sop

= 1.

[What is the volume of the tetrahedron enclosed by these planes ?] 5. If a moveable line, passing through the origin, make with any number of fixed lines angles 0, 01, 02, &c., such that

a cos.0+a, cos.0+...... constant,

=

where a, a,...... are constant scalars, the line describes a right cone. 6. Determine the conditions that

may represent a right cone.

Sppp = 0

7. What property of a cone (or of a spherical conic) is given directly by the following form of its equation,

Χ.ιρκρ =0?

8. What are the conditions that the surfaces represented by