Hence the product of the first and third vectors in expression (2) becomes

scalar + n Vaß,

and the second is mß; therefore expression (2) becomes, by 31. 2,

because Vaß is a vector perpendicular to ß.

Equation (1) is therefore satisfied when p coincides with B.

If Ρ both the second and third vectors are coincide with γ parallel to ẞ (31. 3); therefore their product is a scalar, and equation (1) is satisfied.

The other cases are but repetitions of these.

Hence equation (1) is satisfied if p coincide with any one of the five vectors a, ß, y, d, e; i.e. OA, OB, OC, OD, OE are vectors on the surface of the cone.

3. Let F be the point in which OX cuts the plane ABCDE; then ABCDEF are the angular points of a hexagon inscribed in a conic section.

4. Let the planes OAB, ODE intersect in OP; OBC, OEF in OQ; OCD, OFA in OR; then

therefore

V. VaẞVde = mOP, (31. 4),

V. Vẞy Vep=nOQ,

V. VydVpa=pOR;

S. V (Vaß Vde) V (VßyVep) V (VydVpa) = mnpS(OP.OQ.OR); hence equation (1) gives

S (OP.OQ.OR)=0,

or (31. 2. Cor. 2) OP, OQ, OR are in the same plane.

Hence PQR, the intersection of this plane with the plane ABCDEF is a straight line. But P is the point of intersection of AB, ED, &c.

Therefore, the opposite sides (1st and 4th, 2nd and 5th, 3rd and 6th) of a hexagon inscribed in a conic section being produced meet in the same straight line.

COR. It is evident that the demonstration applies to any six points in the conic, whether the lines which join them form a hexagon or not.

ADDITIONAL EXAMPLES TO CHAP. VIII.

1. Find the locus of a point, the ratio of whose distances from two given straight lines is constant.

2. Find the locus of a point the square of whose distance from a given line is proportional to its distance from a given plane.

3. Prove that the locus of the foot of the perpendicular from the centre on the tangent plane of an ellipsoid is

(ax)2 + (by)2 + (cz)3 = (x2 + y2 +z3)3.

4. The sum of the squares of the reciprocals of any three radii at right angles to one another is constant.

5. If Oy,, Oy, Oy, be perpendiculars from the centre on tangent planes at the extremities of conjugate diameters, and if QQQ be the points where they meet the ellipsoid; then

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6. If tangent planes to an ellipsoid be drawn from points in a plane parallel to that of xy, the curves which contain all the points of contact will lie in planes which all cut the axis of ≈ in the same point.

7. Two similar and similarly situated ellipsoids intersect in a plane curve whose plane is conjugate to the line which joins the centres of the ellipsoids.

8. If points be taken in conjugate semi-diameters produced, at distances from the centre equal to p times those semi-diameters respectively; the sum of the squares of the reciprocals of the

perpendiculars from the centre on their polar planes is equal to p3 times the sum of the squares of the perpendiculars from the centre on tangent planes at the extremities of those diameters.

9. If P be a point on the surface of an ellipsoid, PA, PB, PC any three chords at right angles to each other, the plane ABC will pass through a fixed point, which is in the normal to the ellipsoid at P; and distant from P by

where p is the perpendicular from the centre on the tangent plane at P.

10. Find the equation of the cone which has its vertex in a given point, and which touches and envelopes a given ellipsoid.

CHAPTER IX,

FORMULÆ AND THEIR APPLICATION.

69. PRODUCTS of two or more vectors.

1. Two vectors. The relations which exist between the scalars and vectors of the been exhibited in Art. 22. (a) Saß = Sẞa.

product of two vectors have already We simply extract them :

(c) aß + Ba= 2Saß.

These we shall quote as formulæ (1).

(6) Ταβ== βα.

(α) αββα=2 Γαβ,

2. We may here add a single conclusion for quaternion products.

Any quaternion, such as aß, may be written as the sum of a scalar and a vector. If therefore q and r be quaternions, we

may write

therefore

and

q=Sq + Vq,

r = Sr+ Vr;

gr = SqSr+ Sq Vr + SrVq + VqVr,

S.qr=SqSr+S. VqVr,

V.qr=SqVr+SrVq + V. VqVr,

where S. VqVr is the scalar part, and V. VqVr the vector part of the product of the two vectors Vq, Vr.

If now we transpose q and r, and apply (a) and (b) of formulæ 1, we get

S. qr = S. rq

V.gr + V. rq = 2 (Sq Vr + Sr Vq

3. Three vectors. By observing that S. ySaß is simply the scalar of a vector, and is consequently zero, we may insert or omit such an expression at pleasure. By bearing this in mind the reader will readily apprehend the demonstrations which follow, even in cases where we have studied brevity.

The formulæ marked (3) shew that a change of order amongst three vectors produces no change in the scalar of their product, provided the cyclical order remain unchanged.

This conclusion might have been obtained by a different process, thus:

In (2) let q = aß,. r = y, there results at once

or a cyclical change of order amongst three vectors changes the

sign of the scalar of their product.