1. It is a cone

whose vertex is 0 because it is not altered by writing xp for p. Also it is of the second order in p, occurs in it twice and twice only.

P

since

ρ

2. All the vectors OA, OB, OC, OD, OE lie on its surface. This we shall prove by shewing that if p coincide with any one of them the equation (1) is satisfied.

If Ρ coincide with α, the last term of the left-hand side of the equation, viz. Vpa, becomes Vaa = Va2 = 0, and the equation is satisfied.

If ρ coincide with ß, the left-hand side of the equation be

comes

S. V (VaẞVde) V (Vẞy Veß) V (VydVẞa).........(2).

Now V(VBy Veẞ) = - V (VeẞVẞy), (22. 2), is a vector parallel to B (31. 3), call it mẞ; and

V. {V(VaßVde) V (VydVßa)} = V. {V(VaßVde) V (VaßVyd)}, (22. 2),

= a multiple of Vaß, (31. 3),

= n Vaẞ, say.

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, S. (scalar+nVaß) mß

= mnSB Vaß

= 0,

because Vaß is a vector perpendicular to ẞ.

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

If ρ coincide with γ both the second and third vectors are 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. VẞyVep=nOQ,

V. VydVpa=pOR;

S. V (Vaß Vde) V (VßyVep) V (Vyd√pa) = 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.

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)2 = (x2 + y2 + z2)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 Q1, QQ be the points where they meet the ellipsoid; then

19

3

+

2

1 1 1 = + + a1 b4

OY ̧2. OQ2 + OY2. OQ ̧2* OY3.0Q ̧2

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.

FORMULE AND THEIR APPLICATION.

69.

1.

PRODUCTS of two or more vectors.

Two vectors. The relations which exist between the

scalars and vectors of the

been exhibited in Art. 22.

(a) Saẞ = Sẞa.

(c) aẞ + Ba= 2Saß.

product of two vectors have already

We simply extract them :

These we shall quote as formula (1).

(b) Vaß--Vẞa.

(α) αβ - βα= 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

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