From these, at once, x, since dx and dt are indeterminate. Thus the equation of the envelope is

p = at + f (at)

= £ (at + 2/2),

the hyperbola as before; a, ß 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

p = at (1−x)+ß 2,

dp

= 0 = ad (t (1 − x)) + ßd (†) ·

Hence, as a is not parallel to ß, we must have

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

t and t1 being given, while x and x1 are indeterminate.
At the point of intersection of these lines we have (§ 26),

and thus, for the ultimate intersections, where = 1,

or the intersection lies in the diagonal of the parallelogram on a, ß.

COR. (2). If t1 = mt, where m is constant,

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.

COR. (3). If we take

tt1 (t+t1) = constant the locus is a parabola; and so on.

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 processes of Grassmann's Ausdehnungslehre and Möbius' Barycentrische Calcul. Some very curious investigations connected with curves and surfaces of the second and third orders 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 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, Ce meet in a point, the intersections of AB, ab, of BC, bc, 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 0+√1 sin 0

turns any radius of a given circle through an angle in the positive direction of rotation, without altering its length, deduce the ordinary formulæ for cos (A+B), cos (A−B), sin (A+B), and sin (A–B), 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 tangent at the point where it meets the curves bisects the intercepts of 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.

11. Shew that p = x2 a+ y2 ẞ+(x+y)2 y

is the equation of a cone of the second degree, and that its section by the plane pa + q B + ry

P =

p + q + r

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

12. 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.)

represents a cone of the third order, and that its section by the plane pa + q B+ ry

Ρ

=

p + q + r

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

CHAPTER II.

PRODUCTS AND QUOTIENTS OF VECTORS.

45.] WE now come to the consideration of points in which the Calculus of Quaternions differs entirely from any previous mathematical method; and here we shall get an idea of what a Quaternion is, and whence it derives its name. These points are fundamentally involved in the novel use of the symbols of multiplication and division. And the simplest introduction to the subject seems to be the consideration of the quotient, or ratio, of two vectors.

46.] If the given vectors be parallel to each other, we have already seen (§ 22) that either may be expressed as a numerical multiple of the other; the multiplier being simply the ratio of their lengths, taken positively if they are similarly directed, negatively if they run opposite ways.

47.] If they be not parallel, let OA and OB be drawn parallel and equal to them from any point 0; and the question is reduced to finding the value of the ratio of two vectors drawn from the same point. Let us try to find upon how many distinct numbers this

ratio depends.

We may suppose OA to be changed into OB by the following

processes.

1st. Increase or diminish the length of OA till it becomes

equal to that of OB. For this only one number is required, viz. the ratio of the lengths of the two vectors. As Hamilton remarks, this is a positive, or rather a signless, number.

2nd. Turn OA about O until its direction coincides with that of OB, and (remembering the effect of the first operation)