APPENDIX.

WE have thought it would be acceptable to many students if we should give as an Appendix a brief, and in some cases even a detailed, solution of the most important and most difficult of the ADDITIONAL EXAMPLES. In doing so, we would add as a word of advice, that our solutions be employed simply for the purpose of comparison with those which shall occur to the student himself.

CHAP. II.

Ex. 4. If AB=a, BC=ß, AP = ma, AP' = m'a, BQ =mß, &c.; then

AE AP+xPQ = AP' + x'P'Q'

gives ma+x{(1 − m) a + mß} = m'a + x' { (1 − m') a + m'ß},

whence

x=m', and PE=m'PQ.

Ex. 6. ABCD is a quadrilateral; AB-a, AC=f, AD = y,

gives (1 − m) a + m (ẞ − a) + (1 − m) (y — ß) — my = 0,

or

(1-2m) (a− B + y) = 0;

an equation which is satisfied either when 1-2m = 0, or when a-B+y=0.

The former solution is Ex. 5; the latter gives ABCD a parallelogram.

Ex. 10. Let a, b, c be the points in which the bisectors of the exterior angles at A, B, C meet the opposite sides. Let unit

T. Q.

14

vectors along BC, CA, AB be a, ß, y; then with the usual nota

and also

(b −c) Aa+ (c− a) Ab + (a − b) Ac = 0,

(b−c) + (ca) + (a−b) = 0,

therefore (Art. 13) a, b, c are in a straight line.

Ex. 12. If the figure of Ex. 11, Art. 23, be supposed to represent a parallelepiped; then, with the notation of that example,

1

the vector from 0 to the middle point of OG is (a+ẞ +d),

2

which is the same as the vector to the middle point of AF, viz.

Ex. 13. With the figure and notation of Art. 31, the former part of the enunciation is proved by the equation

Also, if the edges AB, BC, CA be bisected in c, a, b, the mean point of the tetrahedron Oabc is evidently

which proves the latter part of the enunciation.

Ex. 14. Here we have to do with nothing but the triangles on each side of OD.

If OQ = a, QA =pa, AP = ß, PD = qß;

TO=xOD=TQ-OQ=yQP-OQ

Similarly, if OS = a', SB = p'a', BR = f', RD = q'B';

Ex. 15. If AB=a, AC = ß, MN = pa, PQ=qß, RS=r (B− a), we shall have, by making AO=AP+PO = AR + RO,

therefore

(1 − q) a + (1 − p) B = ra + (1 − p) (ß − a) ;

p+q+r=2.

Ex. 17. Let RA = a, RB = ß, AP=ma, AD=pa + qß; then

PD=pa+qẞ-ma,

and RSRP + PS= RQ+ QS gives

(1 +m) a +x (pa + qß − ma) = (1 + m) B + y (pa + qß —mß),

CHAP. III.

Ex. 5. Let ABCD be the quadrilateral; DA, DB, DC, a, ß, y respectively.

Now ẞ (y − a) + (y − a) ß = y (B − a) + (B − a) y

Taking scalars, and applying 22. 3, there results,

SB (y − a) = Sy (B − a) + Sa (y − ß),

which is the proposition.

Ex. 6. If a, ẞ, y be the vectors OA, OB, OC corresponding to the edges a, b, c; we have

the negative square of which is the proposition given.

Ex. 7. If Sa (B-7)=0 and Sẞ (a-7)=0, then, by subtraction, will Sy (a-B)= 0.

Ex. 8. If a2= (B − y)2, ß2 = (y − a)2, y = (a-ẞ); then will

for these are the same equations in another form; and they prove that the corresponding vectors are at right angles to one another.

Ex. 9. If OA, OB, OC, OD are a, ẞ, y, 8;

triangle DAB DAC :: tetrahedron ODAB : ODAC

:

:: Χαβδ : Παγδ

:: triangle OAB : OAC,

because the angles which 8 makes with the planes OAB, OAC are equal.

CHAP. IV.

Ex. 1. Let be the middle point of the common perpendicular to the two given lines; a, a, the vectors from 0 to those lines, unit vectors along which are ẞ, y; p the vector to a point P in a line QR which joins the given lines; P being such that RPmPQ; therefore

Now since a is perpendicular to both ẞ and gives (1+m) Sap = (m-1) a; a plane.

Ex. 2. Retaining what is necessary of the notation of the

last example, let OS = d.

If PR perpendicular on y meet ß in Q, we have

− a + yy + RP = p, which gives yy2 = Syp ;

RQ=2a+xß-yy, which gives yy2 = xSẞy;

which being of the second degree in p shews that the locus is a surface of the second order. See Chap. VI.

Ex. 3. The equation of the plane is

Syp = a,

which, being substituted in the equation of the surface, gives what is obviously the equation of a circle.

Ex. 4. With the notation of Ex. 1, let 8, 8' be the perpendiculars on the lines,

then

p+d=a+xß gives VBS-- VB (p-a),

and the condition given may be written