The inclination of a straight lime 1 to each of the axes : ta cos 1,8 = (1 + a2 + b) +5 cos l3y= (1 + a2 +69)1' +1 cos 1,7 = (1 + a® +62) (cos 1,ix) + (cos 1,y) + (cos 1,x) = 1. If lyxy, &c. denote the angles which I makes with the planes xy, &c. then (sin lyxy)+ (sin lyxz)' + (sin lyx)* =1. The mutual inclination of two lines, l, le: cos ligle = cos 119X.cos lygx + cos lı,y.cos l2,y + cos 11,7.cos l25%. 1+aja, +b, bę + (1 + a,* +6,*)(1 + a, + 6,397 The equations to a line in terms of the angles it makes with the axes are cos lyx cos lny %+ ß. cos 1,7 cos 1,2 (G. G. A. Ch. xvi; H. A. G. 229–51; Biot, 54-69.) (26.) The plane. The most general form of the equation to a plane is A x + By + Cx+D=0, or x= A,X + B y + C1. The equations to the traces of the given plane on the planes of xy, xz, yz, respectively are AX + By+D=0, A x + Cx+ D=0, By + Cx+D=0. If a, b, c, are the distances from the origin at which the plane cuts the axes of x, y, %, respectively, the equation becomes y = If Z = The equation to a plane passing through the point (19913x_), and parallel to the plane Ax + By + Cx+D=0, is A(x — «) + B(y-yl) + C(x– x)=0. y=bx + ß and the plane Ax +By+Cx+D=0, are a(Aa + B6+C) – a(Aa + BB+ D) Aa + Bb + c Aa + Bb +C Aa + Bb C Aa + Bb+C=0. B y-y= -; (A + Bo + C2) The equation to a plane drawn through a point (X1,91,1) perpendicular to a given line, is a (x — «1) +b(y-yı) +%—%=0. The angle contained between the line l, and the plane P; Aa + Bb+C sin 1,P= (1 + a2 + b)(A + B? + C) (27.) The inclination of a plane P to the co-ordinate planes : +C cos Pwy= (A + B2 + C'?), A1 Br cos P,yx= (A® + B? + C)} (cos Py) + (cos P,xz)2 + (cos P,yx)?=1. The mutual inclination of two planes P1, P2; cos P1,P,=cos Powy.cos P29XY + cos P 19xx. cos P 298 % + cos P 1,7x.cos P2Yz. A,A, + B, B, +C,C, (A,? + B,” + C*)(A,? + B +0,2) B C C, (H. A. G. 252—74; Biot, 70–85; G. G. A. Ch. xvii.) THE ORTHOGONAL PROJECTION OF PLANE FIGURES. (28.) Let A represent the area of a plane figure, Ary, Ag, Ayes the areas of its projections on the planes xy, xx, yx respectively, then 4., +42 +4,2 = A. Ay,;, = Ay cos X19X + A, COS XY + A,, cos X1,2, Y1X1, X 1% X 19ı : then X,= X cos X19X + Y cos X1,Y + 2 cos X1,%, Y = X cos Y 1X + Y cos Yı,y + Z cos Y19%, X2 + Y? + Z2 = X + Y + 2%: If the triangle formed by joining any three points in space be represented by a, and the perpendicular on the plane from the origin, by p, the equation to the plane passing through the three points is ays. + arzy + aryX=ap. (H. A. G. 275–90; G. G. A. p. 327–32.) new . . . . OBLIQUE CO-ORDINATES. (29.) The equations to the line ( referred to oblique coordinates are % + a, %+ß. sin 1,298 sin brzyd y The distance of the point (x,y,z) from the origin = x2 + y + x2 + 2(xy cos x,y + x x cos x,x+yx cos y,x). The distance between the two points (27,41,71), (X2392,%,) =(2, - x) + (,- y.) + (2,-2) + (yı- y2)(x1 - x) cos y,x}. sx=0,7 + a, (1.), (12): ty=b3 +8. ly=b2x + ß sin lyx sin x,y% sin 11,4% sin a',y% sin 10,0 % sin lXY cos 11%, cosly + sin %,XY sin lyx x sin llwy cos 1.,%; cosloy + sin Y,& % sin %9X Y cos 1 + dyda+b7b2+1+(a,b2 + a2b)cos x,y+(a + a2)cos x,x+(62 +62)cosy, {a, + b +1+2 (a, b, cos #3y + a, cos x,x+b, cos y,x)} x {a, + b3 +1+2 (a,b,cos x,y + a, cos x,x+ b, cosy,x)} The inclination of a plane P to each of the co-ordinate planes ; let p be the perpendicular from the origin, then the equation to the plane is p=X COS Px + y cosp,y + % CoS pox, 1 cos Y ZNY C, sin % XY COS X 2,8 Y A+ sin Y,8 % COS XY,Y% COS X Y,8 % B. sin y,XX sin x,xy C, sin x,y% sin 2,XY sin 2,4% B + sin Y,8 % -24.5. cos P,7% cos P,xx cos P,wy C=1. sin %,xy Let / xy, = a, xyyyx=ß, 2 X ,yx = y; sin x,yx=d, sin y,xx = e, sin x,xy=f: then A B2 C2 AB AC BC + + 2 COS cos ß + do f2 de df ef The inclination of two planes Ax+By+Ciz=pı, (P) A2X + B,y + C,x=po; (P2) - cos P1,P,= A,A,, B,B2, CC, A,B,+A,B A,C,+ A,C, B,C,+B,C + + ď e de ef B,C 2 COS Y + cosB + ef A.C. B,C, cos y + cos B+ e? f2 de df ef (Whewell, Camb. Phil. Trans. V. 2, Pt. 1; H. A. G. Ch. ix, x.) 1 cos cos y Х d THE TRANSFORMATION OF CO-ORDINATES. 1 (x, sin x 1,4x + yi sin Y14x + x, sin %17%), 1 (x, sin xy + y, sin yıxy + x, sin x wy). X= sin my If the primitive axes are rectangular, and the new ones X = X, COS X 1920 + yu cos Y198 + %1 COS %1,9, (1) %= X 1 COS X 1,7 + Y, cos Y 1,7 + %, COS %19%: |