α

y

с

2abc

√(a2b2 +b2c2 + c2a3) *

24

11. 2-2-25.

α b

(x cos By cos a)2 + (y cos y · z cos ẞ)2 + (≈ cos a x cos y)2 = 0;

Ax+By+Cz-D

A'x+By+C'z-D'

13. 30o.

14.

=

Aa+BB+Cy-D ̄ A'a+BB+C'y-D''

15. D'(Ax+ By + Cz) = D (A'x + By +C'z); the condition is

16. (x-a) {m (c — c') − n (b − b')} + &c. = 0.

17. First obtain the equation to the plane which contains the two given lines; this is

x (m ̧n ̧ — m ̧n ̧) + y (n ̧ ̧ − n ̧ ̧) +z (1 ̧m ̧ — 1 ̧m ̧) = 0;

then find the equation to a plane which is perpendicular to the plane just determined and equally inclined to the two given lines. 22. (x-a) (mn' — m'n) + &c. = 0.

21.5/2.

23. The question is indeterminate, since an unlimited number of lines can be drawn as required.

24. Suppose the given plane to be determined by

and

Ax+ By + Cz=D;

suppose that the line of intersection is to lie in the plane of (x, y); then we may assume for the equation to the required plane Ax+ By +λz = D, and determine A suitably.

27. 60°.

28. Four straight lines.

29. The condition is a2 + b2 + c2 + 2abc = 1.

30.

(p, cos a -p cos a,) x + (p, cos ẞ-p cos B1) y + λz = 0, where (p, cosa-p cosa,)2+(p1cos ß-p cos ẞ ̧)2+λ(p ̧cosy—pcosy,) = 0.

1

31. Let the given point be (a, b, c) and the equations to the given planes

Ax+ By + Cz=D, A'x + B'y+ C′z = D′ ;

41. Let r denote the distance between (a, ẞ, y) and (x, y, z);

then

(A2 + B2 + C12) r2 = (A2 + B2 + C2) {(x − a)3 + (y − ß)3 + (≈ − y)3} ={B (z—y) — C (y — ẞ)}2 + {C'′ (x − a) — A(≈ −y)}2 + {A (y—ẞ) − B (x−a)}2 + {A (x − a) + B (y − ẞ) + C (≈ − y)}3;

The last term is constant if (x, y, z) be in the given plane; hence the least value of r3 is obtained by making the other three terms vanish.

44. The exceptional case is when the line of intersection of two of the planes is perpendicular to the third plane.

46. The condition is al + bm + cn= 0; then the equations may

49. There are two such lines determined by the given equation lx+my+nz= = 0, combined with

x √(l2 + n2) = ± y √(m2 + n2).

50. Let x, y, z be the co-ordinates of any point in the line; let l, m, n be proportional to its direction cosines; then it may be shewn that

(nx-lz) (mx-ly)

{(nx − lz)2 + (ny — mz)2} {(mx − ly)2 + (mz — ny)3}

from this we may shew that tan2 a = · A (mz – ny)1 where A is a symmetrical function of x, y, z, l, m, n.

58. The given equation may be written thus, r2 = (l2 + m2 + n2) p2

where r is the distance of the point (x, y, z) from the origin, and p is

the perpendicular from the point (x, y, z) on the plane lx+my+nz=0. If l2 + m3 + n3 is greater than unity the locus is a right circular 1

cone; the cosine of the semivertical angle is √(l2 + m2 + no)

2

If l2 + m2 + n2 is equal to unity the locus is a right line; see example 12. If l2 + m2 + n3 is less than unity the locus is a point, namely the origin. 59. The locus is a sphere if C be an acute angle, a point if C be a right angle, and impossible if C be an obtuse angle.

63. Let (a, ẞ, y) be the external point; then the required equation is

x2

++++ zy

a

c2

2

66. Let (a, ẞ, y) be the given point; then the required equation is

67. A right circular cone. 70. Take for the equation to

the ellipsoid ++=1, and for the equation to the sphere

α

2

(x − h)2 + y2 + z3 = r2; then the equation to the cylinder will be

71. The circle circumscribed about the opposite face may be determined by the equations

A subcontrary section is a circular section. The equation to the cone may be written thus

hence we see that the plane ax + by + cz − k = 0, cuts the cone in a circle. If the vertex of the cone be at the angular point which is at the other end of the edge c, the equation to the cone is

and the plane ax + by - cz=0 is parallel to the subcontrary sections. 74. An hyperbolic paraboloid.

75. The locus is determined by the equation to the ellipsoid combined with xyz = constant.

76. Suppose the fixed point in the axis of 2; then we have to

Eliminate m from these two equations; thus we obtain a quad

ratic in and the product of the roots becomes known. The

n

92. The condition is amn + bnl + clm = 0.

93. When the given conditions hold, the equation becomes (b'c'x+c'a'y + a'b'z)2 + Qa′b′c′ (a′′x+b′′y + c ̋z) — a'b'c'ƒ = 0;

this represents a parabolic cylinder.