156. To find the volume of a parallelepiped in terms of its edges and their inclinations to one another.

A

B

Let the edges be OA=a, OB=b, OC = c; let the inclinations be BOC=a, COA = ẞ, AOB = y.

plane AOB meeting it at E.

Draw CE perpendicular to the Describe a sphere with O as a

centre, meeting OA, OB, OC, OE at a, b, c, e respectively.

The volume of the parallelepiped is equal to the product of its base and altitude = ab sin y. CE = abc sin y sin cOe. The spherical triangle cae is right-angled at e; thus

sin coesin coa sin cae = sin ẞ sin cab,

and from the spherical triangle cab

√(1- cos2 a cos2 B - cos2 y + 2 cos a cos ẞ cos y)

sin ẞ sin y

therefore the volume of the parallelepiped

= abc √(1 − cos3 a - cos3 B-cos3 y + 2 cos a cos ß cos y).

;

157. To find the diagonal of a parallelepiped in terms of the three edges which it meets and their inclinations to one another.

Let the edges be OA = a, OB=b, OC=c; let the inclinations be BOC=a, COA=ß, AOB=y. Let OD be the diagonal required, and let OE be the diagonal of the face OAB. Then

OD2 = OE2 + ED2 + 20E. ED cos COE

=a+b2+2ab cos y+c+2c0E cos COE.

Describe a sphere with O as centre meeting OA, OB, OC, OE at a, b, c, e respectively; then (see Example 14, Chap. iv.)

therefore OD = a3 + b2 + c2 + 2ab cos y + 2bc cos a + 2ca cos B.

158. To find the volume of a tetrahedron.

A tetrahedron is one-sixth of a parallelepiped which has the same altitude and its base double that of the tetrahedron ; thus if the edges and their inclinations are given we can take one-sixth of the expression for the volume in Art. 156. The volume of a tetrahedron may also be expressed in terms of its six edges; for in the figure of Art. 156 let BC = a', CA=b', AB = c'; then

and if these values are substituted for cos a, cos B, and cos in γ

the expression obtained in Art. 156, the volume of the tetrahedron will be expressed in terms of its six edges.

The following result will be obtained, in which V denotes the volume of the tetrahedron,

144 V2 = — a's b'2 c'2

12

12

+ a2 a'3 (b'2 + c'3 — a'3) + b2b′2 (c'2 + a'2 — b′2) + c3 c22 (a'2 + b′3 — c′2)

12

—- a'3 (a2 — b2) (a2 — c2) — b′3 (b3 — c2) (b3 — a3) — c′3 (c2 — a2) (c2 — b3).

-

Thus for a regular tetrahedron we have 144 V2 = 2ao.

159. If the vertex of a tetrahedron be supposed to be situated at any point in the plane of its base, the volume vanishes; hence if we equate to zero the expression on the right-hand side of the equation just given, we obtain a relation which must hold among the six straight lines which join four points taken arbitrarily in a plane.

Or we may adopt Carnot's method, in which this relation is established independently, and the expression for the volume of a tetrahedron is deduced from it; this we shall now shew, and we shall add some other investigations which are also given by Carnot.

It will be convenient to alter the notation hitherto used, by interchanging the accented and unaccented letters.

160. To find the relation holding among the six straight lines which join four points taken arbitrarily in a plane.

Let A, B, C, D be the four points. Let AB=c, BC = a, CA =6; also let DA a', DB=b', DC = c'.

=

If D falls within the triangle ABC, the sum of the angles ADB, BDC, CDA is equal to four right angles; so that

cos ADB = cos (BDC + CDA).

Hence by ordinary transformations we deduce

1 = cos2 ADB + cos2 BDC + cos2 CDA-2 cos ADB cos BDC cos CDA. If D falls without the triangle ABC, one of the three angles at D is equal to the sum of the other two, and the result just given still holds.

12

Now cos ADB

=

a22 + b22 - c2
2a'b'

and the other cosines may be

expressed in a similar manner; substitute these values in the

above result, and we obtain the required relation, which after reduction may be exhibited thus,

+a'2 a2 (b2 + c2 − a2) + b22 b2 (c2 + a2 − b2) + c'2 c3 (a2 + b2 − c3)

· a2 (a'2 — b′2) (a'2 — c'2) — b2 (b′2 — c'2) (b′2 — a′3) — c2 (c'2 — a'3) (c'3 — b′2).

161. To express the volume of a tetrahedron in terms of its six edges.

Let a, b, c be the lengths of the sides of a triangle ABC forming one face of the tetrahedron, which we may call its base; let a', b', c' be the lengths of the straight lines which join A, B, C respectively to the vertex of the tetrahedron. Let p be the length of the perpendicular from the vertex on the base; then the lengths of the straight lines drawn from the foot of the perpendicular to A, B, C respectively are (a"-p3), √(b' -p3), √(c"-p"). Hence the relation given in Art. 160 will hold if we put √(a'—p3) instead of a', √(b'* -p3) instead of b', and √(c22 -p2) instead of c'. We shall thus obtain

p3 (2a3b2 + 2b3c2 + 2c3 a3 — a* — b* — c1) = — a3b3c2

12

+a'3 a2 (b2 + c2 — a2) + b′2b3 (c2 + a2 − b3) + c23c2 (a2 + b2 — c3)

- a2 (a'2 — b′2) (a'2 — c′2) — b3 (b′2 — c′2) (b'3 — a'3) — c2 (c'2 — a'2) (c'2 — b'2).

The coefficient of p2 in this equation is sixteen times the square of the area of the triangle ABC; so that the left-hand member is 144 V2, where V denotes the volume of the tetrahedron. Hence the required expression is obtained.

162. To find the relation holding among the six arcs of great circles which join four points taken arbitrarily on the surface of a sphere.

Let A, B, C, D be the four points. Let AB=y, BC = a, CA =B; let DA = a', DB = B', DC = y'.

As in Art. 160 we have

1= cos ADB + cos3BDC + cos2 CDA - 2 cos ADB cos BDC cos CDA.

Now cos ADB

cosy-cos a' cos B'
sin a' sin ẞ'

and the other cosines

may be expressed in a similar manner; substitute these values in the above result, and we obtain the required relation, which after reduction may be exhibited thus,

1 = cos3 a + cos3 B+ cos3 y + cos3 a' + cos3 B' + cos3 y'

cos3 a cos2 a' - cos2 ß cos B'-cos y cos3 y' - 2 (cos a cos ẞ cos y + cos a cos ẞ' cos y'

+ cos ẞ cos a' cos y' + cos y cos a' cos B')

+ 2 (cos a cos ẞ cos a' cos ẞ' + cos ẞ cos y cos B'cos y

+ cos y cos a cos y' cos a').

163. To find the radius of the sphere circumscribing a tetrahedron.

Denote the edges of the tetrahedron as in Art. 161. Let the sphere be supposed to be circumscribed about the tetrahedron, and draw on the sphere the six arcs of great circles joining the angular points of the tetrahedron. Then the relation given in Art. 162 holds among the cosines of these six arcs.

Let r denote the radius of the sphere.

Then

and the other cosines may be expressed in a similar manner. Substitute these values in the result of Art. 162, and we obtain, after reduction, with the aid of Art. 161,

4 x 144 V2 =

2a2b2a'b'2 + 2b3c2b′2c'2 + 2c2a2c'2a'2 — a*a'* — b1b'* — c*c'*.

The right-hand member may also be put into factors, as we see by recollecting the mode in which the expression for the area of a triangle is put into factors. Let aa' + bb' + cc' = 20; then

36 Vσ (o-aa') (o-bb') (o- cc').

=