PROBLEM XXXVI. To construct the shortest distance between two given straight lines not in the same plane. The principles on which the construction of this problem depends have been already given in Proposition XXXVI. of the Geometry of Planes. Let AB, CD (fig. XXXVI. a) represent the given straight lines. Through AB draw the plane MN parallel to CD; from the point P, in CD, let fall the perpendicular PR on the plane MN; from R draw RS parallel to CD meeting AB in S; and from S draw ST parallel to PR, and therefore perpendicular to MN (Prop. 18), meeting CD in T: ST, which is perpendicular both to CD and AB, is the shortest distance between these lines. We have, therefore, only to effect these constructions on the planes of projection. Let ab, a'b' (fig. XXXVI.) and cd, c'd' be the projections of the given straight lines. Find the traces a, b' of [ab, a'b'] (Prob. I.). Through o, o the projections of a point in the line [ab, a'b'] draw mn, m'n' parallel to cd, c'd'; from m' and n draw m'm and nn perpendicular to xy; and draw the lines ama, n'b'a: these are the traces of the plane passing through [ab, a'b'] parallel to [cd, cd'] (Prob. XII.). From p, p, the projections of a point in the line [cd, c'd'], draw pq, p'q' perpendicular to aa, b'a: these are the projections of a straight line drawn from the point [p, p'], perpendicular to the plane aab' (Prob. XV.). Determine, by Problem XV., the projections r, r' of the intersection of this perpendicular with the plane aab; and draw rs, r's' parallel to cd, c'd', meeting ab, a'b' in s, s': these points are the projections of the intersection with the line [ab, a'b'] of the straight line drawn in the plane aab' through the point [r, r'], parallel to the line [cd, c'd']. Through s, s' draw st, s't' parallel to rp, rp', meeting cd, c'd' in t, t': st, s't' are the projections of the straight line which is perpendicular to the plane aab', and to the two given lines [ab, a'b'] [cd, c'd'], and which is the shortest distance between these lines. Finally, constructing as in Problem XV., s'u' is the true length of this line. SOLUTION OF THE SEVERAL CASES OF THE TRIHEDRAL ANGLE. In a trihedral angle there are six magnitudes, three faces and three inclinations, that is, three plane angles and three dihedral angles. Any three of these being given, the others may be determined. The data admitting only of six different combinations, there are only six cases for solution; viz. when there are given: 1. The three faces; or the three plane angles forming the trihedral angle. 2. Two faces and their inclination to each other; or two of the plane angles and the dihedral angle formed by their planes. 3. Two of the faces and the inclination of the third face to one of them; or two of the plane angles and the dihedral angle opposite to the plane of one of them. 4. The three inclinations; or the three dihedral angles about the trihedral angle. 5. The inclinations of two of the faces to the third face, and that third face; or two of the dihedral angles and the plane angle in the face common to them. 6. The inclinations of two of the faces to the third face, and one of those two faces; or two of the dihedral angles and the plane angle in the face opposite to one of them. Since the plane angles forming a trihedral angle are the supplements of the opposite dihedral angles in the supplemental trihedral angle, and its dihedral angles are the supplements of the plane angles forming that trihedral angle (Prop. 46), the last three cases may be transformed into the first three in the supplemental trihedral angle, by taking the supplements of the given faces or plane angles as the inclinations or dihedral angles in another trihedral angle, and the given inclinations or dihedral angles as the supplements of the faces or plane angles forming this trihedral angle: the supplements then of the magnitudes to be determined in the supplemental trihedral angle will be those required in the original. We have, therefore, only to consider the first three of these six cases. Their solutions depend upon the methods which have been employed in some of the immediately preceding problems, namely, considering one of the faces of the trihedral angle as the horizontal plane, in turning down upon this plane the other faces of that angle, and likewise other planes which are required in the solution. PROBLEM XXXVII. Given the three faces or plane angles of a trihedral angle, to find the inclinations of these faces, or the dihedral angles formed by them. Taking the plane of one of the faces as the horizontal plane, in that plane make the angle asb (fig. XXXVII.) equal to the given plane angle of that face, and conceiving the other faces to be brought down upon this plane, by turning about sa and sb, let ase and bsc be the given angles thus turned down. It is evident that the trihedral angle would be reconstructed by turning the face asc about the edge sa, and the face bsc' about the edge sb until the two sides sc, sc coincide and form the third edge of that angle. In sc and sc take sf and sf' equal to each other, and draw fgh and f'g'h perpendicular to sa and sb meeting each other in h. During the rotation of the faces asc, bsc', the points ƒ and f' will describe circles about the centres g, g', in vertical planes of which the traces in the fixed plane asb are gh and g'h; and when sc and sc' again coincide, these points will coincide in a point in space. The lines fg and f'g' will then form with gh and g'h angles which are the inclinations of the lateral faces upon the horizontal face asb (Def. 7). In order to determine these inclinations, it is only necessary to turn down the vertical planes described by fg, fg', upon the horizontal plane. Turning the first about gh, the circle described by fg will be turned down upon a circle fk described from the centre g at the distance gf. The point h being common to the two vertical planes, their intersection will be a vertical line drawn from h; and this will be turned down upon hk, perpendicular to hg. The point F, in which the points f, f' unite in space, must be upon this intersection, and must therefore be turned down upon the point k, where the circle fk is intersected by the perpendicular hk. Joining therefore gk, the angle kgh will be the inclination of the faces asb, If the vertical plane described by f'g' be made to turn about g'h, we find, by a similar construction, the angle k'g'h which is equal to the inclination of the faces bsc', asb. asc. The inclination of the faces asc, bsc may be found in a similar manner, by taking one of these planes as the fixed plane, and turning the other down upon this plane; but it may be more conveniently obtained in the following manner. Conceive a plane perpendicular to the third edge in which sc and so unite, to pass through the point F in that edge; it will cut the lateral faces in two straight lines which will contain an angle equal to the inclination sought (Defs. 3, 7). One of these on the face asc, turned down, is fp perpendicular to sc; and the other upon the face bsc', turned down, is fg perpendicular to sc'. The points p and q where these lines meet the edges sa, sb not having changed their position, pq will be the trace upon the plane asb, of the plane containing the angles sought; and this angle will be the angle opposite to the side pq in a triangle of which the other two sides are pf and qf'. This triangle pFq brought down upon the plane asb by turning about pq will have its vertex in m, the intersection of circles described from the centres Ρ and at the distances pf, qf; and joining pm, qm, the angle pmq will be equal to the inclination of the two faces asc, bsc of the trihedral angle. Remark. In this construction we have to notice several verifications. 1. The lines hk, hk', being the same vertical line hF turned down upon the plane asb, ought to be equal. 2. Since pq is the intersection of the planes psq, pFq which are each perpendicular to the projecting plane of the third edge upon the plane asb, it is perpendicular to that projecting plane (Prop. 34), and therefore to shn, the projection of the third edge, meeting it in that plane (Def. 3). And since pq is perpendicular to the projecting plane sFn, it is perpendicular to Fn, the line drawn from F, in the plane pFq, perpendicular to the third edge; consequently when Fn is turned down upon the plane asb, the point F being at m, mn will be perpendicular to pq and be in the same straight line as sn; sh is therefore perpendicular to pq, and being produced passes through m. 3. Producing gh to i, and joining if', ik, these lines are equal, since they are the same line iF brought down, in two different ways, upon the horizontal plane; the first by the triangle iFg', turning about is, and being brought down as if'g'; the second, by the triangle iFh, turning about gi, and being brought down as ikh. Similarly, producing g'h to ', and joining if, i'k', these lines are equal. PROBLEM XXXVIII. Given two faces of a trihedral angle, and their inclination to each other; or two of the plane angles and the dihedral angle contained by their planes; to find the third face and its inclinations to the given faces. Let asb, asc (fig. XXXVIII.) be the two given faces brought down upon the same plane, by turning about as. If fgi be drawn perpendicular to sa, fg and gi will be the intersections of the faces asc, asb with the plane passing through ƒ or F, perpendicular to sa; and the plane asc being in its true position, gf and gi will contain an angle equal to the given inclination. Making then the angle igk equal to the given inclination, if the plane of the angle contained by gf and gi turn about gi, so that this angle be brought down upon the plane asb, it will coincide with igk. Making gk equal to gf, the point for F will be brought down on k; and the perpendicular from F on the plane asb will coincide with kh perpendicular to gi. From h, the foot of this perpendicular, drawing hg'f' perpendicular to sb; describing a circle from the centre s at the distance sf, to intersect g'f' in f'; and drawing sf'c'; bsc will be the third face of the trihedral angle. For this face being in its true position, h is the foot of the perpendicular to the plane asb, from the point F in the edge which is the intersection of the face asc with the third face; and this face being brought down upon the plane of asb, by turning about sb, the point F will describe a circle in a plane perpendicular to sb, and will be in the line g'f', at a distance from s equal to sF or sf. The third face bsc being found, its inclinations to the other faces are found by the last problem. Remark. Joining ik, if', we must have ik equal to if'. PROBLEM XXXIX. Given two faces of a trihedral angle, and the inclination of the third face to one of them, to find the third face and the other inclinations. Let asb, asc (fig. XXXVII.) be the two given faces, turned down upon the same plane, and let the inclination of the third face to asb be given. Conceive the faces of the trihedral angle in their true position. Let a vertical plane pass through any point g in sa, perpendicular to sa, intersecting asb in gi, perpendicular to sa, and asc in a line also perpendicular to sa, and which by the revolution of the face asc about sa will describe a circle of which the radius is gf. Let another vertical plane also pass through g perpendicular to sb, intersecting asb in go perpendicular to sb, and the third face in a line also perpendicular to sb; these perpendiculars to so will contain the given angle of inclination of the third face to the face asb. These two vertical planes passing through g will intersect in a vertical line (Prop. 34), which with go and the other perpendicular to sb will form a rightangled triangle, having the angle at o equal to the given angle of inclination. If this triangle be brought down upon the plane asb, by turning about og, it will be determined by making the angle gom equal to the given inclination, and drawing gm perpendicular to go; so that m is the point of intersection of the plane of the third face with the vertical line passing through g, brought down upon the plane asb. If the other vertical plane, which passes through gi, be brought down upon the plane asb, by turning about gi, the above vertical line passing through g will be brought down upon ga, and its intersection with the plane of the third face will be in n, making gn equal to gm; the line joining this intersection and i, brought down upon asb, will be ni; and the circle described in the vertical plane passing through gi, by the point ƒ in the edge sc, brought down upon asb, will be the circle qpf whose centre is 9. The points n and i being in the plane of the third face, and the circle qpf being described by a point in the edge sc, the intersections p, q, of ni with the circle, are points common to the face asc and the third face. The plane fni being replaced in its true position, at right angles to asb, either of the lines joining s and p, or s and q, in space, will be the third edge of a trihedral angle of which two faces are asb and asc, and the inclination of the third face to asb is the angle mog. Describing therefore circles from i as a centre, at the distances ip, iq, intersecting in p', q' the circle described at the distance sf from the centre s; and joining sp', sq'; either of the angles bsp' or bsq' will be the required third face, brought down upon the plane of asb, by turning about sb. Remarks.-1. When the circle fpq cuts the line ni in two points P, q it appears that there are two solutions to the problem. When VOL. II. G |