Fig. 12. Fig. 13. Find the intersection of the proposed plane with one of the projecting planes of the given line; this line of intersection, being in cach of these planes, will meet the given line at the point at which it pierces the proposed plane. The several steps of this solution may be performed from what was said in article 26. Let (N"O, M'O) (fig. 12) be the given plane, and (QQ", WM') the given line ; N"NM' will be the plane which projects this line upon the horizontal plane: M' and N' are two points in the common section of this plane with the given plane, N"M is the vertical projection of this common section, and P’ is the vertical projection of the point of meeting of this common section with the proposed line; that is, P" is the vertical projection of the point at which the given line pierces the given plane. If then we draw P"P" perpendicular to the ground line, and meeting the horizontal projection of the given line, in the point P', this point will be the horizontal projection of the point sought. 34. PROBLEM. Having the vertical and horizontal traces of a plane, to find, for any point in the horizontal plane, the height of the corresponding point in the proposed plane. Let (GN", GG) (fig. 13) be the given plane, and M’ the given point in the horizontal plane ; through M' pass a vertical plane whose horizontal trace shall be parallel to that of the given plane ; the proposed point will be in this auxiliary plane, and therefore in its intersection with the given plane, which intersection will be parallel to the horizontal plane (27); the point N’ will be a point in this intersection, and consequently the line N"N will measure the height of the required point above the horizontal plane. M" is the vertical projection of this point, and M' its horizontal projection (7). 35. To find the angle which the inclined plane (GG', GN") makes with the horizontal plane BAC; we suppose, from the required point M in the inclined plane, and from its projection M' in the horizontal plane, perpendiculars drawn to the intersection GG' of these planes; these will form the right-angled triangle MMG', of which we know the two sides MM', M'G', and which may therefore be readily constructed. The angle MG'M' will be the inclination required. 36. Having the horizontal trace of the inclined plane, Fig. 13. and the angle of its inclination to the horizontal plane, we may find the height MM’ of the point M above the horizontal plane. Through the point M' draw WM’ parallel to GG' the horizontal trace of the given plane : draw M'G' perpendicular to GG', and make the angle MGM equal to the given angle; this will give us MM’ the height of the point M. 37. Having the horizontal trace of the inclined plane, and its inclination to the horizontal plane, to find the vertical trace. Find the height of this plane above the point M', by the last article ; and through the point N. where M’N, parallel to the horizontal trace, meets the vertical plane, draw NN" perpendicular to the ground line; and, taking WN" equal MM, draw GN", which will be the vertical trace of the proposed plane. 38. ProBLEM. Through a given point to draw a plane parallel to a given plane. We know that the traces of the required plane will be parallel respectively to those of the given plane (29); they are therefore easily drawn, if we can find upon either of the co-ordinate planes one point of the required plane. Suppose, through the proposed point, a straight line parallel to the horizontal trace of the required plane; it will lie wholly in this plane, and will therefore meet the vertical plane in the vertical trace of the required plane, Let (MM', MM") (fig. 14) be the given plane, and Fig. 14. (P', P,") the given point. Through P' draw PE parallel to M'MI; draw EE" perpendicular to the ground line, making EE's equal to PP"; and draw WE" parallel to MM", and NN' parallel to MM’. These two lines WN", NN", are the traces of a plane passing through P (given by its projections, P., P’) parallel to the given plane (MM', MM"). 39. Let (MM', MM") (fig. 15) be an inclined plane, Fig. 15. to which the line LM is supposed perpendicular. Suppose a vertical plane L'EM' to pass through this line cutting the inclined plane in EM'. This auxiliary plane, being perpendicular to the inclined plane and to the horizontal plane, must be perpendicular to their intersection, that is, to the horizontal trace of the given plane. The horizontal projection of the line LM, must , be the line L' Mo ; but M M being perpendicular to the plane L E M', is perpendicular to L'M' (El. 202); that 41. ProBLEM. To draw through a given point a line perpendicular to a given plane ; and to find the point in which it pierces the plane. 42. PROBLEM. To draw through a given point, a plane perpendicular to a given straight line. It is manifest from the constructions which have just been performed, that the traces of the required plane will be perpendicular to the projections of the given line ; if, therefore, we assume a plane which answers this condition, the problem reduces itself to the construction in article 38. 43. Remark. A simple method of constructing the two preceding problems, is to take the vertical plane of the Fig. 17. projection in the proposed line (fig. 17), in each case. In the first problem, the plane might be given by its Fig. 17. horizontal trace and its inclination to the horizontal plane. In the second problem, the line might be given by its horizontal projection and its inclination to the horizontal plane. The details of this construction we shall leave as an exercise for the learner. 44. PROBLEM. To pass a plane through three points, given in space, not in the same straight line. Join the three points by two straight lines; find the points where these lines pierce the co-ordinate planes (18); and the straight line joining the two points in each of the co-ordinate planes will be the trace of the required plane in that plane. Let the three given points be (M', M'), (N',N''), and (P', Po'), (fig. 18); the two lines joining these points and Fig. 18. determining the required plane, will be (M'N', M'N'') (M'P', M/P"); and E, F, will be the points where these lines pierce the horizontal plane. The straight line E'F' joining these two points, will consequently be the horizontal trace of the required plane. To find the vertical trace of the proposed plane, we draw, according to article 38, M/G parallel to E'F', and GG" perpendicular to the ground line and equal to MM"; the straight line IIG” will be the vertical trace of the required plane. 45. We may construct this problem by passing a vertical plane through one of the given points and each of the others, and revolving these two planes about their intersection with the horizontal plane, as in article 20, till they coincide with this plane (fig. 19). If we then draw Fig. 19. MN meeting M'N' in F", and MP meeting M'P' in E, the straight line E'F' will be the horizontal trace of the required plane. It is manifest that, by drawing M'G' perpendicular to this horizontal trace, and constructing the triangleG'M'M" of which the angle at M' is a right-angle and the side M/M" equal to M/M, the angle MG'M' will be the inclination of the required plane to the horizontal plane (35). 46. The lines ME, MF', are the distances in space of the point M from the points E! and F' where these lines pierce the horizontal plane ; we have therefore the three Fig. 19. Fig. 20. sides of the triangle whose vertices are M, E', and F'. This triangle may be conceived to turn about the side E/F till it meets the horizontal plane on the other side of this line ; its vertex M will then be at the point m. 47. Remark. In this construction we have the developement of the entire surface of a tetraedron whose base is M/E/F and whose lateral faces are ME'F', ME/M', and MF'M'. 48. PROBLEM. Two planes being given in position, to find the angle which they make with each other. The angle made by two planes being the plane angle made by two lines drawn in these planes through the same point perpendicular to their line of intersection ; to determine this angle, we may construct a plane perpendicular to the intersection of the given planes. The intersections of this auxiliary plane with the given planes, will form an angle with each other equal to the angle of the proposed planes. This angle may be easily found from the construction in article 46. The following very simple solution of this problem was communicated to Lacroix by Monge. “Suppose the two given planes to be (EH', EF'), (GH', GF"), (fig. 20), and the horizontal projection of their intersection to be FH'. We construct in the vertical plane passing through this straight line, the intersection of the two given planes, by drawing FF perpendicular to FH and equal to I"F"; then through any point M’ taken at pleasure in the line H' F, construct a plane perpendicular to the line F H' (43); and find also the straight line PM’, which is the intersection of this plane with the vertical plane FFH'. But if we revolve the first of these planes about its horizontal trace L'N', the line PM', being perpendicular to M'N', will necessarily fall upon H'M' (20), and the point P will be found in P'; the triangle whose vertices are L', P, N' will not be altered by this revolution ; and the angle L'P'N' will be the angle of the given planes.” 49. PROBLEM. A plane being given, and also a line in that plane, to draw through this line a second plane making with the first a given angle. This problem is easily solved by retracing the several steps of the preceding solution. In this case, the data |