lines which are composed of short lines and dots are the auxiliary lines used in constructing the solution.] SECTION II. Of the Straight Line and the Plane. 19. PROBLEM. The two projections of a line being given, to determine the points in which it pierces the coordinate planes. Let N'N (fig. 3) be the horizontal projection of the Fig. 3. proposed line, and MM" its vertical projection. The point where it pierces the vertical plane must have its horizontal projection in the ground line (8); N must therefore be the horizontal projection of this point. If then we draw, in the vertical plane, NP" perpendicular to the ground line, and meeting the vertical projection in P", this point P" will be the point in which the proposed line meets the vertical plane. And if we draw in the horizontal plane, the line MP' perpendicular to the ground line, the point P' will be the point in which the proposed line pierces the horizontal plane. It is evident that this problem admits of four cases; as the part of the proposed line between the points where it meets the two planes, may be situated in either of the four diedral angles. The solution in figure 3, answers to the case in which this portion of the line is in the first angle. Figure 4 gives us the second case; namely, Fig. 4. that in which the portion of the line between the two points in which it pierces the planes, is in the second angle. The other cases we shall leave to the ingenuity of the learner. 20. PROBLEM. The projections of a line being given, to find the angle which it makes with one of the planes of projection. The inclination which a line has to a plane, is measured in a plane passing through the line, perpendicular to the first plane (El. 207). Suppose that the angle required, is that which the proposed line makes with the horizontal plane (fig. 5). This angle will be in the pro- Fig. 5. jecting plane, and will be contained between the line itself and its horizontal projection. If, therefore, the projecting plane be conceived to revolve round its horizontal trace, till it coincide with the horizontal plane, the required angle will, by this process, be transferred to the horizontal plane, and can be measured. Fig. 5. For this purpose the horizontal projection P'N is considered to be the axis about which the projecting plane P'P'N revolves. In such a rotation, it is perfectly manifest that every point in the revolving plane will be, throughout its revolution, at the same distance from the axis; and that any line in the revolving plane, perpendicular to the axis, will be perpendicular to the axis at the end of this revolution. The line NP" will therefore be perpendicular to the axis when the projecting plane shall be revolved into the horizontal plane. If, therefore, we draw Np perpendicular to P'Ñ, take Np equal NP", and draw P'p, the angle p P'N will be the angle required. 21. PROBLEM. Two points being given by their projections, to find their distance asunder in space. Let the two points be (M', M") (O′, 0′′) (fig. 5). It is obvious from the preceding process, that mo will be the distance required. This might be found at once by drawing through the point M', M'm perpendicular to M'O', and equal M"M; and through the point O ́, the line O'o perpendicular to M'O' and equal to O"0; this gives us m o. 22. Remark. The projection of a line upon either plane, is less than the line itself, unless the line be parallel to this plane. The length of a line is equal to the hypothenuse of a right-angled triangle, of which one side is its horizontal projection, and the other, the difference of the heights of the two extremities of the line above the horizontal plane. The projection of a line parallel to the plane of projection is equal to the line. 23. PROBLEM. Through a given point to draw a line parallel to another line given in position by its projec tions. The projections, upon the same plane, of parallel lines are parallel (16); therefore, if through the projection of the given point in each plane, we draw lines parallel to the projection, in that plane, of the given line, these will be the projections of the required line. 24. PROBLEM. Having the projections of two lines, to ascertain whether they cut each other. The horizontal projections of the two lines may inter sect, and also their vertical projections, and yet the two lines may not come near each other. If the lines intersect in space they will have one point common; and if they are not in the same plane perpendicular to one of the coordinate planes, their projections on each plane will intersect; and these two intersections must be the two projections of the point where the lines actually cut each other. The two projections of the same point are in a straight line perpendicular to the ground line (7); therefore, to ascertain whether two lines given by their projections, actually cut each other, draw from the point where their projections intersect, perpendiculars to the ground line; if these perpendiculars meet the ground line in the same point, the two points where the projections intersect are the projections of the same identical point; and this point must be in each of the given lines; that is, the two given lines cut each other, and are therefore in the same plane. Figure 6 shows a case in which the projections of the Fig. 6. two lines in each of the planes, cut each other, when the lines themselves do not meet. Figure 7 gives us the Fig. 7. projections of two lines which actually intersect in space. 25. Remark. If the two proposed lines cut each other in a plane perpendicular to one of the co-ordinate planes, their projection upon this plane will be but one straight line; if they cut each other in a plane perpendicular to each of the co-ordinate planes, their vertical and horizontal projections will be in one and the same straight line perpendicular to the ground line. 26. PROBLEM. Two planes being given by their traces, to find the projections of their line of intersection. When the traces of the proposed planes upon one of the co-ordinate planes, cut each other, their point of meeting is common to the two planes proposed, and is therefore a point in the required line. Let (NN', NN"), (MM', MM"), be the two given Fig. 8. planes; it is manifest that the point P' is common to the two proposed planes and the horizontal plane ABC; this is therefore one of the points sought. The point Q" evidently belongs to the common section of the two proposed planes with the vertical plane; this then is also a point in the required line. The problem is therefore re Fig. 8. duced to finding the projections of a line passing through the points P', Q". The horizontal projection of Q" is Q (8); P' is its own horizontal projection; P'Q is therefore the horizontal projection of the proposed line. For similar reasons PQ" is its vertical projection. Fig. 9. 27. Remark 1st. If the traces NN', MM', (fig. 9) of the proposed planes upon one of the co-ordinate planes, the horizontal for example, are parallel to each other, the intersection of the proposed planes will be parallel to the horizontal plane; and of this line one point is known, namely, the point P". As this line is in the same plane with each of the horizontal traces of the proposed planes, it will meet them unless it is parallel to them; but it cannot meet them, unless they meet each other, which is impossible as they are parallel; this line of intersection is therefore parallel to each of the horizontal traces of the proposed planes. This question therefore refers itself to the problem in article 23. 28. Remark 2d. It may happen that the line of intersection of the proposed planes meets neither of the coordinate planes; in this case it must be parallel to them Fig. 10. and parallel to the ground line (fig. 10). To find, under these circumstances, the line of intersection of the proposed planes, it will be necessary to refer to a third plane; which, for the sake of simplicity, we take perpendicular to each of the other two co-ordinate planes. We shall not now detain the reader with this process; a little familiarity with constructions will render the solution perfectly easy. 29. If, however, the proposed planes are parallel to each other, their traces upon each of the co-ordinate planes will be parallel; but it is perfectly manifest that they cannot intersect. 30. PROBLEM. To find the projections of a point, when we know three planes in each of which it is situated. As these three planes have only this single point in common; to find the projections of this point, seek first the two projections of the common section of any two of the proposed planes; this line being cut by the third plane, will thus give the point required. We shall come to the same result, by finding the intersection of two of the planes, and then seeking the intersection of one of these with the third. These two intersections will be two straight lines which cut each other. This problem will then refer itself to the preceding (26). 31. The most simple method of determining a point. by means of three planes, is to suppose them perpendicular to each other; and then give the distances of the proposed point from three planes parallel respectively to the three planes in which this point is situated. Suppose the three planes BAC, BAD, and DAC, (fig. 11) to be perpendicular to each other, and that we know Fig. 11. that the point M in space is situated at a distance MM' from the first, MM" from the second, and MM" from the third. As parallel planes are equally distant throughout their whole extent (El. 212), if we draw the planes MM"M", MM' M'"', MM' M", at the given distances respectively from the three planes BAČ, BAD, DAC, and respectively parallel to them, the proposed point will be found at their mutual intersection. 2 These three planes, with the co-ordinate planes, form the rectangular parallelopiped whose diagonal is AM. The square of this diagonal is equal to the sum of the squares of the three edges which meet at the proposed point (El. 245); that is (AM)2= (MM′)2 + (MM”)2 + (MM"). Whence we say-The square of the distance of any point in space from the point of meeting of three rectangular co-ordinate planes, is equal to the sum of the squares of the distances of the proposed point from each of these planes. 32. Remark. It will be perceived that three planes which cut each other form eight triedral angles, in each of which the proposed point may be situated conformably with the above conditions; so that to fix the position in space, of the proposed point, we must not only have its distance from each of the three co-ordinate planes, but must know on which side of each it is situated; or we must in some way particularize the triedral angle in which it is situated. When a point is given by a line and a plane, it is the same thing as if it were given by three planes; for, instead of the given line we must use its two projecting planes. 33. PROBLEM. A plane being given by its traces, and a line, inclined to this plane, being given by its projections, to find the point at which the line pierces the plane. |