PROBLEM 14. From a given point to draw a line perpendicular to a given plane, and then to determine its magnitude 28. We shall first show that the projections of the perpendicular are + to the traces of the given plane. The plane which projects the line on the horizontal plane is - both to the horizontal and the given plane (Prop. XIX., Introduc- . tion); .. conversely, both are I to the projecting plane, and consequently their intersection is also t to it (Prop. XX., Introduction); but this intersection is the horizontal trace of the given plane, :. this trace is 1 to all lines situated in the projecting plane, and :: to the horizontal projection of the perpendicular. A similar method of reasoning applies to the vertical projection. This being the case, draw, through the projections a, a' of the given point, ab, a'b', respectively, – to the traces ym, ym' of the given plane; and it is manifest we shall have the projections of the required perpendicular. We can then determine b and b', (Prob. 3,) the projections of the foot of the perpendicular, and, lastly, we can find its length a'c, (Prob. 5.) PROBLEM 15. Through a given point to draw a line and plane, both per pendicular to a given line. 29. First find the traces of a plane 1 to the given line; these are 1 to the projections of this line, consequently it will be sufficient to know one point of either of these traces to know both. Let a, á' be the projections of the given point, bc, b'ć' those of the given line. Through a, a', suppose a horizontal line to be drawn, parallel to the horizontal trace of the required plane; its horizontal projection being parallel to this trace will be + the horizontal projection bc of the given line; and thus this projection is a known line ad: the line a'd , parallel to xy, will be the vertical projection of this line. And as this horizontal line is altogether in the required plane, therefore d', its intersection with the vertical plane, will be a point of the vertical trace of the required plane. Draw, therefore, through this point ym t to b'c': then ym 1 to bc, and my m' is the plane required. Next, to find the projections of the 1 from the given point upon the given plane. To do so, find p and p', the projections of the intersection of this line with the plane mym': draw the lines ap and a'p': these are the projections required, Problem 16. Given the projections of a line, to find the angles which it makes with the planes of projection. 30. Def. The angle which a line makes with a plane is the angle formed by the line and its projection upon the plane. Let ab and a'b' be the projections of a line, ad' and bb' its traces, determined by a previous proposition; let the plane abb' turn round 66', till it coincide with the vertical plane ; during this revolution the line ab is not altered in magnitude, and as it is always 1 bb', it will take a position as bm along xy. At the same time the line joining a and b' in space will coincide with b'm, and thus the required angle coincides with the angle b'mb. If the plane aa'b' be supposed to revolve round ab', the angle might be found in the horizontal plane; in this case the vertical bb' would be measured along bn, 1 to ab, and the required angle would be nab. To find the angle which the line makes with the vertical plane, a similar construction must be made; it is represented by apa' on the horizontal, or by qb'a' on the vertical plane. When the two points a and b' are very distant, we may use the following method:—Take any two points in the line; let c, c', d and d', be their projections. Through the point dd imagine a line drawn parallel to ab, and find the angle it makes with the line. To do this, let the vertical plane which contains the angle turn round the vertical from c, till the plane is parallel to the vertical plane of projection; then the required angle is projected on this plane in its true magnitude, and is represented by cʻrd. We proceed in the same manner to find the angle with the vertical plane. PROBLEM 17. Given the traces of a plane to find the angles which the plane makes with the two planes of projection, and also the angle between the traces. 31. Let qp and qr be the traces of the plane ; to find first the angle it makes with the horizontal plane. Since the angle between two planes is measured by that between two perpendiculars, drawn, in each of the planes, from the same point of their common intersection ; draw ab 1 to pq, and bc 1 to xy; meeting qr in c; the straight line joining c and a is in the given plané, and is 1 pq; and the < it makes with ab is the < required. (Prop. XI. Introduction.) The 2 is that of a right-angled A, of which ab is the base, and bc the height. To make it coincide with the vertical plane, suppose it to revolve round bc in xy; take ba = ba, and < cab is the angle required ; and for the horizontal plane, draw bc = bc' and I to ba, and cab will be the angle required. To find the 2 the plane makes with the vertical plane draw bd 1 qr, and be I to xy. The line in space, joining d and e, makes with db the required Z; and if this < be made to coincide with the horizontal plane, it is represented by edb; if with the vertical plane, by Edb. Lastly, to find the angle between the traces. Since the line joining the points a and c'is + to pq, if the plane par turn round pq till it coincide with the horizontal plane, ac' will coincide with ac', which is - to pq; the distance aq remaining the same ; the point c'c comes to c', ac' being made equal to ac or to c'a, the real length of the line joining the points ayc'; qr will therefore coincide with 98, drawn through q and c', and pqs is the < of the traces. Problem 18. Given the projections of two lines, to find the angle between the lines. 32. We may remark that, when the points of intersection of the two horizontal projections of the lines, and of their two vertical projections, are not in the same perpendicular to the ground line, the two lines do not really meet. In that case we draw from any point two lines parallel |