PROBLEM XVII. Figs. 43 and 44. From a given point to draw a perpendicular to a given line. Let pp' be the projections of the point, and ab, a'b' the projections of the line; it is required to find the projections of the perpendicular from P on AB. Construction. Draw a plane LMN containing P, and perpendicular to AB (Prob. XVI.), and determine the point of intersection B (Prob. XIV.). PB will be the perpendicular required. Proof. Because AB is perpendicular to the plane LMN, it is also perpendicular to PB (Def. 5). PROBLEM XVIII. Figs. 45 and 46. To determine the common perpendicular to two straight lines which are not in the same plane. Let AB and CD be the two given lines; it is required to find their common perpendicular. Construction. Find the traces of the plane DMC containing CD and parallel to AB (Problem XI.). Find the traces of the plane ANB containing AB and perpendicular to the plane DMC (Prob. xv. Cor. 3). Next determine the projections of the common section EF of the two planes DMC and ANB; and from the point G, where EF intersects CD, draw GH perpendicular to the plane DMC (Prob. xv.). GH will be perpendicular to the two lines AB and CD. Proof. Because GH is perpendicular to the plane DMC it is perpendicular to CD (Def. 5). For the same reason it is perpendicular to EF, and therefore lies in the plane ANB (Theor. v. Cor.). But AB and EF are parallel (Theor. XI.), and since GH is perpendicular to EF it must be also perpendicular to AB. Therefore GH is the common perpendicular to AB and CD. PROBLEM XIX. Figs. 47 and 48. To determine the inclination of a plane to each plane of projection. Let AMD be the traces of a plane, it is required to find its inclination (1) to the horizontal plane; (2) to the vertical plane. Construction. Draw a vertical plane AaB, having its horizontal trace aB at right angles to MD, and find the angle between AB, the common section of the two planes, and aB (Prob. VI.); that will be the inclination of AMD to the horizontal plane. To find the inclination to the vertical plane; draw the traces of a plane CqD, perpendicular to the vertical plane, and having its vertical trace Ca' at right angles to AM; the angle B, between DC and Cq', will be the inclination required. Proof. Since AaB is perpendicular to the horizontal plane, and DB is perpendicular to aB, it is also perpendicular to the plane AaB (Theor. v.), and consequently AB is at right angles to BD (Def. 5). Because AB and aB, one in each plane, are both perpendicular to MD, the common section of the two planes, the angle ABa measures the dihedral angle between the planes (Def. 6). In a similar manner it may be proved that the inclination of AMD to the vertical plane is equal to the angle between DC and Cq'. PROBLEM XX. Figs. 47 and 48. To find the traces of a plane which shall contain a given point, and make given angles with the planes of projection. Let P be the given point, a and ẞ the required inclinations to the horizontal and vertical planes respectively. Construction. Through the point P draw a line PQ, making an angle of (90°-a) with the horizontal and an angle of (90°) with the vertical plane of projection (Prob. VII.). |