Descriptive geometry . ed to find the bisector of the acute angle. InFig. 223, draw first the actual bisector, 0„ in the revolvedposition, bisecting the angle between Ar and Br. Xow CTnecessarily passes through or, and in this case intersects thetrace IIX in the points/. Imagine the plane Xto be counter-revolved (revolved back) about IIX to its original position,taking with it the lines A, B, and C. The lines A and B, andtheir intersection, point o, will resume their former point 83k, being on the axis of revolution HX, will notmove. Hence the //-projection, C* of the line C must
Descriptive geometry . ed to find the bisector of the acute angle. InFig. 223, draw first the actual bisector, 0„ in the revolvedposition, bisecting the angle between Ar and Br. Xow CTnecessarily passes through or, and in this case intersects thetrace IIX in the points/. Imagine the plane Xto be counter-revolved (revolved back) about IIX to its original position,taking with it the lines A, B, and C. The lines A and B, andtheir intersection, point o, will resume their former point 83k, being on the axis of revolution HX, will notmove. Hence the //-projection, C* of the line C must passthrough o* and .%h. To find the ^projection of C, note thato* must be projectedto o, while N,\ being an ff-trace, lies in H, and musl projectto Sj in GL. II ace < \ determined bj and -V- The second example, Fig. 224, cannot be solved, however, inthis way. As soon as the bisector, Cr, i drawn in the revolvedposition, it is seen that the intersection of 0, with the trace 146 DESCRIPTIVE GEOMETRY [XV, § 140. is beyond the limits of the figure. One point of C is o, and one point of Ch is oh. It remains to find a second point in each projection. Assume a point, er, in Cr. Through er draw the revolved position, Lr, of a line lying in the plane of the given lines, by making LT parallel to the revolved position of either of the given lines A or B; in this case LT is drawn parallel to AT. Since the line L is in the plane X, the intersection, t3v, of Lr and VX is the F-trace of the line L, and t3h is in GL. Also, since the line L is parallel to the line A, through tzv draw Lv parallel to A1, and through t3h draw Lh parallel to Ah. Now e is a point in the line L. Hence revolve back from er per- FlG- 22i Orated). pendicular to VX, and find ev on L; jjroject from e and obtain eh on Lh. Since e is a point also in the line C, we have Cv determined by ov and e, and Ch determined by oh and e\ 141. The Angle between a Line and a Plane. The anglewhich a line makes with a plane is e
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