Solution.—The two points A, B in space lie vertically above their plans A1, B1 (fig. 43) and A1A = A0A2, B1B = B0B2. The four points A, B, A1, B1 therefore form a plane quadrilateral on the base A1B1 and having right angles at the base. This plane we rabatt about A1B1 by drawing A1A and B1B perpendicular to A1B1 and making A1A = A0A2, B1B = B0B2. Then AB will give the length required.
The construction might have been performed in the elevation by making A2A = A0A1 and B2B = B0B1 on lines perpendicular to A2B2. Of course AB must have the same length in both cases.
This figure may be turned into a model. Cut the paper along A1A, AB and BB1, and fold the piece A1ABB1 over along A1B1 till it stands upright at right angles to the horizontal plane. The points A, B will then be in their true position in space relative to π1. Similarly if B2BAA2 be cut out and turned along A2B2 through a right angle we shall get AB in its true position relative to the plane π2. Lastly we fold the whole plane of the paper along the axis x till the plane π2 is at right angles to π1. In this position the two sets of points AB will coincide if the drawing has been accurate.
Models of this kind can be made in many cases and their construction cannot be too highly recommended in order to realize orthographic projection.
§ 14. To find the angle between two given lines a, b of which the projections a1, b1 and a2, b2 are given.
Solution.—Let a1, b1 (fig. 44) meet in P1, a2, b2 in T, then if the line P1T is not perpendicular to the axis the two lines will not meet. In this case we draw a line parallel to b to meet the line a. This is easiest done by drawing first the line P1P2 perpendicular to the axis to meet a2 in P2, and then drawing through P2 a line c2 parallel to b2; then b1, c2 will be the projections of a line c which is parallel to b and meets a in P. The plane α which these two lines determine we rabatt to the plan. We determine the traces a′ and c′ of the lines a and c; then a′c′ is the trace α′ of their plane. On rabatting the point P comes to a point S on the line P1Q perpendicular to a′c′, so that QS = QP. But QP is the hypotenuse of a triangle PP1Q with a right angle P1. This we construct by making QR = P0P2; then P1R = PQ. The lines a′S and c′S will therefore include angles equal to those made by the given lines. It is to be remembered that two lines include two angles which are supplementary. Which of these is to be taken in any special case depends upon the circumstances.
To determine the angle between a line and a plane, we draw through any point in the line a perpendicular to the plane (§ 12) and determine the angle between it and the given line. The complement of this angle is the required one.
To determine the angle between two planes, we draw through any point two lines perpendicular to the two planes and determine the angle between the latter as above.
In special cases it is simpler to determine at once the angle between the two planes by taking a plane section perpendicular to the intersection of the two planes and rabatt this. This is especially the case if one of the planes is the horizontal or vertical plane of projection.
Thus in fig. 45 the angle P1QR is the angle which the plane α makes with the horizontal plane.