16. If through a variable point in a plane a line be drawn in a fixed direction and of constant length, find the locus of its extremity. 17. If P and Q be points one on each of two lines which are neither parallel nor concurrent, the locus of the middle point of PQ is a plane. 18. A system of planes drawn through one line meets any other plane in a system of lines which are either concurrent or parallel. 19. The intercepts formed on a system of lines by two parallel planes are divided in a constant ratio. Find the locus of the point of section. 20. A pyramid is described on a given polygon as base. Show that the section of the pyramid made by any plane parallel to the base is a polygon similar to the base. 21. A pyramid is described on a given triangle as base: and the three edges which meet at the vertex are divided in the same ratio. Show that the three points of section determine a plane which is parallel to the base. 22. A pyramid is described on a given polygon as base: and the edges which meet at the vertex are all divided in the same ratio. Show that the points of section all lie in one plane which is parallel to the base. 23. A series of planes are drawn through a point, each containing one of a set of parallel lines. Prove that any plane which does not pass through the given point intersects these planes in a series of parallel or concurrent lines. 24. Three planes intersect in a single line; prove that they divide any two parallel lines proportionately. 25. In Perspective Drawing, if a line is parallel to the Picture Plane, its picture is parallel to itself. 26. In Perspective Drawing, if a given line is not parallel to the Picture Plane, and if the line drawn from the Eye parallel to the given line meets the Picture Plane at e, then the picture of the given line points towards e. CHAPTER XXIX. LINES PERPENDICULAR TO PLANES. 411. THEOREM.-If a given straight line is perpendicular to two straight lines in a given plane which meet it, it will be perpendicular to the given plane. (See Def. § 389.) In Fig. 540, PN meets plane X in N, also NA, NB, are in plane X : Given that PN 1 NA and PN NB, any other straight line, Cons. Produce PN through the plane X to P', making P'N = PN. . In plane X draw any straight line meeting the three lines NA, NB, NC in A, B, C respectively. Join P and P' to each of the points A, B, C.† + The student will find this figure easy to understand, if he realises that it is symmetrical with respect to the plane X: thus PN = P'N, PA P'A, PB P'B. = = A rough model of this figure is easily constructed, and should be of considerable help if the proposition is found difficult without it. Take a piece of thin board for X, with holes bored through at N, A, B, C. Take a piece of wire PP', fasten three pieces of cotton to the end P: drive the wire through the hole W, fixing it at right angles to the board, and so that NP NP': pass the three pieces of cotton through A, B, C : twist them round the end P', and fasten with sealing-wax. = Proof. NA bisects PP' at right angles, .. PA = P'A. NB bisects PP' at right angles, .. PB = P'B. NC. NOTE. If in Fig. 540, NA NB, and ANB = R, then four figures identical with the solid PABP' united along the line PP' would give a solid octahedron. (See Fig. 541.) This octahedron would not be "regular" unless PA = AB, in which case each face would be an equilateral triangle. 412. THEOREM.-If a straight line be perpendicular to each of three concurrent straight lines at their point of intersection, these three straight lines shall lie in one plane. In Fig. 542, PN is perpendicular to each of the three lines NA, NB, NC; Required to prove that NA, NB, NC lie in one plane. B FIG. 542. Cons. Let X be the plane which passes through NB and NC. Let a plane pass through NP and NA; and let ND be the intersection of this plane with the plane X. Proof.-NB, NC, ND lie in the same plane X; .. PN 1 ND. But PNNA, and PN, NA, ND are in one plane. Hence NA lies in plane X. Cons. § 411. COROLLARY 1.—If from a given point in a given straight line a series of lines are drawn perpendicular to the given straight line, the locus of these lines is a plane perpendicular to the given straight line. COROLLARY 2.—Through a given point in a given straight line it is possible to draw one and only one plane perpendicular to the given straight line. 413. THEOREM.-Through a given point in a given plane it is always possible to draw one and only one straight line perpendicular to the given plane. Y In Fig. 543, P is given point, and X given plane. Cons.-In plane X through the point P draw any two perpendicular lines EF, PC. Take any other plane Y through EF, and in it draw PD EF. In plane CPD draw PQ PC. Then PQ is the required perpendicular. Proof (i) To show that PQ plane X : : P X F FIG. 543. PQ lies in plane CPD, also FP1 PC and FP1 PD. Cons. .. FP 1 plane CPD (§ 411); .. FP 1 PQ. that is PQFP (Proved), and PQ 1 PC (Cons.), § 411. (ii) To show that no other line can be drawn through Pi plane X. Suppose that another line PS is drawn through P perpendicular to plane X (Fig. 544). Let plane PQS meet plane X in line AB. Then, since PQ and PS are both plane X, and since AB lies in plane X, Α B .. QPB and SPB are both X FIG. 544. right angles. But these are angles in the same plane, and so cannot be equal unless PS and PQ coincide. That is PQ is the only line through P which is perpendicular to plane X. |