592. Theorem. If a straight line is parallel to a plane, any plane perpendicular to the line is perpendicular to the plane. Given AB || plane M, and plane P 1 AB at C. Proof. From any point B in AB draw BG 1 M. Further Then Therefore § 574 plane M AB, and CD 1 CB. Why? 1. Through a given point pass a plane perpendicular to a given plane. How many such planes can be drawn? 2. If one line is perpendicular to another line, is every plane containing the first line perpendicular to the second? Prove. 3. A plane determined by a line and its projection upon a given plane is perpendicular to the given plane. Illus 4. If three concurrent lines are each perpendicular to the other two, what can be proved concerning the planes determined by the lines? trate by the walls of a room. 5. If a plane is perpendicular to a line in another plane it is perpendicular to that other plane. 6. State the converse of the theorem of § 592. Is it a true theorem? Prove. 7. A circle is divided into eight equal parts by diameters. A plane containing one of these diameters makes an angle of 45° with the plane of the circle. Find the distance of each division point of the circle from the plane. 593. Theorem. If two planes are perpendicular to each other a line drawn in one of them perpendicular to their intersection is perpendicular to the other. Given plane MN 1 plane PQ, AB their intersection, and CD 1 AB in plane MN. To prove CD plane PQ. Proof. Through D and in plane PQ draw DE 1 AB. Then ZCDE is the plane angle of the right diedral ZM-AB-Q. 594. Theorem. If two planes are perpendicular to each other a perpendicular to one of them at any point of their intersection will lie in the other. Given plane MN 1 plane PQ, and intersecting it in AB, also CD plane PQ at any point D in AB. To prove that CD must lie in plane MN. Suggestion. In plane MN draw a line AB at D. Then this line is plane PQ, and must coincide with CD. Why? 595. Theorem. If iwo planes are perpendicular to each other a perpendicular to one from any point in the other will lie in the other. Given plane MN 1 plane PQ, and intersecting it in AB, also CD plane PQ from any point C in plane MN. To prove that CD must lie in plane MN. Suggestion. In plane MN draw a line 1 AB from C. Then this line is plane PQ, and must coincide with CD. Why? 596. Theorem. If each of two intersecting planes is perpendicular to a third plane their line of intersection is perpendicular § 594 Why? Given § 551 If AB is not 1 M, draw a line HB 1 M at B. Hence HB is the intersection of P and Q. But AB is the intersection of P and Q. .. AB 1 M. 597. Theorem. Through any straight line not perpendicular to a plane, one plane, and only one, can be passed perpendicular to the given plane. Given plane P and AB a line not 1 P. To prove that one plane P can be passed through AB, and only one. A B D C P Proof. From any point A in AB, draw AC 1 P. $ 591 .. one plane P can be passed through AB. If another plane would be P by § 596. But AB is not 1 P. .. only one plane IP can be passed through AB. Given EXERCISES 1. Discuss the theorem of § 597 for the case where AB lies in plane P. 2. If a plane is perpendicular to each of two intersecting planes, it is perpendicular to their intersection. 3. Any point in a plane containing the bisector of an angle, and perpendicular to the plane of the angle, is equally distant from the sides of the angle. 598. Theorem. Between two straight lines not in the same plane one common perpendicular, and only one, can be drawn. Given AB and CD two non-coplanar straight lines. To prove that one common perpendicular to AB and CD, and only one, can be drawn. Proof. Through AB pass plane P || CD. § 559 Through CD pass plane EH 1 P, and intersecting P in GH. § 597 Then GHCD. Why? Why? Why? And Why? Why? Why? .. only one common perpendicular to AB and CD can be drawn. 599. Theorem. The common perpendicular between two noncoplanar lines is the shortest line that can be drawn between them. Prove that NO is less than any other line KL, for NO = KR and KR<KŁ. 600. Definition. The length of the shortest line between two lines is called the distance between them. Compare with §§ 187-190, 577. 601. Theorem. Every point in a plane that bisects a diedral angle is equidistant from the faces of the angle. Given the plane R bisecting the diedral angle formed by the planes P and Q, and N any point in plane R. Also NC 1 P and NDIQ. To prove NC=ND. Proof. NC and ND determine plane CD. Why? Plane CD intersects P in CO, Q in DO, and R in NO. Therefore NOC is the plane angle of diedral ZRP, and 602. Theorem. Every point equidistant from the two faces of a diedral angle lies in the plane bisecting the angle. EXERCISES 1. Are both of the preceding theorems included in the following? The locus of a point equidistant from the faces of a diedral angle is the plane bisecting the angle. Explain. 2. From any point within a diedral angle perpendiculars are drawn to the faces. Prove that the angle formed by these perpendiculars is supplementary to the plane angle of the diedral angle. 3. The plane angle of a diedral angle is 120°. A point in the bisector of the diedral angle is 16 in. from the edge of the angle. Find the distance of this point from the faces of the angle. |