The relation of perpendicularity of planes is considered - one of the most important and most used in the geometry of space and its applications.
From all the variety of mutual arrangement
Two planes deserve special attention and study when the planes are perpendicular to each other (for example, the planes of adjacent walls of a room,
fence and plot of land, door and floor, etc. (Fig. 417, a–c).
The above examples allow us to see one of the main properties of the relationship that we will study - the symmetry of the location of each plane relative to the other. Symmetry is ensured by the fact that the planes seem to be “woven” from perpendiculars. Let's try to clarify these observations.
Let us have a plane α and a straight line c on it (Fig. 418, a). Let us draw through each point of the line c straight lines perpendicular to the plane α. All these lines are parallel to each other (why?) and, based on Problem 1 § 8, form a certain plane β (Fig. 418, b). It is natural to call the plane β perpendicular plane α.
In turn, all lines lying in the plane α and perpendicular to the line c form the plane α and are perpendicular to the plane β (Fig. 418, c). Indeed, if a is an arbitrary line, then it intersects the line c at some point M. A line b perpendicular to α passes through the point M in the plane β, therefore b a. Therefore, a c, a b, therefore a β. Thus, the α plane is perpendicular to the β plane, and the straight line c is the line of their intersection.
Two planes are called perpendicular if each of them is formed by straight lines perpendicular to the second plane and passing through the intersection points of these planes.
The perpendicularity of the planes α and β is indicated by the familiar sign: α β.
One illustration of this definition can be imagined if we consider a fragment of a room in a country house (Fig. 419). In it, the floor and wall are made of boards perpendicular to the wall and floor, respectively. Therefore they are perpendicular. On practice
this means the floor is horizontal and the wall is vertical.
The above definition is difficult to use when actually checking the perpendicularity of planes. But if we carefully analyze the reasoning that led to this definition, we see that the perpendicularity of the planes α and β was ensured by the presence in the β plane of a straight line b perpendicular to the α plane (Fig. 418, c). We came to the criterion of perpendicularity of two planes, which is most often used in practice.
406 Perpendicularity of lines and planes
Theorem 1 (test for perpendicularity of planes).
If one of two planes passes through a line perpendicular to the second plane, then these planes are perpendicular.
Let the plane β pass through a line b perpendicular to the plane α and c - the line of intersection of the planes α and β (Fig. 420, a). All straight lines of the plane β, parallel to the line b and intersecting the line c, together with the straight line b form the plane β. By the theorem about two parallel lines, one of which is perpendicular to the plane (Theorem 1 § 19), all of them, together with the line b, are perpendicular to the plane α. That is, plane β consists of straight lines passing through the line of intersection of planes α and β and perpendicular to plane α (Fig. 420, b).
Now in the plane α, through the point A of the intersection of lines b and c, we draw a line a perpendicular to line c (Fig. 420, c). The straight line a is perpendicular to the plane β, based on the perpendicularity of the line and the plane (a c, by construction, and b, since b α). Repeating the previous arguments, we find that the plane α consists of lines perpendicular to the plane β, passing through the line of intersection of the planes. According to the definition, planes α and β are perpendicular. ■
This feature makes it possible to establish the perpendicularity of the planes or ensure it.
Example 1. Attach the shield to the post so that it is positioned vertically.
If the pillar stands vertically, then it is enough to attach a shield at random to the pillar and secure it (Fig. 421, a). According to the feature discussed above, the plane of the shield will be perpendicular to the surface of the earth. In this case, the problem has an infinite number of solutions.
Perpendicularity of planes |
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If the pillar stands obliquely to the ground, then it is enough to attach a vertical rail to the pillar (Fig. 421, b), and then attach the shield to both the rail and the pillar. In this case, the position of the shield will be quite definite, since the post and rail define a single plane. ■
In the previous example, the “technical” task was reduced to a mathematical problem about drawing a plane perpendicular to another plane through a given straight line.
Example 2. From vertex A of square ABCD a segment AK is drawn perpendicular to its plane, AB = AK = a.
1) Determine the relative position of the planes AKC and ABD,
AKD and ABK.
2) Construct a plane passing through line BD perpendicular to plane ABC.
3) Draw a plane perpendicular to the plane KAC through the middle F of the segment KC.
4) Find the area of triangle BDF.
Let’s construct a drawing that corresponds to the conditions of the example (Fig. 422).
1) Planes AKC and ABD are perpendicular, according to the condition of perpendicularity of planes (Theorem 1): AK ABD , according to the condition. Planes AKD and ABK are also perpendicular
are polar, based on the perpendicularity of the planes (Theorem 1). Indeed, the line AB through which the plane ABK passes is perpendicular to the plane AKD, according to the sign of perpendicularity of the line and the plane (Theorem 1 § 18): AB AD are like adjacent sides of a square; AB AK , since
AK ABD.
2) Based on the perpendicularity of the planes, for the desired construction it is enough to draw a straight line BD through some points
408 Perpendicularity of lines and planes
line perpendicular to plane ABC. And to do this, it is enough to draw a line through this point parallel to the line AK.
Indeed, by condition, the line AK is perpendicular to the plane ABC and therefore, according to the theorem about two parallel lines,
our, one of which is perpendicular to the plane (Theorem 1§19), |
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the constructed straight line will be perpendicular to plane ABC. |
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Construction. |
Through the point |
B we conduct |
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BE, |
parallel |
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(Fig. 423). The plane BDE is the desired one. |
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3) Let F be the midpoint of the segment KC. Pro- |
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we lead through the point |
perpendicular- |
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plane |
This straight line |
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children direct |
FO, where |
O - center of the square |
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ABCD (Fig. 424). Indeed, FO || A.K. |
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like average |
triangle line |
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Because the |
perpendicular- |
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on surface |
direct FO |
boo- |
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det is perpendicular to it, according to the theorem about |
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two parallel lines, one of which |
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ry perpendicular to the plane (Theorem 1 |
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§ 19). That's why |
FO DB. And since AC DB, then DB AOF (or |
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KAC). Plane |
BDF passes through a line perpendicular to |
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nal plane KAC, that is, it is the desired one. |
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4) In a triangle |
BDF segment FO |
Height drawn to |
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side BD (see Fig. 424). We have: BD = |
2 a as the diagonal of the quad- |
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rata; FO = 1 |
AK = |
1 a, by the property of the midline of a triangle. |
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Thus, S = 2 BD FO = |
2 2 a |
2 a = |
. ■ |
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Answer: 4) |
a 2 . |
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Study of the properties of the perpendicular- |
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of planes and its applications, let’s start with the simplest |
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that, but very useful theorem. |
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Theorem 2 (about the perpendicular to the line of intersection of perpendicular planes).
If two planes are perpendicular, then a straight line belonging to one plane and perpendicular to the intersection of these planes is perpendicular to the second plane.
Let perpendicular planes
α and β intersect along straight line c, and straight line b in plane β is perpendicular to straight line c and intersects it at point B (Fig. 425). By definition
dividing the perpendicularity of the planes, in the β plane a straight line passes through point B
b 1, perpendicular to the plane α. It is clear that it is perpendicular to the line c. But what-
If you cut a point on a straight line in a plane, you can draw only one straight line perpendicular to the given straight line. That's why
lines b and b 1 coincide. This means that a straight line of one plane, perpendicular to the line of intersection of two perpendicular planes, is perpendicular to the second plane. ■
Let us apply the considered theorem to the substantiation of another sign of the perpendicularity of planes, which is important from the point of view of the subsequent study of the relative position of two planes.
Let the planes α and β be perpendicular, straight line c is the line of their intersection. Through an arbitrary point A we draw a straight line c
in planes α and β, straight lines a and b, perpendicular to straight line c (Fig. 426). According to theory
Me 2, straight lines a and b are perpendicular to the planes β and α, respectively, so they are perpendicular to each other: a b . Straight
a and b define a certain plane γ. Line of intersection with planes α and β
perpendicular to the plane γ, based on the perpendicularity of the line and the plane (Theorem 1 § 18): c a, c b, a γ, b γ. If we take into account the arbitrariness of the choice of point A on line c and the fact that through point A of line c there passes a single plane perpendicular to it, then we can draw the following conclusion.
Theorem 3 (about the plane perpendicular to the line of intersection of perpendicular planes).
A plane perpendicular to the line of intersection of two perpendicular planes intersects these planes along perpendicular straight lines.
Thus, one more property of perpendicular planes has been established. This property is characteristic, that is, if it is true for some two planes, then the planes are perpendicular to each other. We have one more sign of perpendicularity of planes.
Theorem 4 (second criterion for the perpendicularity of planes).
If the direct intersections of two planes by a third plane perpendicular to the line of their intersection are perpendicular, then these planes are also perpendicular.
Let the planes α and β intersect along a straight line with, and the plane γ, perpendicular to the line with, intersects the planes α and β corresponding
respectively along straight lines a and b (Fig. 427). By condition, a b . Since γ c, then a c. And therefore the line a is perpendicular to the plane β, according to the sign of perpendicularity of the line and the plane (Theorem 1 § 18). That's it-
Yes, it follows that the planes α and β are perpendicular, according to the sign of perpendicularity of planes (Theorem 1). ■
Also worthy of attention are theorems on the connections between the perpendicularity of two planes of a third plane and their mutual position.
Theorem 5 (about the line of intersection of two planes perpendicular to the third plane).
If two planes perpendicular to a third plane intersect, then the line of their intersection is perpendicular to this plane.
Let the planes α and β, perpendicular to the plane γ, intersect along the straight line a (a || γ), and A is the point of intersection of the straight line a with
Perpendicularity of planes |
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plane γ (Fig. 428). Point A belongs to |
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lives along the intersection lines of the planes γ and α, γ |
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and β, and, by condition, α γ and β γ. Therefore, according to |
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determining the perpendicularity of the plane |
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tey, through point A you can draw straight lines, |
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lying in the α planes |
and β and perpendicular |
polar planes γ. Because through the point |
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it is possible to draw only one straight line, per- |
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perpendicular to the plane, then the constructed |
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straight lines coincide and coincide with the line |
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intersections of planes α and β. Thus, straight a is a line |
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the intersection of planes α and β is perpendicular to the plane γ. ■ |
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Let us consider a theorem describing the relationship between parallelism and perpendicularity of planes. We already had the corresponding result for straight lines and planes.
Theorem 6 (about parallel planes perpendicular to the third plane).
If one of two parallel planes is perpendicular to the third, then the second plane is perpendicular to it.
Let planes α and β be parallel, and plane γ perpendicular to plane α. Since the plane γ
intersects the plane α, then it must also intersect the plane β parallel to it. Let us take a pro-
an arbitrary straight line m perpendicular to the plane γ and draw through it, as well as through an arbitrary point of the plane β, the plane δ (Fig. 429).
The planes δ and β intersect along a straight line n, and since α ║ β, then m ║ n (Theorem 2 §18). It follows from Theorem 1 that n γ, and therefore the plane β passing through the straight line n will also be perpendicular to the plane γ.
The proved theorem gives another sign of the perpendicularity of planes.
You can draw a plane perpendicular to the given point through a given point using the sign of perpendicularity of planes (Theorem 1). It is enough to draw a straight line through this point perpendicular to the given plane (see Problem 1 § 19). And then draw a plane through the constructed straight line. It will be perpendicular to the given plane according to the specified criterion. It is clear that an infinite number of such planes can be drawn.
More meaningful is the problem of constructing a plane perpendicular to a given one, provided that it passes through a given line. It is clear that if a given line is perpendicular to a given plane, then an infinite number of such planes can be constructed. It remains to consider the case when the given line is not perpendicular to the given plane. The possibility of such a construction is justified at the level of physical models of straight lines and planes in example 1.
Task 1. Prove that through an arbitrary line not perpendicular to a plane, one can draw a plane perpendicular to the given plane.
Let a plane α and a line l, l B\ a be given. Let us take an arbitrary point M on the line l and draw a line m through it, perpendicular to the plane α (Fig. 430, a). Since, by condition, l is not perpendicular to α, then the lines l and m intersect. Through these straight lines it is possible to draw a plane β (Fig. 430, b), which, according to the test for the perpendicularity of planes (Theorem 1), will be perpendicular to the plane α. ■
Example 3. Through vertex A of the regular pyramid SABC with the base ABC, draw a straight line perpendicular to the plane of the side face of SBC.
To solve this problem, we use the theorem about the perpendicular to the line of intersection of perpendicular planes
(Theorem 2). Let K be the midpoint of edge BC (Fig. 431). The planes AKS and BCS are perpendicular, according to the sign of perpendicularity of planes (Theorem 1). Indeed, BC SK and BC AK are like medians drawn to the bases in isosceles triangles. Therefore, according to the criterion of perpendicularity of a line and a plane (Theorem 1 §18), the line BC is perpendicular to the plane AKS. The BCS plane passes through a line perpendicular to the AKS plane.
Construction. Let us draw a line AL in the AKS plane from point A, perpendicular to the KS line - the line of intersection of the AKS and BCS planes (Fig. 432). By the theorem on the perpendicular to the line of intersection of perpendicular planes (Theorem 2), the line AL is perpendicular to the plane BCS. ■
Control questions |
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In Fig. 433 shows the square ABCD, |
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line MD is perpendicular to the plane |
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ABCD. Which of the pairs of planes are not |
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are perpendicular: |
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MAD and MDC; |
MBC and MAV; |
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ABC and MDC; |
MAD and MAV? |
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2. In Fig. 434 is shown correctly- new quadrangular pyramid
SABCD, points P, M, N - middle -
The edges AB, BC, BS, O are the center of the base ABCD. Which of the pairs are flat- bones are perpendicular:
1) ACS and BDS; 2) MOS and POS;
3) COS and MNP; 4) MNP and SOB;
5) CND and ABS?
Perpendicularity of lines and planes |
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3. In Fig. 435 |
depicted rectangular |
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triangle |
with right angle C and |
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straight line BP, perpendicular to the plane |
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ty ABC . Which of the following pairs are flat? |
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bones are perpendicular: |
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1) CBP and ABC; |
2) ABP and ABC; |
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3) PAC and PBC; 4) PAC and PAB?
4. The two planes are perpendicular. Is it possible through an arbitrary point of one of should they draw a straight line in this plane, the second plane?
5. It is impossible to draw a straight line in the α plane, but not in the β plane. Could these planes be mi?
6. Through a certain point on the plane α does a line pass in this plane and is perpendicular to the plane, so that the planes α and β are perpendicular?
A section of fence is attached to a vertical post, is it possible to claim that the plane of the fence is vertical?
How to attach a shield vertically to a rail parallel to the surface of the earth?
Why is the surface of the doors, regardless of whether they are closed or open, vertical to the floor?
Why does a plumb line fit tightly against a vertical wall, but not necessarily against an inclined one?
Is it possible to attach a shield to an inclined post so that it is perpendicular to the surface of the earth?
How to practically determine whether a plane is perpendicular
walls plane floor? perpendicularperpendicularperpendicular- straight, lying down - β.
True 7. . Possible 8.9.10.11.12.
1. Graphic exercises In Fig. 436 shows a cube
1) ABCDA 1 B 1 C 1 D 1 . Specify planes perpendicular to the plane
2) BDD 1.
How are the planes and
Perpendicularity of planes |
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A1 B1 CAB 1 C 1 |
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437 plane squares ABCD and |
ABC1 D1 |
perpendicular. Distance |
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CC1 |
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equals b. Find the length of the segment: |
AB; |
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D1 C; |
D1 D; |
C1 D. |
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Dan- |
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Construct a drawing according to the given |
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1) Planes of equilateral triangles |
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ABC and ABC are perpendicular. |
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Plane ABC is perpendicular to planes BDC and BEA. |
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along the straight line a, the lines of their intersection with the plane γ |
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are straight lines b and c. |
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In a rectangular parallelepiped ABCDA 1 B 1 C 1 D 1 plane |
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bones AB 1 C 1 and ICA 1 are perpendicular. |
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421. The segment OS is drawn from the center O of the square ABCD perpendicular to its plane.
1°) Determine the relative position of the ACS planes
and ABC.
2°) Determine the relative position of the ACS planes
and BDS.
3) Construct a plane passing through the straight line OS perpendicular to the ABS plane.
4) Construct a plane perpendicular to plane ABC and passing through the midpoints of sides AD and CD.
422. From the intersection point O of the diagonals of the rhombus ABCD, a segment OS is drawn perpendicular to the plane of the rhombus; AB=DB=
1°) Determine the relative position of the SDB and
ABC, SDB and ACS.
2°) Construct a plane passing through line BC perpendicular to plane ABD.
3) Draw a plane perpendicular to plane ABC through the middle F of segment CS.
4) Find the area of triangle BDF.
423. Given a cube ABCDA1 B1 C1 D1.
1°) Determine the relative position of the planes AB 1 C 1
and CDD1.
2°) Determine the relative position of the planes AB 1 C 1
and CD1 A1.
3°) Construct a plane passing through point A perpendicular to the plane BB 1 D 1.
4) Construct a section of the cube with a plane passing through the midpoints of the edges A 1 D 1 and B 1 C 1 perpendicular to the plane ABC. 5) Determine the relative position of the plane AA 1 B and the plane passing through the middle of the ribs A 1 B 1, C 1 D 1, CD.
6) Find the cross-sectional area of the cube by the plane passing through the edge BB 1 and the middle of the edge A 1 D 1 (BB 1 = a).
7) Construct a point symmetrical to point A relative to the plane A 1 B 1 C.
424. In a regular tetrahedron ABCD with an edge of 2 cm, point M is the midpoint of DB, and point N is the midpoint of AC.
1°) Prove that straight line DB is perpendicular to the plane
2°) Prove that the plane BDM is perpendicular to the plane AMC.
3) Through point O of the intersection of the medians of triangle ADC, draw a straight line perpendicular to the AMC plane.
4) Find the length of this line segment inside the tetrahedron. 5) In what ratio does the AMC plane divide this segment?
425. Two equilateral triangles ABC and ADC lie in perpendicular planes.
1°) Find the length of segment BD if AC = 1 cm.
2) Prove that the plane BKD (K lies on the line AC) is perpendicular to the plane of each of the triangles if and only if K is the midpoint of side AC.
426. Rectangle ABCD, whose sides are 3 cm and 4 cm, is bent along the diagonal AC so that triangles ABC and ADC are located in perpendicular planes. Determine the distance between points B and D after bending the rectangle ABCD.
427. Through this point draw a plane perpendicular to each of the two given planes.
428°. Prove that the planes of adjacent faces of a cube are perpendicular.
429. Planes α and β are perpendicular to each other. From point A of plane α, a straight line AB is drawn perpendicular to plane β. Prove that line AB lies in the α plane.
430. Prove that if a plane and a line not lying in this plane are perpendicular to the same plane, then they are parallel to each other.
431. Through points A and B lying on the line of intersection p of planes α and β perpendicular to each other, straight lines perpendicular to p are drawn: AA 1 in α, BB 1 in β. Point X lies on straight line AA 1, and point Y lies on BB 1. Prove that straight line ВB 1 is perpendicular to straight line ВХ, and straight line АА 1 is perpendicular to straight line АY.
432*. Through the middle of each side of the triangle a plane is drawn perpendicular to this side. Prove that all three drawn planes intersect along one straight line perpendicular to the plane of the triangle.
Exercises to repeat
433. In an equilateral triangle with side b determine: 1) height; 2) radii of the inscribed and circumscribed circles.
434. From one point a perpendicular and two oblique lines are drawn to a given line. Determine the length of the perpendicular if the inclined ones are 41 cm and 50 cm, and their projections onto this line are in the ratio 3:10.
435. Determine the legs of a right triangle if bis- the sectrix of a right angle divides the hypotenuse into segments of 15 cm and
Basic definition
The two planes are called
are perpendicular , if each of them is formed by straight lines- mi, perpendicular- mi of the second plane and passing through the intersection points of these planes.
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Perpendicular sign |
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perpendicular |
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the second plane, then |
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these planes are per- |
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pendicular. |
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perpend- |
two planes |
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orifice |
are perpendicular, then |
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intersectionsperpen |
direct, belonging to |
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dicular |
flat |
sharing one plane |
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and perpendicular |
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intersections |
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these planes, per- |
α β, b β, c = α ∩β, |
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pendicular to the second |
b c b α |
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plane. |
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The concept of perpendicular planes
When two planes intersect, we get $4$ dihedral angles. Two angles are equal to $\varphi $, and the other two are equal to $(180)^0-\varphi $.
Definition 1
The angle between planes is the minimum of the dihedral angles formed by these planes.
Definition 2
Two intersecting planes are called perpendicular if the angle between these planes is $90^\circ$ (Fig. 1).
Figure 1. Perpendicular planes
Sign of perpendicularity of two planes
Theorem 1
If a straight line of a plane is perpendicular to another plane, then these planes are perpendicular to each other.
Proof.
Let us be given planes $\alpha $ and $\beta $, which intersect along the straight line $AC$. Let the straight line $AB$ lying in the $\alpha $ plane be perpendicular to the $\beta $ plane (Fig. 2).
Figure 2.
Since the line $AB$ is perpendicular to the plane $\beta$, it is also perpendicular to the line $AC$. Let us additionally draw a line $AD$ in the plane $\beta$, perpendicular to the line $AC$.
We find that the angle $BAD$ is the linear angle of the dihedral angle, equal to $90^\circ$. That is, by definition 1, the angle between the planes is $90^\circ$, which means that these planes are perpendicular.
The theorem is proven.
The following theorem follows from this theorem.
Theorem 2
If a plane is perpendicular to the line along which two other planes intersect, then it is also perpendicular to these planes.
Proof.
Let us be given two planes $\alpha $ and $\beta $ intersecting along the straight line $c$. The plane $\gamma $ is perpendicular to the straight line $c$ (Fig. 3)
Figure 3.
Since the line $c$ belongs to the plane $\alpha $ and the plane $\gamma $ is perpendicular to the line $c$, then, by Theorem 1, the planes $\alpha $ and $\gamma $ are perpendicular.
Since the line $c$ belongs to the plane $\beta $ and the plane $\gamma $ is perpendicular to the line $c$, then, by Theorem 1, the planes $\beta $ and $\gamma $ are perpendicular.
The theorem is proven.
For each of these theorems, the converse statements are also true.
Sample problems
Example 1
Let us be given a rectangular parallelepiped $ABCDA_1B_1C_1D_1$. Find all pairs of perpendicular planes (Fig. 5).
Figure 4.
Solution.
By the definition of a rectangular parallelepiped and perpendicular planes, we see the following eight pairs of planes perpendicular to each other: $(ABB_1)$ and $(ADD_1)$, $(ABB_1)$ and $(A_1B_1C_1)$, $(ABB_1)$ and $(BCC_1) $, $(ABB_1)$ and $(ABC)$, $(DCC_1)$ and $(ADD_1)$, $(DCC_1)$ and $(A_1B_1C_1)$, $(DCC_1)$ and $(BCC_1)$, $(DCC_1)$ and $(ABC)$.
Example 2
Let us be given two mutually perpendicular planes. From a point on one plane a perpendicular is drawn to another plane. Prove that this line lies in the given plane.
Proof.
Let us be given perpendicular planes $\alpha $ and $\beta $ intersecting along the straight line $c$. From point $A$ of the plane $\beta $ a perpendicular $AC$ is drawn to the plane $\alpha $. Let us assume that $AC$ does not lie in the $\beta$ plane (Fig. 6).
Figure 5.
Consider triangle $ABC$. It is rectangular with right angle $ACB$. Therefore, $\angle ABC\ne (90)^0$.
But on the other hand, $\angle ABC$ is the linear angle of the dihedral angle formed by these planes. That is, the dihedral angle formed by these planes is not equal to 90 degrees. We find that the angle between the planes is not equal to $90^\circ$. Contradiction. Therefore, $AC$ lies in the $\beta$ plane.
Recall that planes are called perpendicular if the angle between them is right. And this angle is determined like this. Take point O on line C, along which the planes intersect, and draw straight lines through it in the planes (Fig. 1.9a). The angle between a and b is the angle between . When this angle is right, then they say that the planes are mutually perpendicular and write
You, of course, have already noticed that when , then of the three straight lines a, b, c, any two are mutually perpendicular (Fig. 2.28). In particular, . Therefore (based on the perpendicularity of a straight line and a plane). Likewise,
So, each of two mutually perpendicular planes contains a perpendicular to the other plane. Moreover, these perpendiculars fill mutually perpendicular planes. (Fig. 2.29).
Let us prove the last statement. Indeed, if through any point of the plane a we draw a straight line
Then (by Theorem 5 on the parallelism of perpendiculars).
And to indicate the perpendicularity of planes, one perpendicular to the plane is enough.
Theorem 7. (test for perpendicularity of planes). If a plane passes through a perpendicular to another plane, then these planes are mutually perpendicular.
Let the plane a contain a line a perpendicular to the plane P (Fig. 2.28). Then straight line a intersects plane P at point O. Point O lies on line C along which they intersect. Let us draw a straight line in the plane P through point O. Since b also lies in the P plane, then,
This feature has a simple practical meaning: the plane of a door hung on a jamb perpendicular to the floor is perpendicular to the plane of the floor in any position of the door (Fig. 2.1). Another practical application of this feature: when you need to check whether a flat surface (wall, fence, etc.) is installed vertically, this is done using a plumb line - a rope with a load. The plumb line is always directed vertically, and the wall stands vertically if at any point the plumb line, located along it, does not deviate.
When solving problems involving perpendicular planes, the following three sentences are often used.
Proposition 1. A line lying in one of two mutually perpendicular planes and perpendicular to their common line is perpendicular to the other plane.
Let the planes be mutually perpendicular and intersect along straight line C. Let, further, straight line a lie in the plane a and (Fig. 2.28). Line a intersects line C at some point O. Let us draw a line b through point O in the plane P, perpendicular to line c. So that's how it is. Since , then (by Theorem 2).
The second sentence is the opposite of the first.
Proposition 2. A straight line that has a common point with one of two mutually perpendicular planes and is perpendicular to the other plane lies in the first of them.
Let the planes be mutually perpendicular, and let straight line a have a common point A with plane a (Fig. 2.30). Through point A in plane a we draw a straight line perpendicular to straight line C - the line of intersection of the planes. According to the proposal, since in space only one line perpendicular to a given plane passes through each point, then the lines a and coincide. Since a lies in the plane, then a lies in the plane
Proposition 3. If two planes perpendicular to a third plane intersect, then the line of their intersection is perpendicular to the third plane.
Let two planes intersecting along a straight line a be perpendicular to the plane y (Fig. 2.31). Then through any point of line a we draw a line perpendicular to the plane y. According to Proposition 2, this line lies both in the plane a and in the plane P, that is, it coincides with the line a. So,
Lecture on the topic “Test of perpendicularity of two planes”
The idea of a plane in space allows us to obtain, for example, the surface of a table or wall. However, a table or wall has finite dimensions, and the plane extends beyond its boundaries to infinity.Consider two intersecting planes. When they intersect, they form four dihedral angles with a common edge.
Let's remember what a dihedral angle is.
In reality, we encounter objects that have the shape of a dihedral angle: for example, a slightly open door or a half-open folder.
When two planes alpha and beta intersect, we obtain four dihedral angles. Let one of the dihedral angles be equal to (phi), then the second is equal to (180 0 –), third, fourth (180 0 -).
α Andβ, 0°< 90 °
Consider the case when one of the dihedral angles is 90 0 .
Then, all dihedral angles in this case are equal to 90 0 .
dihedral angle between planesα Andβ,
90º
Let us introduce the definition of perpendicular planes:
Two planes are called perpendicular if the dihedral angle between them is 90°.
The angle between the sigma and epsilon planes is 90 degrees, which means the planes are perpendicular
Because =90°
Let us give examples of perpendicular planes.
Wall and ceiling.
Side wall and table top.
Wall and ceiling
Let us formulate a sign of perpendicularity of two planes:
THEOREM:If one of two planes passes through a line perpendicular to the other plane, then these planes are perpendicular.
Let's prove this sign.
By condition it is known that the straight lineAM lies in the α plane, the straight line AM is perpendicular to the β plane,
Prove: planes α and β are perpendicular.
Proof:
1) Planes α andβ intersect along the straight line AR, and AM AR, since AM β by condition, that is, AM is perpendicular to any straight line lying in the β plane.
2) Let us draw a straight line in the β planeAT perpendicularAR.
We get the angle TAM is the linear angle of the dihedral angle. But angle TAM = 90°, since MA is β. So α β.
Q.E.D.
THEOREM:If a plane passes through a line perpendicular to another plane, then these planes are perpendicular.
Given:α, β, AM α, AMβ, AM∩=A
Prove: αβ.
Proof:
1) α∩β = AR, while AM AR, since AM β by condition, that is, AM is perpendicular to any straight line lying in the β plane.
2) ATβ,ATAR.
TAM is the linear angle of the dihedral angle. TAM = 90°, because MA β. So α β.
Q.E.D
From the sign of perpendicularity of two planes we have an important corollary:
IMPACT:A plane perpendicular to a line along which two planes intersect is perpendicular to each of these planes.
Let us prove this corollary: if the gamma plane is perpendicular to the line c, then, based on the parallelism of the two planes, gamma is perpendicular to alpha. Likewise, gamma is perpendicular to beta
That is: if α∩β=с and γс, then γα and γβ.
becauseγс and сα from the sign of perpendicularity γα.
Similar to γ β
Let us reformulate this corollary for a dihedral angle:
The plane passing through the linear angle of a dihedral angle is perpendicular to the edge and faces of this dihedral angle. In other words, if we have constructed a linear angle of a dihedral angle, then the plane passing through it is perpendicular to the edge and faces of this dihedral angle.
Task.
Given: ΔАВС, С = 90°, АС lies in the plane α, the angle between the planes α andABC= 60°, AC = 5 cm, AB = 13 cm.
Find: the distance from point B to plane α.
Solution:
1) Let's construct VC α. Then KS is the projection of the sun onto this plane.
2) BC AC (by condition), which means, according to the theorem of three perpendiculars (TPP), KS AC. Therefore, VSK is the linear angle of the dihedral angle between the plane α and the plane of the triangle ABC. That is, VSK = 60°.
3) From ΔBCA according to the Pythagorean theorem:
From ΔVKS: 
This lesson will help those wishing to gain an understanding of the topic “The sign of perpendicularity of two planes.” At the beginning of it, we will repeat the definition of dihedral and linear angles. Then we will consider which planes are called perpendicular, and prove the sign of perpendicularity of two planes.
Topic: Perpendicularity of lines and planes
Lesson: Sign of perpendicularity of two planes
Definition. A dihedral angle is a figure formed by two half-planes that do not belong to the same plane and their common straight line a (a is an edge).
Rice. 1
Let's consider two half-planes α and β (Fig. 1). Their common border is l. This figure is called a dihedral angle. Two intersecting planes form four dihedral angles with a common edge.
A dihedral angle is measured by its linear angle. We choose an arbitrary point on the common edge l of the dihedral angle. In the half-planes α and β, from this point we draw perpendiculars a and b to the straight line l and obtain the linear angle of the dihedral angle.
Straight lines a and b form four angles equal to φ, 180° - φ, φ, 180° - φ. Recall that the angle between straight lines is the smallest of these angles.
Definition. The angle between planes is the smallest of the dihedral angles formed by these planes. φ is the angle between planes α and β, if
Definition. Two intersecting planes are called perpendicular (mutually perpendicular) if the angle between them is 90°.
Rice. 2
An arbitrary point M is selected on edge l (Fig. 2). Let us draw two perpendicular straight lines MA = a and MB = b to the edge l in the α plane and in the β plane, respectively. We got the angle AMB. Angle AMB is the linear angle of a dihedral angle. If the angle AMB is 90°, then the planes α and β are called perpendicular.
Line b is perpendicular to line l by construction. Line b is perpendicular to line a, since the angle between planes α and β is 90°. We find that line b is perpendicular to two intersecting lines a and l from the plane α. This means that straight line b is perpendicular to plane α.
Similarly, we can prove that straight line a is perpendicular to plane β. Line a is perpendicular to line l by construction. Line a is perpendicular to line b, since the angle between planes α and β is 90°. We find that line a is perpendicular to two intersecting lines b and l from the plane β. This means that straight line a is perpendicular to plane β.
If one of two planes passes through a line perpendicular to the other plane, then such planes are perpendicular.
Prove:
Rice. 3
Proof:
Let planes α and β intersect along straight line AC (Fig. 3). To prove that the planes are mutually perpendicular, you need to construct a linear angle between them and show that this angle is 90°.
The straight line AB is perpendicular to the plane β, and therefore to the straight line AC lying in the plane β.
Let us draw a straight line AD perpendicular to a straight line AC in the β plane. Then BAD is the linear angle of the dihedral angle.
The straight line AB is perpendicular to the plane β, and therefore to the straight line AD lying in the plane β. This means that the linear angle BAD is 90°. This means that the planes α and β are perpendicular, which is what needed to be proven.
The plane perpendicular to the line along which two given planes intersect is perpendicular to each of these planes (Fig. 4).
Prove:
Rice. 4
Proof:
The straight line l is perpendicular to the plane γ, and the plane α passes through the straight line l. This means that, based on the perpendicularity of planes, planes α and γ are perpendicular.
The straight line l is perpendicular to the plane γ, and the plane β passes through the straight line l. This means that according to the perpendicularity of planes, planes β and γ are perpendicular.