The relation of parallelism of planes, its properties and applications is considered.

A visual representation of the location of the two

planes gives modeling using the planes of the surfaces of adjacent walls, ceiling and floor of the room, bunk beds, two sheets of paper fastened

magicians, etc. (Fig. 242–244).

Although there are an infinite number of options for the relative position of various planes, to establish and characterize which measurements of angles and distances will be used in the future, we will first focus on those where the classification (as well as straight lines with planes) is based on the number of their common points.

1. Two planes have at least three common points that do not lie on the same line. Such planes coincide (axiom C 2, §7).

2. The common points of two planes are located on one straight line, which is the line of intersection of these planes (axiom C 3, §7). Such planes intersect.

3. The two planes have no common points.

IN in this case they are called parallel-

Two planes are called parallel if they have no common points.

The parallelism of planes is indicated by the sign ||: α || β.

As always, when introducing geometric concepts arose

There is no problem with their existence. The existence of intersecting-

Xia planes is a characteristic feature of space,

and we have already used this many times. Less obvious is

The existence of parallel planes is revealed. There is no

doubts that, for example, planes of opposite graphs

The cubes are parallel, that is, they do not intersect. But directly

Indeed, by definition, this cannot be established. For solving

understanding of the question posed, as well as other issues related to

parallelism of planes, it is necessary to have a sign of parallelism.

To search for a sign, it is advisable to consider a plane,

“woven” from straight lines. It is obvious that each straight line is one of

parallel planes must be parallel to the other.

Otherwise, the planes will have a common point. Enough

Is the plane β exactly parallel to the same straight line α

so that planes α and β are parallel? Absolutely

but, no (justify this!). Practical experience shows that

two such intersecting lines are sufficient. To secure

on the mast there is a platform parallel to the ground, just place it

If the planes are parallel to the second plane, then these planes are parallel.

 Let the intersecting lines a and b of the plane α be parallel to the plane β. Let us prove that the planes α and β are parallel by contradiction. To do this, let us assume that the planes α and β intersect along a straight line

t (Fig. 246). Lines a and b cannot intersect lines according to the condition. However, then in the plane α two straight lines are drawn through one point that do not intersect with the straight line, that is, parallel to it. This is a contradiction

and completes the proof of the theorem.

The sign of parallelism of planes is used when placing horizontally flat designs (concrete slabs, floor, disk of goniometer devices, etc.) using two levels placed in the plane of the structure on intersecting straight lines. Based on this feature, it is possible to construct a plane parallel to this one.

Problem 1. Through a point lying outside a given plane, draw a plane parallel to the given one.

 Let the plane β and a point M outside the plane be given (Fig. 247, a). Let us draw through point M two intersecting lines a and b, parallel to the plane β. To do this, you need to take two intersecting straight lines c and d in the β plane (Fig. 247, b). Then through point M draw lines a and b parallel to lines c and d, respectively.

but (Fig. 247, c).

Intersecting lines a and b parallel to the plane β, based on the parallelism of the line and the plane (Theorem 1 §11). They uniquely define the plane α. According to the proven criterion, α || β.

Example 1. Given a cube ABCDA 1 B 1 C 1 D 1 , points M , N , P are the midpoints of the edges BC , B 1 C 1 , A 1 D 1 , respectively. Install mutual arrangement planes: 1)ABV 1 and PNM; 2) NMA and A 1 C 1 C ; 3)A 1 NM

and PC 1 C; 4) MAD 1 and DB 1 C.

 1) The planes ABB 1 and РNM (Fig. 248) are parallel, based on the parallelism of the planes (Theorem 1). Indeed, the lines РN and NM intersect and are parallel to the plane ABB 1, based on the parallelism of the line and the plane (Theorem 1 §11), because the segments РN and NM connect the midpoints opposite sides squares, so they are parallel to the sides of the squares:

РN ||A 1 B 1 ,NM ||В 1 B.

2) Planes NMA and A 1 C 1 C intersect along straight line AA 1 (Fig. 249). Indeed, the lines AA 1 and CC 1 are parallel, based on the parallelism of the lines (AA 1 ||ВB 1,ВB 1 ||СC 1). Therefore, straight line AA 1 lies in the plane A 1 C 1 C. The belonging of straight line AA 1 to the plane NMA is similarly justified.

3) Planes A 1 NM and РС 1 C (Fig. 250) are parallel, based on the parallelism of the planes. Indeed, NM ||С 1 C . Therefore, the straight line NM is parallel to the plane PC 1 C. The segments PC 1 and A 1 N are also parallel, since the quadrilateral PC 1 NA 1 is a parallelogram (A 1 P ||NC 1, A 1 P =NC 1). Thus, line A 1 N is parallel to plane PC 1 C. Lines A 1 N and NM intersect.

4) The planes MAD 1 and DB 1 C intersect (Fig. 251). Although the line of their intersection is not easy to construct, it is not difficult to indicate one point of this line. Indeed, the lines A 1 D and B 1 C are parallel, since the quadrilateral A 1 B 1 CD is a parallelogram (A 1 B 1 = AB = CD , A 1 B 1 || AB , AB || CD ). Therefore, line A 1 D belongs to the plane DB 1 C. Lines A 1 D and AD 1 intersect at a point common to the planes MAD 1 and DB 1 C.

If two intersecting lines of one plane are respectively parallel to two lines of another plane, then these planes are parallel.

Using the criterion of parallelism of a line and a plane (Theorem 1 §11), it is easy to establish that the condition of Theorem 1 follows from the conditions of Theorem 1. The application of the theorem inverse to the criterion of parallelism of a line and a plane (Theorem 2 §11) completes the justification for the equivalence of the conditions of Theorems 1 and 1 ′.

Naturally, the question arises about the uniqueness of the construction given in Problem 1. Since we will have to use this property more than once, we will highlight it as a separate theorem. However, let's look at another statement first.

Theorem 2 (about the intersection of two parallel planes with a third).

If two parallel planes are intersected by a third plane, then the lines of intersection of the planes are parallel.

 Let parallel planes α, β and a plane γ intersecting them be given (Fig. 252). Let us denote the intersection lines

through a and b. These lines lie in the γ plane and do not intersect, since the α and β planes do not have common points. Therefore, directly

a and b are parallel.

Theorem 3 (on the existence and uniqueness of a plane parallel to this one).

Through a point located outside a given plane, one can draw a single plane parallel to the given one.

 The construction of such a plane was carried out in problem 1. We will prove the uniqueness of the construction by contradiction. Let us assume that two different planes α and γ are drawn through point M, pa-

parallel planes β (Fig. 253), and straight line t is the line of their intersection. Let us draw a plane δ through the point M, intersecting the line

m and the β plane (how can this be done?). Let us denote by a and b

the line of intersection of the plane δ with the planes α and γ, and through c - the line of intersection of the planes δ and β (Fig. 253). According to Theorem 2,a ||c

and b ||s. That is, in the δ plane through

two straight lines parallel to straight lines pass through point M. A contradiction indicates that the assumption is incorrect.

The relation of parallelism of planes has a number of properties that have analogues in planimetry.

Theorem 4 (on segments of parallel lines between parallel planes).

Segments of parallel lines cut off by parallel planes are equal to each other.

 Let two parallel planes α and β and segments be given AB

and CD of parallel straight lines a and d, cut off by these planes (Fig. 254, a). Let us draw the plane γ through straight lines a and d (Fig. 254, b). It intersects the planes α and β along straight lines AC and BD, which, according to Theorem 2, are parallel. Therefore, quadrilateral ABCD is a parallelogram; its opposite sides AC and BD are equal.

From the above property it follows that if we plot from all points of the plane

on one side of the plane parallel lines of the same length, then the ends of these segments form two parallel planes. It is on this property that the construction of a parallelepiped using the deposition of segments is based (Fig. 255).

Theorem 5 (on the transitivity of the relation of parallelism of planes).

If each of two planes is parallel to a third, then the two planes are parallel to each other.

 Let the planes α and β be parallel to the plane γ. Let's assume that

α and β are not parallel. Then the planes α and β have a common point, and through this point there pass two different planes parallel to the plane γ, which contradicts Theorem 3. Therefore, the planes α and β do not have common points, that is, they are parallel.

Theorem 5 is another sign of parallelism of planes. It is widely used in both geometry and practical activities. For example, in multi-storey building the parallelism of the floor and ceiling planes on each floor guarantees their parallelism on different floors.

Problem 2. Prove that if a straight line intersects the plane α, then it also intersects every plane parallel to the plane α.

 Let planes α and β be parallel, and straight line a intersects plane α at point A. Let us prove that it also intersects the plane

β. Let's assume that this is not the case. Then straight line a is parallel to plane β. Let us draw the plane γ through a straight line and an arbitrary point of the plane β (Fig. 256).

This plane intersects parallel planes α and β along straight lines b is. Co-

according to Theorem 2, b || c, that is, in the plane γ, two lines a and b pass through point A, parallel to line c . This contradiction proves the statement.

Try to prove on your own that if the plane α intersects the plane β, then it also intersects every plane parallel to the plane β.

Example 2. In the tetrahedron ABCD, points K, F, E are the midpoints of edges DA, DC, DB, aM and P - the centers of mass of faces ABD and ВСD, respectively.

1) Establish the relative position of the planes KEF and ABC;

DEF and ABC.

2) Construct the intersection line of the AFB and KEC planes.

3) Find the cross-sectional area of ​​the tetrahedron by a plane parallel to the plane ABD and passing through point P if all the edges of the tetrahedron are equal.

 Let’s construct a drawing that meets the condition (Fig. 257, a). 1) The planes KEF and ABC are parallel, based on the parallelism of the planes (Theorem 1’): the intersecting lines KE and KF of the KEF plane are parallel to the intersecting lines AB and AC of the ABC plane (the midlines of the corresponding

existing triangles).

Planes DEF and ABC intersect along straight line BC, since straight line BC belongs to both planes, and they cannot coincide - points A, B, C, D do not lie in the same plane.

2) Plane AFB intersects with plane KEC along a straight line containing point P, since lines CE and BF lying in these planes are in the plane BCD and intersect at point P. Another point is the point of intersection Q of straight lines AF and CK in the plane ACD (Fig. 257, b). Obviously, this point is the center of mass of the ACD face. The required intersection is the line PQ.

3) Construct the section specified in the condition, using the sign of parallelism of planes. Let us draw lines through points P and Q parallel to lines DB and DA, respectively (Fig. 257, c). These lines intersect the segment CD at point L. The latter follows from the property of the center of mass of a triangle - it divides the medians of the triangle in a ratio of 2: 1, counting from the vertex. It remains to apply Thales' theorem. Thus, the PLQ and BDA planes are parallel. The required section is triangle LSN.

By construction, triangles BCD and SCL are similar with similarity coefficient CE CP =3 2. Therefore LS =3 2 BD . Similar to the established

the following equalities are added: LN =3 2 AD,NS =3 2 AB. It follows that the triangles LSN and ABD are similar with a similarity coefficient of 3 2. According to the properties of the areas of similar triangles,

S LNS =4 9 S ABD . It remains to find the area of ​​triangle ABD. By-

since, by condition, all edges of the tetrahedron are equal to a, then S ABD =4 3 a 2.

The required area is 3 1 3 a 2 .

It is appropriate to note that the answer depends only on the area of ​​the face ABD. Therefore, the equality of all edges is only a means of finding this area. Thus, this task can be significantly generalized.

Answer. 1)KEF ||ABC ; 3)3 1 3 a 2 .

 Test questions

1. Is it true that two planes are parallel if every line lying in one plane is parallel to the other plane?

2. Planes α and β are parallel. Are there skew lines lying in these planes?

3. Two sides of a triangle are parallel to a certain plane. Is the third side of the triangle parallel to this plane?

4. Two sides of a parallelogram are parallel to a certain plane. Is it true that the plane of a parallelogram is parallel to the given plane?

5. Can segments of two straight lines cut off by parallel planes be unequal?

6. Can the cross section of a cube be an isosceles trapezoid? Can the cross section of a cube be a regular pentagon? Is it true that two planes parallel to the same line are parallel to each other?

The lines of intersection of planes α and β with plane γ are parallel to each other. Are planes α and β parallel?

Can three faces of a cube be parallel to the same plane?

Graphic exercises

1. Fig. 258 shows the cube ABCDA 1 B 1 C 1 D 1, points M, N, K, L, P are the midpoints of the corresponding edges. Fill out the table according to the example given, selecting the required location of the α and β planes.

Mutual

location

α || β α = β

α × β α || β α = β

2. In Fig. 259 shows a tetrahedron ABCD, points K, F, M, N, Q are the midpoints of the corresponding edges. Please indicate:

1) a plane passing through point K parallel to plane ABC;

2) a plane passing through the line BD parallel to the plane MNQ.

3. Determine what is the section of a figure by a plane passing through the given three points shown in the figure.

kah 260, a)–e) and 261, a)–d).

4. Construct a drawing based on the given data.

1) From the vertices of a parallelogram ABCD lying in one of two parallel planes, parallel lines are drawn that intersect the second plane at points A 1 , B 1 , C 1 , D 1 , respectively.

2) Triangle A 1 B 1 C 1 is the projection of triangle ABC onto the plane α parallel to it. Point M is the middle of the sun, M 1 is the projection of point M onto the plane α.

207. In the cube ABCDA 1 B 1 C 1 D 1 points O, O 1 are the centers of faces ABCD and A 1 B 1 C 1 D 1, respectively, M is the middle of edge AB.

1°) Determine the relative position of the planes MO 1 O

and ADD 1, ABD 1 and CO 1 C 1.

2°) Construct the point of intersection of the plane DCC 1 and straight line MO 1 and the line of intersection of the planes MCC 1 and A 1 D 1 C 1.

3) Find the cross-sectional area of ​​a cube by a plane parallel to the plane AD 1 C 1 and passing through the point O 1 if the edge of the cube is equal to a.

208. In the tetrahedron ABCD, the points K, L, P are the centers of mass of the faces ABD, BDC, ABC, respectively, and aM is the middle of the edge AD.

1°) Determine the relative position of the ACD planes

and KLP ; MLK and ABC .

2°) Construct the point of intersection of plane ABC and line ML and the line of intersection of planes MKL and ABC.

3) Find the cross-sectional area of ​​the tetrahedron by a plane passing through the points K, L and M parallel to the straight line AD, if all the edges of the tetrahedron are equal.

209. Given a cube ABCDA 1 B 1 C 1 D 1 . Points L, M, M 1 are the midpoints of edges AB, AD and A 1 D 1, respectively.

1°) Determine the relative position of the planes B 1 D 1 D

and LMM1.

2) Construct a plane passing through point M parallel to the plane ACC 1.

3) Construct a section of the cube with a plane passing through point M 1 parallel to the plane CDD 1.

4) Determine the relative position of the planes MA 1 B 1

and CDM1.

5) Construct a plane passing through the line C 1 D 1 parallel to the plane CDM 1.

210. In a regular quadrangular pyramidSABCD, all edges are equal to each other. Points L, M and N are the midpoints of the edges AS, BS, CS, respectively.

1°) Determine the relative position of: straight lines LM and BC; straight line LN and plane ABD; planes LMN and BDC.

2°) Prove that triangles ABC and LMN are similar.

3) Construct a section of the pyramid using the plane AMN; plane LMN; planeLBC.

4*) Which of the sections of the pyramid passing through the vertex S has the largest area?

Parallelism of lines and planes

In the SABC tetrahedron all faces are regular triangles. Points L, M and N are the midpoints of the edges AS, BS, CS, respectively. 1°) Determine the relative position of straight lines LM and BC. 2°) Determine the relative position of straight line LN and plane ABC.

3) Prove that triangles LMN and ABC are similar.

Construct a section of the cube that is: 1°) an equilateral triangle; 2) a pentagon.

217. Construct a section of a tetrahedron that is a parallelogram.

218°. Prove that opposite faces parallelepipeds are parallel.

219. Prove that the set of all lines passing through this point and parallel to a given plane, forms a plane parallel to the given one.

220. Given four points A, B, C, D, not lying in the same plane. Prove that each plane parallel to the lines AB and CD intersects the lines AC, AD, BD, BC at the vertices of the parallelogram.

221. Prove that a plane and a line not belonging to this plane are parallel to each other if both of them are parallel to the same plane.

222. Through the point O of the intersection of the diagonals of the cube ABCDA 1 B 1 C 1 D 1 a plane is drawn parallel to the face ABCD. This plane intersects edges BB 1 and CC 1 at points M and N, respectively. Prove that angle MON is a right angle.

223. Prove that two planes are parallel to each other if and only if every straight line intersecting one of the planes also intersects the second.

224*. In a triangular pyramid SABC, through segments AD and CE, where D is the midpoint SB, and E is the midpoint SA, draw sections of the pyramid parallel to each other.

225. Find geometric places:

1) the midpoints of all segments with ends on two data parallel planes; 2*) midpoints of segments with ends on two given intersecting lines.

226*. Side AB of triangle ABC lying in plane α is parallel to plane β. The equilateral triangle 1 B 1 C 1 is a parallel projection of the triangle ABC onto the plane β; AB = 5, BC = 6, AC = 9.

1) Establish the relative position of straight lines AB and A 1 B 1,

BC and B1 C1, A1 C1 and AC.

2) Find the area of ​​triangle A 1 B 1 C 1.

227*. Given two intersecting lines. Indicate the set of all points in space through which a line can be drawn intersecting each of two given lines.

Basic definition

The two planes are called

are parallel,

if they do not have common points.

Main statements

Parallel sign - If two intersecting straight lines of one plane of the plane are respectively parallel to two straight lines of the second plane, then these planes

the bones are parallel.

Theorem on intersecting If two parallel-intersecting two non-parallel planes are intersected by a third plane, then the lines of the third intersection of the plane

they are parallel.

a α,b α,a ×b ,c β,d β,a ||c ,b ||d α || β

α || β, a = γ∩α,b = γ∩βa ||b

M α

β: α || β,M β

Getting ready for thematic

for assessment on the topic “Parallelism of lines and planes”

Self-control tasks

1. The four points do not belong to the same plane. Can some three of them lie on the same straight line?

2. Can three different planes have exactly two points in common?

3. Can two skew lines be parallel to a third line at the same time?

4. Is it true that straight a and b are not parallel if there is no line c parallel to a and b?

5. Can equal segments have unequal projections?

6. Can a ray be a parallel projection of a line?

7. Can a square be an image of a cube?

8. Is it true that through a given point in space only one plane can be drawn parallel to a given line?

9. Is it always possible to draw a line through a given point parallel to two given planes that do not contain this point?

10. Is it possible to draw parallel planes through two intersecting lines?

Answers to tasks for self-control

Test sample

Two parallelograms ABCD and ABC 1 D 1 lie in different planes.

1°) Determine the relative position of straight lines CD and C 1 D 1.

2°) Determine the relative position of the straight line C 1 D 1 and the plane

3°) Construct the line of intersection of the planes DD 1 C 1 and ВСС 1.

4°) Determine the relative position of the planes ADD 1 and BCC 1.

5) Through point M, dividing segment AB in a ratio of 2:1, counting from point A, draw a plane α parallel to plane C 1 BC. 6) Construct the point of intersection of the straight line AC with the plane α and find the ratio in which this point divides the segment AC.

Relative position of straight lines and planes in space

Trigonometric

You have already dealt with trigonometric functions in geometry lessons. Until now, their applications were mainly limited to solving triangles, that is, we were talking about finding some elements of a triangle from others. From the history of mathematics it is known that the emergence of trigonometry is associated with the measurement of lengths and angles. However, now the sphere

her applications are much wider than in ancient times.

The word "trigonometry" comes from the Greek τριγωνον

(trigonon) – triangle and µετρεω (metreo) – measure, measure-

I bark. Literally it means measuring triangles.

IN This chapter systematizes the material already known to you from the geometry course, and continues the study trigonometric functions and their applications to characterize batch processes, in particular rotational movement, oscillatory processes, etc.

Most applications of trigonometry relate specifically to periodic processes, that is, processes that repeat at regular intervals of time. Sunrise and sunset, changes in seasons, rotation of the wheel - these are the simplest examples of such processes. Mechanical and electromagnetic vibrations are also important examples of periodic processes. Therefore, the study of periodic processes is an important task. And the role of mathematics in its solution is decisive.

getting ready to study the topic “Trigonometric functions”

It is advisable to begin studying the topic “Trigonometric functions” by reviewing the definitions and properties of trigonometric functions of angles of triangles and their applications for solving both right and arbitrary triangles.

Sine, cosine, tangent, cotangent of rectangular angles

triangle

Table 24

The sine of an acute angle is the ratio opposite side to the hypotenuse:

sin α = a c .

The cosine of an acute angle is the ratio of the adjacent leg to the hypotenuse:

cosα = b c .

The tangent of an acute angle is the ratio of the opposite side to the adjacent side:

tg α =a b .

The cotangent of an acute angle is the ratio of the adjacent side to the opposite side:

ctgα = a b .

Sine, cosine, tangent, cotangent of angles from 0° to 180°

Table 25

sin α = R y ; cosα = R x ;

tg α = x y ; cotgα = x y.

(X;at) - point coordinates A located on the upper semicircle, α - the angle formed by the radius OA circle with axis X.

Values ​​of sine, cosine, tangent, cotangent

some corners

Table 26

Corner t

Solving arbitrary triangles

Table 27

Theorem of sines

The sides of a triangle are proportional to the sines of opposite angles:

sin aα = sin bβ = sin cγ .

Cosine theorem

The square of an arbitrary side of a triangle is equal to the sum of the squares of the other two sides without twice the product of these sides by the cosine of the angle between them:

c2 = a2 + b2 − 2 ab cos γ , b2 = a2 + c2 − 2 ac cos β , a2 = b2 + c2 − 2 bc cos α .

The area of ​​a triangle is equal to half the product of its two sides and the sine of the angle between them:

S=1 2 absinγ = 1 2 acsinβ = 1 2 bcsinα .

Basic trigonometric identities

5. Compare the sum of sines sharp corners arbitrary right triangle (we denote it byA) with one.

< 1. B.A= 1.

> 1. G. It's impossible to compare. Arrange the numbers in ascending order: A= sin 30°, b= cos 30°,

= tg 30°.

< b<c.B.a<c<b

15. Specify the coordinates of the pointA, lying on a circle of radius 1 (see figure).

(−1; 0).B.(1; 0).

(0; − 1). G.(0; 1).A.IN.

Two planes in space may be parallel or may intersect, as shown in the following table.

Signs of parallelism of two planes

The first sign of parallelism of two planes. If two intersecting linesintersecting lines, lying in the same plane, respectively parallelparallel two straight lines lying in another plane, then such planes are parallel.

Proof . Consider Figure 1, which shows the planes α and β

Lines a and b lie in the plane α and intersect at point K. Lines c and d lie in the β plane and are parallel to lines a and b, respectively.

We will prove the first sign of parallelism of two planes using the “by contradiction” method. To do this, assume that the planes α and β are not parallel. Consequently, the planes α and β must intersect, and intersect along some straight line. Let us denote the straight line along which the planes α and β intersect with the letter l (Fig. 2) and use the sign of parallelism between the straight line and the plane.

The plane α passes through a line a parallel to the line c and intersects the plane β along the line l. From here, due to , we conclude that the lines a and l are parallel. At the same time, plane α passes through line b, parallel to line d, and intersects plane β along line l. From here, due to the parallelism feature of the line and the plane, we conclude that the lines b and l are parallel. Thus, we have obtained that on the plane α two lines pass through point K, namely, lines a and b , which are parallel to line l. The resulting contradiction with axiom of parallel lines makes it possible to assert that the assumption that the planes α and β intersect is incorrect. The proof of the first sign of parallelism of two planes is completed.

The second sign of parallelism of two planes. If two intersecting lines lying in one plane are parallel to another plane, then such planes are parallel.

Proof . Consider Figure 3, which shows the planes α and β.

This figure also shows lines a and b, which lie in the plane α and intersect at point K. By condition, each of the lines a and b is parallel to the plane β. We need to prove that planes α and β are parallel.

The proof of this statement is similar to the proof of the first criterion for the parallelism of two planes, and we leave it to the reader as a useful exercise.

On our website you can also familiarize yourself with educational materials developed by teachers of the Resolventa training center for preparing for the Unified State Exam in mathematics.

e property of pa parallel lines, called transitiveparallelism:

• If two lines a and b are parallel to a third line c, then they are parallel us to each other.

But it is more difficult to prove this property in stereometry. On a plane, nonparallel lines must intersect and therefore cannot be parallel to a third line at the same time (otherwise the parallel axiom is violated). In proin space there are non-parallel andvolume of disjoint linesif they lie in different planes. Such straight lines are said to cross.

In Fig. 4 shows a cube; straight lines AB and BC intersect, AB and CDare parallel, and AB and B WITH interbreed. In the future, we will often resort to the help of a cube to illustratetriage the concepts and facts of stereometry. Our cube is glued together from six square faces. Based on this, we will derive its other properties. For example, we can say that line AB is parallel to CD,because both of them are parallel to the common side of the CD withsquares holding them.

In stereometry, the parallelism relation is also considered for planes: two planesA line or a line and a plane are parallel if they do not have common points. It is convenient to consider a straight line and a plane to be parallel even when it lies in the plane. For planes and straight lines the following theorems on transitivity are valid:

• If two planes are parallel to a third plane, then they are parallel to each other.
• If a line and a plane are parallel to some line (or plane), then they are parallel to each other.

The most important special case of the second theorem is the sign of parallelism between a line and a plane:

• A line is parallel to a plane if it is parallel to some line in this plane.

And here is a sign of parallel planes:

• If two intersecting lines in one plane are respectively parallel to two intersecting lines in another plane, then the planes are parallel.

The following simple theorem is often used:

• The lines along which two parallel planes intersect with a third are parallel to each other.

Let's look at the cube again (Fig. 4). From the sign of parallelism between a line and a plane it follows, for example, that straight line A IN parallel to plane ABCD (since it is parallel to line AB in this plane), and the opposite faces of the cube, in particular A IN WITH D and ABCD, parallel based on the parallelism of planes: straight lines A B and B WITH in one face are respectively parallel to straight lines AB and BC in the other. And a slightly less simple example. Plane containing parallel lines AA and SS, intersect parallel planes ABCD and A B C D along straight lines AC and A WITH, this means that these lines are parallel: similarly, parallel lines B C and A D. Therefore, parallel planes AB C and A DC intersecting the cube in triangles.

III. Image of spatial figures.

There is such an aphorism Geometryit's a temptationability to reason correctly on an incorrect drawing. Indeed, if we return toBased on the above reasoning, it turns out:

the only benefit we got from the accompanying drawing of the cube was that it saved us some space in explainingNI notations. It could just as easily be depicted as the body in Fig. 4, I, although, obviously, the something represented on it is not only not a cube, but also not a polyhedron. And yet, the above aphorism contains only part of the truth. After all, before discussingpresent a finished proof, it must bethink. And for this you need to clearly imagine the given figure, the relationships between its elements. A good drawing helps to develop such an idea. Moreover, as we will see, in stereometry a successful drawing cancan become not just an illustration, but the basis for solving a problem.

An artist (or rather, a realist artist) ondraws our cube the way we see it (Fig. 5, b), i.e. in perspective, or centralno projection. With a central projection from point O (projection center) onto the plane a,an arbitrary point X is represented by a point X at which a intersects the straight line OX (Fig. 6). Central projection maintains straightnesslinear arrangement of points, but, as a rule, transforms parallel lines into intersectionschanging, not to mention the fact that it changes distances and angles. Studying its properties atled to the emergence of an important section of geometry (see article Projective geometry).

But in geometric drawings a different projection is used. We can say that it is obtained from the central one when the center O moves away to infinity and the straight lines OX become paparallel.

Let us choose a plane a and a straight line l intersecting it. Let's draw a straight line through point X, paparallel l. The point X at which this line meets a is a parallel projection of X onto the plane, a along the straight line l (Fig. 7). Aboutthe projection of a figure consists of the projections of all its points. In geometry, the image of a figure is its parallel projection.

Specifically, the image of a straight lineis it a straight line or (in exceptional cases)tea, when the line is parallel to the direction of projection) point. There is a parallel in the image

This article will study the issues of parallelism of planes. Let us define planes that are parallel to each other; let us denote the signs and sufficient conditions of parallelism; Let's look at the theory with illustrations and practical examples.

Yandex.RTB R-A-339285-1 Definition 1

Parallel planes– planes that do not have common points.

To indicate parallelism, use the following symbol: ∥. If two planes are given: α and β, which are parallel, a short notation about this will look like this: α ‖ β.

In the drawing, as a rule, planes parallel to each other are displayed as two equal parallelograms, offset relative to each other.

In speech, parallelism can be denoted as follows: planes α and β are parallel, and also - plane α is parallel to plane β or plane β is parallel to plane α.

Parallelism of planes: sign and conditions of parallelism

In the process of solving geometric problems, the question often arises: are the given planes parallel to each other? To answer this question, use the parallelism feature, which is also a sufficient condition for the parallelism of planes. Let's write it down as a theorem.

Theorem 1

Planes are parallel if two intersecting lines of one plane are correspondingly parallel to two intersecting lines of another plane.

The proof of this theorem is given in the geometry program for grades 10-11.

In practice, to prove parallelism, the following two theorems are used, among other things.

Theorem 2

If one of the parallel planes is parallel to the third plane, then the other plane is either also parallel to this plane or coincides with it.

Theorem 3

If two divergent planes are perpendicular to a certain line, then they are parallel.

Based on these theorems and the sign of parallelism itself, the fact that any two planes are parallel is proven.

Let us consider in more detail the necessary and sufficient condition for the parallelism of the planes α and β, defined in a rectangular coordinate system of three-dimensional space.

Let us assume that in a certain rectangular coordinate system, a plane α is given, which corresponds to the general equation A 1 x + B 1 y + C 1 z + D 1 = 0, and a plane β is also given, which is determined by a general equation of the form A 2 x + B 2 y + C 2 z + D 2 = 0 .

Theorem 4

For the given planes α and β to be parallel, it is necessary and sufficient that the system of linear equations A 1 x + B 1 y + C 1 z + D 1 = 0 A 2 x + B 2 y + C 2 z + D 2 = 0 has no solution (was incompatible).

Proof

Let us assume that the given planes defined by the equations A 1 x + B 1 y + C 1 z + D 1 = 0 and A 2 x + B 2 y + C 2 z + D 2 = 0 are parallel and therefore have no common points . Thus, there is not a single point in the rectangular coordinate system of three-dimensional space, the coordinates of which would satisfy the conditions of both plane equations simultaneously, i.e. the system A 1 x + B 1 y + C 1 z + D 1 = 0 A 2 x + B 2 y + C 2 z + D 2 = 0 has no solution. If the specified system has no solutions, then there is not a single point in the rectangular coordinate system of three-dimensional space whose coordinates would simultaneously satisfy the conditions of both equations of the system. Consequently, the planes defined by the equations A 1 x + B 1 y + C 1 z + D 1 = 0 and A 2 x + B 2 y + C 2 z + D 2 = 0 do not have a single common point, i.e. they are parallel.

Let us analyze the use of the necessary and sufficient condition for the parallelism of planes.

Example 1

Two planes are given: 2 x + 3 y + z - 1 = 0 and 2 3 x + y + 1 3 z + 4 = 0. It is necessary to determine whether they are parallel.

Solution

Let's write a system of equations from the given conditions:

2 x + 3 y + z - 1 = 0 2 3 x + y + 1 3 z + 4 = 0

Let's check whether it is possible to solve the resulting system of linear equations.

The rank of the matrix 2 3 1 2 3 1 1 3 is equal to one, since the second order minors are equal to zero. The rank of the matrix 2 3 1 1 2 3 1 1 3 - 4 is two, since the minor 2 1 2 3 - 4 is non-zero. Thus, the rank of the main matrix of the system of equations is less than the rank of the extended matrix of the system.

At the same time, from the Kronecker-Capelli theorem it follows: the system of equations 2 x + 3 y + z - 1 = 0 2 3 x + y + 1 3 z + 4 = 0 has no solutions. This fact proves that the planes 2 x + 3 y + z - 1 = 0 and 2 3 x + y + 1 3 z + 4 = 0 are parallel.

Note that if we had used the Gaussian method to solve the system of linear equations, it would have given the same result.

Answer: the given planes are parallel.

The necessary and sufficient condition for the parallelism of planes can be described differently.

Theorem 5

For two non-coinciding planes α and β to be parallel to each other, it is necessary and sufficient that the normal vectors of the planes α and β are collinear.

The proof of the formulated condition is based on the definition of the normal vector of the plane.

Let us assume that n 1 → = (A 1 , B 1 , C 1) and n 2 → = (A 2 , B 2 , C 2) are normal vectors of the planes α and β, respectively. Let us write down the condition for collinearity of these vectors:

n 1 → = t · n 2 ⇀ ⇔ A 1 = t · A 2 B 1 = t · B 2 C 1 = t · C 2 , where t is a real number.

Thus, for the non-coinciding planes α and β with the normal vectors given above to be parallel, it is necessary and sufficient that there be a real number t for which the equality is true:

n 1 → = t n 2 ⇀ ⇔ A 1 = t A 2 B 1 = t B 2 C 1 = t C 2

Example 2

In a rectangular coordinate system of three-dimensional space, planes α and β are specified. The plane α passes through the points: A (0, 1, 0), B (- 3, 1, 1), C (- 2, 2, - 2). The β plane is described by the equation x 12 + y 3 2 + z 4 = 1 It is necessary to prove the parallelism of the given planes.

Solution

Let's make sure that the given planes do not coincide. Indeed, this is so, since the coordinates of point A do not correspond to the equation of the plane β.

The next step is to determine the coordinates of the normal vectors n 1 → and n 2 → corresponding to the planes α and β. We will also check the condition for the collinearity of these vectors.

Vector n 1 → can be specified by taking the vector product of vectors A B → and A C → . Their coordinates are respectively: (- 3, 0, 1) and (- 2, 2, - 2). Then:

n 1 → = A B → × A C → = i → j → k → - 3 0 1 - 2 1 - 2 = - i → - 8 j → - 3 k → ⇔ n 1 → = (- 1 , - 8 , - 3)

To obtain the coordinates of the normal vector of the plane x 12 + y 3 2 + z 4 = 1, we reduce this equation to the general equation of the plane:

x 12 + y 3 2 + z 4 = 1 ⇔ 1 12 x + 2 3 y + 1 4 z - 1 = 0

Thus: n 2 → = 1 12, 2 3, 1 4.

Let's check whether the condition of collinearity of vectors n 1 → = (- 1 , - 8 , - 3) and n 2 → = 1 12 , 2 3 , 1 4 is satisfied

Since - 1 = t · 1 12 - 8 = t · 2 3 - 3 = t · 1 4 ⇔ t = - 12, then the vectors n 1 → and n 2 → are related by the equality n 1 → = - 12 · n 2 → , i.e. are collinear.

Answer: planes α and β do not coincide; their normal vectors are collinear. Thus, the planes α and β are parallel.

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