The angle between planes is a coordinate vector method. Angle between two intersecting planes

Problem 1. Base of a line quadrangular prism ABCDA 1 B 1 C 1 D 1 – rectangle ABCD, in which AB = 5, AD = 11. Find the tangent of the angle between the plane of the base of the prism and the plane passing through the middle of the edge AD perpendicular to the straight line BD 1, if the distance between the straight lines AC and B 1 D 1 equals 12. Solution. Let's introduce a coordinate system. B(0;0;0), A(5;0;0), C(0;11;0), D 1 (5;11;12) Coordinates of the normal to the section plane: Coordinates of the normal to the base plane: – acute angle, then D A B C D1D1 A1A1 B1B1 C1C1 x y z N Angle between planes Answer: 0.5. Nenasheva N.G. mathematics teacher GBOU secondary school 985

Task 2. At the base triangular pyramid SABC is a right triangle ABC. Angle A is straight. AC = 8, BC = 219. The height of the pyramid SA is 6. Point M is taken on the edge AC so that AM = 2. A plane α is drawn through point M, vertex B and point N - the middle of the edge SC. Find the dihedral angle formed by the plane α and the plane of the base of the pyramid. A S x B C M N y z Solution. Let's introduce a coordinate system. Then A (0;0;0), C (0;8;0), M (0;2;0), N (0;4;3), S (0;0;6), Normal to the plane ( ABC) vector Plane normal (PMN) Angle between planes Answer: 60°. Plane equation (ВМN): Nenasheva N.G. mathematics teacher GBOU secondary school 985

Problem 3. The base of the quadrangular pyramid PABCD is a square with side equal to 6, the side edge PD is perpendicular to the plane of the base and equal to 6. Find the angle between the planes (BDP) and (BCP). Solution. 1. Let's take the median DF isosceles triangle CDP (BC = PD = 6) Means DF PC. And from the fact that BC (CDP), it follows that DF BC, which means DF (PCB) A D C B P F 2. Since AC DB and AC DP, then AC (BDP) 3. Thus, the angle between the planes (BDP) and (BCP ) is found from the condition: Angle between planes Nenasheva N.G. mathematics teacher GBOU secondary school 985

Problem 3. The base of the quadrangular pyramid PABCD is a square with side equal to 6, the side edge PD is perpendicular to the plane of the base and equal to 6. Find the angle between the planes (BDP) and (BCP). Solution.4. Let's choose a coordinate system. Coordinates of points: 5. Then the vectors will have the following coordinates: 6. Calculating the values, we find:, then A D C B P F z x y Angle between planes Answer: Nenasheva N.G. mathematics teacher GBOU secondary school 985

Problem 4. In the unit cube ABCDA 1 B 1 C 1 D 1, find the angle between the planes (AD 1 E) and (D 1 FC), where points E and F are the midpoints of the edges A 1 B 1 and B 1 C 1, respectively. Solution: 1. Let's introduce a rectangular coordinate system and determine the coordinates of the points: 2. Let's create an equation of the plane (AD 1 E): 3. Let's create an equation of the plane (D 1 FC): - normal vector plane (AD 1 E). - normal vector of the plane (D 1 FC). Angle between planes x y z Nenasheva N.G. mathematics teacher GBOU secondary school 985

Problem 4. In the unit cube ABCDA 1 B 1 C 1 D 1, find the angle between the planes (AD 1 E) and (D 1 FC), where points E and F are the midpoints of the edges A 1 B 1 and B 1 C 1, respectively. Solution: 4. Find the cosine of the angle between the planes using the formula Answer: Angle between the planes x y z Nenasheva N.G. mathematics teacher GBOU secondary school 985

Problem 5. The segment connecting the center of the base of a regular triangular pyramid with the middle of the side edge is equal to the side of the base. Find the angle between adjacent side faces of the pyramid. Solution: x y z 1. Let's introduce a rectangular coordinate system and determine the coordinates of points A, B, C: K Let the side of the base be equal to 1. For definiteness, consider the faces SAC and SBC 2. Let's find the coordinates of point S: E Angle between planes Nenasheva N.G. . mathematics teacher GBOU secondary school 985

Problem 5. The segment connecting the center of the base of a regular triangular pyramid with the middle of the side edge is equal to the side of the base. Find the angle between adjacent side faces of the pyramid. Solution: x y z K E SO we find from OSB: Angle between planes Nenasheva N.G. mathematics teacher GBOU secondary school 985

Problem 5. The segment connecting the center of the base of a regular triangular pyramid with the middle of the side edge is equal to the side of the base. Find the angle between adjacent side faces of the pyramid. Solution: x y z K E 3. Plane equation (SAC): - plane normal vector (SAC). 4. Plane equation (SBC): - plane normal vector (SBC). Angle between planes Nenasheva N.G. mathematics teacher GBOU secondary school 985

Problem 5. The segment connecting the center of the base of a regular triangular pyramid with the middle of the side edge is equal to the side of the base. Find the angle between adjacent side faces of the pyramid. Solution: x y z K E 5. Find the cosine of the angle between planes using the formula Answer: Angle between planes Nenasheva N.G. mathematics teacher GBOU secondary school 985

The magnitude of the angle between two different planes can be determined for any relative position of the planes.

A trivial case if the planes are parallel. Then the angle between them is considered equal to zero.

A non-trivial case if the planes intersect. This case is the subject of further discussion. First we need the concept of a dihedral angle.

9.1 Dihedral angle

A dihedral angle is two half-planes with a common straight line (which is called the edge of the dihedral angle). In Fig. 50 shows a dihedral angle formed by half-planes and; the edge of this dihedral angle is the straight line a, common to these half-planes.

Rice. 50. Dihedral angle

The dihedral angle can be measured in degrees or radians in a word, enter the angular value of the dihedral angle. This is done as follows.

On the edge of the dihedral angle formed by the half-planes and, we take an arbitrary point M. We draw rays MA and MB, respectively lying in these half-planes and perpendicular to the edge (Fig. 51).

Rice. 51. Linear dihedral angle

The resulting angle AMB is the linear angle of the dihedral angle. The angle " = \AMB is precisely the angular value of our dihedral angle.

Definition. The angular magnitude of a dihedral angle is the magnitude of the linear angle of a given dihedral angle.

All linear angles of a dihedral angle are equal to each other (after all, they are obtained from each other by a parallel shift). That's why this definition correct: the value " does not depend on the specific choice of point M on the edge of the dihedral angle.

9.2 Determining the angle between planes

When two planes intersect, four dihedral angles are obtained. If they all have the same size (90 each), then the planes are called perpendicular; The angle between the planes is then 90.

If not all dihedral angles are the same (that is, there are two acute and two obtuse), then the angle between the planes is the value of the acute dihedral angle (Fig. 52).

Rice. 52. Angle between planes

9.3 Examples of problem solving

Let's look at three problems. The first is simple, the second and third are approximately at level C2 on the Unified State Examination in mathematics.

Problem 1. Find the angle between two faces of a regular tetrahedron.

Solution. Let ABCD regular tetrahedron. Let us draw the medians AM and DM of the corresponding faces, as well as the height of the tetrahedron DH (Fig. 53).

Rice. 53. To task 1

Being medians, AM and DM are also altitudes of equilateral triangles ABC and DBC. Therefore, the angle " = \AMD is the linear angle of the dihedral angle formed by the faces ABC and DBC. We find it from the triangle DHM:

			1 AM

Answer: arccos 1 3 .

Problem 2. In a regular quadrangular pyramid SABCD (with vertex S), the side edge is equal to the side of the base. Point K is the middle of edge SA. Find the angle between the planes

Solution. Line BC is parallel to AD and thus parallel to plane ADS. Therefore, plane KBC intersects plane ADS along straight line KL parallel to BC (Fig. 54).

Rice. 54. To task 2

In this case, KL will also be parallel to line AD; therefore, KL is the midline of triangle ADS, and point L is the midpoint of DS.

Let's find the height of the pyramid SO. Let N be the middle of DO. Then LN is the midline of triangle DOS, and therefore LN k SO. This means LN is perpendicular to plane ABC.

From point N we lower the perpendicular NM to the straight line BC. The straight line NM will be the projection of the inclined LM onto the ABC plane. From the theorem of three perpendiculars it then follows that LM is also perpendicular to BC.

Thus, the angle " = \LMN is the linear angle of the dihedral angle formed by the half-planes KBC and ABC. We will look for this angle from right triangle LMN.

Let the edge of the pyramid be equal to a. First we find the height of the pyramid:

SO=p

Solution. Let L be the intersection point of lines A1 K and AB. Then plane A1 KC intersects plane ABC along straight line CL (Fig.55).

A C

Rice. 55. To problem 3

Triangles A1 B1 K and KBL are equal along the leg and sharp corner. Therefore, the other legs are equal: A1 B1 = BL.

Consider triangle ACL. In it BA = BC = BL. Angle CBL is 120; therefore, \BCL = 30 . Also, \BCA = 60 . Therefore \ACL = \BCA + \BCL = 90 .

So, LC? AC. But line AC serves as a projection of line A1 C onto plane ABC. By the theorem of three perpendiculars we then conclude that LC ? A1 C.

Thus, angle A1 CA is the linear angle of the dihedral angle formed by the half-planes A1 KC and ABC. This is the desired angle. From the isosceles right triangle A1 AC we see that it is equal to 45.

Using the coordinate method when calculating an angle

between planes

Most general method finding the anglebetween planes - the coordinate method (sometimes using vectors). It can be used when all others have been tried. But there are situations in which the coordinate method makes sense to apply immediately, namely when the coordinate system is naturally related to the polyhedron specified in the problem statement, i.e. Three pairwise perpendicular lines are clearly visible, on which coordinate axes can be specified. Such polyhedra are a rectangular parallelepiped and a regular quadrangular pyramid. In the first case, the coordinate system can be specified by edges extending from one vertex (Fig. 1), in the second - by the height and diagonals of the base (Fig. 2)

The application of the coordinate method is as follows.

A rectangular coordinate system in space is introduced. It is advisable to introduce it in a “natural” way - to “link” it to a trio of pairwise perpendicular lines that have a common point.

For each of the planes, the angle between which is sought, an equation is drawn up. The easiest way to create such an equation is to know the coordinates of three points on the plane that do not lie on the same line.

Equation of the plane in general view looks like Ax + By + Cz + D = 0.

Coefficients A, B, The Cs in this equation are the coordinates of the normal vector of the plane (the vector perpendicular to the plane). We then determine the lengths and dot product normal vectors to the planes, the angle between which is sought. If the coordinates of these vectors(A 1, B 1; C 1) and (A 2; B 2; C 2 ), then the desired anglecalculated by the formula

Comment. It must be remembered that the angle between vectors (as opposed to the angle between planes) can be obtuse, and in order to avoid possible uncertainty, the numerator on the right side of the formula contains a modulus.

Solve this problem using the coordinate method.

Problem 1. Given a cube ABCDA 1 B 1 C 1 D 1 . Point K is the middle of edge AD, point L is the middle of edge CD. Why equal to the angle between planes A 1 KL and A 1 AD?

Solution . Let the origin of the coordinate system be at the point A, and the coordinate axes go along the rays AD, AB, AA 1 (Fig. 3). Let’s take the edge of the cube to be equal to 2 (it’s convenient to divide it in half). Then the coordinates of the points A 1 , K, L are as follows: A 1 (0; 0; 2), K(1; 0; 0), L(2; 1; 0).

Rice. 3

Let us write down the equation of the plane A 1 K L in general terms. Then we substitute the coordinates of the selected points of this plane into it. We obtain a system of three equations with four unknowns:

Let's express the coefficients A, B, C through D and we arrive at the equation

Dividing both parts into D (why D = 0?) and then multiplying by -2, we get the equation of the plane A 1 KL: 2x - 2 y + z - 2 = 0. Then the normal vector to this plane has coordinates (2: -2; 1). Plane equation A 1 AD is: y=0, and the coordinates of the normal vector to it, for example, (0; 2: 0). According to the above formula for the cosine of the angle between planes, we obtain:

Goals:

develop the ability to consider different approaches to solving problems and analyze the “effect” of using these methods of solution;
develop the student’s ability to choose a method for solving a problem in accordance with his mathematical preferences, based on more solid knowledge and confident skills;
develop the ability to draw up a plan of successive stages to achieve results;
develop the ability to justify all steps and calculations taken;
repeat and consolidate various topics and issues of stereometry and planimetry, typical stereometric structures related to solving current problems;
develop spatial thinking.

analysis various methods problem solving: coordinate vector method, application of the cosine theorem, application of the theorem of three perpendiculars;
comparison of the advantages and disadvantages of each method;
repetition of the properties of a cube, triangular prism, regular hexagon;
preparation for passing the Unified State Exam;
development of independence in decision making.

Lesson outline

In a cube ABCDA 1 B 1 C 1 D 1 with edge 1 point O – center of face ABCD.

a) the angle between straight lines A 1 D And B.O.;

b) distance from point B to the middle of the segment A 1 D.

Solution to point a).

Let's place our cube in a rectangular coordinate system as shown in the figure, the vertices A 1 (1; 0; 1), D (1; 1; 0), B 1 (0; 0; 1), O (½; ½; 0).

Direction vectors of straight lines A 1 D And B 1 O:

(0; 1; -1) and (½; ½; -1);

we find the desired angle φ between them using the formula:

cos∠φ = ,
whence ∠φ = 30°.

Method 2. We use the cosine theorem.

1) Let's draw a straight line B 1 C parallel to the line A 1 D. Corner CB 1 O will be what you are looking for.

2) From a right triangle BB 1 O according to the Pythagorean theorem:

3) By the theorem of cosines from a triangle CB 1 O calculate the angle CB 1 O:

cos CB 1 O = , the required angle is 30°.

Comment. When solving the problem in the 2nd way, you can notice that according to the theorem of three perpendiculars COB 1 = 90°, therefore from rectangular ∆ CB 1 O It is also easy to calculate the cosine of the desired angle.

Solution to point b).

1 way. Let's use the formula for the distance between two points

Let the point E– middle A 1 D, then the coordinates E (1; 1/2; ½), B (0; 0; 0).

BE = .

Method 2. According to the Pythagorean theorem

From rectangular ∆ B.A.E. with direct B.A.E. we find BE = .

In a regular triangular prism ABCA 1 B 1 C 1 all edges are equal a. Find the angle between lines AB And A 1 C.

1 way. Coordinate vector method

Coordinates of the vertices of the prism in a rectangular system when the prism is positioned as in the figure: A (0; 0; 0), B (a; ; 0), A 1 (0; 0; a), C (0; a; 0).

Direction vectors of straight lines A 1 C And AB:

(0; a; -a) And (a; ; 0} ;

cos φ = ;

Method 2. We use the cosine theorem

We consider ∆ A 1 B 1 C, in which A 1 B 1 || AB. We have

cos φ = .

(From the collection of the Unified State Examination 2012. Mathematics: standard exam options edited by A.L. Semenov, I.V. Yashchenko)

In a regular hexagonal prism ABCDEFA 1 B 1 C 1 D 1 E 1 F 1, all edges of which are equal to 1, find the distance from the point E to a straight line B 1 C 1.

1 way. Coordinate vector method

1) Place the prism in a rectangular coordinate system, placing the coordinate axes as shown in the figure. SS 1, NE And SE are perpendicular in pairs, so you can direct the coordinate axes along them. We get the coordinates:

C 1 (0; 0; 1), E (; 0; 0), B 1 (0;1;1).

2) Find the coordinates of the direction vectors for the lines From 1 to 1 And C 1 E:

(0;1;0), (;0;-1).

3) Find the cosine of the angle between From 1 to 1 And C 1 E, using the scalar product of vectors and :

cos β = = 0 => β = 90° => C 1 E – the required distance.

4)C 1 E = = 2.

Conclusion: knowledge of various approaches to solving stereometric problems allows you to choose the method that is preferable for any student, i.e. one that the student masters confidently, helps to avoid mistakes, leads to successful solution of the problem and getting a good score on the exam. Coordinate method has an advantage over other methods in that it requires less stereometric considerations and vision, and is based on the use of formulas that have many planimetric and algebraic analogies that are more familiar to students.

The form of the lesson is a combination of the teacher’s explanation with the frontal collective work of students.

The polyhedra in question are shown on the screen using a video projector, which allows comparison various ways solutions.

Homework: solve problem 3 in another way, for example using the three perpendicular theorem .

Literature

1. Ershova A.P., Goloborodko V.V. Independent and tests in geometry for grade 11. – M.: ILEKSA, – 2010. – 208 p.

2. Geometry, 10-11: textbook for educational institutions: basic and profile levels / L.S. Atanasyan, V.F. Butuzov, S.B. Kadomtsev et al. - M.: Education, 2007. - 256 p.

3. Unified State Exam-2012. Mathematics: standard exam options: 10 options / ed. A.L. Semenova, I.V. Yashchenko. – M.: National Education, 2011. – 112 p. – (USE-2012. FIPI - school).

Problem 1.6. Given cube. M, N, P - the midpoints of the edges, AB, BC, respectively. Find the angle between planes (MNP) and

a) Let us introduce a rectangular Cartesian coordinate system as indicated in Figure 17. The length of the edge of the cube can be chosen arbitrarily, since with homothety the angle between the planes does not change. It is convenient, for example, to take the length of an edge of a cube equal to 2.

Relative to the selected coordinate system, we find the coordinates of points and vectors:

b) Let be the normal vector of the plane.

In this case, the conditions are met

Similarly, if is a normal vector of the plane, then

c) If, then

Answer:

Problem 1.7. At the base of the regular triangular pyramid SABC lies the regular with side equal to 2. Edge SA is perpendicular to the plane of the base and SA = 1. Points P, Q are the midpoints of edges SB, NE, respectively. The plane is parallel to lines SC and AB, and the plane is parallel to lines AQ and CP. Determine the magnitude of the angle between the planes and.

a) Let us choose a rectangular Cartesian coordinate system as indicated in Figure 18. In the selected coordinate system we have: