I will be brief. The angle between two lines is equal to the angle between their direction vectors. Thus, if you manage to find the coordinates of the direction vectors a \u003d (x 1; y 1; z 1) and b \u003d (x 2; y 2; z 2), you can find the angle. More precisely, the cosine of the angle according to the formula:

Let's see how this formula works on specific examples:

Task. Points E and F are marked in the cube ABCDA 1 B 1 C 1 D 1 - the midpoints of the edges A 1 B 1 and B 1 C 1, respectively. Find the angle between lines AE and BF.

Since the edge of the cube is not specified, we set AB = 1. We introduce a standard coordinate system: the origin is at point A, and the x, y, z axes are directed along AB, AD, and AA 1, respectively. The unit segment is equal to AB = 1. Now let's find the coordinates of the direction vectors for our lines.

Find the coordinates of the vector AE. To do this, we need points A = (0; 0; 0) and E = (0.5; 0; 1). Since the point E is the middle of the segment A 1 B 1 , its coordinates are equal to the arithmetic mean of the coordinates of the ends. Note that the origin of the vector AE coincides with the origin, so AE = (0.5; 0; 1).

Now let's deal with the BF vector. Similarly, we analyze the points B = (1; 0; 0) and F = (1; 0.5; 1), because F - the middle of the segment B 1 C 1 . We have:
BF = (1 - 1; 0.5 - 0; 1 - 0) = (0; 0.5; 1).

So, the direction vectors are ready. The cosine of the angle between the lines is the cosine of the angle between the direction vectors, so we have:

Task. In a regular trihedral prism ABCA 1 B 1 C 1 , all edges of which are equal to 1, points D and E are marked - the midpoints of the edges A 1 B 1 and B 1 C 1, respectively. Find the angle between lines AD and BE.

Let's introduce a standard coordinate system: the origin is at point A, the x-axis is directed along AB, z - along AA 1 . We direct the y axis so that the OXY plane coincides with the ABC plane. The unit segment is equal to AB = 1. Find the coordinates of the direction vectors for the desired lines.

First, let's find the coordinates of the AD vector. Consider the points: A = (0; 0; 0) and D = (0.5; 0; 1), because D - the middle of the segment A 1 B 1 . Since the beginning of the vector AD coincides with the origin, we get AD = (0.5; 0; 1).

Now let's find the coordinates of the vector BE. Point B = (1; 0; 0) is easy to calculate. With point E - the middle of the segment C 1 B 1 - a little more difficult. We have:

It remains to find the cosine of the angle:

Task. 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, the points K and L are marked - the midpoints of the edges A 1 B 1 and B 1 C 1, respectively. Find the angle between lines AK and BL.

We introduce a standard coordinate system for a prism: we place the origin of coordinates at the center of the lower base, direct the x-axis along FC, the y-axis through the midpoints of segments AB and DE, and the z-axis vertically upwards. The unit segment is again equal to AB = 1. Let us write out the coordinates of the points of interest to us:

Points K and L are the midpoints of the segments A 1 B 1 and B 1 C 1, respectively, so their coordinates are found through the arithmetic mean. Knowing the points, we find the coordinates of the direction vectors AK and BL:

Now let's find the cosine of the angle:

Task. In a regular quadrangular pyramid SABCD, all edges of which are equal to 1, points E and F are marked - the midpoints of the sides SB and SC, respectively. Find the angle between lines AE and BF.

We introduce a standard coordinate system: the origin is at point A, the x and y axes are directed along AB and AD, respectively, and the z axis is directed vertically upwards. The unit segment is equal to AB = 1.

Points E and F are the midpoints of the segments SB and SC, respectively, so their coordinates are found as the arithmetic mean of the ends. We write down the coordinates of the points of interest to us:
A = (0; 0; 0); B = (1; 0; 0)

Knowing the points, we find the coordinates of the direction vectors AE and BF:

The coordinates of the vector AE coincide with the coordinates of point E, since point A is the origin of coordinates. It remains to find the cosine of the angle:

Using the coordinate method when calculating an angle

between planes

The most general method for finding an anglebetween planes - the method of coordinates (sometimes - with the involvement of vectors). It can be used when all the others have been tried. But there are situations in which it makes sense to apply the coordinate method 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 the coordinate axes can be set. Such polyhedra are a rectangular parallelepiped and a regular quadrangular pyramid. In the first case, the coordinate system can be set by the edges emerging 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 is introduced in space. It is desirable to introduce it in a "natural" way - "attach" 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 write such an equation is to know the coordinates of three points in the plane that do not lie on one straight line.

The plane equation in general form has the form Ax + By + Cz + D = 0.

Coefficients A, B, C 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 the scalar product of 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 modulus is in the numerator of the right side of the formula.

Solve the following problem using the coordinate method.

Problem 1. A cube ABCDA 1 B 1 C 1 D 1 is given. Point K is the midpoint of edge AD, point L is the midpoint of edge CD. What is the angle between the planes A 1 KL and A 1 AD?

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

Rice. 3

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

We express the coefficients A, B, C through D and come to 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 the planes, we get:

This article is about the angle between planes and how to find it. First, the definition of the angle between two planes is given and a graphic illustration is given. After that, the principle of finding the angle between two intersecting planes by the coordinate method was analyzed, a formula was obtained that allows calculating the angle between intersecting planes using the known coordinates of the normal vectors of these planes. In conclusion, detailed solutions of typical problems are shown.

Page navigation.

Angle between planes - definition.

Let us give arguments that will allow us to gradually approach the definition of the angle between two intersecting planes.

Let us be given two intersecting planes and . These planes intersect in a straight line, which we denote by the letter c. Let's construct a plane passing through the point M of the line c and perpendicular to the line c. In this case, the plane will intersect the planes and . Denote the line along which the planes intersect and as a, and the line along which the planes intersect and as b. Obviously, the lines a and b intersect at the point M.

It is easy to show that the angle between the intersecting lines a and b does not depend on the location of the point M on the line c through which the plane passes.

Let us construct a plane perpendicular to the line c and different from the plane . The plane is intersected by the planes and along straight lines, which we denote by a 1 and b 1, respectively.

From the method of constructing planes and it follows that the lines a and b are perpendicular to the line c, and the lines a 1 and b 1 are perpendicular to the line c. Since the lines a and a 1 lie in the same plane and are perpendicular to the line c, they are parallel. Similarly, lines b and b 1 lie in the same plane and are perpendicular to line c, hence they are parallel. Thus, it is possible to perform a parallel transfer of the plane to the plane, in which the line a 1 coincides with the line a, and the line b with the line b 1. Therefore, the angle between two intersecting lines a 1 and b 1 is equal to the angle between intersecting lines a and b .

This proves that the angle between the intersecting lines a and b lying in the intersecting planes and does not depend on the choice of the point M through which the plane passes. Therefore, it is logical to take this angle as the angle between two intersecting planes.

Now you can voice the definition of the angle between two intersecting planes and .

Definition.

The angle between two planes intersecting in a straight line and is the angle between two intersecting lines a and b, along which the planes and intersect with the plane perpendicular to the line c.

The definition of the angle between two planes can be given a little differently. If on the line c, along which the planes intersect, mark the point M and draw lines through it a and b, perpendicular to the line c and lying in the planes and, respectively, then the angle between the lines a and b is the angle between the planes and. Usually, in practice, such constructions are performed in order to obtain the angle between the planes.

Since the angle between the intersecting lines does not exceed , it follows from the voiced definition that the degree measure of the angle between two intersecting planes is expressed by a real number from the interval . In this case, intersecting planes are called perpendicular if the angle between them is ninety degrees. The angle between parallel planes is either not determined at all, or it is considered equal to zero.

Finding the angle between two intersecting planes.

Usually, when finding the angle between two intersecting planes, you first have to perform additional constructions in order to see the intersecting lines, the angle between which is equal to the desired angle, and then connect this angle with the original data using equal signs, similarity signs, the cosine theorem or the definitions of sine, cosine and the tangent of the angle. In the geometry course of high school, there are similar problems.

For example, let's give a solution to problem C2 from the Unified State Examination in mathematics for 2012 (the condition is intentionally changed, but this does not affect the principle of the solution). In it, it was just necessary to find the angle between two intersecting planes.

Example.

Decision.

First, let's make a drawing.

Let's perform additional constructions to "see" the angle between the planes.

First, let's define a straight line along which the planes ABC and BED 1 intersect. Point B is one of their common points. Find the second common point of these planes. The lines DA and D 1 E lie in the same plane ADD 1, and they are not parallel, and, therefore, intersect. On the other hand, the line DA lies in the plane ABC, and the line D 1 E lies in the plane BED 1, therefore, the intersection point of the lines DA and D 1 E will be a common point of the planes ABC and BED 1. So, we continue the lines DA and D 1 E until they intersect, we denote the point of their intersection with the letter F. Then BF is the straight line along which the planes ABC and BED 1 intersect.

It remains to construct two lines lying in the planes ABC and BED 1, respectively, passing through one point on the line BF and perpendicular to the line BF - the angle between these lines, by definition, will be equal to the desired angle between the planes ABC and BED 1 . Let's do it.

Dot A is the projection of the point E onto the plane ABC. Draw a line that intersects at a right angle the line BF at the point M. Then the line AM is the projection of the line EM onto the plane ABC, and by the three perpendiculars theorem.

Thus, the desired angle between the planes ABC and BED 1 is .

We can determine the sine, cosine or tangent of this angle (and hence the angle itself) from a right triangle AEMif we know the lengths of its two sides. From the condition it is easy to find the length AE: since point E divides side AA 1 in relation to 4 to 3, counting from point A, and the length of side AA 1 is 7, then AE \u003d 4. Let's find the length of AM.

To do this, consider a right triangle ABF with right angle A, where AM is the height. By condition AB=2. We can find the length of the side AF from the similarity of right triangles DD 1 F and AEF :

By the Pythagorean theorem, from the triangle ABF we find . We find the length AM through the area of ​​the triangle ABF: on one side, the area of ​​the triangle ABF is equal to , on the other side , where .

Thus, from the right triangle AEM we have .

Then the desired angle between the planes ABC and BED 1 is (note that ).

Answer:

In some cases, to find the angle between two intersecting planes, it is convenient to specify Oxyz and use the coordinate method. Let's stop on it.

Let's set the task: to find the angle between two intersecting planes and . Let's denote the desired angle as .

We assume that in a given rectangular coordinate system Oxyz we know the coordinates of the normal vectors of the intersecting planes and or it is possible to find them. Let be is the normal vector of the plane, and is the normal vector of the plane . Let us show how to find the angle between intersecting planes and through the coordinates of the normal vectors of these planes.

Let us denote the line along which the planes intersect and as c . Through the point M on the line c we draw a plane perpendicular to the line c. The plane intersects the planes and along the lines a and b, respectively, the lines a and b intersect at the point M. By definition, the angle between intersecting planes and is equal to the angle between intersecting lines a and b.

Let us set aside from the point M in the plane the normal vectors and of the planes and . In this case, the vector lies on a line that is perpendicular to line a, and the vector lies on a line that is perpendicular to line b. Thus, in the plane, the vector is the normal vector of the line a, is the normal vector of the line b.

In the article Finding the angle between intersecting lines, we obtained a formula that allows you to calculate the cosine of the angle between intersecting lines using the coordinates of normal vectors. Thus, the cosine of the angle between the lines a and b, and, consequently, and cosine of the angle between intersecting planes and is found by the formula , where and are the normal vectors of the planes and, respectively. Then it is calculated as .

Let's solve the previous example using the coordinate method.

Example.

A rectangular parallelepiped ABCDA 1 B 1 C 1 D 1 is given, in which AB \u003d 2, AD \u003d 3, AA 1 \u003d 7 and point E divides side AA 1 in a ratio of 4 to 3, counting from point A. Find the angle between the planes ABC and BED 1.

Decision.

Since the sides of a rectangular parallelepiped at one vertex are pairwise perpendicular, it is convenient to introduce a rectangular coordinate system Oxyz as follows: the beginning is aligned with the vertex C, and the coordinate axes Ox, Oy and Oz are directed along the sides CD, CB and CC 1, respectively.

The angle between the planes ABC and BED 1 can be found through the coordinates of the normal vectors of these planes using the formula , where and are the normal vectors of the planes ABC and BED 1, respectively. Let us determine the coordinates of normal vectors.

Problem 1. The base of a straight quadrangular prism ABCDА 1 В 1 С 1 D 1 is a rectangle ABCD, in which AB \u003d 5, AD \u003d 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 line BD 1, if the distance between straight lines AC and B 1 D 1 is 12. Solution. We introduce a coordinate system. В(0;0;0), А(5;0;0), С(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

Problem 2. At the base of the triangular pyramid SABC lies a right triangle ABC. Angle A is straight. AC \u003d 8, BC \u003d 219. The height of the pyramid SA is 6. A point M is taken on the edge AC so that AM \u003d 2. A plane α is drawn through the point M, the vertex B and the 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. We 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 Normal to plane (BMN) Angle between planes Answer: 60°. Equation of the plane (ВМN): N.G. Nenasheva mathematics teacher GBOU secondary school 985

Problem 3. The base of a quadrangular pyramid PABCD is a square with a side equal to 6, the side edge PD is perpendicular to the plane of the base and equals 6. Find the angle between the planes (BDP) and (BCP). Decision. 1. Draw the median DF of an isosceles triangle CDP (BC = PD = 6) So DF PC. And from the fact that BC (CDP), it follows that DF BC 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: The angle between the planes Nenasheva N.G. mathematics teacher GBOU secondary school 985

Problem 3. The base of a quadrangular pyramid PABCD is a square with a side equal to 6, the side edge PD is perpendicular to the plane of the base and equals 6. Find the angle between the planes (BDP) and (BCP). Solution.4. Let's choose a coordinate system. The coordinates of the 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 the planes Answer: Nenasheva N.G. mathematics teacher GBOU secondary school 985

Task 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. Enter a rectangular coordinate system and determine the coordinates of the points: 2. Compose the equation of the plane (AD 1 E): 3. Compose the equation of the plane (D 1 FC): - the normal vector of the plane (AD 1 E). - normal vector of the plane (D 1 FС). Angle between planes x y z Nenasheva N.G. mathematics teacher GBOU secondary school 985

Task 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: The 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 1. For definiteness, consider the faces SAC and SBC 2. Find the coordinates of the point S: E The angle between the 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: The angle between the 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. Equation of the plane (SAC): - normal vector of the plane (SAC). 4. Equation of the plane (SBC): - normal vector of the plane (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 the planes according to the formula Answer: The angle between the planes Nenasheva N.G. mathematics teacher GBOU secondary school 985