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 graphical illustration is given. After this, the principle of finding the angle between two intersecting planes using the coordinate method was analyzed, and a formula was obtained that allows you to calculate the angle between intersecting planes using known coordinates normal vectors of these planes. In conclusion it is shown detailed solutions characteristic tasks.

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Angle between planes - definition.

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

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

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

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

From the method of constructing planes it follows that lines a and b are perpendicular to line c, and lines a 1 and b 1 are perpendicular to line c. Since lines a and a 1 lie in the same plane and are perpendicular to line c, then they are parallel. Similarly, lines b and b 1 lie in the same plane and are perpendicular to line c, therefore, they are parallel. Thus, it is possible to perform a parallel transfer of the plane to the plane, in which straight line a 1 coincides with straight line a, and straight line b with straight 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 intersecting lines a and b lying in intersecting planes and does not depend on the choice of 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– this 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 straight line c along which the planes and intersect, mark a point M and draw straight lines a and b through it, perpendicular to the straight line c and lying in the planes and, respectively, then the angle between the straight lines a and b is the angle between the planes and. Usually in practice, just such constructions are performed in order to obtain the angle between the planes.

Since the angle between intersecting lines does not exceed , it follows from the stated 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 considered equal to zero.

Finding the angle between two intersecting planes.

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

As an example, let’s give the solution to Problem C2 from the Unified State Exam in Mathematics for 2012 (the condition was intentionally changed, but this does not affect the principle of the solution). In it, you just had to find the angle between two intersecting planes.

Example.

Solution.

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 planes ABC and BED 1 intersect. Point B is one of their common points. Let's find the second common point of these planes. 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, line DA lies in the plane ABC, and line D 1 E lies in the plane BED 1, therefore, the intersection point of lines DA and D 1 E will be the common point of planes ABC and BED 1. So, let's continue the lines DA and D 1 E to their intersection, denoting the point of their intersection with the letter F. Then BF is the straight line along which 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 this.

Dot A is the projection of point E onto plane ABC. Let's draw a straight line intersecting line BF at right angles at point M. Then the straight line AM is the projection of the straight line EM onto the plane ABC, and by the theorem of three perpendiculars.

Thus, the required angle between planes ABC and BED 1 is equal to .

We can determine the sine, cosine or tangent of this angle (and therefore the angle itself) from right triangle AEM, if 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 the ratio of 4 to 3, counting from point A, and the length of side AA 1 is 7, then AE = 4. Let's find the length 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 side AF from the similarity of right triangles DD 1 F and AEF:

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

Thus, from the right triangle AEM we have .

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

Answer:

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

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

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

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

Let us plot the normal vectors and planes and from the point M in the plane. 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 received 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 lines a and b, and, consequently, 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.

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

Solution.

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

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

\(\blacktriangleright\) Dihedral angle is an angle formed by two half-planes and a straight line \(a\), which is their common boundary.

\(\blacktriangleright\) To find the angle between the planes \(\xi\) and \(\pi\) , you need to find the linear angle (and spicy or direct) dihedral angle formed by the planes \(\xi\) and \(\pi\) :

Step 1: let \(\xi\cap\pi=a\) (the line of intersection of the planes). In the plane \(\xi\) we mark an arbitrary point \(F\) and draw \(FA\perp a\) ;

Step 2: carry out \(FG\perp \pi\) ;

Step 3: according to TTP (\(FG\) – perpendicular, \(FA\) – oblique, \(AG\) – projection) we have: \(AG\perp a\) ;

Step 4: The angle \(\angle FAG\) is called the linear angle of the dihedral angle formed by the planes \(\xi\) and \(\pi\) .

Note that the triangle \(AG\) is right-angled.
Note also that the plane \(AFG\) constructed in this way is perpendicular to both planes \(\xi\) and \(\pi\) . Therefore, we can say it differently: angle between planes\(\xi\) and \(\pi\) is the angle between two intersecting lines \(c\in \xi\) and \(b\in\pi\) forming a plane perpendicular to and \(\xi\ ) , and \(\pi\) .

Task 1 #2875

Task level: More difficult than the Unified State Exam

Dana quadrangular pyramid, all edges of which are equal, and the base is a square. Find \(6\cos \alpha\) , where \(\alpha\) is the angle between its adjacent side faces.

Let \(SABCD\) be a given pyramid (\(S\) is a vertex) whose edges are equal to \(a\) . Therefore, everything side faces are equal equilateral triangles. Let's find the angle between the faces \(SAD\) and \(SCD\) .

Let's do \(CH\perp SD\) . Because \(\triangle SAD=\triangle SCD\), then \(AH\) will also be the height of \(\triangle SAD\) . Therefore, by definition, \(\angle AHC=\alpha\) is the linear angle of the dihedral angle between the faces \(SAD\) and \(SCD\) .
Since the base is a square, then \(AC=a\sqrt2\) . Note also that \(CH=AH\) is the height of an equilateral triangle with side \(a\), therefore, \(CH=AH=\frac(\sqrt3)2a\) .
Then, by the cosine theorem from \(\triangle AHC\) : \[\cos \alpha=\dfrac(CH^2+AH^2-AC^2)(2CH\cdot AH)=-\dfrac13 \quad\Rightarrow\quad 6\cos\alpha=-2.\]

Answer: -2

Task 2 #2876

Task level: More difficult than the Unified State Exam

The planes \(\pi_1\) and \(\pi_2\) intersect at an angle whose cosine is equal to \(0.2\). The planes \(\pi_2\) and \(\pi_3\) intersect at right angles, and the line of intersection of the planes \(\pi_1\) and \(\pi_2\) is parallel to the line of intersection of the planes \(\pi_2\) and \(\ pi_3\) . Find the sine of the angle between the planes \(\pi_1\) and \(\pi_3\) .

Let the line of intersection of \(\pi_1\) and \(\pi_2\) be a straight line \(a\), the line of intersection of \(\pi_2\) and \(\pi_3\) be a straight line \(b\), and the line of intersection \(\pi_3\) and \(\pi_1\) – straight line \(c\) . Since \(a\parallel b\) , then \(c\parallel a\parallel b\) (according to the theorem from the section of the theoretical reference “Geometry in space” \(\rightarrow\) “Introduction to stereometry, parallelism”).

Let's mark the points \(A\in a, B\in b\) so that \(AB\perp a, AB\perp b\) (this is possible since \(a\parallel b\) ). Let us mark \(C\in c\) so that \(BC\perp c\) , therefore, \(BC\perp b\) . Then \(AC\perp c\) and \(AC\perp a\) .
Indeed, since \(AB\perp b, BC\perp b\) , then \(b\) is perpendicular to the plane \(ABC\) . Since \(c\parallel a\parallel b\), then the lines \(a\) and \(c\) are also perpendicular to the plane \(ABC\), and therefore to any line from this plane, in particular, the line \ (AC\) .

It follows that \(\angle BAC=\angle (\pi_1, \pi_2)\), \(\angle ABC=\angle (\pi_2, \pi_3)=90^\circ\), \(\angle BCA=\angle (\pi_3, \pi_1)\). It turns out that \(\triangle ABC\) is rectangular, which means \[\sin \angle BCA=\cos \angle BAC=0.2.\]

Answer: 0.2

Task 3 #2877

Task level: More difficult than the Unified State Exam

Given straight lines \(a, b, c\) intersecting at one point, and the angle between any two of them is equal to \(60^\circ\) . Find \(\cos^(-1)\alpha\) , where \(\alpha\) is the angle between the plane formed by lines \(a\) and \(c\) and the plane formed by lines \(b\ ) and \(c\) . Give your answer in degrees.

Let the lines intersect at the point \(O\) . Since the angle between any two of them is equal to \(60^\circ\), then all three straight lines cannot lie in the same plane. Let us mark the point \(A\) on the line \(a\) and draw \(AB\perp b\) and \(AC\perp c\) . Then \(\triangle AOB=\triangle AOC\) as rectangular along the hypotenuse and acute angle. Therefore, \(OB=OC\) and \(AB=AC\) .
Let's do \(AH\perp (BOC)\) . Then by the theorem about three perpendiculars \(HC\perp c\) , \(HB\perp b\) . Since \(AB=AC\) , then \(\triangle AHB=\triangle AHC\) as rectangular along the hypotenuse and leg. Therefore, \(HB=HC\) . This means that \(OH\) ​​is the bisector of the angle \(BOC\) (since the point \(H\) is equidistant from the sides of the angle).

Note that in this way we also constructed the linear angle of the dihedral angle formed by the plane formed by the lines \(a\) and \(c\) and the plane formed by the lines \(b\) and \(c\) . This is the angle \(ACH\) .

Let's find this angle. Since we chose the point \(A\) arbitrarily, let us choose it so that \(OA=2\) . Then in rectangular \(\triangle AOC\) : \[\sin 60^\circ=\dfrac(AC)(OA) \quad\Rightarrow\quad AC=\sqrt3 \quad\Rightarrow\quad OC=\sqrt(OA^2-AC^2)=1.\ ] Since \(OH\) ​​is a bisector, then \(\angle HOC=30^\circ\) , therefore, in a rectangular \(\triangle HOC\) : \[\mathrm(tg)\,30^\circ=\dfrac(HC)(OC)\quad\Rightarrow\quad HC=\dfrac1(\sqrt3).\] Then from the rectangular \(\triangle ACH\) : \[\cos\angle \alpha=\cos\angle ACH=\dfrac(HC)(AC)=\dfrac13 \quad\Rightarrow\quad \cos^(-1)\alpha=3.\]

Answer: 3

Task 4 #2910

Task level: More difficult than the Unified State Exam

The planes \(\pi_1\) and \(\pi_2\) intersect along the straight line \(l\) on which the points \(M\) and \(N\) lie. The segments \(MA\) and \(MB\) are perpendicular to the straight line \(l\) and lie in the planes \(\pi_1\) and \(\pi_2\) respectively, and \(MN = 15\) , \(AN = 39\) , \(BN = 17\) , \(AB = 40\) . Find \(3\cos\alpha\) , where \(\alpha\) is the angle between the planes \(\pi_1\) and \(\pi_2\) .

The triangle \(AMN\) is right-angled, \(AN^2 = AM^2 + MN^2\), whence \ The triangle \(BMN\) is rectangular, \(BN^2 = BM^2 + MN^2\), from which \We write the cosine theorem for the triangle \(AMB\): \ Then \ Since the angle \(\alpha\) between the planes is acute angle, and \(\angle AMB\) turned out to be obtuse, then \(\cos\alpha=\dfrac5(12)\) . Then \

Answer: 1.25

Task 5 #2911

Task level: More difficult than the Unified State Exam

\(ABCDA_1B_1C_1D_1\) is a parallelepiped, \(ABCD\) is a square with side \(a\), point \(M\) is the base of the perpendicular dropped from the point \(A_1\) to the plane \((ABCD)\) , in addition, \(M\) is the point of intersection of the diagonals of the square \(ABCD\) . It is known that \(A_1M = \dfrac(\sqrt(3))(2)a\). Find the angle between the planes \((ABCD)\) and \((AA_1B_1B)\) . Give your answer in degrees.

Let's construct \(MN\) perpendicular to \(AB\) as shown in the figure.

Since \(ABCD\) is a square with side \(a\) and \(MN\perp AB\) and \(BC\perp AB\) , then \(MN\parallel BC\) . Since \(M\) is the point of intersection of the diagonals of the square, then \(M\) is the middle of \(AC\), therefore, \(MN\) is the middle line and \(MN =\frac12BC= \frac(1)(2)a\).
\(MN\) is the projection of \(A_1N\) onto the plane \((ABCD)\), and \(MN\) is perpendicular to \(AB\), then, by the theorem of three perpendiculars, \(A_1N\) is perpendicular to \(AB \) and the angle between the planes \((ABCD)\) and \((AA_1B_1B)\) is \(\angle A_1NM\) .
\[\mathrm(tg)\, \angle A_1NM = \dfrac(A_1M)(NM) = \dfrac(\frac(\sqrt(3))(2)a)(\frac(1)(2)a) = \sqrt(3)\qquad\Rightarrow\qquad\angle A_1NM = 60^(\circ)\]

Answer: 60

Task 6 #1854

Task level: More difficult than the Unified State Exam

In a square \(ABCD\) : \(O\) – the point of intersection of the diagonals; \(S\) – does not lie in the plane of the square, \(SO \perp ABC\) . Find the angle between the planes \(ASD\) and \(ABC\) if \(SO = 5\) and \(AB = 10\) .

Right triangles \(\triangle SAO\) and \(\triangle SDO\) are equal in two sides and the angle between them (\(SO \perp ABC\) \(\Rightarrow\) \(\angle SOA = \angle SOD = 90^\circ\); \(AO = DO\) , because \(O\) – the point of intersection of the diagonals of the square, \(SO\) – the common side) \(\Rightarrow\) \(AS = SD\) \(\Rightarrow\) \(\triangle ASD\) – isosceles. Point \(K\) is the middle of \(AD\), then \(SK\) is the height in the triangle \(\triangle ASD\), and \(OK\) is the height in the triangle \(AOD\) \(\ Rightarrow\) plane \(SOK\) is perpendicular to planes \(ASD\) and \(ABC\) \(\Rightarrow\) \(\angle SKO\) – linear angle equal to the desired dihedral angle.

In \(\triangle SKO\) : \(OK = \frac(1)(2)\cdot AB = \frac(1)(2)\cdot 10 = 5 = SO\)\(\Rightarrow\) \(\triangle SOK\) – isosceles right triangle \(\Rightarrow\) \(\angle SKO = 45^\circ\) .

Answer: 45

Task 7 #1855

Task level: More difficult than the Unified State Exam

In a square \(ABCD\) : \(O\) – the point of intersection of the diagonals; \(S\) – does not lie in the plane of the square, \(SO \perp ABC\) . Find the angle between the planes \(ASD\) and \(BSC\) if \(SO = 5\) and \(AB = 10\) .

Right triangles \(\triangle SAO\) , \(\triangle SDO\) , \(\triangle SOB\) and \(\triangle SOC\) are equal in two sides and the angle between them (\(SO \perp ABC\) \(\Rightarrow\) \(\angle SOA = \angle SOD = \angle SOB = \angle SOC = 90^\circ\); \(AO = OD = OB = OC\), because \(O\) – point of intersection of the diagonals of the square, \(SO\) – common side) \(\Rightarrow\) \(AS = DS = BS = CS\) \(\Rightarrow\) \(\triangle ASD\) and \(\triangle BSC\) are isosceles. Point \(K\) is the middle of \(AD\), then \(SK\) is the height in the triangle \(\triangle ASD\), and \(OK\) is the height in the triangle \(AOD\) \(\ Rightarrow\) plane \(SOK\) is perpendicular to plane \(ASD\) . Point \(L\) is the middle of \(BC\), then \(SL\) is the height in the triangle \(\triangle BSC\), and \(OL\) is the height in the triangle \(BOC\) \(\ Rightarrow\) plane \(SOL\) (aka plane \(SOK\)) is perpendicular to the plane \(BSC\) . Thus, we obtain that \(\angle KSL\) is a linear angle equal to the desired dihedral angle.

\(KL = KO + OL = 2\cdot OL = AB = 10\)\(\Rightarrow\) \(OL = 5\) ; \(SK = SL\) – equal heights isosceles triangles, which can be found using the Pythagorean theorem: \(SL^2 = SO^2 + OL^2 = 5^2 + 5^2 = 50\). It can be noticed that \(SK^2 + SL^2 = 50 + 50 = 100 = KL^2\)\(\Rightarrow\) for a triangle \(\triangle KSL\) the inverse Pythagorean theorem holds \(\Rightarrow\) \(\triangle KSL\) – right triangle \(\Rightarrow\) \(\angle KSL = 90^\ circ\) .

Answer: 90

Preparing students for passing the Unified State Exam in mathematics, as a rule, begins with a repetition of basic formulas, including those that allow you to determine the angle between planes. Despite the fact that this section of geometry is covered in sufficient detail within school curriculum, many graduates need repetition base material. Understanding how to find the angle between planes, high school students will be able to quickly calculate the correct answer when solving a problem and count on receiving decent scores on the results of passing the unified state exam.

Main nuances

To ensure that the question of how to find a dihedral angle does not cause difficulties, we recommend following a solution algorithm that will help you cope with Unified State Examination tasks.

First you need to determine the straight line along which the planes intersect.

Then you need to select a point on this line and draw two perpendiculars to it.

Next step- finding trigonometric function dihedral angle formed by perpendiculars. The most convenient way to do this is with the help of the resulting triangle, of which the angle is a part.

The answer will be the value of the angle or its trigonometric function.

Preparing for the exam test with Shkolkovo is the key to your success

During classes on the eve of passing the Unified State Exam, many schoolchildren are faced with the problem of finding definitions and formulas that allow them to calculate the angle between 2 planes. A school textbook is not always at hand exactly when needed. And to find the necessary formulas and examples of them correct application, including finding the angle between planes on the Internet online, sometimes you need to spend a lot of time.

The mathematical portal "Shkolkovo" offers new approach to prepare for the state exam. Classes on our website will help students identify the most difficult sections for themselves and fill gaps in knowledge.

We have prepared and clearly presented everything required material. Basic definitions and formulas are presented in the “Theoretical Information” section.

In order to better understand the material, we also suggest practicing the appropriate exercises. Large selection tasks varying degrees complexity, for example, is presented in the “Catalog” section. All tasks contain detailed algorithm finding the correct answer. The list of exercises on the website is constantly supplemented and updated.

While practicing solving problems that require finding the angle between two planes, students have the opportunity to save any task online as “Favorites.” Thanks to this they will be able to return to him required quantity times and discuss the progress of its solution with a school teacher or tutor.

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 the 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 are, for example, (0; 2: 0). According to the above formula for the cosine of the angle between planes, we obtain:

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 correct triangular pyramid SABC is correct 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:

b) is the normal vector of the plane parallel to the lines SC and AB. then the conditions are met:

c) Let us denote by the plane that is parallel to the lines AQ and CP, and by its normal vector. In this case, we obtain a system of the form

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. The 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).