Angles between planes. How to determine the angle between planes

The measure of the angle between the planes is the acute angle formed by two straight lines lying in these planes and drawn perpendicular to the line of their intersection.

Construction Algorithm

From an arbitrary point K, perpendiculars are drawn to each of the given planes.
Rotation around the level line determines the value of the angle γ ° with the apex at point K.
Calculate the angle between the planes ϕ ° = 180 - γ °, provided that γ °> 90 °. If γ °< 90°, то ∠ϕ° = ∠γ°.

The figure shows the case when the planes α and β are specified by traces. All necessary constructions are performed according to the algorithm and are described below.

Solution

In an arbitrary place of the drawing, mark the point K. From it we drop the perpendiculars m and n, respectively, to the planes α and β. The direction of the projections m and n is as follows: m "" ⊥f 0α, m "⊥h 0α, n" "⊥f 0β, n" ⊥h 0β.
Determine the actual size ∠γ ° between the straight lines m and n. To do this, around the frontal f, we rotate the angle plane with the vertex K to a position parallel to the frontal projection plane. The turning radius R of point K is equal to the value of the hypotenuse right triangle O "" K "" K 0, the leg of which is K "" K 0 = y K - y O.
The sought angle is ϕ ° = ∠γ °, since ∠γ ° is acute.

The figure below shows the solution to the problem in which it is required to find the angle γ ° between the planes α and β, given by parallel and intersecting straight lines, respectively.

Solution

Determine the direction of the projections of the contours h 1, h 2 and the frontals f 1, f 2 belonging to the planes α and β, in the order indicated by the arrows. From an arbitrary point K to pl. α and β we omit the perpendiculars e and k. In this case, e "" ⊥f "" 1, e "⊥h" 1 and k "" ⊥f "" 2, k "⊥h" 2.
Determine ∠γ ° between lines e and k. To do this, draw a horizontal line h 3 and rotate point K around it to position K 1, at which △ CKD will become parallel to the horizontal plane and be reflected on it in full size - △ C "K" 1 D ". The projection of the center of rotation O" is on the drawn to h "3 perpendicular K" O ". The radius R is determined from a right-angled triangle O" K "K 0, whose side is K" K 0 = ZO - ZK.
The value of the sought-for ∠ϕ ° = ∠γ °, since the angle γ ° is acute.

This article is about the angle between the 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 method of coordinates is analyzed, a formula is obtained that allows you to calculate the angle between intersecting planes using the known coordinates of the normal vectors of these planes. The conclusion shows detailed solutions characteristic tasks.

Page navigation.

Angle between planes - definition.

Let us give reasoning that will allow you 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 us construct a plane passing through point M of straight line c and perpendicular to straight line c. In this case, the plane will intersect the planes and. Let's denote the line along which the planes intersect as a, and the line along which the planes intersect as b. Obviously, lines a and b meet at point M.

It is easy to show that the angle between the intersecting straight lines a and b does not depend on the location of the point M on the straight 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 by a 1 and b 1, respectively.

From the method of constructing the planes and it follows that the straight lines a and b are perpendicular to the straight line c, and the straight lines a 1 and b 1 are perpendicular to the straight line c. Since the straight lines a and a 1 lie in the same plane and are perpendicular to the straight line c, 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 carry out 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 straight lines a 1 and b 1 is equal to the angle between intersecting straight lines a and b.

This proves that the angle between the intersecting straight 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 read out 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 straight lines a and b, along which the planes and intersect with the plane perpendicular to the straight 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 the 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 straight lines does not exceed, it follows from the sounded definition that the degree measure of the angle between two intersecting planes is expressed by a real number from the interval. In this case, the intersecting planes are called perpendicular if the angle between them is ninety degrees. The angle between the 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 intersecting straight lines, the angle between which is equal to the desired angle, and then associate this angle with the original data using equality signs, similarity signs, the cosine theorem or definitions of sine, cosine and the tangent of the angle. Similar problems are encountered in the high school geometry course.

For example, we will give the solution to problem C2 from the exam in mathematics for 2012 (the condition was 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.

Solution.

First, let's make a drawing.

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

To begin with, 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 plane ABC, and line D 1 E - in plane BED 1, therefore, the intersection point of lines DA and D 1 E will be common point planes ABC and BED 1. So, we will continue straight lines DA and D 1 E to their intersection, denote the point of their intersection by the letter F. Then BF is the line along which the planes ABC and BED 1 intersect.

It remains to construct two straight lines lying in the planes ABC and BED 1, respectively, passing through one point on the straight line BF and perpendicular to the straight 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 point E onto the ABC plane. Let us draw a straight line intersecting straight line BF at point M. Then the line AM is the projection of the line EM on the plane ABC, and by the three perpendicular 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 the right-angled triangle AEM if we know the lengths of its two sides. From the condition it is easy to find the length of AE: since point E divides side AA 1 in a ratio of 4 to 3, counting from point A, and the length of side AA 1 is 7, then AE = 4. Let us also find the length AM.

To do this, consider a right-angled triangle ABF with a 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 the right-angled 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 triangle ABF: on one side, the area of triangle ABF is equal to , on the other side , where .

Thus, from the right-angled triangle AEM we have .

Then the sought 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 set Oxyz and use the coordinate method. Let's stop at it.

Let's set the task: find the angle between two intersecting planes and. Let's denote the required 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 it is possible to find them. Let is the normal vector of the plane, and is the normal vector of the plane. Let's show how to find the angle between the intersecting planes and through the coordinates of the normal vectors of these planes.

Let us denote the line along which the planes and intersect, as c. Through the point M on the straight line c we draw a plane perpendicular to the straight line c. The plane intersects the plane 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 straight lines a and b.

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

In the article, finding the angle between intersecting straight lines, we obtained a formula that allows us to calculate the cosine of the angle between intersecting straight lines using the coordinates of normal vectors. Thus, the cosine of the angle between straight lines a and b, and, therefore, cosine of the angle between intersecting planes and is found by the formula, where and Are normal vectors of 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 the side AA 1 in a ratio of 4 to 3, counting from point A. Find the angle between the planes ABC and BED 1.

Solution.

Since the sides of the rectangular parallelepiped at one vertex are pairwise perpendicular, it is convenient to introduce the rectangular coordinate system Oxyz as follows: align the origin 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 planes ABC and BED 1 can be found through the coordinates of the normal vectors of these planes by the formula, where and are the normal vectors of the ABC and BED 1 planes, respectively. Let us determine the coordinates of the normal vectors.

Job type: 14
Topic: Angle Between Planes

Condition

Given the correct prism ABCDA_1B_1C_1D_1, M and N are the midpoints of the edges AB and BC, respectively, point K is the midpoint of MN.

a) Prove that lines KD_1 and MN are perpendicular.

b) Find the angle between the planes MND_1 and ABC if AB = 8, AA_1 = 6 \ sqrt 2.

Show solution

Solution

a) In \ triangle DCN and \ triangle MAD we have: \ angle C = \ angle A = 90 ^ (\ circ), CN = AM = \ frac12AB, CD = DA.

Hence \ triangle DCN = \ triangle MAD along two legs. Then MD = DN, \ triangle DMN isosceles. This means that the median DK is also the height. Therefore, DK \ perp MN.

DD_1 \ perp MND by condition, D_1K - oblique, KD - projection, DK \ perp MN.

Hence, by the three perpendicular theorem, MN \ perp D_1K.

b) As proven in a), DK \ perp MN and MN \ perp D_1K, but MN is the intersection line of the MND_1 and ABC planes, so \ angle DKD_1 is the linear angle of the dihedral angle between the MND_1 and ABC planes.

In \ triangle DAM by the Pythagorean theorem DM = \ sqrt (DA ^ 2 + AM ^ 2) = \ sqrt (64 + 16) = 4 \ sqrt 5, MN = \ sqrt (MB ^ 2 + BN ^ 2) = \ sqrt (16 + 16) = 4 \ sqrt 2. Therefore, in \ triangle DKM by the Pythagorean theorem DK = \ sqrt (DM ^ 2-KM ^ 2) = \ sqrt (80-8) = 6 \ sqrt 2. Then in \ triangle DKD_1, tg \ angle DKD_1 = \ frac (DD_1) (DK) = \ frac (6 \ sqrt 2) (6 \ sqrt 2) = 1.

Hence, \ angle DKD_1 = 45 ^ (\ circ).

Answer

45 ^ (\ circ).

Job type: 14
Topic: Angle Between Planes

Condition

In a regular quadrangular prism ABCDA_1B_1C_1D_1, the sides of the base are 4, the side edges are 6. Point M is the midpoint of edge CC_1, point N is marked on edge BB_1, such that BN: NB_1 = 1: 2.

a) In what respect does the AMN plane divide the DD_1 edge?

b) Find the angle between planes ABC and AMN.

Show solution

Solution

a) The plane AMN intersects the edge DD_1 at point K, which is the fourth vertex of the section of this prism by this plane. The section is an ANMK parallelogram, because the opposite faces of a given prism are parallel.

BN = \ frac13BB_1 = 2. Draw KL \ parallel CD, then triangles ABN and KLM are equal, hence ML = BN = 2, LC = MC-ML = 3-2 = 1, KD = LC = 1. Then KD_1 = 6-1 = 5. Now you can find the ratio KD: KD_1 = 1: 5.

b) F is the intersection point of lines CD and KM. Planes ABC and AMN intersect in a straight line AF. Angle \ angle KHD = \ alpha is the linear angle of the dihedral angle (HD \ perp AF, then by the theorem converse to the three perpendicular theorem, KH \ perp AF), and is acute angle right-angled triangle KHD, leg KD = 1.

Triangles FKD and FMC are similar (KD \ parallel MC), therefore FD: FC = KD: MC, solving the proportion FD: (FD + 4) = 1: 3, we get FD = 2. In a right-angled triangle AFD (\ angle D = 90 ^ (\ circ)) with legs 2 and 4, we calculate the hypotenuse AF = \ sqrt (4 ^ 2 + 2 ^ 2) = 2 \ sqrt 5, DH = AD \ cdot FD: AF = \ frac (4 \ cdot 2) (2 \ sqrt 5) = \ frac4 (\ sqrt 5).

In the right-angled triangle KHD we find tg \ alpha = \ frac (KD) (DH) = \ frac (\ sqrt 5) 4, means the required angle \ alpha = arctg \ frac (\ sqrt 5) 4.

Answer

a) 1:5;

b) arctg \ frac (\ sqrt 5) 4.

Source: “Mathematics. Preparation for the exam-2017. Profile level ". Ed. FF Lysenko, S. Yu. Kulabukhova.

Job type: 14
Topic: Angle Between Planes

Condition

You are given a regular quadrangular pyramid KMNPQ with base side MNPQ equal to 6 and a side edge 3 \ sqrt (26).

a) Construct a section of the pyramid with a plane passing through line NF parallel to the diagonal MP, if point F is the midpoint of edge MK.

b) Find the angle between the section plane and the KMP plane.

Show solution

Solution

a) Let KO be the height of the pyramid, F - the middle of MK; FE \ parallel MP (in PKM plane). Since FE - middle line\ triangle PKM, then FE = \ frac (MP) 2.

Let's construct a section of the pyramid with a plane passing through NF and parallel to MP, that is, with the NFE plane. L is the intersection point of EF and KO. Since points L and N belong to the desired section and lie in the plane KQN, the point T obtained as the intersection of LN and KQ is also the intersection point of the desired section and the edge KQ. NETF - the desired section.

b) The planes NFE and MPK intersect in a straight line FE. Hence, the angle between these planes is equal to the linear angle of the dihedral angle OFEN, we construct it: LO \ perp MP, MP \ parallel FE, hence, LO \ perp FE;\ triangle NFE is isosceles (NE = NF as the corresponding medians of equal triangles KPN and KMN), NL is its median (EL = LF, since PO = OM, and \ triangle KEF \ sim \ triangle KPM). Hence NL \ perp FE and \ angle NLO is the desired one.

ON = \ frac12QN = \ frac12MN \ sqrt 2 = 3 \ sqrt 2.

\ triangle KON - rectangular.

The leg KO by the Pythagorean theorem is KO = \ sqrt (KN ^ 2-ON ^ 2).

OL = \ frac12KO = \ frac12 \ sqrt (KN ^ 2-ON ^ 2) = \ frac12 \ sqrt (9 \ cdot 26-9 \ cdot 2) = \ frac12 \ sqrt (9 (26-2)) = \ frac32 \ sqrt (24) = \ frac32 \ cdot 2 \ sqrt 6 = 3 \ sqrt 6.

tg \ angle NLO = \ frac (ON) (OL) = \ frac (3 \ sqrt 2) (3 \ sqrt 6) = \ frac1 (\ sqrt 3),

\ angle NLO = 30 ^ (\ circ).

Answer

Source: “Mathematics. Preparation for the exam-2017. Profile level ". Ed. FF Lysenko, S. Yu. Kulabukhova.

Job type: 14
Topic: Angle Between Planes

Condition

All edges of a regular triangular prism ABCA_ (1) B_ (1) C_ (1) are 6. A cutting plane is drawn through the midpoints of the edges AC and BB_ (1) and the vertex A_ (1).

a) Prove that the edge BC is divided by the cutting plane in a 2: 1 ratio, starting from the vertex C.

b) Find the angle between the section plane and the base plane.

Show solution

Solution

a) Let D and E be the midpoints of the edges AC and BB_ (1), respectively.

In the plane AA_ (1) C_ (1), draw the line A_ (1) D, which intersects the line CC_ (1) at point K, in the plane BB_ (1) C_ (1) - the line KE, which intersects the edge BC at point F ... The connection of the points A_ (1) and E lying in the plane AA_ (1) B_ (1), as well as D and F lying in the plane ABC, we get the section A_ (1) EFD.

\ bigtriangleup AA_ (1) D = \ bigtriangleup CDK along the leg AD = DC and an acute angle.

\ angle ADA_ (1) = \ angle CDK - as vertical, it follows that AA_ (1) = CK = 6. \ bigtriangleup CKF and \ bigtriangleup BFE are similar in two corners \ angle FBE = \ angle KCF = 90 ^ \ circ,\ angle BFE = \ angle CFK - as vertical.

\ frac (CK) (BE) = \ frac (6) (3) = 2, that is, the coefficient of similarity is 2, which implies that CF: FB = 2: 1.

b) Let's draw AH \ perp DF. The angle between the section plane and the base plane is equal to the angle AHA_ (1). Indeed, the segment AH \ perp DF (DF is the line of intersection of these planes) is the projection of the segment A_ (1) H onto the base plane, therefore, by the three perpendicular theorem, A_ (1) H \ perp DF. \ angle AHA_ (1) = arctg \ frac (AA_ (1)) (AH). AA_ (1) = 6.

Find AH. \ angle ADH = \ angle FDC (as vertical).

By the cosine theorem in \ bigtriangleup DFC:

DF ^ 2 = FC ^ 2 + DC ^ 2- 2FC \ cdot DC \ cdot \ cos 60 ^ \ circ,

DF ^ 2 = 4 ^ 2 + 3 ^ 2-2 \ cdot 4 \ cdot 3 \ cdot \ frac (1) (2) = 13.

FC ^ 2 = DF ^ 2 + DC ^ 2- 2DF \ cdot DC \ cdot \ cos \ angle FDC,

4 ^ 2 = 13 + 9-2 \ sqrt (13) \ cdot 3 \ cdot \ cos \ angle FDC,

\ cos \ angle FDC = \ frac (6) (2 \ sqrt (13) \ cdot 3) = \ frac (1) (\ sqrt (13)).

By the corollary of the basic trigonometric identity

\ sin \ angle FDC = \ sqrt (1- \ left (\ frac (1) (\ sqrt (13)) \ right) ^ 2) = \ frac (2 \ sqrt (3)) (\ sqrt (13)) ... Find AH from \ bigtriangleup ADH:

AH = AD \ cdot \ sin \ angle ADH, (\ angle FDC = \ angle ADH). AH = 3 \ cdot \ frac (2 \ sqrt (3)) (\ sqrt (13)) = \ frac (6 \ sqrt (13)) (\ sqrt (13)).

\ angle AHA_ (1) = arctg \ frac (AA_ (1)) (AH) = arctg \ frac (6 \ cdot \ sqrt (13)) (6 \ sqrt (3)) = arctg \ frac (\ sqrt (39)) (3).

Answer

arctg \ frac (\ sqrt (39)) (3).

Source: “Mathematics. Preparation for the exam-2017. Profile level ". Ed. FF Lysenko, S. Yu. Kulabukhova.

Job type: 14
Topic: Angle Between Planes

Condition

The base of the straight prism ABCDA_ (1) B_ (1) C_ (1) D_ (1) is a rhombus with obtuse angle B equal to 120 ^ \ circ. All the edges of this prism are 10. Points P and K are the midpoints of edges CC_ (1) and CD, respectively.

a) Prove that lines PK and PB_ (1) are perpendicular.

b) Find the angle between planes PKB_ (1) and C_ (1) B_ (1) B.

Show solution

Solution

a) We will use the coordinate method. Find the dot product of the vectors \ vec (PK) and \ vec (PB_ (1)), and then the cosine of the angle between these vectors. Let us direct the Oy axis along CD, the Oz axis along CC_ (1), and the Ox \ perp CD axis. C is the origin.

Then C (0; 0; 0); C_ (1) (0; 0; 10); P (0; 0; 5); K (0; 5; 0); B (BC \ cos 30 ^ \ circ; BC \ sin 30 ^ \ circ; 0), that is B (5 \ sqrt (3); 5; 0), B_ (1) (5 \ sqrt (3); 5; 10).

Let's find the coordinates of the vectors: \ vec (PK) = \ (0; 5; -5 \); \ vec (PB_ (1)) = \ (5 \ sqrt (3); 5; 5 \).

Let the angle between \ vec (PK) and \ vec (PB_ (1)) be \ alpha.

\ cos \ alpha = 0, so \ vec (PK) \ perp \ vec (PB_ (1)) and lines PK and PB_ (1) are perpendicular.

b) The angle between the planes is equal to the angle between the nonzero vectors perpendicular to these planes (or, if the angle is obtuse, the angle adjacent to it). Such vectors are called plane normals. Let's find them.

Let \ vec (n_ (1)) = \ (x; y; z \) be perpendicular to the PKB_ (1) plane. Let's find it by solving the system \ begin (cases) \ vec (n_ (1)) \ perp \ vec (PK), \\ \ vec (n_ (1)) \ perp \ vec (PB_ (1)). \ end (cases)

\ begin (cases) \ vec (n_ (1)) \ cdot \ vec (PK) = 0, \\ \ vec (n_ (1)) \ cdot \ vec (PB_ (1)) = 0; \ end (cases)

\ begin (cases) 0x + 5y-5z = 0, \\ 5 \ sqrt (3) x + 5y + 5z = 0; \ end (cases)

\ begin (cases) y = z, \\ x = \ frac (-y-z) (\ sqrt (3)). \ end (cases)

Let's take y = 1; z = 1; x = \ frac (-2) (\ sqrt (3)), \ vec (n_ (1)) = \ left \ (\ frac (-2) (\ sqrt (3)); 1; 1 \ right \).

Let \ vec (n_ (2)) = \ (x; y; z \) be perpendicular to the plane C_ (1) B_ (1) B. Let's find it by solving the system \ begin (cases) \ vec (n_ (2)) \ perp \ vec (CC_ (1)), \\ \ vec (n_ (2)) \ perp \ vec (CB). \ end (cases)

\ vec (CC_ (1)) = \ (0; 0; 10 \), \ vec (CB) = \ (5 \ sqrt (3); 5; 0 \).

\ begin (cases) \ vec (n_ (2)) \ cdot \ vec (CC_ (1)) = 0, \\ \ vec (n_ (2)) \ cdot \ vec (CB) = 0; \ end (cases)

\ begin (cases) 0x + 0y + 10z = 0, \\ 5 \ sqrt (3) x + 5y + 0z = 0; \ end (cases)

\ begin (cases) z = 0, \\ y = - \ sqrt (3) x. \ end (cases)

Let's take x = 1; y = - \ sqrt (3); z = 0, \ vec (n_ (2)) = \ (1; - \ sqrt (3); 0 \).

Find the cosine of the desired angle \ beta (it is equal to the modulus of the cosine of the angle between \ vec (n_ (1)) and \ vec (n_ (2))).

\ cos \ beta = \ frac (\ sqrt (10)) (4), \ beta = \ arccos \ frac (\ sqrt (10)) (4).

Answer

\ arccos \ frac (\ sqrt (10)) (4)

Source: “Mathematics. Preparation for the exam-2017. Profile level ". Ed. FF Lysenko, S. Yu. Kulabukhova.

ABCD - square and side faces - equal rectangles.

Since the section plane passes through the points M and D parallel to the diagonal AC, then to construct it in the plane A_ (1) AC through the point M we draw a segment MN parallel to AC. We get AC \ parallel (MDN) based on the parallelism of a straight line and a plane.

The MDN plane intersects the parallel planes A_ (1) AD and B_ (1) BC, then, by the property of parallel planes, the intersection lines of the faces A_ (1) ADD_ (1) and B_ (1) BCC_ (1) by the MDN plane are parallel.

Let's draw the segment NE parallel to the segment MD.

The quadrilateral DMEN is the required section.

b) Find the angle between the section plane and the base plane. Let the section plane intersect the base plane along some straight line p passing through point D. AC \ parallel MN, hence AC \ parallel p (if a plane passes through a straight line parallel to another plane and intersects this plane, then the line of intersection of the planes is parallel to this straight line). BD \ perp AC as the diagonals of the square, so BD \ perp p. BD is the projection of ED onto the ABC plane, then by the theorem of three perpendiculars ED \ perp p, therefore \ angle EDB is the linear angle of the dihedral angle between the section plane and the base plane.

Let's set the type of the quadrangle DMEN. MD \ parallel EN, similar to ME \ parallel DN, which means DMEN is a parallelogram, and since MD = DN (right-angled triangles MAD and NCD are equal in two legs: AD = DC as sides of a square, AM = CN as distances between parallel lines AC and MN), therefore DMEN is a rhombus. Hence, F is the midpoint of MN.

By the condition AM: MA_ (1) = 2: 3, then AM = \ frac (2) (5) AA_ (1) = \ frac (2) (5) \ cdot 5 \ sqrt (6) = 2 \ sqrt (6).

AMNC - rectangle, F - midpoint of MN, O - midpoint of AC. Means, FO \ parallel MA, FO \ perp AC, FO = MA = 2 \ sqrt (6).

Knowing that the diagonal of the square is a \ sqrt (2), where a is the side of the square, we get BD = 4 \ sqrt (2). OD = \ frac (1) (2) BD = \ frac (1) (2) \ cdot 4 \ sqrt (2) = 2 \ sqrt (2).

In a right triangle FOD \ enspace tg \ angle FDO = \ frac (FO) (OD) = \ frac (2 \ sqrt (6)) (2 \ sqrt (2)) = \ sqrt (3). Therefore, \ angle FDO = 60 ^ \ circ.

Theorem

The angle between the planes is independent of the cut plane selection.

Proof.

Let there be two planes α and β, which intersect in a straight line with. draw the plane γ perpendicular to the straight line с. Then the plane γ intersects the planes α and β along the lines a and b, respectively. The angle between planes α and β is equal to the angle between straight lines a and b.
Take another cut plane γ ', perpendicular to c. Then the plane γ` intersects the planes α and β along the straight lines a` and b`, respectively.
In case of parallel transfer, the point of intersection of the plane γ with the straight line с will go over to the point of intersection of the plane γ` with the straight line с. in this case, by the property of parallel transfer, the straight line a goes over into the straight line a`, b - into the straight line b`. hence the angles between straight lines a and b, a` and b` are equal. The theorem is proved.

This article is about the angle between the 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 method of coordinates is analyzed, a formula is obtained that allows you to calculate the angle between intersecting planes using the known coordinates of the normal vectors of these planes. In the conclusion, detailed solutions to typical problems are shown.

Page navigation.

Angle between planes - definition.

When presenting the material, we will use the definitions and concepts given in the articles, a plane in space and a straight line in space.

Let us give reasoning that will allow you 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 build a plane passing through the point M straight c and perpendicular to the straight line c... In this case, the plane will intersect the planes and. Let us denote the straight line along which the planes intersect and as a, and the straight line along which the planes intersect and how b... Obviously straight a and b intersect at the point M.

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

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

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

This proves that the angle between intersecting straight lines a and b lying in 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 read out the definition of the angle between two intersecting planes and.

Definition.

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

The definition of the angle between two planes can be given a little differently. If on a straight line With, along which the planes intersect and, mark the point M and through it draw straight a and b perpendicular to the straight line c and lying in 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.

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Finding the angle between two intersecting planes.

ABCDA 1 B 1 C 1 D 1, in which AB = 3, AD = 2, AA 1 = 7 and point E divides the side AA 1 in a relationship 4 To 3 counting from the point A ABC and BU 1.

First, let's make a drawing.

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

First, let's define a straight line along which the planes intersect ABC and BED 1... Dot V Is one of their common points. Let's find the second common point of these planes. Direct DA and D 1 E lie in the same plane ADD 1, and they are not parallel, and therefore intersect. On the other hand, straight DA lies in the plane ABC and straight D 1 E- in the plane BED 1, therefore, the intersection point of the lines DA and D 1 E will be the common point of the planes ABC and BED 1... So let's continue straight DA and D 1 E before their intersection, denote the point of their intersection with the letter F... Then Bf- the straight line along which the planes intersect ABC and BED 1.

It remains to construct two lines lying in planes ABC and BED 1 respectively, passing through one point on a straight line Bf and perpendicular to the straight line Bf, - the angle between these straight 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 on the plane ABC... Let's draw a straight line intersecting a straight line at a right angle BF at the point M... Then the straight line AM is a projection of a straight line EAT on the plane ABC, and by the three perpendicular theorem.

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

The sine, cosine or tangent of this angle (and hence the angle itself) we can determine from a right-angled triangle AEM if we know the lengths of its two sides. It is easy to find the length from the condition AE: since the point E divides the side AA 1 in a relationship 4 To 3 counting from the point A and the side length AA 1 is equal to 7 , then AE = 4... Let's find another length AM.

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

By the Pythagorean theorem from a triangle ABF we find. Length AM we find through the area of the triangle ABF: on one side, the area of the triangle ABF equal, on the other hand, whence.

So from a right triangle AEM we have.

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

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

Let's set the task: find the angle between two intersecting planes and. Let's denote the required 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 it is possible to find them. Let be the normal vector of the plane, and be the normal vector of the plane. Let's show how to find the angle between the intersecting planes and through the coordinates of the normal vectors of these planes.

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

Set aside from the point M in the plane are normal vectors and planes and. In this case, the vector lies on a straight line that is perpendicular to the straight line a, and the vector - on a straight line, which is perpendicular to the straight line b... Thus, in the plane, the vector is the normal vector of the straight line a, is the normal vector of the straight line b.

In the article, finding the angle between intersecting straight lines, we obtained a formula that allows us to calculate the cosine of the angle between intersecting straight lines using the coordinates of normal vectors. Thus, the cosine of the angle between the straight lines a and b and, therefore, 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 angle between intersecting planes is calculated as.

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

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

Since the sides of a rectangular parallelepiped at one vertex are pairwise perpendicular, it is convenient to introduce a rectangular coordinate system Oxyz like this: start to align with the top WITH, and the coordinate axes Ox, Oy and Oz send to the sides CD, CB and CC 1 respectively.

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

Since the plane ABC coincides with the coordinate plane Oxy, then its normal vector is the coordinate vector, that is,.

As the normal vector of the plane BED 1 we can take the vector product of vectors and, in turn, the coordinates of the vectors and can be found through the coordinates of the points V, E and D 1(as described in the article, the coordinates of the vector through the coordinates of the points of its beginning and end), and the coordinates of the points V, E and D 1 in the introduced coordinate system will be determined from the condition of the problem.

Obviously, . Since, we find by the coordinates of the points (if necessary, see the article dividing a segment in a given ratio). Then and Oxyz equations and.

When we studied general equation straight line, we found out that the coefficients A, V and WITH represent the corresponding coordinates of the normal vector of the plane. Thus, and are normal vectors of the planes and, respectively.

Substitute the coordinates of the normal vectors of the planes into the formula for calculating the angle between two intersecting planes:

Then . Since the angle between two intersecting planes is not obtuse, then using the basic trigonometric identity we find the sine of the angle:.