The angle between two intersecting planes: definition, examples of finding. Angle between straight lines Find an acute angle between straight lines

\(\blacktriangleright\) A dihedral angle is the angle formed by two half-planes and the 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 spicy or straight) of the 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: draw \(FG\perp \pi\) ;

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 a right triangle.
Note also that the plane \(AFG\) constructed in this way is perpendicular to both the planes \(\xi\) and \(\pi\) . Therefore, it can be said in another way: 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 \(\xi\ ) , and \(\pi\) .

Task 1 #2875

Task level: More difficult than the exam

Given a 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, all side faces are equal equilateral triangles. Find the angle between the faces \(SAD\) and \(SCD\) .

Let's draw \(CH\perp SD\) . As \(\triangle SAD=\triangle SCD\), then \(AH\) will also be a height of \(\triangle SAD\) . Therefore, by definition, \(\angle AHC=\alpha\) is the linear 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 \(a\) , hence \(CH=AH=\frac(\sqrt3)2a\) .
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 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 a right angle, 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 the line \(a\) , the line of intersection of \(\pi_2\) and \(\pi_3\) be the line \(b\) , and the line of intersection \(\pi_3\) and \(\pi_1\) are the 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”).

Mark the points \(A\in a, B\in b\) so that \(AB\perp a, AB\perp b\) (this is possible because \(a\parallel b\) ). Note \(C\in c\) so that \(BC\perp c\) , hence \(BC\perp b\) . 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 hence any line from this plane, in particular, the line \ (AC\) .

Hence 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 exam

Given 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 the lines \(a\) and \(c\) and the plane formed by the 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 lines cannot lie in the same plane. Let us mark a point \(A\) on the line \(a\) and draw \(AB\perp b\) and \(AC\perp c\) . Then \(\triangle AOB=\triangle AOC\) as rectangular in hypotenuse and acute angle. Hence \(OB=OC\) and \(AB=AC\) .
Let's do \(AH\perp (BOC)\) . Then by the three perpendiculars theorem \(HC\perp c\) , \(HB\perp b\) . Since \(AB=AC\) , then \(\triangle AHB=\triangle AHC\) as rectangular along the hypotenuse and leg. Therefore, \(HB=HC\) . Hence, \(OH\) 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 have 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 corner. Since we chose the point \(A\) arbitrarily, then 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 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 exam

The planes \(\pi_1\) and \(\pi_2\) intersect along the line \(l\) , which contains the points \(M\) and \(N\) . The segments \(MA\) and \(MB\) are perpendicular to the 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 right-angled, \(BN^2 = BM^2 + MN^2\) , whence \ We write the cosine theorem for the triangle \(AMB\): \ Then \ Since the angle \(\alpha\) between the planes is an 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 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 \((ABCD)\) , moreover, \(M\) is the intersection point 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.

We 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 intersection point of the diagonals of the square, then \(M\) is the midpoint of \(AC\) , therefore, \(MN\) is 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 three perpendiculars theorem, \(A_1N\) 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 exam

In the square \(ABCD\) : \(O\) is the intersection point of the diagonals; \(S\) is not 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 \(O\) is the point of intersection of the diagonals of the square, \(SO\) is the common side) \(\Rightarrow\) \(AS = SD\) \(\Rightarrow\) \(\triangle ASD\) is isosceles. The point \(K\) is the midpoint 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 the planes \(ASD\) and \(ABC\) \(\Rightarrow\) \(\angle SKO\) is a linear angle equal to the required dihedral angle.

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

Answer: 45

Task 7 #1855

Task level: More difficult than the exam

In the square \(ABCD\) : \(O\) is the intersection point of the diagonals; \(S\) is not 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 \(O\) is the point of intersection of the diagonals of the square, \(SO\) is the common side) \(\Rightarrow\) \(AS = DS = BS = CS\) \(\Rightarrow\) \(\triangle ASD\) and \(\triangle BSC\) are isosceles. The point \(K\) is the midpoint of \(AD\) , then \(SK\) is the height in the triangle \(\triangle ASD\) , and \(OK\) is the height in the triangle \(AOD\) \(\ Rightarrow\) the plane \(SOK\) is perpendicular to the plane \(ASD\) . The point \(L\) is the midpoint of \(BC\) , then \(SL\) is the height in the triangle \(\triangle BSC\) , and \(OL\) is the height in the triangle \(BOC\) \(\ Rightarrow\) the plane \(SOL\) (aka the plane \(SOK\) ) is perpendicular to the plane \(BSC\) . Thus, we obtain that \(\angle KSL\) is a linear angle equal to the desired dihedral angle.

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

Answer: 90

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

Main nuances

So that the question of how to find the dihedral angle does not cause difficulties, we recommend that you follow the solution algorithm that will help you cope with the tasks of the exam.

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

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

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

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

Preparation for the exam test together with Shkolkovo is the key to your success

During class the day before passing the exam many students are faced with the problem of finding definitions and formulas that allow you to calculate the angle between 2 planes. A school textbook is not always at hand exactly when it is needed. And in order to find the necessary formulas and examples of their correct application, including for finding the angle between planes on the Internet online, sometimes you need to spend a lot of time.

Mathematical portal "Shkolkovo" offers a new approach to preparing 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 all the necessary material. Basic definitions and formulas are presented in the "Theoretical Reference" section.

In order to better assimilate the material, we also suggest practicing the corresponding exercises. A large selection of tasks of varying degrees of complexity, for example, on, is presented in the Catalog section. All tasks contain a detailed algorithm for finding the correct answer. The list of exercises on the site is constantly supplemented and updated.

Practicing in solving problems in which it is required to find the angle between two planes, students have the opportunity to save any task online to "Favorites". Thanks to this, they will be able to return to him the necessary number of times and discuss the progress of his solution with a school teacher or tutor.

The article talks about finding the angle between planes. After bringing the definition, we will set a graphic illustration, consider a detailed method for finding coordinates by the method. We obtain a formula for intersecting planes, which includes the coordinates of normal vectors.

The material will use data and concepts that were previously studied in articles about the plane and the line in space. To begin with, it is necessary to move on to reasoning that allows one to have a certain approach to determining the angle between two intersecting planes.

Two intersecting planes γ 1 and γ 2 are given. Their intersection will take the designation c . The construction of the χ plane is connected with the intersection of these planes. The plane χ passes through the point M as a straight line c. The planes γ 1 and γ 2 will be intersected using the χ plane. We accept the designations of the line intersecting γ 1 and χ for the line a, and intersecting γ 2 and χ for the line b. We get that the intersection of lines a and b gives the point M .

The location of the point M does not affect the angle between the intersecting lines a and b, and the point M is located on the line c through which the plane χ passes.

It is necessary to construct a plane χ 1 perpendicular to the line c and different from the plane χ . The intersection of the planes γ 1 and γ 2 with the help of χ 1 will take the designation of lines a 1 and b 1 .

It can be seen that when constructing χ and χ 1, the lines a and b are perpendicular to the line c, then a 1, b 1 are perpendicular to the line c. Finding lines a and a 1 in the plane γ 1 with perpendicularity to the line c, then they can be considered parallel. In the same way, the location of b and b 1 in the plane γ 2 with the perpendicularity of the line c indicates their parallelism. This means that it is necessary to make a parallel transfer of the plane χ 1 to χ, where we get two coinciding lines a and a 1 , b and b 1 . We get that the angle between the intersecting lines a and b 1 equal to the angle intersecting lines a and b.

Consider the figure below.

This judgment is proved by the fact that between the intersecting lines a and b there is an angle that does not depend on the location of the point M, that is, the point of intersection. These lines are located in the planes γ 1 and γ 2 . In fact, the resulting angle can be thought of as the angle between two intersecting planes.

Let's move on to determining the angle between the existing intersecting planes γ 1 and γ 2 .

Definition 1

The angle between two intersecting planes γ 1 and γ 2 call the angle formed by the intersection of lines a and b, where the planes γ 1 and γ 2 intersect with the plane χ perpendicular to the line c.

Consider the figure below.

The definition may be submitted in another form. At the intersection of the planes γ 1 and γ 2, where c is the line on which they intersect, mark the point M, through which draw lines a and b, perpendicular to the line c and lying in the planes γ 1 and γ 2, then the angle between the lines a and b will be the angle between the planes. In practice, this is applicable to constructing an angle between planes.

At the intersection, an angle is formed that is less than 90 degrees in value, that is, the degree measure of the angle is valid on an interval of this type (0, 90] . At the same time, these planes are called perpendicular if a right angle is formed at the intersection. The angle between parallel planes is considered equal to zero.

The usual way to find the angle between intersecting planes is to perform additional constructions. This helps to determine it with accuracy, and this can be done using the signs of equality or similarity of the triangle, sines, cosines of the angle.

Consider solving problems using an example from the problems of the Unified State Examination of block C 2.

Example 1

A rectangular parallelepiped A B C D A 1 B 1 C 1 D 1 is given, where side A B \u003d 2, A D \u003d 3, A A 1 \u003d 7, point E separates side A A 1 in a ratio of 4: 3. Find the angle between planes A B C and B E D 1 .

Decision

For clarity, you need to make a drawing. We get that

A visual representation is necessary in order to make it more convenient to work with the angle between the planes.

We make the definition of a straight line along which the planes A B C and B E D 1 intersect. Point B is a common point. Gotta find another one common point intersections. Consider the lines D A and D 1 E , which are located in the same plane A D D 1 . Their location does not indicate parallelism, which means they have a common intersection point.

However, the line D A is located in the plane A B C, and D 1 E in B E D 1 . Hence we get that the lines D A and D 1 E have a common point of intersection, which is also common for planes A B C and B E D 1 . Indicates the point of intersection of lines D A and D 1 E letter F. From here we get that B F is a straight line along which the planes A B C and B E D 1 intersect.

Consider the figure below.

To obtain an answer, it is necessary to construct straight lines located in the planes A B C and B E D 1 with the passage through a point located on the line B F and perpendicular to it. Then the resulting angle between these lines is considered the desired angle between the planes A B C and B E D 1.

From this it can be seen that the point A is the projection of the point E onto the plane A B C. It is necessary to draw a line intersecting the line B F at a right angle at the point M. It can be seen that the line A M is the projection of the line E M onto the plane A B C, based on the theorem about those perpendiculars A M ⊥ B F . Consider the figure below.

∠ A M E is the desired angle formed by the planes A B C and B E D 1 . From the resulting triangle A E M we can find the sine, cosine or tangent of the angle, after which the angle itself, only with its two known sides. By condition, we have that the length of A E is found in this way: the line A A 1 is divided by the point E in a ratio of 4: 3, which means the total length of the line is 7 parts, then A E \u003d 4 parts. We find A.M.

It is necessary to consider a right triangle A B F. We have a right angle A with height A M. From the condition A B \u003d 2, then we can find the length A F by the similarity of triangles D D 1 F and A E F. We get that A E D D 1 = A F D F ⇔ A E D D 1 = A F D A + A F ⇒ 4 7 = A F 3 + A F ⇔ A F = 4

It is necessary to find the length of the side B F from the triangle A B F using the Pythagorean theorem. We get that B F = A B 2 + A F 2 = 2 2 + 4 2 = 2 5 . The length of the side A M is found through the area of the triangle A B F. We have that the area can be equal to both S A B C = 1 2 · A B · A F , and S A B C = 1 2 · B F · A M .

We get that A M = A B A F B F = 2 4 2 5 = 4 5 5

Then we can find the value of the tangent of the angle of the triangle A E M. We get:

t g ∠ A M E = A E A M = 4 4 5 5 = 5

The desired angle obtained by the intersection of the planes A B C and B E D 1 is equal to a r c t g 5, then, when simplified, we get a r c t g 5 = a r c sin 30 6 = a r c cos 6 6 .

Answer: a r c t g 5 = a r c sin 30 6 = a r c cos 6 6 .

Some cases of finding the angle between intersecting lines are given using coordinate plane About x y z and the coordinate method. Let's consider in more detail.

If a problem is given where it is necessary to find the angle between the intersecting planes γ 1 and γ 2, we denote the desired angle by α.

Then the given coordinate system shows that we have the coordinates of the normal vectors of the intersecting planes γ 1 and γ 2 . Then we denote that n 1 → = n 1 x , n 1 y , n 1 z is a normal vector of the plane γ 1 , and n 2 → = (n 2 x , n 2 y , n 2 z) - for the plane γ 2 . Consider a detailed finding of the angle located between these planes according to the coordinates of the vectors.

It is necessary to designate the straight line along which the planes γ 1 and γ 2 intersect with the letter c. On the line with we have a point M, through which we draw a plane χ, perpendicular to c. The plane χ along the lines a and b intersects the planes γ 1 and γ 2 at the point M . it follows from the definition that the angle between the intersecting planes γ 1 and γ 2 is equal to the angle of the intersecting lines a and b belonging to these planes, respectively.

In the χ plane, we set aside the normal vectors from the point M and denote them n 1 → and n 2 →. Vector n 1 → is located on a line perpendicular to line a, and vector n 2 → on a line perpendicular to line b. Hence we get that given planeχ has a normal vector of the line a equal to n 1 → and for the straight line b equal to n 2 → . Consider the figure below.

From here we obtain a formula by which we can calculate the sine of the angle of intersecting lines using the coordinates of the vectors. We found that the cosine of the angle between the straight lines a and b is the same as the cosine between the intersecting planes γ 1 and γ 2 is derived from the formula cos α = cos n 1 → , n 2 → ^ = n 1 x n 2 x + n 1 y n 2 y + n 1 z n 2 z n 1 x 2 + n 1 y 2 + n 1 z 2 n 2 x 2 + n 2 y 2 + n 2 z 2 , where we have that n 1 → = (n 1 x , n 1 y , n 1 z) and n 2 → = (n 2 x , n 2 y , n 2 z) are the coordinates of the vectors of the represented planes.

The angle between intersecting lines is calculated using the formula

α = a r c cos n 1 x n 2 x + n 1 y n 2 y + n 1 z n 2 z n 1 x 2 + n 1 y 2 + n 1 z 2 n 2 x 2 + n 2 y 2 + n 2 z 2

Example 2

By condition, a parallelepiped А В С D A 1 B 1 C 1 D 1 is given , where A B \u003d 2, A D \u003d 3, A A 1 \u003d 7, and point E separates the side A A 1 4: 3. Find the angle between planes A B C and B E D 1 .

Decision

It can be seen from the condition that its sides are pairwise perpendicular. This means that it is necessary to introduce a coordinate system O x y z with a vertex at point C and coordinate axes O x, O y, O z. It is necessary to put the direction on the appropriate sides. Consider the figure below.

Intersecting planes A B C and B E D 1 form an angle, which can be found by the formula 2 x 2 + n 2 y 2 + n 2 z 2 , where n 1 → = (n 1 x , n 1 y , n 1 z) and n 2 → = (n 2 x , n 2 y , n 2 z ) are normal vectors of these planes. It is necessary to determine the coordinates. From the figure, we see that the coordinate axis O x y coincides in the plane A B C, which means that the coordinates of the normal vector k → equal to the value n 1 → = k → = (0, 0, 1) .

The normal vector of the plane B E D 1 is the vector product of B E → and B D 1 → , where their coordinates are found by the coordinates extreme points B, E, D 1 , which are determined based on the condition of the problem.

We get that B (0 , 3 , 0) , D 1 (2 , 0 , 7) . Because A E E A 1 = 4 3 , from the coordinates of the points A 2 , 3 , 0 , A 1 2 , 3 , 7 we find E 2 , 3 , 4 . We get that B E → = (2 , 0 , 4) , B D 1 → = 2 , - 3 , 7 n 2 → = B E → × B D 1 = i → j → k → 2 0 4 2 - 3 7 = 12 i → - 6 j → - 6 k → ⇔ n 2 → = (12, - 6, - 6)

It is necessary to substitute the found coordinates into the formula for calculating the angle through the arc cosine. We get

α = a r c cos 0 12 + 0 (- 6) + 1 (- 6) 0 2 + 0 2 + 1 2 12 2 + (- 6) 2 + (- 6) 2 = a r c cos 6 6 6 = a r c cos 6 6

The coordinate method gives a similar result.

Answer: a r c cos 6 6 .

The final problem is considered in order to find the angle between the intersecting planes with the available known equations of the planes.

Example 3

Calculate the sine, cosine of the angle, and the value of the angle formed by two intersecting lines, which are defined in the O x y z coordinate system and given by the equations 2 x - 4 y + z + 1 = 0 and 3 y - z - 1 = 0 .

Decision

When studying the topic of the general equation of a straight line of the form A x + B y + C z + D = 0, it was revealed that A, B, C are coefficients equal to the coordinates of the normal vector. Hence, n 1 → = 2 , - 4 , 1 and n 2 → = 0 , 3 , - 1 are normal vectors of given lines.

It is necessary to substitute the coordinates of the normal vectors of the planes into the formula for calculating the desired angle of intersecting planes. Then we get that

α = a r c cos 2 0 + - 4 3 + 1 (- 1) 2 2 + - 4 2 + 1 2 = a r c cos 13 210

Hence we have that the cosine of the angle takes the form cos α = 13 210 . Then the angle of the intersecting lines is not obtuse. Substituting into the trigonometric identity, we get that the value of the sine of the angle is equal to the expression. We calculate and get that

sin α = 1 - cos 2 α = 1 - 13 210 = 41 210

Answer: sin α = 41 210 , cos α = 13 210 , α = a r c cos 13 210 = a r c sin 41 210 .

If you notice a mistake in the text, please highlight it and press Ctrl+Enter

Injection φ general equations A 1 x + B 1 y + C 1 = 0 and A 2 x + B 2 y + C 2 = 0, is calculated by the formula:

Injection φ between two straight lines canonical equations(x-x 1) / m 1 \u003d (y-y 1) / n 1 and (x-x 2) / m 2 \u003d (y-y 2) / n 2, is calculated by the formula:

Distance from point to line

Each plane in space can be represented as linear equation called general equation plane

Special cases.

o If in equation (8), then the plane passes through the origin.

o With (,) the plane is parallel to the axis(axis, axis), respectively.

o When (,) the plane is parallel to the plane(plane, plane).

Solution: use (7)

Answer: the general equation of the plane.

Example.

The plane in the rectangular coordinate system Oxyz is given by the general equation of the plane . Write down the coordinates of all normal vectors in this plane.

We know that the coefficients of the variables x, y, and z in the general equation of the plane are the corresponding coordinates of the normal vector of that plane. Therefore, the normal vector of the given plane has coordinates. The set of all normal vectors can be given as.

Write the equation of a plane if in a rectangular coordinate system Oxyz in space it passes through a point , a is the normal vector of this plane.

We present two solutions to this problem.

From the condition we have . We substitute these data into the general equation of the plane passing through the point:

Write the general equation for a plane parallel to the coordinate plane Oyz and passing through the point .

A plane that is parallel to the coordinate plane Oyz can be given by a general incomplete equation of the plane of the form . Since the point belongs to the plane by condition, then the coordinates of this point must satisfy the equation of the plane, that is, equality must be true. From here we find. Thus, the desired equation has the form.

Decision. The vector product, by definition 10.26, is orthogonal to the vectors p and q. Therefore, it is orthogonal to the desired plane and the vector can be taken as its normal vector. Find the coordinates of the vector n:

i.e . Using formula (11.1), we obtain

Opening the brackets in this equation, we arrive at the final answer.

Answer: .

Let's rewrite the normal vector in the form and find its length:

According to the above:

Answer:

Parallel planes have the same normal vector. 1) From the equation we find the normal vector of the plane:.

2) We compose the equation of the plane according to the point and the normal vector:

Answer:

Vector equation of a plane in space

Parametric equation of a plane in space

Equation of a plane passing through a given point perpendicular to a given vector

Let in three-dimensional space a rectangular Cartesian coordinate system is specified. Let's formulate the following problem:

Write an equation for a plane passing through a given point M(x 0, y 0, z 0) perpendicular to the given vector n = ( A, B, C} .

Decision. Let be P(x, y, z) is an arbitrary point in space. Dot P belongs to the plane if and only if the vector MP = {x − x 0, y − y 0, z − z 0) orthogonal to vector n = {A, B, C) (Fig. 1).

Having written the orthogonality condition for these vectors (n, MP) = 0 in coordinate form, we get:

A(x − x 0) + B(y − y 0) + C(z − z 0) = 0

Equation of a plane by three points

In vector form

In coordinates

Mutual arrangement of planes in space

are general equations of two planes. Then:

1) if , then the planes coincide;

2) if , then the planes are parallel;

3) if or , then the planes intersect and the system of equations

(6)

are the equations of the line of intersection of the given planes.

Decision: We compose the canonical equations of the straight line by the formula:

Answer:

We take the resulting equations and mentally “pin off”, for example, the left piece: . Now we equate this piece to any number(remember that there was already a zero), for example, to one: . Since , then the other two "pieces" must also be equal to one. Essentially, you need to solve the system:

Write parametric equations for the following lines:

Decision: The lines are given by canonical equations and at the first stage one should find some point belonging to the line and its direction vector.

a) From the equations remove the point and the direction vector: . You can choose another point (how to do this is described above), but it is better to take the most obvious one. By the way, to avoid mistakes, always substitute its coordinates into the equations.

Let us compose the parametric equations of this straight line:

The convenience of parametric equations is that with their help it is very easy to find other points of the line. For example, let's find a point whose coordinates, say, correspond to the value of the parameter :

Thus: b) Consider the canonical equations . The choice of a point here is simple, but insidious: (be careful not to mix up the coordinates!!!). How to pull out a guide vector? You can speculate what this line is parallel to, or you can use a simple formal trick: the proportion is “y” and “Z”, so we write the direction vector , and put zero in the remaining space: .

We compose the parametric equations of the straight line:

c) Let's rewrite the equations in the form , that is, "Z" can be anything. And if any, then let, for example, . Thus, the point belongs to this line. To find the direction vector, we use the following formal technique: in the initial equations there are "x" and "y", and in the direction vector at these places we write zeros: . In the remaining place we put unit: . Instead of one, any number, except zero, will do.

We write the parametric equations of the straight line:

corner between straight lines in space we will call any of the adjacent angles formed by two straight lines drawn through an arbitrary point parallel to the data.

Let two straight lines be given in space:

Obviously, the angle φ between the lines can be taken as the angle between their direction vectors and . Since , then according to the formula for the cosine of the angle between the vectors we get

The conditions of parallelism and perpendicularity of two lines are equivalent to the conditions of parallelism and perpendicularity of their direction vectors and:

Two straight are parallel if and only if their respective coefficients are proportional, i.e. l 1 parallel l 2 if and only if parallel .

Two straight perpendicular if and only if the sum of the products of the corresponding coefficients is equal to zero: .

At goal between line and plane

Let the line d- not perpendicular to the plane θ;
d′− projection of a straight line d to the plane θ;
The smallest of the angles between straight lines d and d′ we will call angle between line and plane.
Let's denote it as φ=( d,θ)
If a d⊥θ , then ( d,θ)=π/2

Oi→j→k→− rectangular coordinate system.
Plane equation:

θ: Ax+By+cz+D=0

We consider that the line is given by a point and a direction vector: d[M 0,p→]
Vector n→(A,B,C)⊥θ
Then it remains to find out the angle between the vectors n→ and p→, denote it as γ=( n→,p→).

If the angle γ<π/2 , то искомый угол φ=π/2−γ .

If the angle γ>π/2 , then the required angle φ=γ−π/2

sinφ=sin(2π−γ)=cosγ

sinφ=sin(γ−2π)=−cosγ

Then, angle between line and plane can be calculated using the formula:

sinφ=∣cosγ∣=∣ ∣ Ap 1+bp 2+cp 3∣ ∣ √A 2+B 2+C 2√p 21+p 22+p 23

Question 29. The concept of a quadratic form. The sign-definiteness of quadratic forms.

Quadratic form j (x 1, x 2, ..., x n) n real variables x 1, x 2, ..., x n is called a sum of the form
, (1)

where aij are some numbers called coefficients. Without loss of generality, we can assume that aij = a ji.

The quadratic form is called valid, if aij О GR. Matrix of quadratic form is called the matrix composed of its coefficients. Quadratic form (1) corresponds to a unique symmetric matrix
i.e. A T = A. Therefore, quadratic form (1) can be written in matrix form j ( X) = x T Ah, where x T = (X 1 X 2 … x n). (2)

And vice versa, any symmetric matrix (2) corresponds to a unique quadratic form up to the notation of variables.

The rank of the quadratic form is called the rank of its matrix. The quadratic form is called non-degenerate, if its matrix is nonsingular BUT. (recall that the matrix BUT is called non-degenerate if its determinant is non-zero). Otherwise, the quadratic form is degenerate.

positive definite(or strictly positive) if

j ( X) > 0 , for anyone X = (X 1 , X 2 , …, x n), Besides X = (0, 0, …, 0).

Matrix BUT positive definite quadratic form j ( X) is also called positive definite. Therefore, a positive definite quadratic form corresponds to a unique positive definite matrix and vice versa.

The quadratic form (1) is called negative definite(or strictly negative) if

j ( X) < 0, для любого X = (X 1 , X 2 , …, x n), Besides X = (0, 0, …, 0).

Similarly as above, a negative-definite quadratic matrix is also called negative-definite.

Therefore, a positively (negatively) definite quadratic form j ( X) reaches the minimum (maximum) value j ( X*) = 0 for X* = (0, 0, …, 0).

Note that most of the quadratic forms are not sign-definite, that is, they are neither positive nor negative. Such quadratic forms vanish not only at the origin of the coordinate system, but also at other points.

When n> 2, special criteria are required to check the sign-definiteness of a quadratic form. Let's consider them.

Major Minors quadratic form are called minors:

that is, these are minors of order 1, 2, …, n matrices BUT, located in the upper left corner, the last of them coincides with the determinant of the matrix BUT.

Criterion for positive definiteness (Sylvester criterion)

X) = x T Ah is positive definite, it is necessary and sufficient that all principal minors of the matrix BUT were positive, that is: M 1 > 0, M 2 > 0, …, M n > 0. Criterion of negative certainty In order for the quadratic form j ( X) = x T Ah is negative definite, it is necessary and sufficient that its principal minors of even order be positive, and those of odd order are negative, i.e.: M 1 < 0, M 2 > 0, M 3 < 0, …, (–1)n

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.

We 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 the right 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. It remains to find the cosine of the angle: