Find the angle between two. Angle between lines on a plane

Oh-oh-oh-oh-oh ... well, it's tinny, as if you read the sentence to yourself =) However, then relaxation will help, especially since today I bought suitable accessories. Therefore, let's proceed to the first section, I hope, by the end of the article I will keep a cheerful mood.

Mutual arrangement of two straight lines

The case when the hall sings along in chorus. Two lines can:

1) match;

2) be parallel: ;

3) or intersect at a single point: .

Help for dummies : please remember the mathematical sign of the intersection , it will occur very often. The entry means that the line intersects with the line at the point.

How to determine the relative position of two lines?

Let's start with the first case:

Two lines coincide if and only if their respective coefficients are proportional, that is, there is such a number "lambda" that the equalities

Let's consider straight lines and compose three equations from the corresponding coefficients: . From each equation it follows that, therefore, these lines coincide.

Indeed, if all the coefficients of the equation multiply by -1 (change signs), and reduce all the coefficients of the equation by 2, you get the same equation: .

The second case when the lines are parallel:

Two lines are parallel if and only if their coefficients at the variables are proportional: , but.

As an example, consider two straight lines. We check the proportionality of the corresponding coefficients for the variables :

However, it is clear that .

And the third case, when the lines intersect:

Two lines intersect if and only if their coefficients of the variables are NOT proportional, that is, there is NOT such a value of "lambda" that the equalities are fulfilled

So, for straight lines we will compose a system:

From the first equation it follows that , and from the second equation: , hence, the system is inconsistent(no solutions). Thus, the coefficients at the variables are not proportional.

Conclusion: lines intersect

In practical problems, the solution scheme just considered can be used. By the way, it is very similar to the algorithm for checking vectors for collinearity, which we considered in the lesson. The concept of linear (non) dependence of vectors. Vector basis. But there is a more civilized package:

Example 1

Find out the relative position of the lines:

Solution based on the study of directing vectors of straight lines:

a) From the equations we find the direction vectors of the lines: .

, so the vectors are not collinear and the lines intersect.

Just in case, I will put a stone with pointers at the crossroads:

The rest jump over the stone and follow on, straight to Kashchei the Deathless =)

b) Find the direction vectors of the lines:

The lines have the same direction vector, which means they are either parallel or the same. Here the determinant is not necessary.

Obviously, the coefficients of the unknowns are proportional, while .

Let's find out if the equality is true:

In this way,

c) Find the direction vectors of the lines:

Let's calculate the determinant, composed of the coordinates of these vectors:
, therefore, the direction vectors are collinear. The lines are either parallel or coincide.

The proportionality factor "lambda" is easy to see directly from the ratio of collinear direction vectors. However, it can also be found through the coefficients of the equations themselves: .

Now let's find out if the equality is true. Both free terms are zero, so:

The resulting value satisfies this equation (any number generally satisfies it).

Thus, the lines coincide.

Answer:

Very soon you will learn (or even have already learned) to solve the considered problem verbally literally in a matter of seconds. In this regard, I see no reason to offer something for an independent solution, it is better to lay one more important brick in the geometric foundation:

How to draw a line parallel to a given one?

For ignorance of this simplest task, the Nightingale the Robber severely punishes.

Example 2

The straight line is given by the equation . Write an equation for a parallel line that passes through the point.

Solution: Denote the unknown line by the letter . What does the condition say about it? The line passes through the point. And if the lines are parallel, then it is obvious that the directing vector of the line "ce" is also suitable for constructing the line "te".

We take out the direction vector from the equation:

Answer:

The geometry of the example looks simple:

Analytical verification consists of the following steps:

1) We check that the lines have the same direction vector (if the equation of the line is not properly simplified, then the vectors will be collinear).

2) Check if the point satisfies the resulting equation.

Analytical verification in most cases is easy to perform orally. Look at the two equations and many of you will quickly figure out how the lines are parallel without any drawing.

Examples for self-solving today will be creative. Because you still have to compete with Baba Yaga, and she, you know, is a lover of all kinds of riddles.

Example 3

Write an equation for a line passing through a point parallel to the line if

There is a rational and not very rational way to solve. The shortest way is at the end of the lesson.

We did a little work with parallel lines and will return to them later. The case of coinciding lines is of little interest, so let's consider a problem that is well known to you from the school curriculum:

How to find the point of intersection of two lines?

If straight intersect at the point , then its coordinates are the solution systems of linear equations

How to find the point of intersection of lines? Solve the system.

Here's to you geometric meaning of a system of two linear equations with two unknowns are two intersecting (most often) straight lines on a plane.

Example 4

Find the point of intersection of lines

Solution: There are two ways to solve - graphical and analytical.

The graphical way is to simply draw the given lines and find out the point of intersection directly from the drawing:

Here is our point: . To check, you should substitute its coordinates into each equation of a straight line, they should fit both there and there. In other words, the coordinates of a point are the solution of the system . In fact, we considered a graphical way to solve systems of linear equations with two equations, two unknowns.

The graphical method, of course, is not bad, but there are noticeable disadvantages. No, the point is not that seventh graders decide this way, the point is that it will take time to make a correct and EXACT drawing. In addition, some lines are not so easy to construct, and the intersection point itself can be somewhere in the thirtieth kingdom outside the notebook sheet.

Therefore, it is more expedient to search for the intersection point by the analytical method. Let's solve the system:

To solve the system, the method of termwise addition of equations was used. To develop the relevant skills, visit the lesson How to solve a system of equations?

Answer:

The verification is trivial - the coordinates of the intersection point must satisfy each equation of the system.

Example 5

Find the point of intersection of the lines if they intersect.

This is a do-it-yourself example. The task can be conveniently divided into several stages. Analysis of the condition suggests that it is necessary:
1) Write the equation of a straight line.
2) Write the equation of a straight line.
3) Find out the relative position of the lines.
4) If the lines intersect, then find the point of intersection.

The development of an action algorithm is typical for many geometric problems, and I will repeatedly focus on this.

Full solution and answer at the end of the tutorial:

A pair of shoes has not yet been worn out, as we got to the second section of the lesson:

Perpendicular lines. The distance from a point to a line.
Angle between lines

Let's start with a typical and very important task. In the first part, we learned how to build a straight line parallel to the given one, and now the hut on chicken legs will turn 90 degrees:

How to draw a line perpendicular to a given one?

Example 6

The straight line is given by the equation . Write an equation for a perpendicular line passing through a point.

Solution: It is known by assumption that . It would be nice to find the direction vector of the straight line. Since the lines are perpendicular, the trick is simple:

From the equation we “remove” the normal vector: , which will be the directing vector of the straight line.

We compose the equation of a straight line by a point and a directing vector:

Answer:

Let's unfold the geometric sketch:

Hmmm... Orange sky, orange sea, orange camel.

Analytical verification of the solution:

1) Extract the direction vectors from the equations and with the help dot product of vectors we conclude that the lines are indeed perpendicular: .

By the way, you can use normal vectors, it's even easier.

2) Check if the point satisfies the resulting equation .

Verification, again, is easy to perform verbally.

Example 7

Find the point of intersection of perpendicular lines, if the equation is known and dot.

This is a do-it-yourself example. There are several actions in the task, so it is convenient to arrange the solution point by point.

Our exciting journey continues:

Distance from point to line

Before us is a straight strip of the river and our task is to reach it in the shortest way. There are no obstacles, and the most optimal route will be movement along the perpendicular. That is, the distance from a point to a line is the length of the perpendicular segment.

The distance in geometry is traditionally denoted by the Greek letter "ro", for example: - the distance from the point "em" to the straight line "de".

Distance from point to line is expressed by the formula

Example 8

Find the distance from a point to a line

Solution: all you need is to carefully substitute the numbers into the formula and do the calculations:

Answer:

Let's execute the drawing:

The distance found from the point to the line is exactly the length of the red segment. If you make a drawing on checkered paper on a scale of 1 unit. \u003d 1 cm (2 cells), then the distance can be measured with an ordinary ruler.

Consider another task according to the same drawing:

The task is to find the coordinates of the point , which is symmetrical to the point with respect to the line . I propose to perform the actions on your own, however, I will outline the solution algorithm with intermediate results:

1) Find a line that is perpendicular to a line.

2) Find the point of intersection of the lines: .

Both actions are discussed in detail in this lesson.

3) The point is the midpoint of the segment. We know the coordinates of the middle and one of the ends. By formulas for the coordinates of the middle of the segment find .

It will not be superfluous to check that the distance is also equal to 2.2 units.

Difficulties here may arise in calculations, but in the tower a microcalculator helps out a lot, allowing you to count ordinary fractions. Have advised many times and will recommend again.

How to find the distance between two parallel lines?

Example 9

Find the distance between two parallel lines

This is another example for an independent solution. A little hint: there are infinitely many ways to solve. Debriefing at the end of the lesson, but better try to guess for yourself, I think you managed to disperse your ingenuity well.

Angle between two lines

Whatever the corner, then the jamb:


In geometry, the angle between two straight lines is taken as the SMALLER angle, from which it automatically follows that it cannot be obtuse. In the figure, the angle indicated by the red arc is not considered to be the angle between intersecting lines. And its “green” neighbor or oppositely oriented crimson corner.

If the lines are perpendicular, then any of the 4 angles can be taken as the angle between them.

How are the angles different? Orientation. First, the direction of "scrolling" the corner is fundamentally important. Secondly, a negatively oriented angle is written with a minus sign, for example, if .

Why did I say this? It seems that you can get by with the usual concept of an angle. The fact is that in the formulas by which we will find the angles, a negative result can easily be obtained, and this should not take you by surprise. An angle with a minus sign is no worse, and has a very specific geometric meaning. In the drawing for a negative angle, it is imperative to indicate its orientation (clockwise) with an arrow.

How to find the angle between two lines? There are two working formulas:

Example 10

Find the angle between lines

Solution and Method one

Consider two straight lines given by equations in general form:

If straight not perpendicular, then oriented the angle between them can be calculated using the formula:

Let's pay close attention to the denominator - this is exactly scalar product direction vectors of straight lines:

If , then the denominator of the formula vanishes, and the vectors will be orthogonal and the lines will be perpendicular. That is why a reservation was made about the non-perpendicularity of the lines in the formulation.

Based on the foregoing, the solution is conveniently formalized in two steps:

1) Calculate the scalar product of directing vectors of straight lines:
so the lines are not perpendicular.

2) We find the angle between the lines by the formula:

Using the inverse function, it is easy to find the angle itself. In this case, we use the oddness of the arc tangent (see Fig. Graphs and properties of elementary functions):

Answer:

In the answer, we indicate the exact value, as well as the approximate value (preferably both in degrees and in radians), calculated using a calculator.

Well, minus, so minus, it's okay. Here is a geometric illustration:

It is not surprising that the angle turned out to be of a negative orientation, because in the condition of the problem the first number is a straight line and the “twisting” of the angle began precisely from it.

If you really want to get a positive angle, you need to swap the straight lines, that is, take the coefficients from the second equation , and take the coefficients from the first equation . In short, you need to start with a direct .

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 a 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.

Solution. 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:

that is . 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 a rectangular Cartesian coordinate system be given in three-dimensional space. 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} .

Solution. Let 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.

Solution: 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:

Solution: 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:

a. Let two lines be given. These lines, as it was indicated in Chapter 1, form various positive and negative angles, which can be either acute or obtuse. Knowing one of these angles, we can easily find any other.

By the way, for all these angles, the numerical value of the tangent is the same, the difference can only be in the sign

Equations of lines. The numbers are the projections of the directing vectors of the first and second lines. The angle between these vectors is equal to one of the angles formed by straight lines. Therefore, the problem is reduced to determining the angle between the vectors, We get

For simplicity, we can agree on an angle between two straight lines to understand an acute positive angle (as, for example, in Fig. 53).

Then the tangent of this angle will always be positive. Thus, if a minus sign is obtained on the right side of formula (1), then we must discard it, i.e., keep only the absolute value.

Example. Determine the angle between lines

By formula (1) we have

With. If it is indicated which of the sides of the angle is its beginning and which is its end, then, counting always the direction of the angle counterclockwise, we can extract something more from formulas (1). As is easy to see from Fig. 53 the sign obtained on the right side of the formula (1) will indicate which one - acute or obtuse - the angle forms the second line with the first.

(Indeed, from Fig. 53 we see that the angle between the first and second direction vectors is either equal to the desired angle between the lines, or differs from it by ±180°.)

d. If the lines are parallel, then their direction vectors are also parallel. Applying the condition of parallelism of two vectors, we get!

This is a necessary and sufficient condition for two lines to be parallel.

Example. Direct

are parallel because

e. If the lines are perpendicular, then their direction vectors are also perpendicular. Applying the condition of perpendicularity of two vectors, we obtain the condition of perpendicularity of two lines, namely

Example. Direct

perpendicular because

In connection with the conditions of parallelism and perpendicularity, we will solve the following two problems.

f. Draw a line parallel to a given line through a point

The decision is made like this. Since the desired line is parallel to the given one, then for its directing vector we can take the same one as that of the given line, i.e., a vector with projections A and B. And then the equation of the desired line will be written in the form (§ 1)

Example. Equation of a straight line passing through a point (1; 3) parallel to a straight line

will be next!

g. Draw a line through a point perpendicular to the given line

Here, it is no longer suitable to take a vector with projections A and as a directing vector, but it is necessary to win a vector perpendicular to it. The projections of this vector must therefore be chosen according to the condition that both vectors are perpendicular, i.e., according to the condition

This condition can be fulfilled in an infinite number of ways, since here there is one equation with two unknowns. But the easiest way is to take it. Then the equation of the desired straight line will be written in the form

Example. Equation of a line passing through a point (-7; 2) in a perpendicular line

will be the following (according to the second formula)!

h. In the case when the lines are given by equations of the form

rewriting these equations differently, we have

Definition. If two lines are given y = k 1 x + b 1 , y = k 2 x + b 2 , then the acute angle between these lines will be defined as

Two lines are parallel if k 1 = k 2 . Two lines are perpendicular if k 1 = -1/ k 2 .

Theorem. The straight lines Ax + Vy + C \u003d 0 and A 1 x + B 1 y + C 1 \u003d 0 are parallel when the coefficients A 1 \u003d λA, B 1 \u003d λB are proportional. If also С 1 = λС, then the lines coincide. The coordinates of the point of intersection of two lines are found as a solution to the system of equations of these lines.

Equation of a line passing through a given point

Perpendicular to this line

Definition. The line passing through the point M 1 (x 1, y 1) and perpendicular to the line y \u003d kx + b is represented by the equation:

Distance from point to line

Theorem. If a point M(x 0, y 0) is given, then the distance to the line Ax + Vy + C \u003d 0 is defined as

.

Proof. Let the point M 1 (x 1, y 1) be the base of the perpendicular dropped from the point M to the given line. Then the distance between points M and M 1:

(1)

The x 1 and y 1 coordinates can be found as a solution to the system of equations:

The second equation of the system is the equation of a straight line passing through a given point M 0 perpendicular to a given straight line. If we transform the first equation of the system to the form:

A(x - x 0) + B(y - y 0) + Ax 0 + By 0 + C = 0,

then, solving, we get:

Substituting these expressions into equation (1), we find:

The theorem has been proven.

Example. Determine the angle between the lines: y = -3 x + 7; y = 2 x + 1.

k 1 \u003d -3; k2 = 2; tgφ = ; φ= p /4.

Example. Show that the lines 3x - 5y + 7 = 0 and 10x + 6y - 3 = 0 are perpendicular.

Solution. We find: k 1 \u003d 3/5, k 2 \u003d -5/3, k 1 * k 2 \u003d -1, therefore, the lines are perpendicular.

Example. The vertices of the triangle A(0; 1), B (6; 5), C (12; -1) are given. Find the equation for the height drawn from vertex C.

Solution. We find the equation of the side AB: ; 4 x = 6 y - 6;

2x – 3y + 3 = 0;

The desired height equation is: Ax + By + C = 0 or y = kx + b. k = . Then y = . Because the height passes through point C, then its coordinates satisfy this equation: whence b = 17. Total: .

Answer: 3x + 2y - 34 = 0.

Equation of a line passing through a given point in a given direction. Equation of a straight line passing through two given points. Angle between two lines. Condition of parallelism and perpendicularity of two lines. Determining the point of intersection of two lines

1. Equation of a line passing through a given point A(x 1 , y 1) in a given direction, determined by the slope k,

y - y 1 = k(x - x 1). (1)

This equation defines a pencil of lines passing through a point A(x 1 , y 1), which is called the center of the beam.

2. Equation of a straight line passing through two points: A(x 1 , y 1) and B(x 2 , y 2) is written like this:

The slope of a straight line passing through two given points is determined by the formula

3. Angle between straight lines A and B is the angle by which the first straight line must be rotated A around the point of intersection of these lines counterclockwise until it coincides with the second line B. If two lines are given by slope equations

y = k 1 x + B 1 ,

y = k 2 x + B 2 , (4)

then the angle between them is determined by the formula

It should be noted that in the numerator of the fraction, the slope of the first straight line is subtracted from the slope of the second straight line.

If the equations of a straight line are given in general form

A 1 x + B 1 y + C 1 = 0,

A 2 x + B 2 y + C 2 = 0, (6)

the angle between them is determined by the formula

4. Conditions for parallelism of two lines:

a) If the lines are given by equations (4) with a slope, then the necessary and sufficient condition for their parallelism is the equality of their slopes:

k 1 = k 2 . (8)

b) For the case when the lines are given by equations in general form (6), the necessary and sufficient condition for their parallelism is that the coefficients at the corresponding current coordinates in their equations are proportional, i.e.

5. Conditions for perpendicularity of two lines:

a) In the case when the lines are given by equations (4) with a slope, the necessary and sufficient condition for their perpendicularity is that their slopes are reciprocal in magnitude and opposite in sign, i.e.

This condition can also be written in the form

k 1 k 2 = -1. (11)

b) If the equations of straight lines are given in general form (6), then the condition for their perpendicularity (necessary and sufficient) is to fulfill the equality

A 1 A 2 + B 1 B 2 = 0. (12)

6. The coordinates of the point of intersection of two lines are found by solving the system of equations (6). Lines (6) intersect if and only if

1. Write the equations of the lines passing through the point M, one of which is parallel and the other is perpendicular to the given line l.

\(\blacktriangleright\) 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 \(CH\perp SD\) . Because \(\triangle SAD=\triangle SCD\), 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 the midline 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 a triangle \(\triangle KSL\) the inverse Pythagorean theorem holds \(\Rightarrow\) \(\triangle KSL\) is a 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 within the framework of the 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 to find the trigonometric function of the dihedral angle, which is formed by the 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

In the process of studying on the eve of 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.