How to find the angle between straight lines. Angle between straight lines on a plane

Oh-oh-oh-oh-oh... well, it’s tough, as if he was reading out a sentence to himself =) However, relaxation will help later, especially since today I bought the appropriate accessories. Therefore, let's proceed to the first section, I hope that by the end of the article I will maintain a cheerful mood.

The relative position of two straight lines

This is the case when the audience sings along in chorus. Two straight lines can:

1) match;

2) be parallel: ;

3) or intersect at a single point: .

Help for dummies : Please remember the mathematical intersection sign, it will appear very often. The notation means that the line intersects with the line at 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 corresponding coefficients are proportional, that is, there is a number “lambda” such that the equalities are satisfied

Let's consider the straight lines and create 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 all coefficients of the equation cut 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 of 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 quite obvious 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 NO such value of “lambda” that the equalities are satisfied

So, for straight lines we will create a system:

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

Conclusion: lines intersect

In practical problems, you can use the solution scheme just discussed. By the way, it is very reminiscent of the algorithm for checking vectors for collinearity, which we looked at in class The concept of linear (in)dependence of vectors. Basis of vectors. But there is a more civilized packaging:

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: .

, which means that the vectors are not collinear and the lines intersect.

Just in case, I’ll put a stone with signs at the crossroads:

The rest jump over the stone and follow further, straight to Kashchei the Immortal =)

b) Find the direction vectors of the lines:

The lines have the same direction vector, which means they are either parallel or coincident. There is no need to count the determinant here.

It is obvious that the coefficients of the unknowns are proportional, and .

Let's find out whether the equality is true:

Thus,

c) Find the direction vectors of the lines:

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

The proportionality coefficient “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 whether the equality is true. Both free terms are zero, so:

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

Thus, the lines coincide.

Answer:

Very soon you will learn (or even have already learned) to solve the problem discussed verbally literally in a matter of seconds. In this regard, I don’t see any point in offering anything for an independent solution; it’s better to lay another important brick in the geometric foundation:

How to construct 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: Let's denote the unknown line by the letter . What does the condition say about her? The straight line passes through the point. And if the lines are parallel, then it is obvious that the direction vector of the straight line “tse” is also suitable for constructing the straight line “de”.

We take the direction vector out of the equation:

Answer:

The example geometry looks simple:

Analytical testing consists of the following steps:

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

2) Check whether the point satisfies the resulting equation.

In most cases, analytical testing can be easily performed orally. Look at the two equations, and many of you will quickly determine the parallelism of the lines without any drawing.

Examples for independent solutions today will be creative. Because you will still have to compete with Baba Yaga, and she, you know, is a lover of all sorts 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 so rational way to solve it. The shortest way is at the end of the lesson.

We worked a little 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 very familiar to you from the school curriculum:

How to find the point of intersection of two lines?

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

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

Here you go geometric meaning of a system of two linear equations with two unknowns- these are two intersecting (most often) 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 method is to simply draw the given lines and find out the intersection point directly from the drawing:

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

The graphical method is, of course, 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 create a correct and ACCURATE drawing. In addition, some straight lines are not so easy to construct, and the point of intersection itself may be located somewhere in the thirtieth kingdom outside the notebook sheet.

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

To solve the system, the method of term-by-term addition of equations was used. To develop relevant skills, take a lesson How to solve a system of equations?

Answer:

The check 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 an example for you to solve on your own. It is convenient to split the task into several stages. Analysis of the condition suggests that it is necessary:
1) Write down the equation of the straight line.
2) Write down the equation of the 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 lesson:

Not even a pair of shoes were worn out before we got to the second section of the lesson:

Perpendicular lines. Distance from a point to a line.
Angle between straight 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 this one, and now the hut on chicken legs will turn 90 degrees:

How to construct a line perpendicular to a given one?

Example 6

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

Solution: By condition it is known that . It would be nice to find the directing vector of the 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.

Let's compose the equation of a straight line using a point and a direction vector:

Answer:

Let's expand the geometric sketch:

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

Analytical verification of the solution:

1) We take out the direction vectors from the equations and with the help scalar product of vectors we come to the conclusion that the lines are indeed perpendicular: .

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

2) Check whether the point satisfies the resulting equation .

The test, again, is easy to perform orally.

Example 7

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

This is an example for you to solve on your own. There are several actions in the problem, so it is convenient to formulate the solution point by point.

Our exciting journey continues:

Distance from point to line

In front of us is a straight strip of the river and our task is to get to it by the shortest route. There are no obstacles, and the most optimal route will be to move along the perpendicular. That is, the distance from a point to a line is the length of the perpendicular segment.

Distance in geometry is traditionally denoted by the Greek letter “rho”, for example: – the distance from the point “em” to the straight line “de”.

Distance from point to line expressed by the formula

Example 8

Find the distance from a point to a line

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

Answer:

Let's make the drawing:

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

Let's consider another task based on the same drawing:

The task is to find the coordinates of a point that is symmetrical to the point relative to the straight line . I suggest performing the steps yourself, but I will outline the solution algorithm with intermediate results:

1) Find a line that is perpendicular to the 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 midpoint of a segment we find .

It would be a good idea to check that the distance is also 2.2 units.

Difficulties may arise in calculations here, but a microcalculator is a great help in the tower, allowing you to calculate ordinary fractions. I have advised you many times and will recommend you again.

How to find the distance between two parallel lines?

Example 9

Find the distance between two parallel lines

This is another example for you to decide on your own. I’ll give you a little hint: there are infinitely many ways to solve this. Debriefing at the end of the lesson, but it’s better to try to guess for yourself, I think your ingenuity was well developed.

Angle between two straight lines

Every corner is a jamb:


In geometry, the angle between two straight lines is taken to be 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 the angle between intersecting lines. And his “green” neighbor or oppositely oriented"raspberry" 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. Firstly, the direction in which the angle is “scrolled” is fundamentally important. Secondly, a negatively oriented angle is written with a minus sign, for example if .

Why did I tell you this? It seems that we can get by with the usual concept of an angle. The fact is that the formulas by which we will find angles can easily result in a negative result, 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, be sure to indicate its orientation with an arrow (clockwise).

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

Example 10

Find the angle between lines

Solution And Method one

Let's consider two straight lines defined by equations in general form:

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

Let us pay close attention to the denominator - this is exactly scalar product directing vectors of straight lines:

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

Based on the above, it is convenient to formalize the solution in two steps:

1) Let's calculate the scalar product of the direction vectors of the lines:
, which means the lines are not perpendicular.

2) Find the angle between straight lines using the formula:

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

Answer:

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

Well, minus, minus, no big deal. Here is a geometric illustration:

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

If you really want to get a positive angle, you need to swap the 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 .

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 lines Ax + Bу + C = 0 and A 1 x + B 1 y + C 1 = 0 are parallel when the coefficients A 1 = λA, B 1 = λB are proportional. If also C 1 = λC, 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 a given line

Definition. A straight line passing through the point M 1 (x 1, y 1) and perpendicular to the straight line y = 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 + Bу + C = 0 is determined as

.

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

(1)

The coordinates x 1 and y 1 can be found by solving the system of equations:

The second equation of the system is the equation of a line passing through a given point M 0 perpendicular to a given 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 = -3; k 2 = 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 = 3/5, k 2 = -5/3, k 1* k 2 = -1, therefore, the lines are perpendicular.

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

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

2 x – 3 y + 3 = 0;

The required height equation has the form: Ax + By + C = 0 or y = kx + b. k = . Then y = . Because the height passes through point C, then its coordinates satisfy this equation: from where b = 17. Total: .

Answer: 3 x + 2 y – 34 = 0.

The equation of a line passing through a given point in a given direction. Equation of a line passing through two given points. The angle between two straight lines. The condition of parallelism and perpendicularity of two straight 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 beam center.

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

The angular coefficient 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 straight lines are given by equations with a slope

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 line is subtracted from the slope of the second line.

If the equations of a 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 an angular coefficient, then the necessary and sufficient condition for their parallelism is the equality of their angular coefficients:

k 1 = k 2 . (8)

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

5. Conditions for perpendicularity of two straight lines:

a) In the case when the lines are given by equations (4) with an angular coefficient, a necessary and sufficient condition for their perpendicularity is that their angular coefficients are inverse 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 lines are given in general form (6), then the condition for their perpendicularity (necessary and sufficient) is to satisfy 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 lines passing through the point M, one of which is parallel and the other perpendicular to the given line l.

Angle 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 lines be given in space:

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

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

Two straight parallel if and only if their corresponding 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: .

U goal between line and plane

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

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

θ: Ax+By+Cz+D=0

We assume that the straight line is defined 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→, let us denote it as γ=( n→,p→).

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

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

sinφ=sin(2π−γ)=cosγ

sinφ=sin(γ−2π)=−cosγ

Then, angle between straight 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

Question29. The concept of quadratic form. 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 a ij – some numbers called coefficients. Without loss of generality, we can assume that a ij = a ji.

The quadratic form is called valid, If a ij Î GR. Matrix of quadratic form is called a matrix made up of its coefficients. The quadratic form (1) corresponds to the only symmetric matrix
That is A T = A. Consequently, 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, conversely, every symmetric matrix (2) corresponds to a unique quadratic form up to the notation of variables.

Rank of quadratic form is called the rank of its matrix. The quadratic form is called non-degenerate, if its matrix is ​​non-singular A. (recall that the matrix A is called non-degenerate if its determinant is not equal to 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), except X = (0, 0, …, 0).

Matrix A 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 negatively defined(or strictly negative) if

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

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

Consequently, the positive (negative) definite quadratic form j ( X) reaches the minimum (maximum) value j ( X*) = 0 at X* = (0, 0, …, 0).

Note that most 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 of a quadratic form. Let's look at them.

Major minors quadratic form are called minors:

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

Positive Definiteness Criterion (Sylvester criterion)

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

A. Let two straight lines be given. These straight lines, as 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 direction vectors of the first and second straight lines. The angle between these vectors is equal to one of the angles formed by straight lines. Therefore, the problem comes down to determining the angle between the vectors. We get

For simplicity, we can agree that the angle between two straight lines is an acute positive angle (as, for example, in Fig. 53).

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

Example. Determine the angle between straight lines

According to 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, always counting the direction of the angle counterclockwise, we can extract something more from formula (1). As is easy to see from Fig. 53, the sign obtained on the right side of formula (1) will indicate what kind of angle - acute or obtuse - the second straight line forms 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 straight lines, or differs from it by ±180°.)

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

This is a necessary and sufficient condition for the parallelism of two lines.

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 straight lines, namely

Example. Direct

are perpendicular due to the fact that

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

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

The solution is carried out like this. Since the desired line is parallel to this one, then for its direction 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 line passing through the point (1; 3) parallel to the line

there will be next!

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

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

This condition can be fulfilled in countless ways, since here is one equation with two unknowns. But the easiest way is to take or Then the equation of the desired line will be written in the form

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

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

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

Instructions

note

The period of the trigonometric tangent function is equal to 180 degrees, which means that the slope angles of straight lines cannot, in absolute value, exceed this value.

Helpful advice

If the angular coefficients are equal to each other, then the angle between such lines is 0, since such lines either coincide or are parallel.

To determine the value of the angle between intersecting lines, it is necessary to move both lines (or one of them) to a new position using the parallel translation method until they intersect. After this, you should find the angle between the resulting intersecting lines.

You will need

• Ruler, right triangle, pencil, protractor.

Instructions

So, let the vector V = (a, b, c) and the plane A x + B y + C z = 0 be given, where A, B and C are the coordinates of the normal N. Then the cosine of the angle α between the vectors V and N is equal to: cos α = (a A + b B + c C)/(√(a² + b² + c²) √(A² + B² + C²)).

To calculate the angle in degrees or radians, you need to calculate the inverse to cosine function from the resulting expression, i.e. arccosine:α = аrsсos ((a A + b B + c C)/(√(a² + b² + c²) √(A² + B² + C²))).

Example: find corner between vector(5, -3, 8) and plane, given by the general equation 2 x – 5 y + 3 z = 0. Solution: write down the coordinates of the normal vector of the plane N = (2, -5, 3). Substitute all known values ​​into the given formula: cos α = (10 + 15 + 24)/√3724 ≈ 0.8 → α = 36.87°.

Video on the topic

A straight line that has one common point with a circle is tangent to the circle. Another feature of the tangent is that it is always perpendicular to the radius drawn to the point of contact, that is, the tangent and radius form a straight line corner. If two tangents to a circle AB and AC are drawn from one point A, then they are always equal to each other. Determining the angle between tangents ( corner ABC) is made using the Pythagorean theorem.

Instructions

To determine the angle, you need to know the radius of the circle OB and OS and the distance of the starting point of the tangent from the center of the circle - O. So, angles ABO and ACO are equal, the radius OB is, for example, 10 cm, and the distance to the center of the circle AO is 15 cm. Determine the length of the tangent using formula in accordance with the Pythagorean theorem: AB = square root of AO2 – OB2 or 152 - 102 = 225 – 100 = 125;