It will be useful for every student who is preparing for the exam in mathematics to repeat the topic “Finding the angle between lines”. As statistics show, when passing a certification test, tasks in this section of stereometry cause difficulties for a large number students. At the same time, tasks requiring finding the angle between straight lines are found in the USE at both the basic and profile levels. This means that everyone should be able to solve them.
There are 4 types in space relative position direct. They can coincide, intersect, be parallel or intersecting. The angle between them can be acute or straight.
To find the angle between the lines in the Unified State Examination or, for example, in the solution, schoolchildren in Moscow and other cities can use several methods for solving problems in this section of stereometry. You can complete the task by classical constructions. To do this, it is worth learning the basic axioms and theorems of stereometry. The student needs to be able to logically build reasoning and create drawings in order to bring the task to a planimetric problem.
You can also use the vector-coordinate method by applying simple formulas, rules and algorithms. The main thing in this case is to correctly perform all the calculations. Hone your problem solving skills in stereometry and other topics school course will help you educational project"Shkolkovo".
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 view
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.
Let two lines l and m on a plane in a Cartesian coordinate system be given by the general equations: l: A 1 x + B 1 y + C 1 = 0, m: A 2 x + B 2 y + C 2 = 0
The vectors of normals to these lines: = (A 1 , B 1) - to the line l,
= (A 2 , B 2) to the line m.
Let j be the angle between lines l and m.
Since angles with mutually perpendicular sides are either equal or add up to p, then 
, i.e. cos j = .
So, we have proved the following theorem.
Theorem. Let j be the angle between two straight lines in the plane, and let these straight lines be given in the Cartesian coordinate system by the general equations A 1 x + B 1 y + C 1 = 0 and A 2 x + B 2 y + C 2 = 0. Then cos j = 
.
Exercises.
1) Derive a formula for calculating the angle between lines if:
(1) both lines are given parametrically; (2) both lines are given canonical equations; (3) one straight line is given parametrically, the other straight line – by the general equation; (4) both lines are given by the slope equation.
2) Let j be the angle between two straight lines in the plane, and let these straight lines be given to the Cartesian coordinate system by the equations y = k 1 x + b 1 and y =k 2 x + b 2 .
Then tan j = .
3) Explore the relative position of two lines given by general equations in the Cartesian coordinate system and fill in the table:
The distance from a point to a line in a plane.
Let the line l on the plane in the Cartesian coordinate system be given by the general equation Ax + By + C = 0. Find the distance from the point M(x 0 , y 0) to the line l.
The distance from the point M to the line l is the length of the perpendicular HM (H н l, HM ^ l).
The vector and the normal vector to the line l are collinear, so that | | = | | | | and | | = .
Let the coordinates of the point H be (x,y).
Since the point H belongs to the line l, then Ax + By + C = 0 (*).
The coordinates of the vectors and: = (x 0 - x, y 0 - y), = (A, B).
| | = 
= 
= 
(C = -Ax - By , see (*))
Theorem. Let the line l be given in the Cartesian coordinate system by the general equation Ax + By + C = 0. Then the distance from the point M(x 0 , y 0) to this line is calculated by the formula: r (M; l) = .
Exercises.
1) Derive a formula for calculating the distance from a point to a line if: (1) the line is given parametrically; (2) the line is given by the canonical equations; (3) the straight line is given by the slope equation.
2) Write the equation of a circle tangent to the line 3x - y = 0 centered at Q(-2,4).
3) Write the equations of the lines dividing the angles formed by the intersection of the lines 2x + y - 1 = 0 and x + y + 1 = 0 in half.
§ 27. Analytical definition of a plane in space
Definition. The normal vector to the plane we will call a non-zero vector, any representative of which is perpendicular to the given plane.
Comment. It is clear that if at least one representative of the vector is perpendicular to the plane, then all other representatives of the vector are perpendicular to this plane.
Let a Cartesian coordinate system be given in space.
Let the plane a be given, = (A, B, C) – the normal vector to this plane, the point M (x 0 , y 0 , z 0) belongs to the plane a.
For any point N(x, y, z) of the plane a, the vectors and are orthogonal, that is, their scalar product equals zero: = 0. Let's write the last equality in coordinates: A(x - x 0) + B(y - y 0) + C(z - z 0) = 0.
Let -Ax 0 - By 0 - Cz 0 = D, then Ax + By + Cz + D = 0.
Take a point K (x, y) such that Ax + By + Cz + D \u003d 0. Since D \u003d -Ax 0 - By 0 - Cz 0, then A(x - x 0) + B(y - y 0) + C(z - z 0) = 0. Since the coordinates of the directed segment = (x - x 0 , y - y 0 , z - z 0), the last equality means that ^ , and, therefore, K н a.
So, we have proved the following theorem:
Theorem. Any plane in space in the Cartesian coordinate system can be defined by an equation of the form Ax + By + Cz + D = 0 (A 2 + B 2 + C 2 ≠ 0), where (A, B, C) are the coordinates of the normal vector to this plane.
The reverse is also true.
Theorem. Any equation of the form Ax + By + Cz + D \u003d 0 (A 2 + B 2 + C 2 ≠ 0) in the Cartesian coordinate system defines a certain plane, while (A, B, C) are the coordinates of the normal vector to this plane.
Proof.
Take a point M (x 0 , y 0 , z 0) such that Ax 0 + By 0 + Cz 0 + D = 0 and vector = (A, B, C) ( ≠ q).
A plane (and only one) passes through the point M perpendicular to the vector. According to the previous theorem, this plane is given by the equation Ax + By + Cz + D = 0.
Definition. An equation of the form Ax + By + Cz + D = 0 (A 2 + B 2 + C 2 ≠ 0) is called the general equation of the plane.
Example.
Let's write the equation of the plane passing through the points M (0.2.4), N (1,-1.0) and K (-1.0.5).
1. Find the coordinates of the normal vector to the plane (MNK). Since the vector product ´ is orthogonal to non-collinear vectors and , the vector is collinear to ´ .
= (1, -3, -4), = (-1, -2, 1);
´ = 
,
´ = (-11, 3, -5).
So, as a normal vector, take the vector = (-11, 3, -5).
2. Let us now use the results of the first theorem:
the equation of this plane A(x - x 0) + B(y - y 0) + C(z - z 0) = 0, where (A, B, C) are the coordinates of the normal vector, (x 0 , y 0 , z 0) – coordinates of a point lying in the plane (for example, point M).
11(x - 0) + 3(y - 2) - 5(z - 4) = 0
11x + 3y - 5z + 14 = 0
Answer: -11x + 3y - 5z + 14 = 0.
Exercises.
1) Write the equation of the plane if
(1) the plane passes through the point M (-2,3,0) parallel to the plane 3x + y + z = 0;
(2) the plane contains the (Ox) axis and is perpendicular to the x + 2y – 5z + 7 = 0 plane.
2) Write the equation for a plane passing through three given points.
§ 28. Analytical specification of a half-space*
Comment*. Let some plane be fixed. Under half-space we will understand the set of points lying on one side of a given plane, that is, two points lie in the same half-space if the segment connecting them does not intersect the given plane. This plane is called boundary of this half-space. The union of a given plane and a half-space will be called closed half-space.
Let a Cartesian coordinate system be fixed in space.
Theorem. Let the plane a be given by the general equation Ax + By + Cz + D = 0. Then one of the two half-spaces into which the plane a divides the space is given by the inequality Ax + By + Cz + D > 0, and the second half-space is given by the inequality Ax + By + Cz + D< 0.
Proof.
Let us plot the normal vector = (A, B, С) to the plane a from the point M (x 0 , y 0 , z 0) lying on this plane: = , M н a, MN ^ a. The plane divides the space into two half-spaces: b 1 and b 2 . It is clear that the point N belongs to one of these half-spaces. Without loss of generality, we assume that N н b 1 .
Let us prove that the half-space b 1 is defined by the inequality Ax + By + Cz + D > 0.
1) Take a point K(x,y,z) in the half-space b 1 . The angle Ð NMK is the angle between the vectors and is acute, therefore the scalar product of these vectors is positive: > 0. Let's write this inequality in coordinates: A(x - x 0) + B(y - y 0) + C(z - z 0) > 0, i.e. Ax + By + Cy - Ax 0 - By 0 - C z 0 > 0.
Since M н b 1 , then Ax 0 + By 0 + C z 0 + D = 0, therefore -Ax 0 - By 0 - C z 0 = D. Therefore, the last inequality can be written as follows: Ax + By + Cz + D > 0.
2) Take a point L(x,y) such that Ax + By + Cz + D > 0.
Let us rewrite the inequality, replacing D with (-Ax 0 - By 0 - C z 0) (since M н b 1, then Ax 0 + By 0 + C z 0 + D = 0): A(x - x 0) + B(y - y 0) + C(z - z 0) > 0.
The vector with coordinates (x - x 0 ,y - y 0 , z - z 0) is a vector , so the expression A(x - x 0) + B(y - y 0) + C(z - z 0) can be understood , as the scalar product of the vectors and . Since the scalar product of the vectors and is positive, the angle between them is acute and the point L н b 1 .
Similarly, one can prove that the half-space b 2 is given by the inequality Ax + By + Cz + D< 0.
Remarks.
1) It is clear that the above proof does not depend on the choice of the point M in the plane a.
2) It is clear that the same half-space can be defined by different inequalities.
The reverse is also true.
Theorem. Any linear inequality of the form Ax + By + Cz + D > 0 (or Ax + By + Cz + D< 0) (A 2 + B 2 + C 2 ≠ 0) задает в пространстве в декартовой системе координат полупространство с границей Ax + By + Cz + D = 0.
Proof.
The equation Ax + By + Cz + D = 0 (A 2 + B 2 + C 2 ≠ 0) in space defines some plane a (see § ...). As was proved in the previous theorem, one of the two half-spaces into which the plane divides the space is given by the inequality Ax Ax + By + Cz + D > 0.
Remarks.
1) It is clear that a closed half-space can be defined by a non-strict linear inequality, and any non-strict linear inequality in the Cartesian coordinate system defines a closed half-space.
2) Any convex polyhedron can be defined as the intersection of closed half-spaces (the boundaries of which are planes containing the faces of the polyhedron), that is, analytically, by a system of linear non-strict inequalities.
Exercises.
1) Prove the two theorems presented for an arbitrary affine coordinate system.
2) Is the converse true that any system of nonstrict linear inequalities defines a convex polygon?
An exercise.1) Explore the relative position of two planes given by general equations in the Cartesian coordinate system and fill in the table.
Oh-oh-oh-oh-oh ... well, it's tinny, as if you read the sentence to yourself =) However, then relaxation will help, especially since I bought suitable accessories today. Therefore, let's proceed to the first section, I hope, by the end of the article I will keep a cheerful mood.
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.
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 all the coefficients of the equation 
reduce 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:
For ignorance of this the simplest task severely punishes the Nightingale the Robber.
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 "de".
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:
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 the 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 look for the intersection point 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:
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:
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 an amusing trip continues:
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 common fractions. Have advised many times and will recommend again.
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.
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:
By using inverse function easy to find the corner itself. In this case, we use the oddness of the arc tangent (see Fig. Graphs and properties of elementary functions):
Answer: 
In the answer, indicate exact value, as well as an approximate value (preferably in both degrees and 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 
.
This material is devoted to such a concept as the angle between two intersecting straight lines. In the first paragraph, we will explain what it is and show it in illustrations. Then we will analyze how you can find the sine, cosine of this angle and the angle itself (we will separately consider cases with a plane and three-dimensional space), we will give the necessary formulas and show with examples how exactly they are applied in practice.
Yandex.RTB R-A-339285-1
In order to understand what an angle formed at the intersection of two lines is, we need to recall the very definition of an angle, perpendicularity and an intersection point.
Definition 1
We call two lines intersecting if they have one common point. This point is called the point of intersection of the two lines.
Each line is divided by the point of intersection into rays. In this case, both lines form 4 angles, of which two are vertical and two are adjacent. If we know the measure of one of them, then we can determine the other remaining ones.
Let's say we know that one of the angles is equal to α. In such a case, the angle that is vertical to it will also be equal to α. To find the remaining angles, we need to calculate the difference 180 ° - α . If α is equal to 90 degrees, then all angles will be right. Lines intersecting at right angles are called perpendicular (a separate article is devoted to the concept of perpendicularity).
Take a look at the picture:
Let us proceed to the formulation of the main definition.
Definition 2
The angle formed by two intersecting lines is the measure of the smaller of the 4 angles that form these two lines.
An important conclusion must be drawn from the definition: the size of the angle in this case will be expressed by any real number in the interval (0 , 90 ] . If the lines are perpendicular, then the angle between them will in any case be equal to 90 degrees.
The ability to find the measure of the angle between two intersecting lines is useful for solving many practical problems. The solution method can be selected from several options.
For starters, we can take geometric methods. If we know something about additional angles, then we can connect them to the angle we need using the properties of equal or similar shapes. For example, if we know the sides of a triangle and need to calculate the angle between the lines on which these sides are located, then the cosine theorem is suitable for solving. If we have a right triangle in the condition, then for calculations we will also need to know the sine, cosine and tangent of the angle.
The coordinate method is also very convenient for solving problems of this type. Let's explain how to use it correctly.
We have a rectangular (cartesian) coordinate system O x y with two straight lines. Let's denote them by letters a and b. In this case, straight lines can be described using any equations. The original lines have an intersection point M . How to determine the desired angle (let's denote it α) between these lines?
Let's start with the formulation of the basic principle of finding an angle under given conditions.
We know that such concepts as directing and normal vector are closely related to the concept of a straight line. If we have the equation of some straight line, we can take the coordinates of these vectors from it. We can do this for two intersecting lines at once.
The angle formed by two intersecting lines can be found using:
Now let's look at each method separately.
1. Suppose we have a line a with direction vector a → = (a x , a y) and a line b with direction vector b → (b x , b y) . Now let's set aside two vectors a → and b → from the intersection point. After that, we will see that they will each be located on their own line. Then we have four options for their relative position. See illustration:
If the angle between two vectors is not obtuse, then it will be the angle we need between the intersecting lines a and b. If it is obtuse, then the desired angle will be equal to the angle adjacent to the angle a → , b → ^ . Thus, α = a → , b → ^ if a → , b → ^ ≤ 90 ° , and α = 180 ° - a → , b → ^ if a → , b → ^ > 90 ° .
Based on the fact that the cosines of equal angles are equal, we can rewrite the resulting equalities as follows: cos α = cos a → , b → ^ if a → , b → ^ ≤ 90 ° ; cos α = cos 180 ° - a → , b → ^ = - cos a → , b → ^ if a → , b → ^ > 90 ° .
In the second case, reduction formulas were used. In this way,
cos α cos a → , b → ^ , cos a → , b → ^ ≥ 0 - cos a → , b → ^ , cos a → , b → ^< 0 ⇔ cos α = cos a → , b → ^
Let's write the last formula in words:
Definition 3
The cosine of the angle formed by two intersecting lines will be equal to the modulus of the cosine of the angle between its direction vectors.
The general form of the formula for the cosine of the angle between two vectors a → = (a x, a y) and b → = (b x, b y) looks like this:
cos a → , b → ^ = a → , b → ^ a → b → = a x b x + a y + b y a x 2 + a y 2 b x 2 + b y 2
From it we can derive the formula for the cosine of the angle between two given lines:
cos α = a x b x + a y + b y a x 2 + a y 2 b x 2 + b y 2 = a x b x + a y + b y a x 2 + a y 2 b x 2 + b y 2
Then the angle itself can be found using the following formula:
α = a r c cos a x b x + a y + b y a x 2 + a y 2 b x 2 + b y 2
Here a → = (a x , a y) and b → = (b x , b y) are the direction vectors of the given lines.
Let us give an example of solving the problem.
Example 1
In a rectangular coordinate system, two intersecting lines a and b are given on the plane. They can be described by parametric equations x = 1 + 4 · λ y = 2 + λ λ ∈ R and x 5 = y - 6 - 3 . Calculate the angle between these lines.
Solution
We have a parametric equation in the condition, which means that for this straight line we can immediately write down the coordinates of its direction vector. To do this, we need to take the values of the coefficients at the parameter, i.e. the line x = 1 + 4 λ y = 2 + λ λ ∈ R will have a direction vector a → = (4 , 1) .
The second straight line is described using the canonical equation x 5 = y - 6 - 3 . Here we can take the coordinates from the denominators. Thus, this line has a direction vector b → = (5 , - 3) .
Next, we proceed directly to finding the angle. To do this, simply substitute the available coordinates of the two vectors into the above formula α = a r c cos a x b x + a y + b y a x 2 + a y 2 b x 2 + b y 2 . We get the following:
α = a r c cos 4 5 + 1 (- 3) 4 2 + 1 2 5 2 + (- 3) 2 = a r c cos 17 17 34 = a r c cos 1 2 = 45°
Answer: These lines form an angle of 45 degrees.
We can solve a similar problem by finding the angle between normal vectors. If we have a line a with a normal vector n a → = (n a x , n a y) and a line b with a normal vector n b → = (n b x , n b y) , then the angle between them will be equal to the angle between n a → and n b → or the angle that will be adjacent to n a → , n b → ^ . This method is shown in the picture:
The formulas for calculating the cosine of the angle between intersecting lines and this angle itself using the coordinates of normal vectors look like this:
cos α = cos n a → , n b → ^ = n a x n b x + n a y + n b y n a x 2 + n a y 2 n b x 2 + n b y 2
Here n a → and n b → denote the normal vectors of two given lines.
Example 2
Two straight lines are given in a rectangular coordinate system using the equations 3 x + 5 y - 30 = 0 and x + 4 y - 17 = 0 . Find the sine, cosine of the angle between them, and the magnitude of that angle itself.
Solution
The original straight lines are given using normal straight line equations of the form A x + B y + C = 0 . Denote the normal vector n → = (A , B) . Let's find the coordinates of the first normal vector for one straight line and write them down: n a → = (3 , 5) . For the second line x + 4 y - 17 = 0 the normal vector will have coordinates n b → = (1 , 4) . Now add the obtained values to the formula and calculate the total:
cos α = cos n a → , n b → ^ = 3 1 + 5 4 3 2 + 5 2 1 2 + 4 2 = 23 34 17 = 23 2 34
If we know the cosine of an angle, then we can calculate its sine using the basic trigonometric identity. Since the angle α formed by straight lines is not obtuse, then sin α \u003d 1 - cos 2 α \u003d 1 - 23 2 34 2 \u003d 7 2 34.
In this case, α = a r c cos 23 2 34 = a r c sin 7 2 34 .
Answer: cos α = 23 2 34 , sin α = 7 2 34 , α = a r c cos 23 2 34 = a r c sin 7 2 34
Let's analyze the last case - finding the angle between the lines, if we know the coordinates of the directing vector of one line and the normal vector of the other.
Assume that line a has a direction vector a → = (a x , a y) , and line b has a normal vector n b → = (n b x , n b y) . We need to postpone these vectors from the intersection point and consider all options for their relative position. See picture:
If the angle between the given vectors is no more than 90 degrees, it turns out that it will complement the angle between a and b to a right angle.
a → , n b → ^ = 90 ° - α if a → , n b → ^ ≤ 90 ° .
If it is less than 90 degrees, then we get the following:
a → , n b → ^ > 90 ° , then a → , n b → ^ = 90 ° + α
Using the rule of equality of cosines of equal angles, we write:
cos a → , n b → ^ = cos (90 ° - α) = sin α for a → , n b → ^ ≤ 90 ° .
cos a → , n b → ^ = cos 90 ° + α = - sin α at a → , n b → ^ > 90 ° .
In this way,
sin α = cos a → , n b → ^ , a → , n b → ^ ≤ 90 ° - cos a → , n b → ^ , a → , n b → ^ > 90 ° ⇔ sin α = cos a → , n b → ^ , a → , n b → ^ > 0 - cos a → , n b → ^ , a → , n b → ^< 0 ⇔ ⇔ sin α = cos a → , n b → ^
Let's formulate a conclusion.
Definition 4
To find the sine of the angle between two lines intersecting in a plane, you need to calculate the modulus of the cosine of the angle between the direction vector of the first line and the normal vector of the second.
Let's write down the necessary formulas. Finding the sine of an angle:
sin α = cos a → , n b → ^ = a x n b x + a y n b y a x 2 + a y 2 n b x 2 + n b y 2
Finding the corner itself:
α = a r c sin = a x n b x + a y n b y a x 2 + a y 2 n b x 2 + n b y 2
Here a → is the direction vector of the first line, and n b → is the normal vector of the second.
Example 3
Two intersecting lines are given by the equations x - 5 = y - 6 3 and x + 4 y - 17 = 0 . Find the angle of intersection.
Solution
We take the coordinates of the directing and normal vector from the given equations. It turns out a → = (- 5 , 3) and n → b = (1 , 4) . We take the formula α \u003d a r c sin \u003d a x n b x + a y n b y a x 2 + a y 2 n b x 2 + n b y 2 and consider:
α = a r c sin = - 5 1 + 3 4 (- 5) 2 + 3 2 1 2 + 4 2 = a r c sin 7 2 34
Note that we took the equations from the previous problem and got exactly the same result, but in a different way.
Answer:α = a r c sin 7 2 34
Here is another way to find the desired angle using the slope coefficients of given lines.
We have a line a , which is defined in a rectangular coordinate system using the equation y = k 1 · x + b 1 , and a line b , defined as y = k 2 · x + b 2 . These are equations of lines with a slope. To find the angle of intersection, use the formula:
α = a r c cos k 1 k 2 + 1 k 1 2 + 1 k 2 2 + 1 , where k 1 and k 2 are slope factors given lines. To obtain this record, formulas for determining the angle through the coordinates of normal vectors were used.
Example 4
There are two straight lines intersecting in the plane, given by the equations y = - 3 5 x + 6 and y = - 1 4 x + 17 4 . Calculate the angle of intersection.
Solution
The slopes of our lines are equal to k 1 = - 3 5 and k 2 = - 1 4 . Let's add them to the formula α = a r c cos k 1 k 2 + 1 k 1 2 + 1 k 2 2 + 1 and calculate:
α = a r c cos - 3 5 - 1 4 + 1 - 3 5 2 + 1 - 1 4 2 + 1 = a r c cos 23 20 34 24 17 16 = a r c cos 23 2 34
Answer:α = a r c cos 23 2 34
In the conclusions of this paragraph, it should be noted that the formulas for finding the angle given here do not have to be learned by heart. To do this, it is sufficient to know the coordinates of the guides and/or normal vectors of the given lines and be able to determine them from different types equations. But the formulas for calculating the cosine of an angle are better to remember or write down.
The calculation of such an angle can be reduced to the calculation of the coordinates of the direction vectors and the determination of the magnitude of the angle formed by these vectors. For such examples, we use the same reasoning that we have given before.
Let's say we have rectangular system coordinates located in three-dimensional space. It contains two lines a and b with the intersection point M . To calculate the coordinates of the direction vectors, we need to know the equations of these lines. Denote the direction vectors a → = (a x , a y , a z) and b → = (b x , b y , b z) . To calculate the cosine of the angle between them, we use the formula:
cos α = cos a → , b → ^ = a → , b → a → b → = a x b x + a y b y + a z b z a x 2 + a y 2 + a z 2 b x 2 + b y 2 + b z 2
To find the angle itself, we need this formula:
α = a r c cos a x b x + a y b y + a z b z a x 2 + a y 2 + a z 2 b x 2 + b y 2 + b z 2
Example 5
We have a straight line defined in 3D space using the equation x 1 = y - 3 = z + 3 - 2 . It is known that it intersects with the O z axis. Calculate the angle of intersection and the cosine of that angle.
Solution
Let's denote the angle to be calculated by the letter α. Let's write down the coordinates of the direction vector for the first straight line - a → = (1 , - 3 , - 2) . For the applicate axis, we can take the coordinate vector k → = (0 , 0 , 1) as a guide. We have received the necessary data and can add it to the desired formula:
cos α = cos a → , k → ^ = a → , k → a → k → = 1 0 - 3 0 - 2 1 1 2 + (- 3) 2 + (- 2) 2 0 2 + 0 2 + 1 2 = 2 8 = 1 2
As a result, we got that the angle we need will be equal to a r c cos 1 2 = 45 °.
Answer: cos α = 1 2 , α = 45 ° .
If you notice a mistake in the text, please highlight it and press Ctrl+Enter
