corner between straight lines in space we will call any of the adjacent angles formed by two straight lines drawn through an arbitrary point parallel to the data.
Let two straight lines be given in space:
Obviously, the angle φ between the lines can be taken as the angle between their direction vectors and . Since , then according to the formula for the cosine of the angle between the vectors we get
The conditions of parallelism and perpendicularity of two lines are equivalent to the conditions of parallelism and perpendicularity of their direction vectors and:
Two straight parallel if and only if their respective coefficients are proportional, i.e. l 1 parallel l 2 if and only if parallel 
.
Two straight perpendicular if and only if the sum of the products of the corresponding coefficients is equal to zero: .
At goal between line and plane
Let the line d- not perpendicular to the plane θ;
d′− projection of a straight line d to the plane θ;
The smallest of the angles between straight lines d and d′ we will call angle between line and plane.
Let's denote it as φ=( d,θ)
If a d⊥θ , then ( d,θ)=π/2
Oi→j→k→− rectangular coordinate system.
Plane equation:
θ: Ax+By+cz+D=0
We consider that the line is given by a point and a direction vector: d[M 0,p→]
Vector n→(A,B,C)⊥θ
Then it remains to find out the angle between the vectors n→ and p→, denote it as γ=( n→,p→).
If the angle γ<π/2 , то искомый угол φ=π/2−γ .
If the angle γ>π/2 , then the required angle φ=γ−π/2
sinφ=sin(2π−γ)=cosγ
sinφ=sin(γ−2π)=−cosγ
Then, angle between line and plane can be calculated using the formula:
sinφ=∣cosγ∣=∣ ∣ Ap 1+bp 2+cp 3∣ ∣ √A 2+B 2+C 2√p 21+p 22+p 23
Question 29. The concept of a quadratic form. The sign-definiteness of quadratic forms.
Quadratic form j (x 1, x 2, ..., x n) n real variables x 1, x 2, ..., x n is called a sum of the form 
, (1)
where aij are some numbers called coefficients. Without loss of generality, we can assume that aij = a ji.
The quadratic form is called valid, if aij
О GR. Matrix of quadratic form is called the matrix composed of its coefficients. Quadratic form (1) corresponds to a unique symmetric matrix 
i.e. A T = A. Therefore, quadratic form (1) can be written in matrix form j ( X) = x T Ah, where x T = (X 1 X 2 … x n). (2)
And vice versa, any symmetric matrix (2) corresponds to a unique quadratic form up to the notation of variables.
The rank of the quadratic form is called the rank of its matrix. The quadratic form is called non-degenerate, if its matrix is nonsingular BUT. (recall that the matrix BUT is called non-degenerate if its determinant is non-zero). Otherwise, the quadratic form is degenerate.
positive definite(or strictly positive) if
j ( X) > 0 , for anyone X = (X 1 , X 2 , …, x n), Besides X = (0, 0, …, 0).
Matrix BUT positive definite quadratic form j ( X) is also called positive definite. Therefore, a positive definite quadratic form corresponds to a unique positive definite matrix and vice versa.
The quadratic form (1) is called negative definite(or strictly negative) if
j ( X) < 0, для любого X = (X 1 , X 2 , …, x n), Besides X = (0, 0, …, 0).
Similarly as above, a negative-definite quadratic matrix is also called negative-definite.
Therefore, a positively (negatively) definite quadratic form j ( X) reaches the minimum (maximum) value j ( X*) = 0 for X* = (0, 0, …, 0).
Note that most of the quadratic forms are not sign-definite, that is, they are neither positive nor negative. Such quadratic forms vanish not only at the origin of the coordinate system, but also at other points.
When n> 2, special criteria are required to check the sign-definiteness of a quadratic form. Let's consider them.
Major Minors quadratic form are called minors:
that is, these are minors of order 1, 2, …, n matrices BUT, located in the upper left corner, the last of them coincides with the determinant of the matrix BUT.
Criterion for positive definiteness (Sylvester criterion)
X) = x T Ah is positive definite, it is necessary and sufficient that all principal minors of the matrix BUT were positive, that is: M 1 > 0, M 2 > 0, …, M n > 0. Criterion of negative certainty In order for the quadratic form j ( X) = x T Ah is negative definite, it is necessary and sufficient that its principal minors of even order are positive, and those of odd order are negative, i.e.: M 1 < 0, M 2 > 0, M 3 < 0, …, (–1)n
ANGLE BETWEEN PLANES
Let's consider two planes α 1 and α 2 given respectively by the equations:
Under corner between two planes we mean one of the dihedral angles formed by these planes. It is obvious that the angle between the normal vectors and the planes α 1 and α 2 is equal to one of the indicated adjacent dihedral angles or 
. That's why 
. Because 
and 
, then
.
Example. Determine the angle between planes x+2y-3z+4=0 and 2 x+3y+z+8=0.
Condition of parallelism of two planes.
Two planes α 1 and α 2 are parallel if and only if their normal vectors and are parallel, and hence 
.
So, two planes are parallel to each other if and only if the coefficients at the corresponding coordinates are proportional:
or
Condition of perpendicularity of planes.
It is clear that two planes are perpendicular if and only if their normal vectors are perpendicular, and therefore, or .
In this way, .
Examples.
DIRECT IN SPACE.
VECTOR EQUATION DIRECT.
PARAMETRIC EQUATIONS DIRECT
The position of a straight line in space is completely determined by specifying any of its fixed points M 1 and a vector parallel to this line.
A vector parallel to a straight line is called guiding the vector of this line.
So let the straight l passes through a point M 1 (x 1 , y 1 , z 1) lying on a straight line parallel to the vector .
Consider an arbitrary point M(x,y,z) on a straight line. It can be seen from the figure that 
.
The vectors and are collinear, so there is such a number t, what , where is the multiplier t can take any numeric value depending on the position of the point M on a straight line. Factor t is called a parameter. Denoting the radius vectors of points M 1 and M respectively, through and , we obtain . This equation is called vector straight line equation. It shows that each parameter value t corresponds to the radius vector of some point M lying on a straight line.
We write this equation in coordinate form. Notice, that , 
and from here
The resulting equations are called parametric straight line equations.
When changing the parameter t coordinates change x, y and z and dot M moves in a straight line.
CANONICAL EQUATIONS DIRECT
Let M 1 (x 1 , y 1 , z 1) - a point lying on a straight line l, and 
is its direction vector. Again, take an arbitrary point on a straight line M(x,y,z) and consider the vector .
It is clear that the vectors and are collinear, so their respective coordinates must be proportional, hence
– canonical straight line equations.
Remark 1. Note that the canonical equations of the line could be obtained from the parametric equations by eliminating the parameter t. Indeed, from the parametric equations we obtain 
or .
Example. Write the equation of a straight line 
in a parametric way.
Denote 
, hence x = 2 + 3t, y = –1 + 2t, z = 1 –t.
Remark 2. Let the line be perpendicular to one of the coordinate axes, for example, the axis Ox. Then the direction vector of the line is perpendicular Ox, Consequently, m=0. Consequently, the parametric equations of the straight line take the form
Eliminating the parameter from the equations t, we obtain the equations of the straight line in the form
However, in this case too, we agree to formally write the canonical equations of the straight line in the form 
. Thus, if the denominator of one of the fractions is zero, then this means that the line is perpendicular to the corresponding coordinate axis.
Similarly, the canonical equations 
corresponds to a straight line perpendicular to the axes Ox and Oy or parallel axis Oz.
Examples.
GENERAL EQUATIONS A DIRECT LINE AS A LINE OF INTERCEPTION OF TWO PLANES
Through each straight line in space passes an infinite number of planes. Any two of them, intersecting, define it in space. Therefore, the equations of any two such planes, considered together, are the equations of this line.
In general, any two non-parallel planes given by the general equations
determine their line of intersection. These equations are called general equations straight.
Examples.
Construct a straight line given by equations 
To construct a line, it is enough to find any two of its points. The easiest way is to choose the points of intersection of the line with the coordinate planes. For example, the point of intersection with the plane xOy we obtain from the equations of a straight line, assuming z= 0:
Solving this system, we find the point M 1 (1;2;0).
Similarly, assuming y= 0, we get the point of intersection of the line with the plane xOz:
From the general equations of a straight line, one can proceed to its canonical or parametric equations. To do this, you need to find some point M 1 on the line and the direction vector of the line.
Point coordinates M 1 we obtain from this system of equations, giving one of the coordinates an arbitrary value. To find the direction vector, note that this vector must be perpendicular to both normal vectors 
and 
. Therefore, for the direction vector of the straight line l you can take the cross product of normal vectors:
.
Example. Give the general equations of the straight line 
to the canonical form.
Find a point on a straight line. To do this, we choose arbitrarily one of the coordinates, for example, y= 0 and solve the system of equations:
The normal vectors of the planes defining the line have coordinates 
Therefore, the direction vector will be straight
. Consequently, l: 
.
ANGLE BETWEEN RIGHTS
corner between straight lines in space we will call any of the adjacent angles formed by two straight lines drawn through an arbitrary point parallel to the data.
Let two straight lines be given in space:
Obviously, the angle φ between the lines can be taken as the angle between their direction vectors and . Since , then according to the formula for the cosine of the angle between the vectors we get
This material is devoted to such a concept as the angle between two intersecting 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 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 we 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:
- angle between direction vectors;
- angle between normal vectors;
- the angle between the normal vector of one line and the direction vector of the other.
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 straight 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 the slopes of the 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 enough to know the coordinates of the guides and/or normal vectors of the given lines and be able to determine them using different types of equations. But the formulas for calculating the cosine of an angle are better to remember or write down.
How to calculate the angle between intersecting lines in space
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 a rectangular coordinate system located in 3D 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 ° .
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Oh-oh-oh-oh-oh ... well, it's tough, 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.
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 math sign intersections, 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 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.
It is obvious that 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 "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 verbally. 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 may 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. It is convenient to divide the problem 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 straight 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 
.