The angle between two straight lines. Angle between straight lines on a plane
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 the line passing through given point M 0 is 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 = -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 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 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.
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 see no point in offering anything for independent decision, we’d better lay down another one important brick into a geometric foundation:
How to construct a line parallel to a given one?
For ignorance of this simplest task 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 geometry of the example 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 familiar to you from 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 systems 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 look for the intersection point 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.
Complete solution and the 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
We have a straight strip of river in front of us 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 common 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 scalar product directing vectors of straight lines:
, which means the lines are not perpendicular.
2) Find the angle between straight lines using the formula:
By using inverse function It's easy to find the corner itself. In this case, we use the oddness of the arctangent (see. Graphs and properties of elementary functions):
Answer: 
In the answer we indicate 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 
.
Angle between lines in space we will call any of adjacent corners, 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 system coordinates
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. Hence, 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 of 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 left top 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
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 look at 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 present necessary formulas and show with examples exactly how they are used in practice.
Yandex.RTB R-A-339285-1
In order to understand what the angle formed when two lines intersect is, we need to remember the very definition of angle, perpendicularity and point of intersection.
Definition 1
We call two lines intersecting if they have one common point. This point is called the point of intersection of two lines.
Each straight line is divided by an intersection point into rays. Both straight lines form 4 angles, two of which are vertical, and two are adjacent. If we know the measure of one of them, then we can determine the remaining ones.
Let's say we know that one of the angles is equal to α. In this case, the angle that is vertical with respect 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 angles. 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's move on to formulating 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 chosen from several options.
To begin with, we can take geometric methods. If we know something about supplementary angles, then we can relate them to the angle we need using the properties of equal or similar figures. 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 it. If we have the condition right triangle, then for calculations we will also need knowledge of sine, cosine and tangent of an angle.
The coordinate method is also very convenient for solving problems of this type. Let us explain how to use it correctly.
We have a rectangular (Cartesian) coordinate system O x y, in which two straight lines are given. Let's denote them by letters a and b. The straight lines can be described using some equations. The original lines have an intersection point M. How to determine the required angle (let's denote it α) between these straight lines?
Let's start by formulating the basic principle of finding an angle under given conditions.
We know that the concept of a straight line is closely related to such concepts as a direction vector and a normal vector. If we have an equation of a certain line, we can take the coordinates of these vectors from it. We can do this for two intersecting lines at once.
The angle subtended 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. Let us assume that we have a line a with a direction vector a → = (a x, a y) and a line b with a direction vector b → (b x, b y). Now let’s plot two vectors a → and b → from the intersection point. After this we will see that they will each be located on their own straight line. Then we have four options for their relative arrangement. 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 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. Thus,
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 straight 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 straight 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's give an example of solving the problem.
Example 1
In a rectangular coordinate system on a plane, two intersecting lines a and b are given. They can be described by the 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 our condition, which means that for this line we can immediately write down the coordinates of its direction vector. To do this, we need to take the values of the coefficients for the parameter, i.e. the straight line x = 1 + 4 · λ y = 2 + λ λ ∈ R will have a direction vector a → = (4, 1).
The second 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 move directly to finding the angle. To do this, simply substitute the existing 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 straight 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:
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 α = a r c cos 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
In a rectangular coordinate system, two straight lines are given using the equations 3 x + 5 y - 30 = 0 and x + 4 y - 17 = 0. Find the sine and cosine of the angle between them and the magnitude of this angle itself.
Solution
The original lines are specified using normal line equations of the form A x + B y + C = 0. We denote the normal vector as n → = (A, B). Let's find the coordinates of the first normal vector for one line and write them: n a → = (3, 5) . For the second line x + 4 y - 17 = 0, the normal vector will have coordinates n b → = (1, 4). Now let’s 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 = 1 - cos 2 α = 1 - 23 2 34 2 = 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 us analyze the last case - finding the angle between straight lines if we know the coordinates of the direction vector of one straight line and the normal vector of the other.
Let us assume that straight line a has a direction vector a → = (a x , a y) , and straight line b has a normal vector n b → = (n b x , n b y) . We need to set these vectors aside from the intersection point and consider all options for their relative positions. See in the 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 for a → , n b → ^ > 90 ° .
Thus,
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 us formulate a conclusion.
Definition 4
To find the sine of the angle between two lines intersecting on 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 angle 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 guide and normal vector from the given equations. It turns out a → = (- 5, 3) and n → b = (1, 4). We take the formula 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 and calculate:
α = a r c sin = - 5 1 + 3 4 (- 5) 2 + 3 2 1 2 + 4 2 = a r c sin 7 2 34
Please note that we took the equations from the previous problem and obtained exactly the same result, but in a different way.
Answer:α = a r c sin 7 2 34
Let us present another way to find the desired angle using the angular coefficients of given straight 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 slopes. To find the angle of intersection, we 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 angle coefficients given straight lines. To obtain this record, formulas for determining the angle through the coordinates of normal vectors were used.
Example 4
There are two lines intersecting in a plane, given by the equations y = - 3 5 x + 6 and y = - 1 4 x + 17 4. Calculate the value of the intersection angle.
Solution
The angular coefficients 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 given lines and be able to determine them by different types equations. But it’s better to remember or write down the formulas for calculating the cosine of an angle.
How to calculate the angle between intersecting lines in space
The calculation of such an angle can be reduced to calculating the coordinates of the direction vectors and determining the magnitude of the angle formed by these vectors. For such examples, the same reasoning that we gave before is used.
Let's assume that we have a rectangular coordinate system located in three-dimensional space. It contains two straight lines a and b with an intersection point M. To calculate the coordinates of the direction vectors, we need to know the equations of these lines. Let us 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 line defined in three-dimensional space using the equation x 1 = y - 3 = z + 3 - 2. It is known that it intersects with the O z axis. Calculate the intercept angle and the cosine of that angle.
Solution
Let us denote the angle that needs 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 found 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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I'll be brief. The angle between two straight lines is equal to the angle between their direction vectors. Thus, if you manage to find the coordinates of the direction vectors a = (x 1 ; y 1 ; z 1) and b = (x 2 ; y 2 ; z 2), then you can find the angle. More precisely, the cosine of the angle according to the formula:
Let's see how this formula works using specific examples:
Task. In the cube ABCDA 1 B 1 C 1 D 1, points E and F are marked - the midpoints of the edges A 1 B 1 and B 1 C 1, respectively. Find the angle between lines AE and BF.
Since the edge of the cube is not specified, we set AB = 1. We introduce standard system coordinates: the origin is at point A, the x, y, z axes are directed along AB, AD and AA 1, respectively. The unit segment is equal to AB = 1. Now let's find the coordinates of the direction vectors for our lines.
Let's find the coordinates of vector AE. For this we need points A = (0; 0; 0) and E = (0.5; 0; 1). Since point E is the middle of the segment A 1 B 1, its coordinates are equal to the arithmetic mean of the coordinates of the ends. Note that the origin of the vector AE coincides with the origin of coordinates, so AE = (0.5; 0; 1).
Now let's look at the BF vector. Similarly, we analyze the points B = (1; 0; 0) and F = (1; 0.5; 1), because F is the middle of the segment B 1 C 1. We have:
BF = (1 − 1; 0.5 − 0; 1 − 0) = (0; 0.5; 1).
So, the direction vectors are ready. The cosine of the angle between straight lines is the cosine of the angle between the direction vectors, so we have:
Task. In a regular triangular prism ABCA 1 B 1 C 1, all edges of which are equal to 1, points D and E are marked - the midpoints of the edges A 1 B 1 and B 1 C 1, respectively. Find the angle between lines AD and BE.
Let's introduce a standard coordinate system: the origin is at point A, the x axis is directed along AB, z - along AA 1. Let's direct the y-axis so that the OXY plane coincides with the ABC plane. The unit segment is equal to AB = 1. Let us find the coordinates of the direction vectors for the required lines.
First, let's find the coordinates of the vector AD. Consider the points: A = (0; 0; 0) and D = (0.5; 0; 1), because D - the middle of the segment A 1 B 1. Since the beginning of the vector AD coincides with the origin of coordinates, we obtain AD = (0.5; 0; 1).
Now let's find the coordinates of vector BE. Point B = (1; 0; 0) is easy to calculate. With point E - the middle of the segment C 1 B 1 - it is a little more complicated. We have:
It remains to find the cosine of the angle:
Task. In a regular hexagonal prism ABCDEFA 1 B 1 C 1 D 1 E 1 F 1 , all edges of which are equal to 1, points K and L are marked - the midpoints of the edges A 1 B 1 and B 1 C 1, respectively. Find the angle between lines AK and BL.
Let us introduce a standard coordinate system for a prism: we place the origin of coordinates at the center of the lower base, the x axis is directed along FC, the y axis is directed through the midpoints of segments AB and DE, and the z axis is directed vertically upward. The unit segment is again equal to AB = 1. Let’s write down the coordinates of the points of interest to us:
Points K and L are the midpoints of the segments A 1 B 1 and B 1 C 1, respectively, so their coordinates are found through the arithmetic mean. Knowing the points, we find the coordinates of the direction vectors AK and BL:
Now let's find the cosine of the angle:
Task. In a regular quadrangular pyramid SABCD, all edges of which are equal to 1, points E and F are marked - the midpoints of sides SB and SC, respectively. Find the angle between lines AE and BF.
Let's introduce a standard coordinate system: the origin is at point A, the x and y axes are directed along AB and AD, respectively, and the z axis is directed vertically upward. The unit segment is equal to AB = 1.
Points E and F are the midpoints of the segments SB and SC, respectively, so their coordinates are found as the arithmetic mean of the ends. Let's write down the coordinates of the points of interest to us:
A = (0; 0; 0); B = (1; 0; 0)
Knowing the points, we find the coordinates of the direction vectors AE and BF:
The coordinates of vector AE coincide with the coordinates of point E, since point A is the origin. It remains to find the cosine of the angle:
