Find the angle between straight lines formula. Angle between two straight lines
A. Let two straight lines be given. These straight lines, as indicated in Chapter 1, form various positive and negative angles, which can be either acute or obtuse. Knowing one of these angles, we can easily find any other.
By the way, for all these angles the numerical value of the tangent is the same, the difference can only be in the sign
Equations of lines. The numbers are the projections of the direction vectors of the first and second straight lines. The angle between these vectors is equal to one of the angles formed by straight lines. Therefore, the problem comes down to determining the angle between the vectors. We get
For simplicity, we can agree that the angle between two straight lines is an acute positive angle (as, for example, in Fig. 53).
Then the tangent of this angle will always be positive. Thus, if there is a minus sign on the right side of formula (1), then we must discard it, i.e., save only the absolute value.
Example. Determine the angle between straight lines
According to formula (1) we have
With. If it is indicated which of the sides of the angle is its beginning and which is its end, then, always counting the direction of the angle counterclockwise, we can extract something more from formula (1). As is easy to see from Fig. 53, the sign obtained on the right side of formula (1) will indicate what kind of angle - acute or obtuse - the second straight line forms with the first.
(Indeed, from Fig. 53 we see that the angle between the first and second direction vectors is either equal to the desired angle between the straight lines, or differs from it by ±180°.)
d. If the lines are parallel, then their direction vectors are parallel. Applying the condition of parallelism of two vectors, we get!
This is a necessary and sufficient condition for the parallelism of two lines.
Example. Direct
are parallel because
e. If the lines are perpendicular then their direction vectors are also perpendicular. Applying the condition of perpendicularity of two vectors, we obtain the condition of perpendicularity of two straight lines, namely
Example. Direct
are perpendicular due to the fact that
In connection with the conditions of parallelism and perpendicularity, we will solve the following two problems.
f. Draw a line through a point parallel to the given line
The solution is carried out like this. Since the desired line is parallel to this one, then for its direction vector we can take the same one as that of the given line, i.e., a vector with projections A and B. And then the equation of the desired line will be written in the form (§ 1)
Example. Equation of a line passing through the point (1; 3) parallel to the line
there will be next!
g. Draw a line through a point perpendicular to the given line
Here it is no longer suitable to take the vector with projections A and as the guiding vector, but it is necessary to take the vector perpendicular to it. The projections of this vector must therefore be chosen according to the condition of perpendicularity of both vectors, i.e. according to the condition
This condition can be fulfilled in countless ways, since here is one equation with two unknowns. But the easiest way is to take or Then the equation of the desired line will be written in the form
Example. Equation of a line passing through the point (-7; 2) in a perpendicular line
there will be the following (according to the second formula)!
h. In the case when the lines are given by equations of the form
Instructions
note
The period of the trigonometric function tangent is equal to 180 degrees, which means that the slope angles of straight lines cannot, in absolute value, exceed this value.
Helpful advice
If the angular coefficients are equal to each other, then the angle between such lines is 0, since such lines either coincide or are parallel.
To determine the value of the angle between intersecting lines, it is necessary to move both lines (or one of them) to a new position using the parallel translation method until they intersect. After this, you should find the angle between the resulting intersecting lines.
You will need
- Ruler, right triangle, pencil, protractor.
Instructions
So, let the vector V = (a, b, c) and the plane A x + B y + C z = 0 be given, where A, B and C are the coordinates of the normal N. Then the cosine of the angle α between the vectors V and N is equal to: cos α = (a A + b B + c C)/(√(a² + b² + c²) √(A² + B² + C²)).
To calculate the angle in degrees or radians, you need to calculate the inverse to cosine function from the resulting expression, i.e. arccosine:α = аrsсos ((a A + b B + c C)/(√(a² + b² + c²) √(A² + B² + C²))).
Example: find corner between vector(5, -3, 8) and plane, given by the general equation 2 x – 5 y + 3 z = 0. Solution: write down the coordinates of the normal vector of the plane N = (2, -5, 3). Substitute all known values into the given formula: cos α = (10 + 15 + 24)/√3724 ≈ 0.8 → α = 36.87°.
Video on the topic
A straight line that has one common point with a circle is tangent to the circle. Another feature of the tangent is that it is always perpendicular to the radius drawn to the point of contact, that is, the tangent and radius form a straight line corner. If two tangents to a circle AB and AC are drawn from one point A, then they are always equal to each other. Determining the angle between tangents ( corner ABC) is made using the Pythagorean theorem.
Instructions
To determine the angle, you need to know the radius of the circle OB and OS and the distance of the starting point of the tangent from the center of the circle - O. So, angles ABO and ACO are equal, the radius OB is, for example, 10 cm, and the distance to the center of the circle AO is 15 cm. Determine the length of the tangent using formula in accordance with the Pythagorean theorem: AB = square root of AO2 – OB2 or 152 - 102 = 225 – 100 = 125;
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), 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, let us set AB = 1. We introduce a standard coordinate system: 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:
Definition. If two lines are given y = k 1 x + b 1, y = k 2 x + b 2, then the acute angle between these lines will be defined as
Two lines are parallel if k 1 = k 2. Two lines are perpendicular if k 1 = -1/ k 2.
Theorem. The lines Ax + Bу + C = 0 and A 1 x + B 1 y + C 1 = 0 are parallel when the coefficients A 1 = λA, B 1 = λB are proportional. If also C 1 = λC, then the lines coincide. The coordinates of the point of intersection of two lines are found as a solution to the system of equations of these lines.
Equation of a line passing through a given point
Perpendicular to a given line
Definition. A straight line passing through the point M 1 (x 1, y 1) and perpendicular to the straight line y = kx + b is represented by the equation:
Distance from point to line
Theorem. If a point M(x 0, y 0) is given, then the distance to the line Ax + Bу + C = 0 is determined as
.
Proof. Let point M 1 (x 1, y 1) be the base of the perpendicular dropped from point M to a given straight line. Then the distance between points M and M 1:
(1)
The coordinates x 1 and y 1 can be found by solving the system of equations:
The second equation of the system is the equation of a line passing through a given point M 0 perpendicular to a given line. If we transform the first equation of the system to the form:
A(x – x 0) + B(y – y 0) + Ax 0 + By 0 + C = 0,
then, solving, we get:
Substituting these expressions into equation (1), we find:
The theorem has been proven.
Example. Determine the angle between the lines: y = -3 x + 7; y = 2 x + 1.
k 1 = -3; k 2 = 2; tgφ = 
; φ= p /4.
Example. Show that the lines 3x – 5y + 7 = 0 and 10x + 6y – 3 = 0 are perpendicular.
Solution. We find: k 1 = 3/5, k 2 = -5/3, k 1* k 2 = -1, therefore, the lines are perpendicular.
Example. Given are the vertices of the triangle A(0; 1), B (6; 5), C (12; -1). Find the equation of the height drawn from vertex C.
Solution. We find the equation of side AB: 
; 4 x = 6 y – 6;
2 x – 3 y + 3 = 0;
The required height equation has the form: Ax + By + C = 0 or y = kx + b. k = . Then y = . Because the height passes through point C, then its coordinates satisfy this equation: 
from where b = 17. Total: . 
Answer: 3 x + 2 y – 34 = 0.
The equation of a line passing through a given point in a given direction. Equation of a line passing through two given points. The angle between two straight lines. The condition of parallelism and perpendicularity of two straight lines. Determining the point of intersection of two lines
1. Equation of a line passing through a given point A(x 1 , y 1) in a given direction, determined by the slope k,
y - y 1 = k(x - x 1). (1)
This equation defines a pencil of lines passing through a point A(x 1 , y 1), which is called the beam center.
2. Equation of a line passing through two points: A(x 1 , y 1) and B(x 2 , y 2), written like this:
The angular coefficient of a straight line passing through two given points is determined by the formula
3. Angle between straight lines A And B is the angle by which the first straight line must be rotated A around the point of intersection of these lines counterclockwise until it coincides with the second line B. If two straight lines are given by equations with a slope
y = k 1 x + B 1 ,
y = k 2 x + B 2 , (4)
then the angle between them is determined by the formula
It should be noted that in the numerator of the fraction, the slope of the first line is subtracted from the slope of the second line.
If the equations of a line are given in general form
A 1 x + B 1 y + C 1 = 0,
A 2 x + B 2 y + C 2 = 0, (6)
the angle between them is determined by the formula
4. Conditions for parallelism of two lines:
a) If the lines are given by equations (4) with an angular coefficient, then the necessary and sufficient condition for their parallelism is the equality of their angular coefficients:
k 1 = k 2 . (8)
b) For the case when the lines are given by equations in general form (6), a necessary and sufficient condition for their parallelism is that the coefficients for the corresponding current coordinates in their equations are proportional, i.e.
5. Conditions for perpendicularity of two straight lines:
a) In the case when the lines are given by equations (4) with an angular coefficient, a necessary and sufficient condition for their perpendicularity is that their angular coefficients are inverse in magnitude and opposite in sign, i.e.
This condition can also be written in the form
k 1 k 2 = -1. (11)
b) If the equations of lines are given in general form (6), then the condition for their perpendicularity (necessary and sufficient) is to satisfy the equality
A 1 A 2 + B 1 B 2 = 0. (12)
6. The coordinates of the point of intersection of two lines are found by solving the system of equations (6). Lines (6) intersect if and only if
1. Write the equations of lines passing through the point M, one of which is parallel and the other perpendicular to the given line l.
Angle between straight lines in space we will call any of the adjacent angles formed by two straight lines drawn through an arbitrary point parallel to the data.
Let two lines be given in space:
Obviously, the angle φ between straight lines can be taken as the angle between their direction vectors and . Since , then using the formula for the cosine of the angle between vectors we get
The conditions of parallelism and perpendicularity of two straight lines are equivalent to the conditions of parallelism and perpendicularity of their direction vectors and:
Two straight parallel if and only if their corresponding coefficients are proportional, i.e. l 1 parallel l 2 if and only if parallel 
.
Two straight perpendicular if and only if the sum of the products of the corresponding coefficients is equal to zero: .
U goal between line and plane
Let it be straight d- not perpendicular to the θ plane;
d′− projection of a line d to the θ plane;
The smallest angle between straight lines d And d′ we will call angle between a straight line and a plane.
Let us denote it as φ=( d,θ)
If d⊥θ, then ( d,θ)=π/2
Oi→j→k→− rectangular coordinate system.
Plane equation:
θ: Ax+By+Cz+D=0
We assume that the straight line is defined by a point and a direction vector: d[M 0,p→]
Vector n→(A,B,C)⊥θ
Then it remains to find out the angle between the vectors n→ and p→, let us denote it as γ=( n→,p→).
If the angle γ<π/2 , то искомый угол φ=π/2−γ .
If the angle is γ>π/2, then the desired angle is φ=γ−π/2
sinφ=sin(2π−γ)=cosγ
sinφ=sin(γ−2π)=−cosγ
Then, angle between straight line and plane can be calculated using the formula:
sinφ=∣cosγ∣=∣ ∣ Ap 1+Bp 2+Cp 3∣ ∣ √A 2+B 2+C 2√p 21+p 22+p 23
Question29. The concept of quadratic form. Sign definiteness of quadratic forms.
Quadratic form j (x 1, x 2, …, x n) n real variables x 1, x 2, …, x n is called a sum of the form 
, (1)
Where a ij – some numbers called coefficients. Without loss of generality, we can assume that a ij = a ji.
The quadratic form is called valid, If a ij
Î GR. Matrix of quadratic form is called a matrix made up of its coefficients. The quadratic form (1) corresponds to the only symmetric matrix 
That is A T = A. Consequently, quadratic form (1) can be written in matrix form j ( X) = x T Ah, Where x T = (X 1 X 2 … x n). (2)
And, conversely, every symmetric matrix (2) corresponds to a unique quadratic form up to the notation of variables.
Rank of quadratic form is called the rank of its matrix. The quadratic form is called non-degenerate, if its matrix is non-singular A. (recall that the matrix A is called non-degenerate if its determinant is not equal to zero). Otherwise, the quadratic form is degenerate.
positive definite(or strictly positive) if
j ( X) > 0 , for anyone X = (X 1 , X 2 , …, x n), except X = (0, 0, …, 0).
Matrix A positive definite quadratic form j ( X) is also called positive definite. Therefore, a positive definite quadratic form corresponds to a unique positive definite matrix and vice versa.
The quadratic form (1) is called negatively defined(or strictly negative) if
j ( X) < 0, для любого X = (X 1 , X 2 , …, x n), except X = (0, 0, …, 0).
Similarly as above, a matrix of negative definite quadratic form is also called negative definite.
Consequently, the positive (negative) definite quadratic form j ( X) reaches the minimum (maximum) value j ( X*) = 0 at X* = (0, 0, …, 0).
Note that most quadratic forms are not sign-definite, that is, they are neither positive nor negative. Such quadratic forms vanish not only at the origin of the coordinate system, but also at other points.
When n> 2, special criteria are required to check the sign of a quadratic form. Let's look at them.
Major minors quadratic form are called minors:
that is, these are minors of the order of 1, 2, ..., n matrices A, located in the upper left corner, the last of them coincides with the determinant of the matrix A.
Positive Definiteness Criterion (Sylvester criterion)
X) = x T Ah was positive definite, it is necessary and sufficient that all major minors of the matrix A were positive, that is: M 1 > 0, M 2 > 0, …, M n > 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