Let the line pass through the point M1 (x1, y1, z1) and be parallel to the vector (m ,n, l). Let's write an equation for this line.

Let's take an arbitrary point M (x, y, z) on this line and find the relationship between x, y, z. Let's build a vector

The vectors are collinear.

- canonical equation of a straight line in space.

44 Parametric equations of a straight line

Because this equation is satisfied by the coordinates of any point on the line, then the resulting equation is a parametric equation of the line.

This vector equation can be represented in coordinate form:

Transforming this system and equating the values ​​of the parameter t, we obtain the canonical equations of a straight line in space:

Definition. The direction cosines of the straight line are the direction cosines of the vector, which can be calculated by the formulas:

From here we get: m: n: p = cosa: cosb: cosg.

The numbers m, n, p are called the slope of the line. Since is a non-zero vector, then m, n and p cannot be equal to zero at the same time, but one or two of these numbers can be equal to zero. In this case, in the equation of a straight line, the corresponding numerators should be equated to zero.

45 Equation of a straight line in space passing through two different given points.

Analytic geometry

Equation of a straight line passing through two given points.

Let M1(x1y1) and M2(x2y2) be given on the plane. Let us compose the canonical equation of the straight line passing through these two points, as the direction vector S we take M1M2

troika.

This is the equation of a straight line passing through two given points (x1 y1) and (x2, y2)

Let us now turn to the equations of the straight line and the plane in space.

Analytical geometry in 3-dimensional space

Similarly to the two-dimensional case, any equation of the first degree with respect to three variables x, y, z is an equation of a plane in the space Оxyz. planes. The canonical equation of the plane passing through the point M(x0,y0,z0) and having the normal N(A,B,C) A(x – x0) + B(y – y0) + C(z – z0)=0 – which is this equation?

The values ​​x-x0, y-y0 and z-z0 are the differences between the coordinates of the current point and the fixed point. Therefore, the vector a (x-x 0, y-y0, z-z0) is a vector lying in the described plane, and the vector N is a vector perpendicular to the plane, which means they are perpendicular to each other.

Then their scalar product must be equal to zero.

In coordinate form (N,a)=0 looks like this:

A (x-x0)+B (y-y0)+C (z-z0)=0

In space, right and left triples of vectors are distinguished. A triple of non-coplanar vectors a, b, c is called right if, from their common origin, the traversal of the ends of the vectors a, b, c in the specified order seems to be going clockwise. Otherwise a,b,c are left.

46 Angle between lines in space

An angle between straight lines in space is any of adjacent corners, 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 lines are parallel if and only if their respective coefficients are proportional, i.e. l1 is parallel to l2 if and only if it is parallel .

Two lines are perpendicular if and only if the sum of the products of the corresponding coefficients is equal to zero: .

Find the equations of the line passing through the point М1(1;2;3) parallel to the line l1:

Since the desired line l is parallel to l1, then as the direction vector of the desired line l, we can take the direction vector of the line l1.

One of the sub-items of the topic “The equation of a straight line on a plane” is the issue of compiling parametric equations of a straight line on a plane in a rectangular coordinate system. The article below discusses the principle of compiling such equations for certain known data. Let us show how to pass from parametric equations to equations of a different form; Let's analyze the solution of typical problems.

A particular line can be defined by specifying a point that belongs to that line and a direction vector for the line.

Suppose we are given a rectangular coordinate system O x y . And also the straight line a is given, indicating the point M 1 lying on it (x 1, y 1) and the direction vector of the given straight line a → = (a x , a y) . We give a description of the given line a using equations.

We use an arbitrary point M (x, y) and get a vector M 1 M →; calculate its coordinates from the coordinates of the start and end points: M 1 M → = (x - x 1 , y - y 1) . Let's describe the result: the line is given by a set of points M (x, y), passes through the point M 1 (x 1, y 1) and has a direction vector a → = (a x , a y) . The specified set defines a straight line only when the vectors M 1 M → = (x - x 1 , y - y 1) and a → = (a x , a y) are collinear.

There is a necessary and sufficient condition for the collinearity of vectors, which in this case for the vectors M 1 M → = (x - x 1 , y - y 1) and a → = (a x , a y) can be written as an equation:

M 1 M → = λ · a → , where λ is some real number.

Definition 1

The equation M 1 M → = λ · a → is called the vector-parametric equation of the line.

In coordinate form, it looks like:

M 1 M → = λ a → ⇔ x - x 1 = λ a x y - y 1 = λ a y ⇔ x = x 1 + a x λ y = y 1 + a y λ

The equations of the resulting system x = x 1 + a x · λ y = y 1 + a y · λ are called parametric equations of a straight line on a plane in a rectangular coordinate system. The essence of the name is as follows: the coordinates of all points of the line can be determined by parametric equations on the plane of the form x = x 1 + a x λ y = y 1 + a y λ when iterating over all real values ​​of the parameter λ

According to the above, the parametric equations of a straight line on the plane x \u003d x 1 + a x λ y \u003d y 1 + a y λ determine a straight line that is given in a rectangular coordinate system, passes through the point M 1 (x 1, y 1) and has a guide vector a → = (a x , a y) . Therefore, if the coordinates of a certain point of the straight line and the coordinates of its directing vector are given, then it is possible to immediately write down the parametric equations of the given straight line.

Example 1

It is necessary to compose parametric equations of a straight line on a plane in a rectangular coordinate system, if the point M 1 (2, 3) belonging to it and its direction vector are given a → = (3 , 1) .

Solution

Based on the initial data, we get: x 1 \u003d 2, y 1 \u003d 3, a x \u003d 3, a y \u003d 1. The parametric equations will look like:

x = x 1 + a x λ y = y 1 + a y λ ⇔ x = 2 + 3 λ y = 3 + 1 λ ⇔ x = 2 + 3 λ y = 3 + λ

Let's clearly illustrate:

Answer: x = 2 + 3 λ y = 3 + λ

It should be noted: if the vector a → = (a x , a y) serves as a directing vector of the straight line a, and the points M 1 (x 1, y 1) and M 2 (x 2, y 2) belong to this line, then it can be determined by setting parametric equations of the form: x = x 1 + a x λ y = y 1 + a y λ , as well as this option: x = x 2 + a x λ y = y 2 + a y λ .

For example, we are given a directing vector of a straight line a → \u003d (2, - 1), as well as points M 1 (1, - 2) and M 2 (3, - 3) belonging to this line. Then the straight line is determined by parametric equations: x = 1 + 2 · λ y = - 2 - λ or x = 3 + 2 · λ y = - 3 - λ .

Attention should also be paid to the following fact: if a → = (a x , a y) is the directing vector of the straight line a , then any of the vectors will also be its directing vector μ a → = (μ a x , μ a y) , where μ ϵ R , μ ≠ 0 .

Thus, a straight line a on a plane in a rectangular coordinate system can be defined by parametric equations: x = x 1 + μ a x λ y = y 1 + μ a y λ for any value of μ that is different from zero.

Suppose the line a is given by the parametric equations x = 3 + 2 λ y = - 2 - 5 λ . Then a → = (2 , - 5) - direction vector of this line. And also any of the vectors μ · a → = (μ · 2 , μ · - 5) = 2 μ , - 5 μ , μ ∈ R , μ ≠ 0 will become the direction vector for the given straight line. For clarity, consider a specific vector - 2 · a → = (- 4 , 10) , it corresponds to the value μ = - 2 . In this case, the given straight line can also be determined by the parametric equations x = 3 - 4 · λ y = - 2 + 10 · λ .

Transition from parametric equations of a straight line on a plane to other equations of a given straight line and vice versa

In solving some problems, the use of parametric equations is not the most optimal option, then it becomes necessary to translate the parametric equations of a straight line into equations of a straight line of a different type. Let's see how to do it.

Parametric equations of the straight line x = x 1 + a x · λ y = y 1 + a y · λ will correspond to the canonical equation of the straight line on the plane x - x 1 a x = y - y 1 a y .

We solve each of the parametric equations with respect to the parameter λ, equate the right parts of the obtained equalities and obtain the canonical equation of the given straight line:

x = x 1 + a x λ y = y 1 + a y λ ⇔ λ = x - x 1 a x λ = y - y 1 a y ⇔ x - x 1 a x = y - y 1 a y

In this case, it should not be embarrassing if a x or a y will be equal to zero.

Example 2

It is necessary to carry out the transition from the parametric equations of the straight line x = 3 y = - 2 - 4 · λ to the canonical equation.

Solution

We write the given parametric equations in the following form: x = 3 + 0 λ y = - 2 - 4 λ

We express the parameter λ in each of the equations: x = 3 + 0 λ y = - 2 - 4 λ ⇔ λ = x - 3 0 λ = y + 2 - 4

We equate the right parts of the system of equations and obtain the required canonical equation of a straight line in the plane:

x - 3 0 = y + 2 - 4

Answer: x - 3 0 = y + 2 - 4

In the case when it is necessary to write down the equation of the straight line of the form A x + B y + C = 0 , while the parametric equations of the straight line on the plane are given, it is necessary first to make the transition to the canonical equation, and then to the general equation of the straight line. Let's write down the whole sequence of actions:

x = x 1 + a x λ y = y 1 + a y λ ⇔ λ = x - x 1 a x λ = y - y 1 a y ⇔ x - x 1 a x = y - y 1 a y ⇔ ⇔ a y (x - x 1) = a x (y - y 1) ⇔ A x + B y + C = 0

Example 3

It is necessary to write down the general equation of a straight line if the parametric equations defining it are given: x = - 1 + 2 λ y = - 3 λ

Solution

First, let's make the transition to the canonical equation:

x = - 1 + 2 λ y = - 3 λ ⇔ λ = x + 1 2 λ = y - 3 ⇔ x + 1 2 = y - 3

The resulting proportion is identical to the equality - 3 · (x + 1) = 2 · y. Let's open the brackets and get the general equation of the straight line: - 3 x + 1 = 2 y ⇔ 3 x + 2 y + 3 = 0 .

Answer: 3x + 2y + 3 = 0

Following the above logic of actions, to obtain the equation of a straight line with slope factor, the equation of a straight line in segments or the normal equation of a straight line, it is necessary to obtain the general equation of a straight line, and from it to carry out a further transition.

Now consider the reverse action: writing the parametric equations of a straight line for a different given form of the equations of this straight line.

The easiest transition: from the canonical equation to the parametric ones. Let the canonical equation of the form be given: x - x 1 a x = y - y 1 a y . We take each of the relations of this equality equal to the parameter λ:

x - x 1 a x = y - y 1 a y = λ ⇔ λ = x - x 1 a x λ = y - y 1 a y

Let us solve the resulting equations for the variables x and y:

x = x 1 + a x λ y = y 1 + a y λ

Example 4

It is necessary to write down the parametric equations of the straight line if the canonical equation of the straight line on the plane is known: x - 2 5 = y - 2 2

Solution

Let's equate the parts of the known equation to the parameter λ: x - 2 5 = y - 2 2 = λ . From the obtained equality we obtain the parametric equations of the straight line: x - 2 5 = y - 2 2 = λ ⇔ λ = x - 2 5 λ = y - 2 5 ⇔ x = 2 + 5 λ y = 2 + 2 λ

Answer: x = 2 + 5 λ y = 2 + 2 λ

When it is necessary to make a transition to parametric equations from a given general equation of a straight line, an equation of a straight line with a slope or an equation of a straight line in segments, it is necessary to bring the original equation to the canonical one, and then make the transition to parametric equations.

Example 5

It is necessary to write down the parametric equations of the straight line with the known general equation of this straight line: 4 x - 3 y - 3 = 0 .

Solution

We transform the given general equation into an equation of the canonical form:

4 x - 3 y - 3 = 0 ⇔ 4 x = 3 y + 3 ⇔ ⇔ 4 x = 3 y + 1 3 ⇔ x 3 = y + 1 3 4

We equate both parts of the equality to the parameter λ and obtain the required parametric equations of the straight line:

x 3 = y + 1 3 4 = λ ⇔ x 3 = λ y + 1 3 4 = λ ⇔ x = 3 λ y = - 1 3 + 4 λ

Answer: x = 3 λ y = - 1 3 + 4 λ

Examples and problems with parametric equations of a straight line on a plane

Let us consider the most common types of problems using parametric equations of a straight line on a plane in a rectangular coordinate system.

1. In problems of the first type, the coordinates of points are given, whether or not they belong to a straight line described by parametric equations.

The solution of such problems is based on the following fact: the numbers (x, y) determined from the parametric equations x \u003d x 1 + a x λ y \u003d y 1 + a y λ for some real value λ are the coordinates of a point belonging to the straight line, which is described these parametric equations.

Example 6

It is necessary to determine the coordinates of a point that lies on a straight line given by the parametric equations x = 2 - 1 6 · λ y = - 1 + 2 · λ for λ = 3 .

Solution

We substitute the known value λ = 3 into the given parametric equations and calculate the required coordinates: x = 2 - 1 6 3 y = - 1 + 2 3 ⇔ x = 1 1 2 y = 5

Answer: 1 1 2 , 5

The following problem is also possible: let some point M 0 (x 0, y 0) be given on a plane in a rectangular coordinate system and it is necessary to determine whether this point belongs to the line described by the parametric equations x = x 1 + a x λ y = y 1 + a y λ .

To solve such a problem, it is necessary to substitute the coordinates of a given point into the known parametric equations of a straight line. If it is determined that such a value of the parameter λ = λ 0 is possible, in which both parametric equations are true, then the given point belongs to the given straight line.

Example 7

Points M 0 (4, - 2) and N 0 (- 2, 1) are given. It is necessary to determine whether they belong to the straight line defined by the parametric equations x = 2 · λ y = - 1 - 1 2 · λ .

Solution

We substitute the coordinates of the point M 0 (4, - 2) into the given parametric equations:

4 = 2 λ - 2 = - 1 - 1 2 λ ⇔ λ = 2 λ = 2 ⇔ λ = 2

We conclude that the point M 0 belongs to a given line, because corresponds to the value λ = 2 .

2 = 2 λ 1 = - 1 - 1 2 λ ⇔ λ = - 1 λ = - 4

It is obvious that there is no such parameter λ to which the point N 0 will correspond. In other words, the given line does not pass through the point N 0 (- 2 , 1) .

Answer: point M 0 belongs to a given line; the point N 0 does not belong to the given line.

1. In problems of the second type, it is required to compose parametric equations of a straight line on a plane in a rectangular coordinate system. The simplest example of such a problem (with known coordinates of the point of the line and direction vector) was considered above. Now let's look at examples in which you first need to find the coordinates of the direction vector, and then write down the parametric equations.

Point M 1 1 2 , 2 3 is given. It is necessary to compose parametric equations of a straight line passing through this point and a parallel straight line x 2 \u003d y - 3 - 1.

Solution

According to the condition of the problem, the straight line, the equation of which we have to get ahead of, is parallel to the straight line x 2 \u003d y - 3 - 1. Then, as a direction vector, the straight line passing through given point, it is possible to use the direction vector of the straight line x 2 = y - 3 - 1 , which we write in the form: a → = (2 , - 1) . Now all the necessary data are known in order to compose the desired parametric equations:

x = x 1 + a x λ y = y 1 + a y λ ⇔ x = 1 2 + 2 λ y = 2 3 + (- 1) λ ⇔ x = 1 2 + x λ y = 2 3 - λ

Answer: x = 1 2 + x λ y = 2 3 - λ .

Example 9

Point M 1 (0, - 7) is given. It is necessary to write the parametric equations of a straight line passing through this point perpendicular to the straight line 3 x – 2 y – 5 = 0 .

Solution

As the directing vector of the straight line, the equation of which must be composed, it is possible to take the normal vector of the straight line 3 x - 2 y - 5 = 0 . Its coordinates are (3 , - 2) . We write the required parametric equations of the straight line:

x = x 1 + a x λ y = y 1 + a y λ ⇔ x = 0 + 3 λ y = - 7 + (- 2) λ ⇔ x = 3 λ y = - 7 - 2 λ

Answer: x = 3 λ y = - 7 - 2 λ

1. In problems of the third type, it is required to make a transition from parametric equations of a given straight line to other types of equations that determine it. Solution similar examples we considered above, we will give one more.

Given a straight line on a plane in a rectangular coordinate system, defined by the parametric equations x = 1 - 3 4 · λ y = - 1 + λ . It is necessary to find the coordinates of some normal vector of this line.

Solution

To determine the desired coordinates of the normal vector, we will make the transition from parametric equations to the general equation:

x = 1 - 3 4 λ y = - 1 + λ ⇔ λ = x - 1 - 3 4 λ = y + 1 1 ⇔ x - 1 - 3 4 = y + 1 1 ⇔ ⇔ 1 x - 1 = - 3 4 y + 1 ⇔ x + 3 4 y - 1 4 = 0

The coefficients of the variables x and y give us the required coordinates of the normal vector. Thus, the normal vector of the line x = 1 - 3 4 · λ y = - 1 + λ has coordinates 1 , 3 4 .

Answer: 1 , 3 4 .

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Equating in the canonical equations of the straight line each of the fractions to some parameter t:

We obtain equations expressing the current coordinates of each point of the straight line through the parameter t.

thus, the parametric equations of the straight line have the form:

Equations of a straight line passing through two given points.

Let two points M 1 (x1,y1,z1) and M 2 (x2,y2,z2). The equations of a straight line passing through two given points are obtained in the same way as a similar equation on a plane. Therefore, we immediately give the form of this equation.

A straight line at the intersection of two planes. General equation of a straight line in space.

If we consider two non-parallel planes, then their intersection will be a straight line.

If normal vectors and non-collinear.

Below, when considering examples, we will show a way to transform such straight line equations to canonical equations.

5.4 Angle between two straight lines. Condition of parallelism and perpendicularity of two lines.

An angle between two straight lines in space is any of the angles formed by two straight lines drawn through an arbitrary point parallel to the data.

Let two lines be given by their canonical equations.

For the angle between two straight lines we will take the angle between the direction vectors.

And

The condition of perpendicularity of two straight lines is reduced to the condition of perpendicularity of their direction vectors and , that is, to the equality to zero of the scalar product: or in coordinate form: .

The condition of parallelism of two lines is reduced to the condition of parallelism of their direction vectors and

5.5 Mutual arrangement straight and plane.

Let the equations of the straight line be given:

and planes. The angle between the line and the plane will be any of the two adjacent angles formed by the line and its projection onto the plane (Figure 5.5).

Figure 5.5

If the line is perpendicular to the plane, the directing vector of the line and the normal vector to the plane are collinear. Thus, the condition of perpendicularity of a straight line and a plane is reduced to the condition of collinear vectors

In the case of parallelism of a straight line and a plane, their vectors indicated above are mutually perpendicular. Therefore, the condition of parallelism of a straight line and a plane is reduced to the condition of perpendicularity of the vectors; those. their dot product is zero or in coordinate form: .

Below are examples of solving problems related to the topic of Chapter 5.

Example 1:

Write an equation for a plane passing through point A (1,2,4) perpendicular to the straight line given by the equation:

Solution:

We use the equation of a plane passing through a given point perpendicular to a given vector.

A(x-x 0)+B(y-y 0)+C(z-z 0)=0

As a point, we take the point A (1,2,4), through which the plane passes by the condition.

Knowing the canonical equations of the line, we know the vector parallel to the line.

Due to the fact that, by the condition, the line is perpendicular to the desired plane, the direction vector can be taken as the normal vector of the plane.

Thus, we obtain the equation of the plane in the form:

2(x-1)+1(y-2)+4(z-4)=0

2x+y+4z-16=0

2x+y+4z-20=0

Example 2:

Find on the plane 4x-7y+5z-20=0 a point P for which OP makes equal angles with the coordinate axes.

Solution:

Let's make a schematic drawing. (Figure 5.6)

at

Figure 5.6

The empty point Р has coordinates . Since the vector makes the same angles with the coordinate axes, the direction cosines of this vector are equal to each other

Let's find the projections of the vector:

then the direction cosines of this vector are easily found.

From the equality of the direction cosines, the equality follows:

x p \u003d y p \u003d z p

since the point P lies on the plane, substituting the coordinates of this point into the equation of the plane turns it into an identity.

4x p -7x p +5x p -20=0

2x p \u003d 20

x p \u003d 10

Respectively: y r=10; z p=10.

Thus, the desired point P has coordinates P (10; 10; 10)

Example 3:

Given two points A (2, -1, -2) and B (8, -7.5). Find the equation of the plane passing through the point B, perpendicular to the segment AB.

Solution:

To solve the problem, we use the equation of a plane passing through a given point perpendicular to a given vector.

A(x-x 0)+B(y-y 0)+C(z-z 0)=0

As a point, we use point B (8, -7.5), and as a vector perpendicular to the plane, vector. Let's find the projections of the vector:

then we get the equation of the plane in the form:

6(x-8)-6(y+7)+7(z-5)=0

6x-48-6y-42+7z-35=0

6x-6y+7z-35=0

6x-6y+7z-125=0

Example 4:

Find the equation of a plane parallel to the OY axis and passing through the points K(1,-5,1) and M(3,2,-2).

Solution:

Since the plane is parallel to the OY axis, we will use the incomplete equation of the plane.

Ax+Cz+D=0

Due to the fact that the points K and M lie on the plane, we obtain two conditions.

Let us express from these conditions the coefficients A and C in terms of D.

We substitute the found coefficients into the incomplete equation of the plane:

since , then we reduce D:

Example 5:

Find the equation of a plane passing through three points M(7,6,7), K(5,10,5), R(-1,8,9)

Solution:

Let's use the equation of a plane passing through 3 given points.

substituting the coordinates points M, K, R as the first, second and third we get:

expand the determinant along the 1st line.

Example 6:

Find the equation of the plane passing through the points M 1 (8, -3,1); M 2 (4,7,2) and perpendicular to the plane 3x+5y-7z-21=0

Solution:

Let's make a schematic drawing (Figure 5.7)

Figure 5.7

We denote the given plane P 2 and the desired plane P 2. . From the equation of a given plane Р 1 we determine the projections of the vector perpendicular to the plane Р 1.

The vector can be moved to the plane P 2 by means of parallel translation, since, according to the condition of the problem, the plane P 2 is perpendicular to the plane P 1, which means that the vector is parallel to the plane P 2.

Let's find the projections of the vector lying in the plane Р 2:

now we have two vectors and lying in the plane R 2 . obviously vector , equal to the vector product of vectors and will be perpendicular to the plane R 2, since it is perpendicular to and, therefore, its normal vector to the plane R 2.

The vectors and are given by their projections, therefore:

Next, we use the equation of a plane passing through a given point perpendicular to the vector. As a point, you can take any of the points M 1 or M 2, for example M 1 (8, -3.1); As a normal vector to the plane Р 2 we take .

74(x-8)+25(y+3)+50(z-1)=0

3(x-8)+(y-3)+2(z-1)=0

3x-24+y+3+27-2=0

3x+y+2z-23=0

Example 7:

A straight line is defined by the intersection of two planes. Find the canonical equations of the line.

Solution:

We have an equation in the form:

Need to find a point x 0, y 0, z 0) through which the straight line and direction vector pass.

We choose one of the coordinates arbitrarily. For example, z=1, then we get a system of two equations with two unknowns:

Thus, we have found a point lying on the desired line (2,0,1).

As a directing vector of the desired straight line, we take the cross product of vectors and , which are normal vectors since , which means parallel to the desired line.

Thus, the direction vector of the straight line has projections . Using the equation of a straight line passing through a given point parallel to a given vector:

So the desired canonical equation has the form:

Example 8:

Find the coordinates of the point of intersection of a line and a plane 2x+3y+3z-8=0

Solution:

Let us write the given equation of a straight line in a parametric form.

x=3t-2; y=-t+2; z=2t-1

each point of the straight line corresponds to a single value of the parameter t. To find the parameter t corresponding to the point of intersection of the line and the plane, we substitute the expression into the equation of the plane x, y, z via parameter t.

2(3t-2)+(-t+2)+3(2t-1)-8=0

6t-4-3t+6+6t-3-8=0

t=1

then the coordinates of the desired point

the desired intersection point has coordinates (1;1;1).

Example 9:

Find the equation of a plane passing through parallel lines.

Let's make a schematic drawing (Figure 5.9)

Figure 5.9

From given equations lines and determine the projections of the direction vectors of these lines. We find the projections of the vector lying in the plane P, and take the points and from the canonical equations of the lines M 1 (1, -1,2) and M 2 (0,1, -2).

1. Parametric equation of a straight line

Let a point M 0 (x 0 , y 0 , z 0 ) be given on a straight line and a vector s = (l ,m ,n ) lying on

this line (or parallel to it). The vector s is also called guide vector straight.

These conditions uniquely define a straight line in space. Let's find her

the equation. Take an arbitrary point M (x, y, z) on the line. It is clear that the vectors

M 0 M (x − x 0 , y − y 0 , z − z 0 ) and s are collinear.

Therefore, M 0 M = t s − is a vector equation of a straight line.

In coordinate notation, the last equation has the following parametric representation

two angles formed by two straight lines, respectively, parallel to the given one and passing through one point (which may require parallel translation of one of the straight lines).

It follows from the definition that one of the angles is equal to the angle ϕ between

The lines are perpendicular to s 1 s 2 l 1 l 2 + m 1 m 2 + n 1 n 2 = 0 .

4. The angle between a line and a plane. Conditions « » and « » direct and

plane

Let the line L be given by its canonical equation x − l x 0 = y − m y 0 = z − n z 0 ,

and the plane P by the equation

Ax + By + Cz + D = 0.

Definition. Angle between line L

and the plane p is called sharp corner between the line L and its projection onto the plane.

It follows from the definition (and figure) that the required angle ϕ is additional (up to right angle) to the angle between the normal vector n (A , B ,C ) and

direction vector s (l ,m ,n ) .

(. is taken to get an acute angle).

If L Р, then s n (s, n) = 0

Now, if we substitute the found "t" into the parametric equations of the straight line, then we will find the desired intersection point

One of the main operations of mathematical analysis is the operation of passage to the limit, which occurs in the course in various forms. We start with the simplest form of the passage to the limit operation, based on the concept of the limit of the so-called number sequence. This will facilitate the introduction of another very important form of the passage to the limit operation, the limit of a function. In what follows, the constructions of passages to the limit will be used in the construction of the differential and integral calculus.



 Infinitesimal and infinitely large sequences

 Relationship between infinitely large and infinitely small sequences.

 The simplest properties of infinitesimal sequences

 Sequence limit.

 Properties of convergent sequences

 Arithmetic operations on convergent sequences

 Monotonic sequences

 Cauchy Convergence Criterion

 The number e and its economic illustration.

 Application of limits in economic calculations

§ 1. Numerical sequences and simple properties

1. The concept of a numerical sequence. Arithmetic operations on sequences

Number sequences are infinite sets of numbers. Example sequences are known from school:

1) the sequence of all members of an infinite arithmetic and geometric progression;

2) sequence of regular perimeters n-gons inscribed in a given circle;

will be called the number sequence (or just a sequence).

Separate numbers x 3 , x 5 , x n will be called elements or members of the sequence (1). The symbol x n is called the common or n-th member of this sequence. Giving the value n = 1, 2, … in the common term x n we get, respectively, the first x 1 , the second x 2 and so on. members.

A sequence is considered given (see Def.) if a method for obtaining any of its elements is specified. Often a sequence is given by a formula for the common term of the sequence.

To shorten the notation, the sequence (1) is sometimes written as

The structure of the common term (its formula) can be complex. For example,

Sometimes the sequence is given by the so-called recurrent formulas, i.e. formulas that allow you to find subsequent members of the sequence from known previous ones.

Example (Fibonacci numbers). Let x 1 = x 2 = 1 and the recurrent formula x n = x n − 1 + x n − 2 for n = 3, 4, … is given. Then we have the sequence 1, 1,

2, 3, 5, 8, ... (the numbers of Leonardo from Pisa, nicknamed Fibonacci). Geometrically, a numerical sequence can be depicted on a numerical

axis in the form of a sequence of points whose coordinates are equal to the corresponding

corresponding members of the sequence. For example, ( x n ) = 1 n .

Lecture № 8-9 Fundamentals of mathematical analysis prof. Dymkov M.P. 66

Consider along with the sequence ( x n ) another sequence ( y n ) : y 1 , y 2 , y ,n (2).

Definition. The sum (difference, product, quotient) of the sequence

values ​​( xn ) and ( yn ) is called a sequence ( zn ) whose members are

The product of a sequence ( xn ) and a number c R is a sequence ( c xn ) .

Definition. The sequence ( xn ) is called bounded

from above (from below), if there is a real number M (m) such that each element of this sequence xn satisfies the unequal

xn ≤ M (xn ≥ m) . A sequence is called bounded if it is bounded both above and below m ≤ xn ≤ M . The sequence xn is called

is unbounded if for a positive number A (arbitrarily large) there is at least one element of the sequence xn , satisfies

which gives the inequality xn > A.

( x n ) = ( 1n ) 0 ≤ x n ≤ 1.

( x n ) = ( n ) − is bounded from below by 1, but is unbounded.

( x n ) = ( − n ) − bounded from above (–1), but also unbounded.

Definition. The sequence ( x n ) is called infinitesimal,

if for any positive real number ε (no matter how small it is taken) there exists a number N depending, generally speaking, on ε , (N = N (ε )) such that for all n ≥ N the inequality x n< ε .

Example. ( x n ) = 1 n .

Definition. The sequence ( xn ) is called endless pain-

shoy if for a positive real number A (no matter how large it is) there is a number N (N = N(A)) such that for all n ≥ N

the inequality xn > A is obtained.

ANGLE BETWEEN PLANES

Let's consider two planes α 1 and α 2 given respectively by the equations:

Under angle 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 vector product normal vectors:

.

Example. Lead general equations straight 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