A straight line in space can always be defined as the line of intersection of two non-parallel planes. If the equation of one plane is the equation of the second plane, then the equation of the line is given as

Here non-collinear
. These equations are called general equations straight in space.

Canonical equations of the line

Any non-zero vector lying on a given line or parallel to it is called the direction vector of this line.

If the point is known
straight line and its direction vector
, then the canonical equations of the line have the form:

. (9)

Parametric equations of a line

Let the canonical equations of the line be given

.

From here, we obtain the parametric equations of the line:

(10)

These equations are useful for finding the intersection point of a line and a plane.

Equation of a line passing through two points
And
has the form:

.

Angle between straight lines

Angle between straight lines

And

equal to the angle between their direction vectors. Therefore, it can be calculated using formula (4):

Condition for parallel lines:

.

Condition for planes to be perpendicular:

Distance of a point from a line

P let's say the point is given
and straight

.

From the canonical equations of the line we know the point
, belonging to a line, and its direction vector
. Then the distance of the point
from a straight line is equal to the height of a parallelogram built on vectors And
. Hence,

.

Condition for the intersection of lines

Two non-parallel lines

,

intersect if and only if

.

The relative position of a straight line and a plane.

Let the straight line be given
and plane. Corner between them can be found by the formula

.

Problem 73. Write the canonical equations of the line

(11)

Solution. In order to write down the canonical equations of the line (9), it is necessary to know any point belonging to the line and the direction vector of the line.

Let's find the vector , parallel to this line. Since it must be perpendicular to the normal vectors of these planes, i.e.

,
, That

.

From the general equations of the straight line we have that
,
. Then

.

Since the point
any point on a line, then its coordinates must satisfy the equations of the line and one of them can be specified, for example,
, we find the other two coordinates from system (11):

From here,
.

Thus, the canonical equations of the desired line have the form:

or
.

Problem 74.

And
.

Solution. From canonical equations the first line knows the coordinates of the point
belonging to the line, and the coordinates of the direction vector
. From the canonical equations of the second line the coordinates of the point are also known
and coordinates of the direction vector
.

The distance between parallel lines is equal to the distance of the point
from the second straight line. This distance is calculated by the formula

.

Let's find the coordinates of the vector
.

Let's calculate the vector product
:

.

Problem 75. Find a point symmetrical point
relatively straight

.

Solution. Let us write down the equation of a plane perpendicular to a given line and passing through a point . As its normal vector you can take the directing vector of a straight line. Then
. Hence,

Let's find a point
the point of intersection of this line and plane P. To do this, we write down the parametric equations of the line using equations (10), we get

Hence,
.

Let
point symmetrical to point
relative to this line. Then point
midpoint
. To find the coordinates of a point we use the formulas for the coordinates of the midpoint of the segment:

,
,
.

So,
.

Problem 76. Write the equation of a plane passing through a line
And

a) through a point
;

b) perpendicular to the plane.

Solution. Let's write it down general equations this line. To do this, consider two equalities:

This means that the desired plane belongs to a bundle of planes with generators and its equation can be written in the form (8):

a) Let's find
And from the condition that the plane passes through the point
, therefore, its coordinates must satisfy the equation of the plane. Let's substitute the coordinates of the point
into the equation of a bunch of planes:

Found value
Let's substitute it into equation (12). we obtain the equation of the desired plane:

b) Let's find
And from the condition that the desired plane is perpendicular to the plane. The normal vector of a given plane
, normal vector of the desired plane (see equation of a bunch of planes (12).

Two vectors are perpendicular if and only if they are scalar product equals zero. Hence,

Let's substitute the found value
into the equation of a bunch of planes (12). We obtain the equation of the desired plane:

Tasks for independent decision

Task 77. Lead to canonical form equations of lines:

1)
2)

Problem 78. Write the parametric equations of the line
, If:

1)
,
; 2)
,
.

Problem 79. Write the equation of the plane passing through the point
perpendicular to a straight line

Problem 80. Write the equations of a line passing a point
perpendicular to the plane.

Problem 81. Find the angle between the lines:

1)
And
;

2)
And

Problem 82. Prove parallelism of lines:

And
.

Problem 83. Prove the perpendicularity of the lines:

And

Problem 84. Calculate the distance of a point
from straight line:

1)
; 2)
.

Problem 85. Calculate the distance between parallel lines:

And
.

Problem 86. In the equations of the line
define parameter so that this line intersects with the line and find the point of their intersection.

Problem 87. Show that it is straight
parallel to the plane
, and the straight line
lies in this plane.

Problem 88. Find a point symmetrical point relative to the plane
, If:

1)
, ;

2)
, ;.

Problem 89. Write the equation of a perpendicular dropped from a point
directly
.

Problem 90. Find a point symmetrical point
relatively straight
.

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 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 hold

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 for the variables are NOT proportional, that is, there is NO such “lambda” value that the equalities hold

So, for straight lines we will create a system:

From the first equation it follows that , and from the second equation: , which means that the system is inconsistent (there are 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 discussed in the lesson The concept of linear (non) dependence of vectors. Basis of vectors. But there is a more civilized packaging:

Example 1

Find out the relative position of the lines:

The solution is 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 don’t see any point in offering anything for an independent solution; it’s better to lay another important brick in the geometric foundation:

How to construct a line parallel to a given one?

For ignorance of this simplest task, the Nightingale the Robber severely punishes.

Example 2

The straight line is given by the equation. Write an equation for a parallel line that passes through the point.

Solution: 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 example geometry 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 a solution to the system 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 these 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 way to solve a system 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 search for the intersection point using an 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, visit the 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.

Full solution and 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 using the 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

In front of us is a straight strip of the river 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. Using the formulas for the coordinates of the midpoint of the 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 count 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 Consider two straight lines given by the equations in:

general view If the lines are not perpendicular, then oriented

The angle between them can be calculated using the formula:

Let us pay close attention to the denominator - this is exactly the scalar product of the direction vectors of the 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 the scalar product of the direction vectors of the lines:

, which means the lines are not perpendicular.

2) Find the angle between straight lines using the formula:

Answer :

Using the inverse function, it is easy to find the angle itself. In this case, we use the oddness of the arctangent (see Graphs and properties of elementary functions):

In your answer, we indicate the 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


In July 2020, NASA launches an expedition to Mars. The spacecraft will deliver to Mars an electronic medium with the names of all registered expedition participants.

One of these code options needs to be copied and pasted into the code of your web page, preferably between tags and or immediately after the tag. According to the first option, MathJax loads faster and slows down the page less. But the second option automatically monitors and loads the latest versions of MathJax. If you insert the first code, it will need to be updated periodically. If you insert the second code, the pages will load more slowly, but you will not need to constantly monitor MathJax updates.

The easiest way to connect MathJax is in Blogger or WordPress: in the site control panel, add a widget designed to insert third-party JavaScript code, copy the first or second version of the download code presented above into it, and place the widget closer to the beginning of the template (by the way, this is not at all necessary , since the MathJax script is loaded asynchronously). That's all. Now learn the markup syntax of MathML, LaTeX, and ASCIIMathML, and you are ready to insert mathematical formulas into your site's web pages.

Another New Year's Eve... frosty weather and snowflakes on the window glass... All this prompted me to write again about... fractals, and what Wolfram Alpha knows about it. There is an interesting article on this subject, which contains examples of two-dimensional fractal structures. Here we will look at more complex examples three-dimensional fractals.

A fractal can be visually represented (described) as a geometric figure or body (meaning that both are a set, in this case, a set of points), the details of which have the same shape as the original figure itself. That is, this is a self-similar structure, examining the details of which when magnified, we will see the same shape as without magnification. Whereas in the case of ordinary geometric figure(not a fractal), when zoomed in we will see details that have more simple form than the original figure itself. For example, at a high enough magnification, part of an ellipse looks like a straight line segment. This does not happen with fractals: with any increase in them, we will again see the same complex shape, which will be repeated again and again with each increase.

Benoit Mandelbrot, the founder of the science of fractals, wrote in his article Fractals and Art in the Name of Science: “Fractals are geometric shapes that are as complex in their details as they are in their general form. That is, if part of a fractal is enlarged to the size of the whole, it will appear as the whole, either exactly, or perhaps with a slight deformation."