A point symmetrical to a point with respect to a straight plane. The simplest problems with a straight line on a plane. Mutual arrangement of lines. Angle between lines. How to find the distance between two parallel lines

A straight line in space can always be defined as a 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 straight line is given as

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

Canonical equations of the straight line

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

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

. (9)

Parametric equations of a straight line

Let the canonical equations of the line be given

.

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

(10)

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

Equation of a line passing through two points
And
looks like:

.

Angle between lines

Angle between lines

And

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

Condition of parallel lines:

.

Condition of perpendicularity of planes:

Distance of a point from a straight line

P given point
and direct

.

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

.

Line intersection condition

Two non-parallel lines

,

intersect if and only if

.

Mutual arrangement of a straight line and a plane.

Let the straight line
and flat. 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 directing vector of the line.

Let's find the vector parallel to the given 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 of the 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 the system (11):

From here,
.

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

or
.

Problem 74.

And
.

Solution. From the canonical equations of the first line, the coordinates of the point are known
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 direction vector coordinates
.

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

.

Let's find the coordinates of the vector
.

Compute the vector product
:

.

Problem 75. Find a point symmetrical point
relatively straight

.

Solution. We write the equation of the plane perpendicular to the given line and passing through the point . As its normal vector we can take the directing vector as a straight line. Then
. Hence,

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

Hence,
.

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

,
,
.

So,
.

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

a) through a dot
;

b) perpendicular to the plane.

Solution. Let us write down the general equations of this straight line. To do this, consider two equalities:

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

a) find
And from the condition that the plane passes through the point
, therefore, its coordinates must satisfy the equation of the plane. Substitute the coordinates of the point
into the equation of a beam of planes:

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

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

Two vectors are perpendicular if and only if their dot product is zero. Hence,

Substitute the found value
into the equation of a beam of planes (12). We obtain the equation of the desired plane:

Tasks for independent solution

Problem 77. Bring to the canonical form the equations of lines:

1)
2)

Problem 78. Write parametric equations of a straight line
, If:

1)
,
; 2)
,
.

Problem 79. Write an equation for a plane passing through a point
perpendicular to the line

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

Problem 81. Find the angle between lines:

1)
And
;

2)
And

Problem 82. Prove parallel lines:

And
.

Problem 83. Prove perpendicularity of lines:

And

Problem 84. Calculate point distance
from straight:

1)
; 2)
.

Problem 85. Calculate the distance between parallel lines:

And
.

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

Problem 87. Show it's 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 for a perpendicular dropped from a point
directly
.

Problem 90. Find a point symmetrical point
relatively straight
.

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

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One of these code options needs to be copied and pasted into the code of your web page, preferably between the tags And or right after the tag . According to the first option, MathJax loads faster and slows down the page less. But the second option automatically tracks and loads the latest versions of MathJax. If you insert the first code, then it will need to be updated periodically. If you paste the second code, then 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 load code above into it, and place the widget closer to the beginning of the template (by the way, this is not necessary at all , since the MathJax script is loaded asynchronously). That's all. Now learn the MathML, LaTeX, and ASCIIMathML markup syntax and you're ready to embed math formulas into your 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. On this occasion, there is an interesting article in which there are examples of two-dimensional fractal structures. Here we will consider more complex examples of 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, it is a self-similar structure, considering the details of which, when magnified, we will see the same shape as without magnification. Whereas in the case of an ordinary geometric figure (not a fractal), when zoomed in, we will see details that have a simpler shape than the original figure itself. For example, at a sufficiently high 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 with each increase will be repeated again and again.

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

Oh-oh-oh-oh-oh ... well, it's tinny, as if you read the sentence to yourself =) However, then relaxation will help, especially since I bought suitable accessories today. Therefore, let's proceed to the first section, I hope, by the end of the article I will keep a cheerful mood.

Mutual arrangement of two straight lines

The case when the hall sings along in chorus. Two lines can:

1) match;

2) be parallel: ;

3) or intersect at a single point: .

Help for dummies : please remember the math sign intersections, it will occur very often. The entry means that the line intersects with the line at the point.

How to determine the relative position of two lines?

Let's start with the first case:

Two lines coincide if and only if their respective coefficients are proportional, that is, there is such a number "lambda" that the equalities

Let's consider straight lines and compose three equations from the corresponding coefficients: . From each equation it follows that, therefore, these lines coincide.

Indeed, if all the coefficients of the equation multiply by -1 (change signs), and all the coefficients of the equation reduce by 2, you get the same equation: .

The second case when the lines are parallel:

Two lines are parallel if and only if their coefficients at the variables are proportional: , But.

As an example, consider two straight lines. We check the proportionality of the corresponding coefficients for the variables :

However, it is clear that .

And the third case, when the lines intersect:

Two lines intersect if and only if their coefficients of the variables are NOT proportional, that is, there is NOT such a value of "lambda" that the equalities are fulfilled

So, for straight lines we will compose a system:

From the first equation it follows that , and from the second equation: , hence, the system is inconsistent (no solutions). Thus, the coefficients at the variables are not proportional.

Conclusion: lines intersect

In practical problems, the solution scheme just considered can be used. By the way, it is very similar to the algorithm for checking vectors for collinearity, which we considered in the lesson. The concept of linear (non) dependence of vectors. Vector basis . But there is a more civilized package:

Example 1

Find out the relative position of the lines:

Solution based on the study of directing vectors of straight lines:

a) From the equations we find the direction vectors of the lines: .

, so the vectors are not collinear and the lines intersect.

Just in case, I will put a stone with pointers at the crossroads:

The rest jump over the stone and follow on, straight to Kashchei the Deathless =)

b) Find the direction vectors of the lines:

The lines have the same direction vector, which means they are either parallel or the same. Here the determinant is not necessary.

Obviously, the coefficients of the unknowns are proportional, while .

Let's find out if the equality is true:

Thus,

c) Find the direction vectors of the lines:

Let's calculate the determinant, composed of the coordinates of these vectors:
, therefore, the direction vectors are collinear. The lines are either parallel or coincide.

The proportionality factor "lambda" is easy to see directly from the ratio of collinear direction vectors. However, it can also be found through the coefficients of the equations themselves: .

Now let's find out if the equality is true. Both free terms are zero, so:

The resulting value satisfies this equation (any number generally satisfies it).

Thus, the lines coincide.

Answer:

Very soon you will learn (or even have already learned) to solve the considered problem verbally literally in a matter of seconds. In this regard, I see no reason to offer something for an independent solution, it is better to lay one more important brick in the geometric foundation:

How to draw a line parallel to a given one?

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

Example 2

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

Solution: Denote the unknown line by the letter . What does the condition say about it? The line passes through the point. And if the lines are parallel, then it is obvious that the directing vector of the line "ce" is also suitable for constructing the line "te".

We take out the direction vector from the equation:

Answer:

The geometry of the example looks simple:

Analytical verification consists of the following steps:

1) We check that the lines have the same direction vector (if the equation of the line is not properly simplified, then the vectors will be collinear).

2) Check if the point satisfies the resulting equation.

Analytical verification in most cases is easy to perform verbally. Look at the two equations and many of you will quickly figure out how the lines are parallel without any drawing.

Examples for self-solving today will be creative. Because you still have to compete with Baba Yaga, and she, you know, is a lover of all kinds of riddles.

Example 3

Write an equation for a line passing through a point parallel to the line if

There is a rational and not very rational way to solve. The shortest way is at the end of the lesson.

We did a little work with parallel lines and will return to them later. The case of coinciding lines is of little interest, so let's consider a problem that is well known to you from the school curriculum:

How to find the point of intersection of two lines?

If straight intersect at the point , then its coordinates are the solution systems of linear equations

How to find the point of intersection of lines? Solve the system.

Here's to you geometric meaning of a system of two linear equations with two unknowns are two intersecting (most often) straight lines on a plane.

Example 4

Find the point of intersection of lines

Solution: There are two ways to solve - graphical and analytical.

The graphical way is to simply draw the given lines and find out the point of intersection directly from the drawing:

Here is our point: . To check, you should substitute its coordinates into each equation of a straight line, they should fit both there and there. In other words, the coordinates of a point are the solution of the system . In fact, we considered a graphical way to solve systems of linear equations with two equations, two unknowns.

The graphical method, of course, is not bad, but there are noticeable disadvantages. No, the point is not that seventh graders decide this way, the point is that it will take time to make a correct and EXACT drawing. In addition, some lines are not so easy to construct, and the intersection point itself can be somewhere in the thirtieth kingdom outside the notebook sheet.

Therefore, it is more expedient to search for the intersection point by the analytical method. Let's solve the system:

To solve the system, the method of termwise addition of equations was used. To develop the relevant skills, visit the lesson How to solve a system of equations?

Answer:

The verification is trivial - the coordinates of the intersection point must satisfy each equation of the system.

Example 5

Find the point of intersection of the lines if they intersect.

This is a do-it-yourself example. The task can be conveniently divided into several stages. Analysis of the condition suggests that it is necessary:
1) Write the equation of a straight line.
2) Write the equation of a straight line.
3) Find out the relative position of the lines.
4) If the lines intersect, then find the point of intersection.

The development of an action algorithm is typical for many geometric problems, and I will repeatedly focus on this.

Full solution and answer at the end of the tutorial:

A pair of shoes has not yet been worn out, as we got to the second section of the lesson:

Perpendicular lines. The distance from a point to a line.
Angle between lines

Let's start with a typical and very important task. In the first part, we learned how to build a straight line parallel to the given one, and now the hut on chicken legs will turn 90 degrees:

How to draw a line perpendicular to a given one?

Example 6

The straight line is given by the equation . Write an equation for a perpendicular line passing through a point.

Solution: It is known by assumption that . It would be nice to find the direction vector of the straight line. Since the lines are perpendicular, the trick is simple:

From the equation we “remove” the normal vector: , which will be the directing vector of the straight line.

We compose the equation of a straight line by a point and a directing vector:

Answer:

Let's unfold the geometric sketch:

Hmmm... Orange sky, orange sea, orange camel.

Analytical verification of the solution:

1) Extract the direction vectors from the equations and with the help dot product of vectors we conclude that the lines are indeed perpendicular: .

By the way, you can use normal vectors, it's even easier.

2) Check if the point satisfies the resulting equation .

Verification, again, is easy to perform verbally.

Example 7

Find the point of intersection of perpendicular lines, if the equation is known and dot.

This is a do-it-yourself example. There are several actions in the task, so it is convenient to arrange the solution point by point.

Our exciting journey continues:

Distance from point to line

Before us is a straight strip of the river and our task is to reach it in the shortest way. There are no obstacles, and the most optimal route will be movement along the perpendicular. That is, the distance from a point to a line is the length of the perpendicular segment.

The distance in geometry is traditionally denoted by the Greek letter "ro", for example: - the distance from the point "em" to the straight line "de".

Distance from point to line is expressed by the formula

Example 8

Find the distance from a point to a line

Solution: all you need is to carefully substitute the numbers into the formula and do the calculations:

Answer:

Let's execute the drawing:

The distance found from the point to the line is exactly the length of the red segment. If you make a drawing on checkered paper on a scale of 1 unit. \u003d 1 cm (2 cells), then the distance can be measured with an ordinary ruler.

Consider another task according to the same drawing:

The task is to find the coordinates of the point , which is symmetrical to the point with respect to the line . I propose to perform the actions on your own, however, I will outline the solution algorithm with intermediate results:

1) Find a line that is perpendicular to a line.

2) Find the point of intersection of the lines: .

Both actions are discussed in detail in this lesson.

3) The point is the midpoint of the segment. We know the coordinates of the middle and one of the ends. By formulas for the coordinates of the middle of the segment find .

It will not be superfluous to check that the distance is also equal to 2.2 units.

Difficulties here may arise in calculations, but in the tower a microcalculator helps out a lot, allowing you to count ordinary fractions. Have advised many times and will recommend again.

How to find the distance between two parallel lines?

Example 9

Find the distance between two parallel lines

This is another example for an independent solution. A little hint: there are infinitely many ways to solve. Debriefing at the end of the lesson, but better try to guess for yourself, I think you managed to disperse your ingenuity well.

Angle between two lines

Whatever the corner, then the jamb:


In geometry, the angle between two straight lines is taken as the SMALLER angle, from which it automatically follows that it cannot be obtuse. In the figure, the angle indicated by the red arc is not considered to be the angle between intersecting lines. And its “green” neighbor or oppositely oriented crimson corner.

If the lines are perpendicular, then any of the 4 angles can be taken as the angle between them.

How are the angles different? Orientation. First, the direction of "scrolling" the corner is fundamentally important. Secondly, a negatively oriented angle is written with a minus sign, for example, if .

Why did I say this? It seems that you can get by with the usual concept of an angle. The fact is that in the formulas by which we will find the angles, a negative result can easily be obtained, and this should not take you by surprise. An angle with a minus sign is no worse, and has a very specific geometric meaning. In the drawing for a negative angle, it is imperative to indicate its orientation (clockwise) with an arrow.

How to find the angle between two lines? There are two working formulas:

Example 10

Find the angle between lines

Solution And Method one

Consider two straight lines given by equations in general form:

If straight not perpendicular, That oriented the angle between them can be calculated using the formula:

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

If , then the denominator of the formula vanishes, and the vectors will be orthogonal and the lines will be perpendicular. That is why a reservation was made about the non-perpendicularity of the lines in the formulation.

Based on the foregoing, the solution is conveniently formalized in two steps:

1) Calculate the scalar product of directing vectors of straight lines:
so the lines are not perpendicular.

2) We find the angle between the lines by the formula:

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

Answer:

In the answer, we indicate the exact value, as well as the approximate value (preferably both in degrees and in radians), calculated using a calculator.

Well, minus, so minus, it's okay. Here is a geometric illustration:

It is not surprising that the angle turned out to be of a negative orientation, because in the condition of the problem the first number is a straight line and the “twisting” of the angle began precisely from it.

If you really want to get a positive angle, you need to swap the straight lines, that is, take the coefficients from the second equation , and take the coefficients from the first equation . In short, you need to start with a direct .