Forward respect

1. The stereometry has geometric bodies and spacious figures, not all points of which lie in the same plane. Expansive figures are depicted on the armchair for the help of little ones, as if roaring at the eye is approximately the same as the figure itself. Qi little ones follow the singing rules, which are grounded on the geometric power of the figures.
One of the ways of depicting spacious figures on the flat will be indicated far away (§ 54-66).

ROZDIL FIRST STRAIGHT AND FLAT

I. PLANE POSITION

2. Image of the area. In everyday life, there are a lot of objects, on top of some guessing geometrical plane, to form a rectangular shape: a palette of a book, a mistake, on top of a writing table, etc. draw the shape of a parallelogram. Therefore, it is customary to depict the plane on the armchair as a parallelogram 1. Tsyu area sound means one letter, for example, "area M" (Chart 1).

1 The order of the assigned images of the area is possible and the same, like on armchairs 15-17 and іn.
(Ed. note)

3. The main characteristics of the surface. Let's say that the power of the flat, which is accepted without proof, is to be an axiom:

1) If two points of a straight line lie on a plane, then a skin point on a straight line lies on a plane.

2) If two flats make a burning point, then the stench tumbles in a straight line through that point.

3) Through whether there are three points that do not lie on one straight line, it is possible to draw a plane, and before that there is only one.

4. Heritage. From the rest of the proposition, you can enter the following:

1) Through a straight line and a point behind it, you can draw a plane (and more than one). Indeed, the point of the pose is straight at once with some two points along the straight line, three points are added, through the yak it is possible to draw a plane (and before that, one).

2) Through two straight lines that intertwine, it is possible to draw a flat (and only one). Effectively, taking the point of the crossbar and one more point on the skin straight line, or three points, through the yaki it is possible to draw a plane (and before that, one).

3) Through two parallel straight lines it is possible to draw only one plane. Indeed, parallel to the straight lines behind the appointments, lie in the same plane; This plane is one, so that through one of the parallel and as a point of another it is possible to draw no more than one plane.

5. Wrap the area around the straight line. Through the skin straight in the open space it is possible to draw an infinite area.

Really, let it be straight A (devil 2).

Let's take point A behind it. Through point A and straight A pass through a single plane (§ 4). We call it the plane M. Take a new point after the plane M. Through the point B and the straight line A to pass through the flat with his blackness. It’s called the plane N. It can sp_vpada z M, the rocks at it lie the point Y, so that the plane M lies. A pass a new surface. It’s called її R. Vaughn does not escape from M, N from N, because there is a point C in it, so it does not lie either up to the M area, or up to the N area. take away all the new and new planes that pass through the straight line A . There will be no such areas. All these planes can be seen as different positions of one and the same plane, which wraps around straight A .

We can show one more power of the plane: the plane can wrap itself around, be it straight, which lies near this plane.

6. Appointment for a visit to the open space. Mustaches, who scrambled in planimetry, victorious in the same plane behind the help of armchair tools. For pobudov near the open space, the armchair tools become already unacceptable, so it’s impossible to chair the figures near the open space. In addition, when there is a new element in the space, there is a new element - the flat, which, if it is in the space, cannot be laid on the floor with simple locks, like a straight line on the flat.

Therefore, when pobudov near the open space, it is necessary to accurately determine what it means to vikonate that chi іnsha pobudovu that, zokrema, which means to rouse the flat near the open space. At all occasions in space, we allow:

1) that the plane can be induced, as the elements are found that signify these positions in space (§ 3 and 4), so that we can induce the plane to pass through three given points, through a straight line and a point behind it, or through two two parallel lines;

2) if there are two planes that overlap, then a line of their lane is given, so that we can know the line of the lane of two planes;

3) as a plane is given in space, then we can win over in it, stay, as if we were beaten by planimetry.

Vikonati yak-nebudova pobudova in space - tse means to call yogo until the end of the day of the main appointments. With the help of these main tasks, you can untie the folding tasks.

In these speeches, there are problems with the need for stereometry.

7. The butt of a task to stay at the open space.
Manager. Find the crossing point of a given straight line A (Chart 3) from the center of the R.

Take on the plane P as a point A. Through the point A and a straight line A conductively the plane Q. She crosses the plane P along the acting straight line b . At the plane Q, we know the point 3 of the span of straight lines A і b . Tsya dot and be a shukana. How straight A і b appear parallel, then the task is not a solution.

Alignment of a straight line like a line of a peretina of two planes:

Through the skin straight in the open space to pass an impersonal area. Be-yakі from them, changing, they signify її in space. Otzhe, if there were two equal flats, which are looked at together, they are equal to the straight lines.

Vzagali be-like two not parallel planes

designate a straight line. Qi equal are called wild jealousy straight.

Alignment of a straight line that passes through two points:

Give the given points A(x 1 ;y 1) and B(x 2 ;y 2). Alignment of a straight line to pass through the points A (x 1; y 1) and B (x 2; y 2) can look like:

If given points A and B lie on a straight line, parallel to the axis O x (y 2 -y 1 \u003d 0) or the axis O y (x 2 - x 1 \u003d 0), then the alignment of the straight line will be similar to the mother looking at \u003d y 1 or x = x 1

Example 4. Lay straight lines to pass through points A(1;2) and B(-1;1).

Solution: Substituting alignment (8) x 1 =1, y 1 =2, x 2 =-1; y 2 \u003d 1
stars either 2y-4=x-1, or else x-2y+3=0

Canonically straight lines:

Let the Cartesian coordinate system be fixed on the plane Oxy. Let's set our own goals: take a straight line a, yakscho - Deyak point of a straight line a i - direct vector a.

Nehai - floating point is straight a. Then the vector is a direct vector of a straight line a and maє coordinates (if necessary, marvel at the status of the coordinates of the vector through the coordinate points). It is obvious that an impersonal point on a plane is assigned a straight line, so that a direct vector can pass through a point i can only and only if the vectors are collinear.

Let's write down the necessary and sufficient collinarity of vectors for the mind: . The rest of the equality of the coordinate form can be seen.

Yakscho i , then we can write down

Otrimane equal to mind is called canonical lines straight on the flat in a rectangular coordinate system Oxy. Rivnyannia is also called equal to the straight line at the canonical look.

Again, the canonical alignment of a straight line on a plane of mind is given by a rectangular coordinate system Oxy a straight line that passes through a point and can be a direct vector.

We will direct the butt of the canonical straight line on the flat.

For example, equal to the straight line of the canonical look. The straight line, which makes it possible to pass through the point , and - її is a direct vector. Below is a graphic illustration.

Significantly important facts:

· yakscho-straight vector straight lines and straight lines pass like through a point, so і through a point, then її canonically equal can be written like, so і;

· if it is a direct vector of a straight line, then whether any of the vectors is also a direct vector of a given straight line, then, be it equal to a straight line in the canonical view of a straight line.

Parametric straight line alignment:

Theorem. The system of straight lines is advancing with parametric straight lines:

de – coordinates of a fairly fixed point of a given straight line, – general coordinates of a fairly direct vector of a given straight line, t – parameter.

Proof. Vidpovidno up to vyznachennya evenness, whether it be multiplying the points of the coordinate space, we are responsible to bring that equals (7) satisfy all points of the straight line L i, from the other side, do not satisfy the coordinates of the point, which do not lie on a straight line.

Let's have a good point. The same vectors and є for the purpose of collinear and theorems about the collinearness of two vectors following, which are linearly expressed through the other, that. there is such a number, what. The equality of the vectors and the exactness of the coordinates:

Ch.t.d.

Back, come on point. Then, according to the theorem about the collinearity of vectors, there can be linear expressions through another, then. I want one of the equalities (7) not to win. In this order, the equals (7) are satisfied with the coordinates of less quiet points, like lying on the straight line L and only a little stench, etc.

The theorem has been completed.

Normal alignment of the area:

IN vector form flatness of the area may look

Likewise, the normal vector of the area is single,

even the flatness of the area can be recorded as

(normal flatness).

– move from the cob of coordinates to the plane, , , – direct cosine of the normal

de - cut between the normal of the plane and the coordinate axes are vodpovidno.

The vertical plane of the plane (8) can be brought to normal form by multiplication by the normalizing factor, the sign in front of the fraction is opposite to the sign of the free term (8).

Vіdstan vіd point to the plane(8) be behind the formula, taken by substituting the point in normal alignment

The deep flatness of the plane, after the deep flatness of the flat:

As for the trivial space, a rectangular coordinate system is given Oxyz, then the equal planes in the trivi-world coordinate system are called the same equal to the triple x, yі z, I am satisfied with the coordinates of all points of the plane and are not satisfied with the coordinates of any other points. In other words, when substantiating the coordinates of the first point of the plane, the equalness of the plane is taken away, and when substituting the equal plane of the coordinates, whether it is the other point, the equalness is incorrect.

First of all, write down the central plane of the plane, guessing the straight line perpendicular to the plane: the straight line is perpendicular to the plane, as if it is perpendicular to the straight line that lies at this plane. From which designation it is clear whether there is any normal vector of the plane of perpendiculars to any non-zero vector that lies near this plane. This fact mimics the proof of the attacking theorem, as it sets the appearance of the wild flatness of the area.

Theorem.

Be like equal to the mind, de A, B, Cі D- Deyakі dіysnі numbers, moreover A, INі C not equal to zero at once, designating the area in a given rectangular coordinate system Oxyz near the trivial space, and whether it be a plane near a rectangular coordinate system Oxyz in a trivial space they are equal to the mind with a certain set of numbers A, B, Cі D.

Proof.

Like Bachite, the theorem is composed of two parts. At the first part, we have been given a level and we need to bring it to the surface. At the other part, we were given a deuce of flatness and it is necessary to bring what we can assign to equals with a simple choice of numbers A, IN, Zі D.

Let's just confirm the first part of the theorem.

Oskіlki numbers A, INі Z overnight not equal to zero, then є point , the coordinates of which are satisfied with the equivalence, so equality is fair. Vіdnіmemo lvu and right part of the otrimanoї rivnostі vіdpovіdno vіd lіvoї and pravaі ї parts іvnyannja, tsomu otmáєmo ravnyannja vіdnіvnіnya vіdіvalentno vihіdnomu іvnyannju. Now, as we know, that equalizes the plane, then it will be brought, that is equivalent to it, equal also marks the plane of the given rectangular coordinate system for the trivi- merous space.

Equity is the necessary and sufficient mental perpendicularity of the vectors and . In other words, the coordinates of a floating point are satisfied evenly, and only once, if the vectors are perpendicular. Then, violating the fact, inductions before the theorem, we can confirm that equality is true, then an impersonal point defines a plane, a normal vector like є, moreover, this plane passes through a point. In other words, the alignment is indicative of a rectangular coordinate system Oxyz near the trivimir expanse, a larger area is assigned. Otzhe, equivalently, equalizes the area itself. The first part of the theorem has been completed.

Let's proceed to the confirmation of the other part.

Let us be given a plane to pass through a point, with a normal vector, which is є . Let us know that a rectangular coordinate system Oxyzїї setting the level of the mind.

For which we take a sufficient point of the plane. Let me be the point. Then the vectors i will be perpendicular, then, їх scalar twіr will be equal to zero: . Accepting, looking forward to seeing. Tse equal and signify our area. Again, the theorem is confirmed again. (for the first values ​​of the numbers A, IN, Zі D);

Let's aim a butt to illustrate the rest of the phrase.

Marvel at the little ones from the images of the area near the trivial expanse near the fixed rectangular coordinate system Oxyz. Tsіy ploshchinі vіdpovіdaє rіvnyannya, to that scho you are pleased with the coordinates of any point of the plaza. On the other side, the alignment is determined by the given coordinate system Oxyz an impersonal point, the image of which is a small flat.

Flatness of the area at the windrows:

Let the trivial space have a right-angled coordinate system Oxyz.

For a rectangular coordinate system Oxyz in a trivial expanse equal to the mind, de a, bі c– in the form of zero the current number is called equal to the area at the windbreaks. Such a name is not vipadkova. Absolute values ​​of numbers a, bі c equal to vіdrіzkіv vіdzhina, yakі vіdsіkaє the plane on the coordinate axes Ox, Ohі Oz vіdpovіdno, rahuyuchi vіd cob of coordinates. Number sign a, bі c shows, in a straight line (positive and negative) there are brackets on the coordinate axes. Definitely, the coordinate points satisfy the plane of the winds:

Look at the little ones, which explains the moment.

The level of the plane passing through the point is perpendicular to the vector: Let the trivial space have a rectangular Cartesian coordinate system. We formulate the following task:

Fold flat planes to pass through this point
M(x 0 , y 0 , z 0) perpendicular to the given vector →n = {A, B, C} .

Solution. Come on P(x, y, z) - enough point to space. Dot, mottled P overlie flat todі і less todі, if the vector
MP = {x − x 0 , y − y 0 , z − z 0 ) orthogonal vector → n = {A, B, C) (Fig. 1).

Having written the mental orthogonality of these vectors (→ n, MP) = 0 for the coordinate form, optional.

AREA.

Appointment. Any non-zero vector, perpendicular to the plane, is called її normal vector, and is indicated.

Appointment. Equal to the surface of the mind de coefficient - sufficient effective numbers, not equal to zero at the same time, is called zagalnym flats of the area.

Theorem. Leveling determines the area that the normal vector can pass through the point.

Appointment. Rivnyannia mind

de - sufficient, not equal to zero, actual numbers, called equal to the area at the windbreaks.

Theorem. Come on - the flatness of the flat at the windbreaks. Todi - coordinate point її crossbar with coordinate axes.

Appointment. The deep flatness of the area is called rationing or normal equal to the area, like

that .

Theorem. Normally, the alignment of the plane can be written in the form of - in the column of coordinates to the given plane, - the direct cosine of the normal vector ).

Appointment. Normalizing multiplier the number of the flat area is called de sign is chosen by the opposite sign of the free member D.

Theorem. Come on - a multiplier that normalizes, a wild flatness of the area. Todi rivnyannya є rationing rivnyannyam given area.

Theorem. Vіdstan d type of specks up to the flat .

Mutually rotashuvannya two flats.

Two planes either run, or they are parallel, or they intertwine with a straight line.

Theorem. Let the superficial assignments be overhead: . Todi:

1) yakscho then the flats zbіgayutsya;

2) yakscho then the planes are parallel;

3) otherwise the planes are tinted along straight lines, equal to which the equal system serves: .

Theorem. Come on - normal vectors of two planes, then one of the two cuts between given planes is more:.

Last. Come on ,- Normal vectors of two given areas. As a scalar addition, given areas are perpendicular.

Theorem. Give the given coordinates of three different points of the coordinate space:

Todi river –є equal planes that pass through qi three points.

Theorem. Let the data of the gale of the two flats, which overlap: moreover. Todi:

– leveling of the bi-sectoral area of ​​the gostry duhedral kut, studded with peratin of these flats;

– alignment of the bi-sectoral area of ​​a blunt dihedral cut.

Zv'yazuvannya that beam of flats.

Appointment. Zv'yazuvannyam flats the impersonality of all planes is called, that one can see one bright point, as it is called linkage center.

Theorem. Let's go - three flats that make a single bright point equalization of zv'yazuvannya of flats.

Theorem. Rivnyannya, de dovilnі deisnі parameters, not equal to zero at the same time, є equal to the linking of the planes with the center of the linking at the point

Theorem. Let me give you the data of the glacial level of three flats:

-їх vidpovіdnі normal vectors. In order for three given planes to overlap into a single point, it is necessary and sufficient, so that the difference between the two normal vectors does not reach zero:

In this way, the coordinates of the single central point are the single solutions of the equalization system:

Appointment. A bunch of flats the impersonal planes are called, which are intertwined along the same straight line, the title of the entire beam.

Theorem. Let two flats that intertwine in a straight line. Todі vnyannja, de dovіlnі dіisnі parameters at once not equal to zero, є alignment of a beam of planes from the top of the beam

DIRECT.

Appointment. Whether it be a non-zero vector, a collinear given straight line is called її direct vector, and is indicated

Theorem. parametric straight lines in space: de coordinates of a fairly fixed point of a given straight line, and general coordinates of a fairly direct vector of a given straight line parameter.

Last. The system of equals is advancing, equals straight in the open space and is called canonical equals straight in space: de - coordinates of a fairly fixed point of a given straight line, - general coordinates of a fairly direct vector of a given straight line.

Appointment. Canonical equivalent of direct view - called canonical alignments of straight lines that pass through two different given points

Mutually roztashuvannya two straight lines in the open space.

It is possible to have 4 slopes of rotting two straight lines near the open space. They can straighten up, be parallel, cross at one point or cross.

Theorem. Let me give the canonical equalization of two straight lines:

de - їх straight vectors, - enough fixed points that lie on straight lines. Todi:

і ;

and do not win if only one of the equalities

;

, then.

4) directly cross, like , then.

Theorem. Come on

– two rather straight lines near the open space, set by parametric alignments. Todi:

1) how the system is equal

if there is a single solution, then they are directly intertwined in one point;

2) if the system is equal there is no solution, then it is directly crossed in parallel.

3) if the system is equal to more than one rozvyazku, then straight zbіgayutsya.

Stand between two straight lines in the open space.

Theorem.(Formula between two parallel lines.): Move between two parallel lines

De - їх overhead direct vector, - points of ціх straight lines, can be calculated by the formula:

or

Theorem.(Formula between two straight lines to cross.): Stand between two straight lines to cross.

can be calculated using this formula:

de – module of mixed creation of direct vectors і і vector, module of vector creation of direct vectors.

Theorem. Come on - the alignment of two flats that overlap. Then comes the system of alignment and alignment of straight lines, which are intertwined with planes: . Direct vector can serve as a vector , de ,- Normal vectors of given areas.

Theorem. Let it be given a canonical straight line: de . Then the system of equals comes along, and the equals are given straight lines, given by the span of two planes: .

Theorem. Alignment of a perpendicular dropped from a point straight may view de - coordinates of the vector creation, - coordinates of the direct vector given to the straight line. The length of the perpendicular can be known by the formula:

Theorem. Alignment of the perpendicular perpendicular of two straight lines, which can be crossed, can be seen: de.

Mutually roztashuvannya straight lines and flats near the open space.

There are three possible ways of mutual expansion of a straight line near the expanse of that area:

Theorem. Let the plane be given to the straight lines, and the straight line is given to the canonical or parametric lines abo, de vector is the normal vector of the area – coordinates of a fairly fixed point of a straight line, – general coordinates of a fairly direct vector of a straight line. Todi:

1) yakscho, then we directly cross the plane of the point, the coordinates of which can be known from the equalization system

2) if i, then lie straight on the flat;

3) if i, then the line is parallel to the plane.

Last. If the system (*) has a single solution, then it is directly overflowing from the flat; if the system (*) has no solution, then it is straight parallel to the plane; if the system (*) can be an impersonal decision, then it is straight to lie on the plane.

Virishennya typical tasks.

manager №1 :

Fold flat planes to pass through the point parallel to vectors.

We know the normal vector of the area:

= =

As a normal vector of the area, you can take the vector of the same globally equal area in the future when you look:

In order to know, it is necessary to replace the coordinates of the points with which the plane lies.

manager №2 :

Two faces of the cube lie on the planes and calculate the total number of the cube.

Obviously, the planes are parallel. Dovzhina edge of the cube є vіdstan mіzh flats. Vibero on the first plane to a point: let's not know.

We know how to walk between planes, how to walk from a point to another plane:

Otzhe, the volume of the cube is good ()

manager №3 :

Know the cut between the faces and piramid peaks

Cut between planes – ce cut between normal vectors up to these planes. We know the normal vector of the area: [,];

, or

Similarly

manager №4 :

Lay canonically equal straight lines .

Otzhe,

The vector is perpendicular to the straight line, to

Otzhe, canonically equal, I will look straight ahead.

manager №5 :

Know the difference between straight lines

і .

Directly parallel, because їх direct vectors and ірівні. Come on point lie on the first line, and the point lies on the other line. We know the area of ​​the parallelogram based on vectors.

[,];

Shukanoi vіdstannyu є the height of the parallelogram, omitted from the points:

manager №6 :

Calculate the shortest distance between straight lines:

It will be shown that it is straight to cross, tobto. vectors, and not lie on the same plane: ≠ 0.

1 way:

Through another straight line we draw a plane parallel to the first straight line. For shukano's area v_domі tі, scho to lie їй vectorії. Normal area vector є vector tvir vectorіv, .

Also, as a normal vector of the area, you can take the vector of that area’s alignment in the future, if you know that the point to lie on the area can be found and write down the alignment:

Shukana v_dstan - tsya vіdstan from the point of the first straight line to the plane is known by the formula:

13.

2 way:

On the vectors i we will create a parallelepiped.

Shukana vіdstan' – the height of the parallelepiped, omitted from the points on the yogo basis, based on vectors.

Result: 13 singles.

manager №7 :

Know the projection of a point onto a plane

The normal vector of the area is the direct vector of the straight line:

We know the crossing point of the straight line

that area:

.

Substituting in a flat plane, we know, and then

Respect. In order to know a point that is symmetrical to a point similar to the plane, it is necessary (similarly to the forward tasks) to know the projection of the point onto the plane, then look at the vіdіzok with the vіdomimikobkami middle, squirming with formulas,,.

manager №8 :

Find the alignment of a perpendicular dropped from a point on a straight line .

1 way:

2 way:

Tasks are written in a different way:

The area is perpendicular to the given line, so the direct vector of the line is the normal vector of the area. Knowing the normal vector of the plane and a point on the plane, we write її equal:

We know the cross point of the plane and the straight line, written parametrically:

,

Let's make a straight line to pass through the points i:

.

Suggestion: .

In the same way, you can virishity and the same task:

manager №9 :

Find a point that is symmetrical to a point like a straight line .

manager №10 :

Danish tricot with tops Know the level of height, lowered from the top to the back.

The heading is absolutely analogous to the previous tasks.

Suggestion: .

manager №11 :

Designate the alignment of the perpendicular perpendicular to two straight lines: .

0.

Vrakhovuchi, scho pass through a point, we write down the alignment of the plane:

The point is to lay down, I will see that the equal of the area will look:.

Suggestion:

manager №12 :

Fold straight lines to pass through a point and cross straight lines .

The first straight line to pass through the point that can be a direct vector; other - to pass through points and may direct vector

It is shown that qi lines are such that they can be crossed, for which we fold the arbitrator, the rows of which are the coordinates of vectors ,, ,vectors do not overlap to the same plane.

Let's draw a plane through the specks and move straight ahead:

Come on - a sufficient point of the plane of the same vectors, and complanar. Flatness of the area may look:.

Similarly, we can fold the flatness of the plane, which can pass through the specks and the other straight: 0.

Shukana is straight є span of flats, tobto.

The illuminated result of the aftermath given by those is the formation of components, statements at the entrance, the totality of competencies (nobility, mind, power) on two levels: threshold and prosunuty. The threshold rіven gives the assessment “presumably”, the sticking rіven gives the assessments “good” or “remarkable”, fallow in the results of the case-task.

For self-diagnostics of these components, you will be shown the next steps.

INSTUP

Chapter 1

1 Cross point of a straight line with a plane

1 Variations in the position of the straight line in space

2 Kut mizh straight and flat

WISNOVOK

LIST OF DZHEREL VICTORIES

INSTUP

Be-yaké equalization of the first stage of coordinates x, y, z

By + Cz + D = 0

sets the area and now: be it the area can be represented by equals, as it is called the equals of the area.

The vector n (A, B, C), orthogonal to the plane, is called the normal vector of the plane. The equal coefficients A, B, C do not equal 0 at the same time.

D = 0, Ax+By+Cz = 0 – the plane passes through the cob of coordinates.

C \u003d 0, Ax + By + D \u003d 0 - the plane is parallel to the Oz axis.

C = D = 0, Ax + By = 0 – the area to pass through all Oz.

B = C = 0, Ax + D = 0 – the plane is parallel to the Oyz plane.

Alignment of coordinate planes: x=0, y=0, z=0.

A straight line in space can be given:

) as a line to cross two planes, tobto. rivnyan system:

A 1 x+B 1 y+C 1 z+D 1= 0, A 2 x+B 2 y+C 2z + D 2 = 0;

) with its two points M 1(x 1,y 1, z 1) and M 2(x 2,y 2, z 2), even if it is straight, what to pass through them, is given by equals:

=;

) point M 1(x 1,y 1, z 1), which ї th lie, that vector a (m, n, р), ї th is collinear. Todi is directly attributed to equals:

Equations are called canonical straight lines.

The vector a is called the direct vector of the straight line.

The parametric alignment of the straight line is taken away, equating the skin from the eye to the parameter t:

X 1+mt, y = y 1+ nt, z = z1 + Pt.

Razv'azyuchi system like a system of linear alignments, where unknown x and y, come to straight lines in projections or to pointing straight lines:

Mz + a, y = nz + b

You can go to the canonical ranks, knowing the z from the dermal rank and adding the value:

In the upper levels (3.2), one can pass to the canonical one in another way, in order to know whether the line point of the straight line and the direct vector n = , de n 1(A 1, B 1, C 1) and n 2(A 2, B 2, C 2) are normal vectors of given areas. If one of the signs m, n and r in equals (3.4) is equal to zero, then the number of the double fraction must be set equal to zero, that is. system

equal system ; such a line is perpendicular to the axis Ox.

System system is equally strong x = x 1,y=y 1; straight line parallel to the Oz axis.

Purpose of course work:straight up that flat area by the open space.

Head of course work:look at the area near the open space, її equal, and look at the flat near the open space.

The structure of the course work:entry, 2 chapters, visnovok, list of vikoristanih dzherel.

Chapter 1

.1 Crossing point of a straight line with a plane

Let the area Q be given to the curved type: Ax+By+Cz+D=0, and the line L to the parametric type: x=x 1+mt, y=y 1+nt, z=z 1+pt, otherwise, to know the cross point of the line L and the plane Q, it is necessary to know the value of the parameter t, for which the point of the line lies on the plane. By substituting the values ​​of x, y, z, the plane is equal, and by deriving t, we subtract

The value of t will be the same, since the plane is not straight and parallel.

Wash the parallelism and perpendicularity of the straight line and the plane

Looking directly at L:

and flatness?

Straight line L and plane? :

a) perpendicular to one to one or less to one, if the direct vector is straight and normal vector collinear planes, tobto.

b) parallel to one to the same and less to the same, if the vectors і perpendicular, that is.

i Am + Bn + Ср = 0.

.2 Kut mizh straight and flat

Kut ?between the normal vector of the area i by a direct vector be calculated according to the following formula:

Beam of flats

The totality of all planes that pass through a given straight line L is called a bundle of planes, and the straight line L is called the entire bundle. Let the whole beam is given by equals

We multiply the other system's rank by term by term and we store it with the first ranks:

A 1x+B 1y+C 1z+D 1+ ?(A 2x+B 2y+C2 z+D 2)=0.

Tse equal may the first step should be x, y, z i, then, for any numerical value ?define the area. So, as a given equalization is the last of two equals, then the coordinates of the point, which are satisfied with these equals, are satisfied with this equal. Father, for whatever numerical value ?given the alignment of the planes that pass through the given straight line. Otrimane rivnyannia є alignment of a beam of planes.

butt.Write the plane plane passing through the point M 1(2, -3, 4) parallel to lines

Solution.We write down the alignment of the connection of the planes that pass through the point M1 :

A (x - 2) + B (y + 3) + C (z - 4) = 0.

Since the plane is needed but is parallel to these lines, then the normal vector is due to both perpendicular to the lines. tsikh straight lines. Therefore, as a vector N, you can take a vector tv_r vector_v:

Also, A \u003d 4, B \u003d 30, C \u003d - 8. Substituting the known values ​​\u200b\u200bof A, B, Z

4(x-2)+30(y + 3) -8(z-4) = 0 or 2x + 15y - 4z + 57 = 0.

butt.Find the point of the line that area 2x + 3y-2z + 2 = 0.

Solution.Let's write down the alignment of this straight line with the parametric view:

Let's imagine qi vrazi for x, y, z equalization of the plane:

(2t+1)+3(3t-1)-2(2t+5)+2=0 Þ t=1.

Imagine t = 1 parametric alignment of the straight line. Take away

Also, the straight line intersects at the point M(3, 2, 7).

butt.Know kut ?between a straight line that area is 4x-2y-2z+7=0. Solution.Let us fix the formula (3.20). so yak

That

Father, = 30°.

The straight line in the open space is not narrow, so you can ask it more easily with the help of a friend. From the school course of Euclidean geometry, there is an axiom, “through two points in the space you can draw a straight line i, before that, only one.” Also, on the diagram, the straight line can be given by two frontal and two horizontal projections of points. But if it’s a straight line - it’s a straight line (and not a curve), then with a full basis we can combine points in a straight line and take a frontal and horizontal projection of a straight line (Fig. 13).

The proof is inverted: in the planes of the projections V and H, two projections a "b" and ab are given (Fig. 14). We draw through them the plane, perpendicular to the planes of the projections V and H (Fig. 14), the line of the peretina of the planes will be straight line AB.

.1 Different slopes

At the slopes we looked at, the straight lines were neither parallel, nor perpendicular to the projection planes V, H, W. The stench can be vishіdnimi or low (rozіbratisya independently).

On fig. 17 shows the straight line of the stave, set by three projections. Let's look at the family of straight lines, which may be important authorities - straight lines, be parallel to the projection plane.

On fig. 17 shows the straight line of the stave, set by three projections.

Let's look at the family of straight lines, which may be important authorities - straight lines, be parallel to the projection plane.

a) Horizontal straight line (nakshe - horizontal, straight horizontal line). This is the name of the straight line, parallel to the horizontal plane of the projections. Її the image near the space on the plot is shown in fig. 18.

The horizontal is easy to recognize on the plot "in the guise": її frontal projection is always parallel to the axis ОХ. As a whole, the most important horizontal power is formulated as follows:

At the horizontal - the frontal projection is parallel to the axis ОХ, and the horizontal projection is natural size. Preferably a horizontal projection of the horizontal on the diagram allows you to designate the cut її to the plane V (cut b) and to the plane W (y) - fig.18.

b) Frontal straight line (frontal, straight line of the frontal alignment) - not straight, parallel to the frontal plane of the projections. We are not illustrative of the actual images, but are shown by the epures (Fig. 19).

The frontal diagram is characteristic, which is horizontal and profile projections parallel to the X and Z axes, and the frontal projection expands fairly and shows the natural size of the frontal. Preferably on the diagram, cut straight ahead to horizontal (a) and profile (flat) projections. Otzhe, one more time:

At the frontal, the horizontal projection is parallel to the axis ОХ, and the frontal projection is natural size

c) Profile straight line. Obviously, it is straight, parallel to the profile plane of the projections (Fig. 20). It is also obvious that the natural value of the profile straight line є on the profile plane of the projections (projection a "b" - Fig. 20) and here you can bachiti kuti її nahilu to the planes H (a) and V (b).

A family of straight lines is coming, wanting and important laying, like straight lines - not projecting straight lines.

Straight lines, perpendicular to the planes of projections, are called projecting (by analogy with projecting changes - Fig. 21).

AV sq. H - straight horizontally projecting; sq. V - straight frontally projecting; sq. W - straight profile-projecting.

2.2 Kut mizh straight and flat

flat square cut trikutnik

Rectangular tricout method

Straight zagalnogo camp, as we said, heeled to the planes of the projections under a kind of full kut.

The cut between the straight line and that plane is projected by the cut, we add the straight line to that projection on the plane (Fig. 22). Kut a vyznaє kut nakhily vіdrіzka AB to pl. H. W fig. 22: Ab1 | 1pl. H; Bb1 = Bb – Aa = Z 22

In a straight-cut tricot ABb1, the leg Ab1 has a normal horizontal projection ab; and the other leg Bb1 is the most expensive retail point A and B in the square. N. Since the points on the horizontal projection of the straight line ab are drawn perpendicular and put on the new value Z, then, having taken the point a with the taken point b0, we take the hypotenuse ab0, equal to the natural value of the AB. On the diagram it looks like this (mal. 23):

In a similar way, the straight line extends to the frontal plane of the projections (b) - fig. 24.

To give respect: in case of pobudov on the horizontal direct projection, we add to the additional direct value Z; when on the anterior projection - the value of Y.

The look-ahead method is called a straight-cut tricutnik. With the help of yoga, you can determine the natural size of any kind of crack that cries us, as well as cut yoga sickly to the planes of the projections.

Mutually becoming straight

Previously, we looked at the nutritional value of a point of a straight line: if a point lies on a straight line, then projections lie on one-dimensional projections of a straight line (the rule of belonging, div. Fig. 14). From the high school course of geometry, it is possible to guess: two straight lines intertwine in one point (otherwise: if two straight lines make one double point, then the stinks intertwine in the other point).

The projections of the straight lines, which are intertwined, on the diagram may have a clearly pronounced sign: the projections of the tread point lie on the same line of the link (Fig. 25). It is clear: point K lie down і AB and CD; on the plot, point k lies on the same line linking with point k.

Straight AB and CD - reshape

Coming from the possible mutual roztashuvannyah two straight lines in the open - straight crossed. It is possible to have a fall, if the straight lines are not parallel, but they do not overlap. Such straight lines can be laid at two parallel planes (Fig. 26). It doesn't even mean that two are straight, that they cross, lie ob'yazkovo at two parallel planes; and even less those that two parallel planes can be drawn through them.

The projections of two lines that intersect can overlap, but the points of their overlap do not lie on the same line of the link (Fig. 27).

It is important to see the nutrition of competing points (Fig. 27). On the horizontal projection there are two dots (e, f), but in the frontal stench they turn into one (e "f"), moreover, it is unreasonable, as the point is visible, as it is not visible (competing points).

Two points, the frontal projections of which are collapsing, are called frontal-competing.

We saw such a twist earlier (Fig. 11), but by those “mutually stowing two points”. Therefore, the rule is stagnant:

From two competing points, the one that is visible is the one whose coordinate is larger.

3 fig. 27 it can be seen that the horizontal projection of the point E (e) is far from the OX axis, the lower point is f. Again, the Y coordinate of the point "e" is larger, lower at the point f; later, point E will be visible. On the front projection, point f "is placed in the arches as invisible.

One more thing: the point e lies on the projection of the straight line ab, and tse means that on the frontal projection the straight line a "b" is drawn "on top" of the straight line c "d".

Parallel lines

Parallel lines on the plot are easy to recognize in appearance, but one-dimensional projections of two parallel lines are parallel.

To give respect: the same! Tobto. frontal projections are parallel between themselves, and horizontal projections are between themselves (Fig. 29).

Proof: in Figure 28, two parallel lines AB and CD are given in space. Let us draw through them the projecting planes Q and T - they will appear parallel (for as two straight lines that overlap, one plane parallel to two overlaps with a straight line, the other plane, then such planes are parallel).

On diagram 30a, the tasks are parallel to straight lines, on diagram 30b, straight lines are crossed, although in that, and in a different direction, the frontal and horizontal projections are mutually parallel.

I use, however, a trick, for the help of which it is possible to assign mutually the positions of two profile straight lines, without going into the third projections. For which it is enough to have two projections with additional lines, as shown in Fig. 30. It will appear that the points of intersection of these lines lie on the same line of connection - the profile lines are parallel to each other - fig. 30a. Yakshcho nі - profile straight lines cross (Fig. 306).

Features of the fall of the straight line:

Projections of a direct cut

As if two straight lines of the stave are tucked under a straight cut, their projections make a cut not equal to 90° (Fig. 31).

Shards at the crossbar of two parallel planes of the third in the crossbar appear parallel to straight lines, then the horizontal projections ab and cd are parallel.

To repeat the operation and project straight lines AB and CD onto the frontal plane of the projections, we will take the same result.

A special slope is two profile straight lines, set by frontal and horizontal projections (Fig. 30). As it was said, in the profile lines, the frontal and horizontal projections are mutually parallel, prote, for this sign it is impossible to judge the parallelism of the two profile lines without inducing the third projection.

Manager. Try a right-angled tricut ABC with leg BC lying on straight MN (Fig. 34).

Solution. It can be seen from the diagram that the straight line MN is the horizontal. And behind the mind, the tricutnik is straight-cut.

The speed of the power of the projection of the direct kut is omitted from the point "a" perpendicular to the projection mn (on the square H, our direct kuta is projected without creation) - fig. 35.

As an additional straight line, which is to be carried out from the end of the cut under the direct cut to this point, we win the part of the horizontal projection of the straight line, and bm itself (Fig. 36). Let's look at the value of the difference in the Z coordinates, taken from the front projection, and take the point “a” from the end of the removed windrow. We take the actual size of the leg AB (ab ; ab).

Figures 31 and 32 show two straight lines with a corner position, which make 90 ° cut between them (in Fig. 32, straight lines lie in the same plane P). Yak bachimo, on diagrams of kut, projections of straight lines, not up to 90 °.

Let's cherish the power of the world by looking at the projection of the direct kuta from the offensive cause:

Since one side of the straight kuta is parallel to the projection plane, then the straight kuta is projected onto this plane without a hitch (Fig. 33).

We do not lead to the same position (produce it independently), but we can look at the odds, as if you can follow this rule.

Nasampered, it is significant that behind the mind one side of the direct kuta is parallel to whether it be the projection plane, then, one side will be either frontal, or horizontal (maybe a profile straight line) - fig. 33.

And the frontal and horizontal on the diagram are easy to recognize “in disguise” (one of the projections is parallel to the axis ОХ), or you can easily induce it as needed. In addition, the frontal line has the most important power: one of the projections of the obov's language looks like

According to the rule of moisture, we know the frontal projection of the point b "behind the auxiliary line link. We have the leg AB (a" b "; ab).

In order to place the leg BC on the side MN, on the back it is necessary to designate the natural size of the AB arm (a d ; ab). For which one is fast, we already have the rule of a straight-cut tricutnik.

WISNOVOK

Zagalni rіvnyannya straight ahead

The alignment of the straight line can be seen as the alignment of the line of the peretina of two planes. As it was seen more, the area of ​​the vector form can be set equal:

× + D = 0, de

Normal surface; - radius - the vector of a small point of the plane.

Let the space set two planes: × + D 1= 0 and × + D 2= 0, normal vectors and coordinates: (A 1, B 1, C 1), (A 2, B 2, C 2); (x, y, z). Similar curves for straight lines in vector form:

Zagalnі vnyannya straight line in the coordinate form:

For which you need to know the full point of the straight line of the numbers m, n, p. If so, a direct vector can be known as a vectorial extension of a vector in the normal to given planes.

Flatness of the area near the space

Send data points and non-zero vector (tobto , de

wash away is the normal vector.

Yakscho , , , ..., then equal can be changed to look . Numbers , і , і

Come on - as a point of the plane, - Vector perpendicular to the plane. Todi river є equalization of the surface.

Coefficient , ; near the equal area є coordinates of the vector perpendicular to the plane.

How to divide the plane by a number that equals the length of the vector , then we take away the flatness of the area of ​​the normal form.

Flatness of the plane, like passing through a point i is perpendicular to a non-zero vector, .

Be-yak equal to the first step sets the coordinate space to a single plane, which is perpendicular to the vector with coordinates .

Rivnyannia є equal to the plane that passes through the point i perpendicular to a nonzero vector.

Skin area set in a system of rectangular coordinates , , equal to the mind.

mind you, what are the average coefficients , , є non-zero, sets the space for the area of ​​the system of rectangular coordinates. The area near the space is set in the system of rectangular coordinates , , equal to mind , mind you, sho.

Correct is that return of firmness: equal to the mind wash away set the space for the system of rectangular coordinates.

De , , , , ,

The area near the space is assigned to equals , de , , , - decimal numbers, moreover , , do not equal 0 and set the coordinates of the vector at once , perpendicular to this plane and called the normal vector.

Send data points and non-zero vector (tobto ). Same vector flat area , de - enough point of the plane) looks like - alignment of the area behind the point and the normal vector.

Skin leveling of the first stage wash away put in a rectangular coordinate system a single plane, for which the vector is the normal vector.

Yakscho , , , , then equal can be changed to look . Numbers , і rivnі vіdzhina vіdrіzkіv, yakі vіdsіkayut flat on the axes , і obviously. To that equal called equal to the area "at the winds".

LIST OF DZHEREL VICTORIES

1.Stereometry. Geometry in space. Aleksandrov A.D., Werner A.L., Rizhik V.I.

2.Alexandrov P. S. Course of Analytical Geometry and Linear Algebra. - Head edition of physical and mathematical literature, 2000. - 512 p.

.Beklemishev D.V. Course of Analytical Geometry and Linear Algebra, 2005. – 304 p.

.Ilyin V. A., Poznyak E. G. Analytical geometry: Navch. for universities. - 7th view., Sr., 2004. - 224 p. - (Course of advanced mathematics and mathematical physics.)

.Efimov N. V. A short course in analytical geometry: Navch. help. - 13th view., stereo. –, 2005. – 240 p.

.Kanatnikov O.M., Krishchenko O.P. Analytical geometry. -2nd view. -, 2000, 388 s (Ser. Mathematics at the Technical University

.Kadomtsev SB. Analytic geometry and linear algebra, 2003. – 160 p.

.Fedorchuk V. V. A course of analytical geometry and linear algebra: Navch. posibnik, 2000. - 328 p.

.Analytical Geometry (lecture notes by Y.V. Troitsky, 1st year, 1999/2000) – 118 p.

.Bortakovsky, A.S. Analytical geometry in applications and tasks: Navch. Posibnik/O.S. Bortakovsky, A.V. Panteliev. - Vishch. school, 2005. - 496 s: il. - (Series "Applied Mathematics").

.Morozova E.A., Sklyarenko E.G. Analytical geometry. Methodological guide 2004. - 103 p.

.Methodical instructions and working program for the course "Vishcha Mathematics" - 55 p.

Two are straight at the parallel expanse, as if the stench of lying in one flat does not overlap.

Two straight lines cross in space, as if there is no such area, in which stink to lie.

Signs to cross straight lines. If one of the two straight lines lies at the deakіy and lagnosti, and another straight line crosses the plane at a point, if the first straight line does not overlap, then the straight lines cross.

The plane is straight, so that the plane, parallel, do not overlap, so that the stench does not muffle the sleeping points.

The sign of parallelism of the straight line and the plane. If it’s straight, if it doesn’t overlap the plane, if it’s parallel, if it’s straight, if it overlaps the plane, it’s parallel to the plane.

Power of the plane and the straight line, parallel to the plane:

1) if the plane is to be moved straight, parallel to the other plane, and if it crosses the plane, then the line of the planes is parallel to this straight line;

2) if through the skin from two parallel straight lines, the planes that overlap, then the line of their line is parallel to these straight lines.

Two planes are parallel, as if the stench could not be sleeping points.

Signs of parallelism of planes, as if two straight planes of the same plane, which overlap, seem to be parallel to two straight planes, then two planes are parallel.

The straight line is perpendicular to the plane, as if it were perpendicular to the straight line, so that the plane lies.

Signs of perpendicularity of a straight line and a plane: if a straight line is perpendicular to two straight lines that overlap, lie near the plane, then it is perpendicular to the plane.

The dominance of a straight line, perpendicular to the plane.

1) if one of the two parallel lines is perpendicular to the plane, then the other line is perpendicular to the center of the plane;

2) straight, perpendicular to one of two parallel planes, perpendicular to the other plane.

Sign of perpendicularity of planes. If the plane is to move perpendicular to the other plane, it is perpendicular to that plane.

A straight line that crosses the plane, but is not perpendicular to it, is called a frailty to the plane.

Theorem about three perpendiculars. In order for it to be straight, which lies near the flat, the bula is perpendicular to the sickness, it is necessary and sufficient, so that it is perpendicular to the projection of the sickly on the flat.

On the baby 1 straight b- khila to the flat, straight c- projection A┴ h, That a┴ b

Kutom between the frailty and flatness is called the cut between the frailty and the projection onto the flat. On the little one 2 straight b- pokhila to the flat, straight a- projection of the chiloi on the flat, α - cut between the chilo and the flat.

The two-faced kutvoryutsya in the past the peretina of two planes. The straight line, cut off by the end of the span of two planes, is called the edge of the two-sided cut. Two pіvploschini іz zagalny rib are called the faces of the two-faced kut.

Napіvnі the area, between which zbіgaєtsya with the edge of the two-faced kuta and how to divide the two-sided kut into two equal kuti, is called a bi-sector flat.

The two-faced cut is reduced to a similar linear cut. The linear cut of a two-sided cut is called a cut between perpendiculars drawn from the skin face to the edge.

Prism

Bagatohedron, two sides of a river n- kosinci, which lie near parallel flats, and reshta n faces - parallelograms, called n- Vugіlnoy prism.

Two n- kosintsya є podstavami prism, parallelograms - bіchnymi faces. The sides of the faces are called the edges of the prism, and the ends of the edges are called the tops of the prism.

The height of the prism is called the height of the perpendicular, the arrangements between the bases of the prism.

The diagonal of the prism is called a cross, which connects two vertices of the bases, which does not lie on the same face.

A straight prism is a prism, the ribs of which are perpendicular to the planes of the bases (Fig. 3).

A frail prism is called a prism, the ribs of which are frail to the flatness of the foundations (Fig. 4).

Obsyag and area of ​​the surface of the height prism are known by the formulas:

The area of ​​the lateral surface of the straight prism can be calculated using the formula.

Volume that surface area frail prisms (Fig. 4) can also be calculated in the same way: de ΔPNK - overcut, perpendicular to the edge l.

A right prism is called a straight prism, the basis of which is a regular bagatokutnik.

A prism is called a parallelepiped, and all of its faces are called parallelograms.

A straight parallelepiped is a parallelepiped, the ribs of which are perpendicular to the planes of the bases.

A straight parallelepiped is called a straight parallelepiped, the basis of which is a straight cut.

Power of the diagonal of a rectangular parallelepiped

The square of the diagonal of a rectangular parallelepiped is the sum of the squares of three yogo vimiriv: d² = a² + b² + c², de a,b,c- Dozhina ribs that come out of one vertex, d- diagonal of the parallelepiped (Fig. 3).

The volume of a rectangular parallelepiped is known from the formula V=abc.

A cube is called a rectangular parallelepiped with equal ribs. All faces of a cube are squares.

The volume, surface area and diagonal of a cube with an edge are known by the formulas:

V = a³, S = 6a² d² = 3 a².

pyramid

A bagatohedron, one facet of which is a bagatokutnik, and the other faces are tricutniks from a steep peak, are called a pyramid. The bagatokutnik is called the base of the pyramid, and the trikutniks are called the bichny faces.

The height of the pyramid is called the height of the perpendicular drawn from the top of the pyramid to the surface of the base.

If all the side ribs of the pyramid are even, or they are heeled to the flatness of the base under the same kut, then the height descends to the center of the described stake.

If the sides of the pyramid are heeled to the plane of the base under the same kut (double-faced kuti when standing equal), then the height descends to the center of the inscribed stake.

The pyramid is called correct, as its basis is the correct bagatokutnik, and the height falls into the center of the inscribed and described stake of the bagatokutnik, which lies at the base of the pyramid. The height of the bіchnі facet of the right pyramid, drawn from the її peaks, is called an apothem.

For example, on the little 5 is depicted the correct tricot pyramid SABC(tetrahedron): AB= BC= AC= a, OD=r- the radius of the stake inscribed in the tricutnik ABC, OA=R- the radius of the stake, the described white of the tricot ABC, SO=h- Visota

pyramids, SD = l- apothem, - kut nakhily

ribs SA to the plane of the base, - cut of the slope of the side face SBC up to the base of the pyramid.

The tricot pyramid is called a tetrahedron. The tetrahedron is called regular, as if its edges were equal.

Obsyag piramidi that area її superficially know for the formulas:

De h- The height of the pyramid.

The area of ​​the square of the surface of the regular pyramid to know behind the formula, de - the apothem of the pyramid.

A truncated pyramid is called a bagatohedron, the vertices of which serve as the tops of the base of the pyramid and the tops of the її cut across the plane, parallel to the base of the pyramid. Submit a truncated pyramid - like a bagatokutniki.

Obsyag truncated piramidi know behind the formula , de - the area of ​​the base, h - the height of the truncated pyramid.

Correct bagatohedrons

A regular bagatohedron is called a puffy bagatohedron, which has all the faces - regular bagatokutniki with the same number of sides and the same number of ribs converge in the skin vertex of the bagatohedron.

The faces of a regular bagatohedron can be either equal-sided tricutniks, or squares, or regular pyatikutniks.

Just as a regular bagatohedron has faces - regular tricuts, then regular bagatohedrons have a regular tetrahedron (vin maє 4 faces), a regular octahedron (vin maє 8 faces), a regular icosahedron (vin maє 20 faces).

If a regular bagatohedron has square faces, then the bagatohedron is called a cube or a hexahedron (there may be 6 faces).

If a regular bagatohedron has faces with regular p'yatikutniks, then the bagatohedron is called a dodecahedron (there may be 12 faces).

cylinder

A figure is called a cylinder, and as a result, a rectangle is wrapped around one side.

On the little one 6 is straight - all the wrapping; - Visota, l- Satisfying; ABCD- axial section of the cylinder, cut off by the wrapping of the rectangle on the side. The volume of that surface area of ​​the cylinder is known by the formulas:

, , , , de R- base radius, h- Visota, l- fix the cylinder.

Cone

A figure is called a cone, and the wrapping of a straight-cut tricot is cut off next to one of the catheters. On the little one 7 straight OB- all wrapping; OB = h- Visota, l- satisfying;Δ ABC- axial cut of the cone, cut off the wrappings of the straight-cut tricutnik OBC next to the leg OB.