Rivnyannia is right next to the open space. Flattening of the plane: steep, through three points, normal.

The canonical alignments of a straight line in space are called a straight line, which signifies a straight line, which passes through a given point colinearly to a direct vector.

Let a point and a direct vector be given. A sufficient point to lie on a straight line l only in that way, as vectors and collenears, so that for them the mind wins:

Directed more straight lines and canonical straight lines.

Numbers m , nі pє projections of the direct vector of the coordinate axis. Since the vector is non-zero, then all numbers m , nі p cannot reach zero overnight. But one or two of them can appear equal to zero. In analytic geometry, for example, the following notation is allowed:

which means that the vector projections on the axis Ouchі Oz equal to zero. Therefore, i is a vector, i is a straight line, given by canonical lines, perpendicular to the axes. Ouchі Oz, i.e. flat yOz .

example 1. Lay straight lines in the space, perpendicular to the plane i pass through the point of the crossbar tsієї planes Oz .

Solution. We know the cross point Oz. So like a point, what to lie on the axle Oz maє coordinates, then, vvazhayuchi in a given equal plane x=y= 0 , take 4 z- 8 = 0 or z\u003d 2. Later, the crossing point is given by the plane from the line Oz may coordinates (0; 0; 2). Oskolki straight line is perpendicular to the plane, it is parallel to the normal vector її. Therefore, a direct vector can be a normal vector given surface.

Now let's write down the straight lines to pass through the point A= (0; 0; 2) y straight vector:

Alignment of a straight line to pass through two given points

A straight line can be given by two points, which lie on it і For this direct vector, a direct vector can be a vector . Todі canonіchnі rіvnyannya direct nabudu vglyadu

Pointing more straight lines determine a straight line, which will pass through two given points.

butt 2. Lay straight lines in the open space to pass through the points i.

Solution. Let's write down the sound of the straight line at the sight, brought to the surface in the theoretical conclusion:

Oskіlki, then the line is perpendicular to the axis Ouch .

A straight line like a line of flats

A straight line in space can be designated as a line of crossing of two non-parallel planes, so that it is like an impersonal point that satisfies the systems of two linear rivers

The leveling of the system is also called the wild leveling of the straight line in the open space.

example 3. Lay down the canonical straight line in the open space, set by the wild lines

Solution. In order to write the canonical alignment of straight lines, or, that those same straight lines that pass through two given points, it is necessary to know the coordinates of whether there are two points of the straight line. They can be points of the cross line with some sort of two coordinate planes, for example yOzі xOz .

The point of the cross line is straight with the plane yOz abscissa x= 0. To that, vvazhayuchi in my system equals x= 0 we take the system with two changes:

Її decision y = 2 , z= 6 at once x= 0 assigns a point A(0; 2; 6) shukano straight lines. Vvazhayuchi potim at the task of the system rivnyan y= 0 take the system

Її decision x = -2 , z= 0 together y= 0 assigns a point B(-2; 0; 0) straight line with flat xOz .

Now let's write down the alignment of the straight line, which will pass through the specks A(0; 2; 6) B (-2; 0; 0) :

otherwise, I sent the bannermen to -2:

lecture 6-7. Elements of analytical geometry.

The surface is smooth.

example 1.

Sphere

butt 2.

F(x, y, z) = 0(*),

Tse - smooth surfaces

Apply:

x 2 + y 2 - z 2 = 0 (cone)

Flat.

Flatness of a plane that passes through a given point perpendicularly to a given vector.

Let's take a look at the area near the space. Let M 0 (x 0, y 0, z 0) - a point of the plane P is given, and - a vector of perpendiculars to the plane ( normal vector flats).

(1) – vector flatness of the area.

For the coordinate form:

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

We took away the flatness of the area to pass through a given point.

Zagalne flatness of the area.

Opening the arches in (2): Ax + By + Cz + (-Ax 0 - By 0 - Cz 0) = 0 or

Ax + By + Cz + D = 0 (3)

Otrimane leveling of the area linearly, then. leveling 1 step for x, y, z coordinates. Therefore, the area - first order surface .

Confirmation: Whether it's equal, linearly x, y, z sets the plane.

Be-yak flat m. given to equals (3), as it is called zagalnym flats of the area.

Okremi vipadki zagalnogo rivnyannya.

a) D = 0: Ax + By + Cz = 0. Since Since the coordinates of the point O(0, 0, 0) satisfy the same level, then the plane is given to pass through the cob of coordinates.

b) C \u003d 0: Ax + By + D \u003d 0. In this way, the normal vector of the area that square, assigned to equals parallel to the OZ axis.

c) C \u003d D \u003d 0: Ax + By \u003d 0. The plane is parallel to the OZ axis (because C \u003d 0) and pass through the cob of coordinates (because D \u003d 0). Get out, get out through the whole OZ.

d) B \u003d C \u003d 0: Ax + D \u003d 0 or . vector, tobto. that . Also, the plane is parallel to the axes OY and OZ, tobto. parallel to the plane YOZ and pass through the point.

Look at the falls yourself: B = 0, B = D = 0, A = 0, A = D = 0, A = C = 0, A = B = 0/

Alignment of the plane through three given points.

Because If two or more points lie on the plane, then q vectors are coplanar, so. x zmіshany tvіr one zero:

Got a flat plane to pass through three points vector look.

For the coordinate form:

(7)

As soon as the vyznachnik is opened, then we take away the flatness of the area at the sight:

Ax+By+Cz+D=0.

butt. Write the level of the plane to pass through the points M1 (1, -1,0);

M 2 (-2.3.1) and M 3 (0.0.1).

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

3x + 2y + z - 1 = 0.

Leveling of the area near the windrows

Let's give a higher level of the area Ax + By + Cz + D = 0 and D ≠ 0, then. the plane does not pass through the cob of coordinates. Let's divide the offending parts into -D: that is significant: ; ; . Todi

took leveling of the plane at the windrows .

where a, b, c are the values of the vіrіzkіv, which are seen by the plane of the coordinate axes.

example 1. Write the level of the plane so as to pass through the points A(3, 0, 0);

B(0, 2, 0) and C(0, 0, -3).

a=3; b = 2; c = -3, or 2x + 3y - 2z - 6 = 0.

butt 2. Know the magnitude of the vіdrіzkіv, yakі vіdtinaє flat

4x - y - 3z - 12 = 0 on the coordinate axes.

4x - y - 3z = 12 a=3, b=-12, c=-4.

Normal flatness.

Let deyak be given the plane Q. From the cob of coordinates we draw the perpendicular OP to the plane. Let the task |OP|=р that vector: . Let's take a current point M(x, y, z) of the area and a calculably scalar vector augmentation that : .

If you project point M on a straight line, then we can go to point P. So, we take equal

(9).

Installed lines in space.

The line L in space can be set as a peretina of two surfaces. Let the point M(x, y, z) lie on the line L, lie on the surface of P1, and on the surface of P2. So the coordinates of the points of the fault are to be equal to both surfaces. To that pid equal to the line L in space to understand the marriage of two equals, the skin of those equals to the equal surfaces:

Lines L lie between th and th points, the coordinates of which satisfy both levels in (*). Let's look at other ways of setting up lines in space.

A bunch of flats.

Beam of flats- A lot of all the planes that pass through a given straight line - the entire beam.

To set a bunch of planes, it’s enough to install it all. Let the line of straight lines be given to the infamous looker:

Fold the bundle- means to fold the level, from which you can take away for the supplementary mind the level of the beam, cream b.m. alone. Let's multiply the II equal by l and store it by the I equal:

A 1 x + B 1 y + C 1 z + D 1 + l(A 2 x + B 2 y + C 2 z + D 2) = 0 (1) or

(A 1 + lA 2)x + (B 1 + lB 2)y + (C 1 + lC 2)z + (D 1 + lD 2) = 0 (2).

l - parameter - a number, so you can get real values. For whatever the opposite meaning is (1) and (2) linear, that is. tse - leveling of the deykoi area.

1. showy, so that the plane passes through the entire beam L. Take a sufficient point M 0 (x 0, y 0, z 0) L. Later, M 0 R 1 and M 0 R 2 . To mean:

Otzhe, the flat, which is described by the equals (1) or (2), to lie on the beam.

2. You can bring and protilezhne: whether or not the plane that passes through the straight line L is described by equals (1) for a different choice of the parameter l.

butt 1. Lay flat planes to pass through the line of planes x + y + 5z - 1 = 0 and 2x + 3y - z + 2 = 0 and through the point M (3, 2, 1).

Beam alignment is recorded: x + y + 5z - 1 + l (2x + 3y - z + 2) = 0. For the value of l, it is correct that M R:

Be it on the surface in space, you can see it as a geometrically spaced point, which may have power, a sign for all points.

example 1.

Sphere - an impersonal point, which is exactly in the distance from the given point C (to the center). Z (x 0, y 0, z 0). For appointments |CM|=R or . Tse equal vikonuєtsya all points of the sphere and only їm. If x 0 = 0, y 0 = 0, z 0 = 0, then .

By a similar rank, you can fold the alignment, whether it be on the surface, as if the coordinate system has been chosen.

butt 2. x=0 – alignment of the YOZ area.

Virazivshi geometrical design surface through the coordinates of the flow point and picking all the warehouses in one part, we take into account the equality of the mind

F(x, y, z) = 0(*),

Tse - smooth surfaces , so the coordinates of all points on the surface are satisfied with this evenness, and the coordinates of points that do not lie on the surface are not satisfied.

Including, the skin surface of the chosen coordinate system shows its own alignment. However, not skin-like appearance (*) shows the surface in a significant way.

Apply:

2x - y + z - 3 = 0 (flat)

x 2 + y 2 - z 2 = 0 (cone)

x 2 + y 2 +3 = 0 - the coordinates of the same point are not satisfied.

x 2 + y 2 + z 2 = 0 - one point (0,0,0).

x 2 \u003d 3y 2 \u003d 0 - Straight (all OZ).

All the leveling of the area, as if broken in the offensive points, can be removed from the deep level of the area, and also brought to the deep level of the area. In this rank, if we talk about the flatness of the flat, then we can say that the flat is flat on the ground, as it is not designated otherwise.

Leveling of the flat at the windrows.

Rivnyannia mind , de a, b і c – vіdminnі vіd zero dіysnі numbers, called equal to the area at the windbreaks.

Such a name is not vipadkova. The absolute values of the numbers a, b and c are equal to the lengths of the vіdrіzkіv, yakі vіdsіkaє the area on the coordinate axes Ox, Oy and Oz is clearly, fluctuating in the cob of coordinates. The sign of the numbers a, b and c shows, for which direction (positive or negative) there is a trace of a cross on the coordinate axes.

For the buttstock, it will be necessary for the rectangular coordinate system Oxyz to have a plane, which is assigned to the equal planes of the holes . For which point is indicated, it is 5 units away from the cob of coordinates near the negative direction of the abscissa axis, 4 units along the negative direction of the ordinate axis and 4 units away from the positive direction of the applicative axis. Lost some qi points with straight lines. The area of the trimmed tricot and є flat, which shows the flatness of the flat at the sight .

For more information, please go to the article leveling of the plane at the windrows, there is shown a given leveling of the plane at the vіdrіzkakh to the deep leveling of the plane, there you will also know the reports of the characteristic applications and the task.

Normal flatness.

Call the more flat area of the mind normal planes, like dovnyuє loneliness, tobto, , ta .

You can often tell that it is normal for the plane to be recorded as . Here are the direct cosines of the normal vector of a given area of a single plane, tobto , and p is an unknown number that is equal to the cob of coordinates to the plane.

The normal alignment of the plane in the rectangular coordinate system Oxyz defines the plane, as far as the cob of coordinates is on the distance p of the positive direction of the normal vector of the plane . If p=0 then the plane passes through the cob of coordinates.

Let's direct the butt of the normal flatness of the area.

Let the area be given in the rectangular coordinate system Oxyz to the upper planes of the plane in the form . Tse galnee alignment of the area є normal alignment of the area. Correct, i normal vector of the plane may dozhina equal loneliness, shards .

The flatness of the plane at the normal sight allows you to know move from point to plane.

It is recommended to study in more detail with this type of leveling of the plane, to look over the reports of typical applications on that day, and also to learn how to bring the leveling of the plane to a normal look. Tse can be robiti, having turned to the state.

List of literature.

Atanasyan L.S., Butuzov V.F., Kadomtsev S.B., Kiselova L.S., Poznyak E.G. Geometry. Assistant for grades 10–11 of a secondary school.
Bugrov Ya.S., Mikilsky S.M. Great math. Volume One: Elements of Linear Algebra and Analytic Geometry.
Ilyin V.A., Poznyak E.G. Analytical geometry.

Be-yak equal to the first step of the coordinates x, y, z

Ax + By + Cz + D = 0 (3.1)

defines the area, and again: whether or not the area can be represented by equals (3.1), as it is called equal to the area.

Vector n(A, B, C), orthogonal to the plane, called normal vector flats. Equation (3.1) has coefficients A, B, C that do not equal 0 at the same time.

Particular types of equivalence (3.1):

1. D = 0, Ax+By+Cz = 0 – the area to pass through the cob of coordinates.

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

3. C \u003d D \u003d 0, Ax + By \u003d 0 - the area to pass through all Oz.

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

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

The straight line at the space can be given:

1) like 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 2 z + D 2 = 0; (3.2)

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

= ; (3.3)

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

. (3.4)

Rivnyannia (3.4) are called canonical equals straight.

Vector a called direct vector direct.

The parameter is taken away by equating the skin from the wear (3.4) to the parameter t:

x=x1+mt, y=y1+nt, z=z1+rt. (3.5)

Virishuyuchi system (3.2) as a system of linear equalities xі y we come straight to the river projections or until let's make straight lines:

x = mz + a, y = nz + b. (3.6)

Vіd rivnyan (3.6) you can go to the canonical rivnyan, knowing z from the skin level and the difference between the values:

View wild rivers(3.2) it is possible to pass to the canonical and other methods, in order to know whether the line point of the line and the line n= [n 1 , n 2], de n 1 (A 1 , B 1 , C 1) and n 2 (A 2 B 2 C 2) - normal vectors of given planes. Like one of the famous m,n or R equals (3.4) appear to be equal to zero, then the number-book of a similar fraction must be set equal to zero, then. system

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

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

Stock 1.15. Fold the alignment of the plane, knowing that the point A (1, -1,3) is the substava perpendicular drawn from the cob of coordinates to the center of the plane.

Solution. Behind the brains vector OA(1,-1,3)
x-y+3z+D=0. Substituting the coordinates of the point A(1,-1,3), which lie flat, we know D: 1-(-1)+3×3+D = 0, D = -11. Also, x-y+3z-11=0.

Stock 1.16. Lay down the flatness of the area to pass through the entire Oz and make up with the area 2x+y-z-7=0 kut 60 o.

Solution. The area that passes through all Oz is given by Ax + By = 0, while A and do not turn to zero at once. Come on, don't
one 0, A/Bx+y=0. According to the cosine formula kuta between two planes

Virishyuchi square alignment 3m 2 + 8m - 3 \u003d 0, we know the yogo root
m 1 \u003d 1/3, m 2 \u003d -3, stars are taken from two areas 1/3x + y \u003d 0 and -3x + y \u003d 0.

butt 1.17. Fold canonical straight lines:
5x + y + z = 0, 2x + 3y – 2z + 5 = 0.

Solution.Canonical equality straight look:

de m, n, p- coordinates of the direct vector of the straight line, x1, y1, z1- coordinates of any point, which lie straight. The straight line is given as a line between two planes. In order to know the point, to lay straight lines, to fix one of the coordinates (the simplest is to put, for example, x = 0) and I will decompose the system as a system of linear alignments with two unknowns. Then, high x \u003d 0, then y + z \u003d 0, 3y - 2z + 5 \u003d 0, stars y \u003d -1, z \u003d 1. The coordinates of the point M (x 1, y 1, z 1), which lie on this line , we showed: M (0,-1,1). The direct vector is easy to know if you know the normal vectors and outward planes. n 1 (5,1,1) that n 2(2,3,-2). Todi

The canonical alignment of the straight line can be seen: x/(-5) = (y + 1)/12 =
= (z – 1)/13.

Stock 1.18. For a beam that is defined by planes 2x-y+5z-3=0 and x+y+2z+1=0, find two perpendicular planes, one of which passes through the point M(1,0,1).

Solution. The alignment of the beam, which is defined by these planes, may look like u (2x-y + 5z-3) + v (x + y + 2z + 1) \u003d 0, de u and v do not turn to zero overnight. Let's rewrite the alignment of the beam with an offensive rank:

(2u + v) x + (-u + v) y + (5u + 2v) z - 3u + v = 0.

In order to see a plane from the beam that passes through the point M, we represent the coordinates of the point M of the alignment of the beam. We take:

(2u+v)×1 + (-u + v) ×0 + (5u + 2v)×1 -3u + v = 0, or v = - u.

We know that we know the equalization of the plane, which avenues M, by substituting v = - u the equalization of the beam:

u(2x-y+5z - 3) - u(x+y+2z+1) = 0.

Because u ¹0 (now v=0, otherwise supercalculate the beam direction), maybe even the plane x-2y+3z-4=0. Another plane, which lies on the beam, may be perpendicular. Let's write down the mental orthogonality of the planes:

(2u + v) ×1 + (v - u) ×(-2) + (5u +2v)×3 = 0, or v = - 19/5u.

Otzhe, equal to another area may look:

u(2x -y+5z - 3) - 19/5 u(x + y +2z +1) = 0 or 9x +24y + 13z + 34 = 0.