1.7.1. Plane.

Consider in a Cartesian basis an arbitrary plane P and a normal vector (perpendicular) to it `n (A, B, C). Let us take an arbitrary fixed point M0(x0, y0, z0) and a current point M(x, y, z) in this plane.

It is obvious that ?`n = 0 (1.53)

(see (1.20) for j = p /2). This is the equation of a plane in vector form. Moving on to the coordinates, we obtain the general equation of the plane

A(x – x0) + B(y – y0) + C(z – z0) = 0 ?Ах + Ву + Сz + D = 0 (1.54).

(D = –Ах0– Ву0 – Сz0; А2 + В2 + С2 ? 0).

It can be shown that in Cartesian coordinates, every plane is determined by an equation of the first degree and, conversely, every equation of the first degree determines a plane (i.e., a plane is a surface of the first order and a surface of the first order is a plane).

Let's consider some special cases of the location of the plane specified by the general equation:

A = 0 – parallel to the Ox axis; B = 0 – parallel to the Oy axis; C = 0 – parallel to the Oz axis. (Such planes perpendicular to one of the coordinate planes are called projecting planes); D = 0 – passes through the origin; A = B = 0 – perpendicular to the Oz axis (parallel to the xOy plane); A = B = D = 0 – coincides with the xOy plane (z = 0). All other cases are analyzed similarly.

If D? 0, then by dividing both sides of (1.54) by -D, we can bring the equation of the plane to the form: (1.55),

a = – D /A, b = –D/ B, c = –D /C. Relationship (1.55) is called the equation of the plane in segments; a, b, c – abscissa, ordinate and applicate of the points of intersection of the plane with the Ox, Oy, Oz axes, and |a|, |b|, |c| – the lengths of the segments cut off by the plane on the corresponding axes from the origin of coordinates.

Multiplying both sides (1.54) by a normalizing factor (mD xcosa + ycosb + zcosg – p = 0 (1.56)

where cosa = Am, cosb = Bm, cosg = Cm are the direction cosines of the normal to the plane, p is the distance to the plane from the origin.

Let's consider the basic relationships used in the calculations. The angle between the planes A1x + B1y + C1z + D1 = 0 and A2x + B2y + C2z + D2 = 0 can be easily defined as the angle between the normals of these planes `n1 (A1, B1, C1) and

`n2 (A2, B2, C2): (1.57)

From (1.57) it is easy to obtain the perpendicularity condition

A1A2 + B1 B2 + C1 C2 = 0 (1.58)

and parallelism (1.59) planes and their normals.

Distance from an arbitrary point M0(x0, y0, z0) to the plane (1.54)

is determined by the expression: (1.60)

The equation of a plane passing through three given points M1(x1, y1, z1), M2(x2, y2, z2), M3(x3, y3, z3) is most conveniently written using the coplanarity condition (1.25) of the vectors where M(x, y , z) – current point of the plane.

(1.61)

Let us present the equation of a bundle of planes (i.e.

Sets of planes passing through one straight line) - it is convenient to use in a number of problems.

(A1x + B1y + C1z + D1) + l(A2x + B2y + C2z + D2) = 0 (1.62)

Where l О R, and in brackets are the equations of any two planes of the beam.

Control questions.

1) How to check that a given point lies on the surface defined by this equation?

2) What is the characteristic feature that distinguishes the equation of a plane in the Cartesian coordinate system from the equation of other surfaces?

3) How is the plane located relative to the coordinate system if its equation does not contain: a) a free term; b) one of the coordinates; c) two coordinates; d) one of the coordinates and a free term; d) two coordinates and a free term?

1) Given points M1(0,-1,3) and M2(1,3,5). Write the equation of a plane passing through point M1 and perpendicular to the vector Choose the correct answer:

A) ; b) .

2) Find the angle between the planes and . Choose the correct answer:

a) 135o, b) 45o

1.7.2. Straight. Planes whose normals are not collinear or intersect, unambiguously defining the straight line as the line of their intersection, which is written as follows:

An infinite number of planes can be drawn through this line (the bundle of planes (1.62)), including those that project it onto coordinate planes. To obtain their equations, it is enough to transform (1.63), eliminating one unknown from each equation and reducing them, for example, to the form (1.63`).

Let us set the task - to draw through the point M0(x0,y0,z0) a straight line parallel to the vector `S (l, m, n) (it is called a directing line). Let us take an arbitrary point M(x,y,z) on the desired line. Vectors and must be collinear, from which we obtain the canonical equations of the line.

(1.64) or (1.64`)

where cosa, cosb, cosg are the direction cosines of the vector `S. From (1.64) it is easy to obtain the equation of a straight line passing through given points M1(x1, y1, z1) and M2(x2, y2, z2) (it is parallel )

Or (1.64``)

(The values ​​of the fractions in (1.64) are equal for each point on the line and can be denoted by t, where t R. This allows you to enter the parametric equations of the line

Each value of the parameter t corresponds to a set of coordinates x, y, z of a point on a line or (otherwise) - values ​​of unknowns satisfying the equations of a line).

Using the already known properties of vectors and operations on them and the canonical equations of the straight line, it is easy to obtain the following formulas:

Angle between straight lines: (1.65)

Parallelism condition (1.66).

perpendicularity l1l2 + m1m2 + n1n2 = 0 (1.67) straight lines.

The angle between the straight line and the plane (easily obtained by finding the angle between the straight line and the normal to the plane, which adds up to the desired p/2)

(1.68)

From (1.66) we obtain the parallelism condition Al + Bm + Cn = 0 (1.69)

and perpendicularity (1.70) of a straight line and a plane. The necessary and sufficient condition for two lines to be in the same plane can be easily obtained from the coplanarity condition (1.25).

(1.71)

Control questions.

1) What are the ways to define a straight line in space?

1) Write the equations of a line passing through point A(4,3,0) and parallel to the vector Indicate the correct answer:

A) ; b) .

2) Write the equations of a straight line passing through points A(2,-1,3) and B(2,3,3). Indicate the correct answer.

A) ; b) .

3) Find the point of intersection of the line with the plane: , . Indicate the correct answer:

a) (6,4,5); b) (6,-4,5).

1.7.3. Surfaces of the second order. If a linear equation in a three-dimensional Cartesian basis uniquely defines a plane, any nonlinear equation containing x, y, z describes some other surface. If the equation is of the form

Ax2 + Ву2 + Cz2 + 2Dxy + 2Exz + 2Fyz + 2Gx + 2Hy + 2Kz + L = 0, then it describes a second-order surface (general equation of a second-order surface). By choosing or transforming Cartesian coordinates, the equation can be simplified as much as possible, leading to one of the following forms describing the corresponding surface.

1. Canonical equations of second-order cylinders, the generators of which are parallel to the Oz axis, and the corresponding second-order curves lying in the xOy plane serve as the guides:

(1.72), (1.73), y2 = 2px (1.74)

elliptic, hyperbolic and parabolic cylinders respectively.

(Recall that a cylindrical surface is a surface obtained by moving a straight line, called a generatrix, parallel to itself. The line of intersection of this surface with a plane perpendicular to the generatrix is ​​called a guide - it determines the shape of the surface).

By analogy, we can write down the equations of the same cylindrical surfaces with generatrices parallel to the Oy axis and the Ox axis. The guide can be defined as the line of intersection of the surface of the cylinder and the corresponding coordinate plane, i.e. system of equations of the form:

2. Equations of a second order cone with a vertex at the origin:

(1.75)

(the axes of the cone are the Oz, Oy and Ox axes, respectively)

3. Canonical equation of the ellipsoid: (1.76);

Special cases are ellipsoids of revolution, for example – surface obtained by rotating an ellipse around the Oz axis (At

a > c the ellipsoid is compressed, with a x2 + y2+ z2 + = r2 – the equation of a sphere of radius r with center at the origin).

4. Canonical equation of a one-sheet hyperboloid

(the “–” sign can appear in front of any of the three terms on the left side - this only changes the position of the surface in space). Special cases are single-sheet hyperboloids of revolution, for example – surface obtained by rotating a hyperbola around the Oz axis (the imaginary axis of the hyperbola).

5. Canonical equation of a two-sheet hyperboloid

(the “–” sign can appear in front of any of the three terms on the left side).

Special cases are two-sheet hyperboloids of revolution, for example, a surface obtained by rotating a hyperbola around the Oz axis (the real axis of the hyperbola).

6. Canonical equation of an elliptic paraboloid

(p >0, q >0) (1.79)

7. Canonical equation of a hyperbolic paraboloid

(p >0, q >0) (1.80)

(the variable z can change places with any of the variables x and y - the position of the surface in space will change).

Note that an idea of ​​the features (shape) of these surfaces can be easily obtained by considering sections of these surfaces by planes perpendicular to the coordinate axes.

Control questions.

1) What set of points in space determines the equation?

2) What are the canonical equations of second order cylinders; second order cone; ellipsoid; single-sheet hyperboloid; two-sheet hyperboloid; elliptical paraboloid; hyperbolic paraboloid?

1) Find the center and radius of the sphere and indicate the correct answer:

a) C(1.5;-2.5;2), ; b) C(1.5;2.5;2), ;

2) Determine the type of surface given by the equations: . Indicate the correct answer:

a) single-sheet hyperboloid; hyperbolic paraboloid; elliptical paraboloid; cone.

b) two-sheet hyperboloid; hyperbolic paraboloid; elliptical paraboloid; cone.

§7. Plane as a surface of the first order. General equation of the plane. Equation of a plane passing through a given point perpendicular to a given vector. Let us introduce a rectangular Cartesian coordinate system Oxyz in space and consider an equation of the first degree (or linear equation) for x, y, z: (7.1) Ax  By  Cz  D  0, A2  B2  C 2  0 . Theorem 7.1. Any plane can be specified in an arbitrary rectangular Cartesian coordinate system by an equation of the form (7.1). In exactly the same way as in the case of a line on a plane, the converse of Theorem 7.1 is valid. Theorem 7.2. Any equation of the form (7.1) defines a plane in space. The proof of Theorems 7.1 and 7.2 can be carried out similarly to the proof of Theorems 2.1, 2.2. From Theorems 7.1 and 7.2 it follows that the plane and only it is a surface of the first order. Equation (7.1) is called the general plane equation. Its  coefficients A, B, C are interpreted geometrically as the coordinates of the vector n perpendicular to the plane defined by this equation. This vector  n(A, B, C) is called the normal vector to the given plane. Equation (7.2) A(x  x0)  B(y  y0)  C (z  z0)  0 for all possible values ​​of the coefficients A, B, C defines all planes passing through the point M 0 (x0 , y0 , z0) . It is called the equation of a bunch of planes. The choice of specific values ​​of A, B, C in (7.2) means the choice of the plane P from the link passing through the point M 0 perpendicular to the given vector n(A, B, C) (Fig. 7.1). Example 7.1. Write the equation of the plane P passing through the point   A(1, 2, 0) parallel to the vectors a  (1, 2,–1), b  (2, 0, 1) .    The normal vector n to P is orthogonal to the given vectors a and b (Fig. 7.2),   therefore for n we can take their vector n product: A    P i j k    2 1  1 1   2 n  a  b  1 2  1  i  j 2 1  k 12 0  0 1 2 0 1 n   a    b 2i  3 j  4k . Let's substitute the coordinates of Fig. 7.2. For example, 7.1 P M0  point M 0 and vector n into equation (7.2), we obtain Fig. 7.1. To the equation of the plane of a bundle of planes P: 2(x  1)  3(y  2)  4z  0 or P: 2x  3y  4z  4  0 .◄ 1 If two of the coefficients A, B, C of the equation (7.1) are equal to zero, it specifies a plane parallel to one of the coordinate planes. For example, when A  B  0, C  0 – plane P1: Cz  D  0 or P1: z   D / C (Fig. 7.3). It is parallel to the Oxy plane, because its normal vector  n1(0, 0, C) is perpendicular to this plane. For A  C  0, B  0 or B  C  0, A  0, equation (7. 1) defines the planes P2: By  D  0 and P3: Ax  D  0, parallel to the coordinate planes Oxz and Oyz, since   their normal vectors n2(0, B, 0) and n3(A, 0, 0 ) are perpendicular to them (Fig. 7.3). If only one of the coefficients A, B, C of equation (7.1) is equal to zero, then it specifies a plane parallel to one of the coordinate axes (or containing it if D  0). Thus, plane P: Ax  By  D  0 is parallel to the Oz axis, z z  n1  n  n2 P1 L P O  n3 x y O P2 y P3 x Fig. 7.4. Plane P: Ax  B y  D  0, parallel to the Oz axis Fig. 7.3. The planes are parallel to the coordinate planes  since its normal vector n(A, B, 0) is perpendicular to the Oz axis. Note that it passes through the straight line L: Ax  By  D  0 lying in the Oxy plane (Fig. 7.4). For D  0, equation (7.1) specifies a plane passing through the origin. Example 7.2. Find the values ​​of the parameter  for which the equation x  (2  2) y  (2    2)z    3  0 defines the plane P: a) parallel to one of the coordinate planes; b) parallel to one of the coordinate axes; c) passing through the origin of coordinates. Let us write this equation in the form x  (  2) y  (  2)(  1) z    3  0 . (7.3) For any value , equation (7.3) defines a certain plane, since the coefficients of x, y, z in (7.3) do not simultaneously vanish. a) For   0, equation (7.3) defines a plane P parallel to the plane Oxy, P: z  3 / 2, and for   2 it defines a plane P 2 parallel to the plane Oyz, P: x  5/ 2. For no values ​​of  the plane P defined by equation (7.3) is parallel to the plane Oxz, since the coefficients of x, z in (7.3) do not simultaneously vanish. b) For   1, equation (7.3) defines a plane P parallel to the Oz axis, P: x  3y  2  0. For other values ​​of the parameter , it does not define a plane parallel to only one of the coordinate axes. c) For   3, equation (7.3) defines the plane P passing through the origin, P: 3x  15 y  10 z  0 . ◄ Example 7.3. Write the equation of the plane P passing through: a) point M (1,  3, 2) parallel to the plane axis Oxy; b) the Ox axis and point M (2, – 1, 3).   a) For the normal vector n to P here we can take the vector k (0, 0,1) - the unit vector of the Oz axis, since it is perpendicular to the Oxy plane. Substitute the coordinates of the point  M (1,  3, 2) and the vector n into equation (7.2), we obtain the equation of the plane P: z 3  0.   b) The normal vector n to P is orthogonal to the vectors i (1, 0, 0) and OM (2,  1, 3) ,  therefore we can take their vector product as n:    i j k       n  i  OM  1 0 0   j 12 03  k 12 01   3 j  k . 2 1 3  Substitute the coordinates of point O and vector n into equation (7.2), we obtain the equation of the plane P:  3(y  0)  (z  0)  0 or P: 3 y  z  0 .◄ 3

With the difference that instead of “flat” graphs, we will consider the most common spatial surfaces, and also learn how to competently build them by hand. I spent quite a long time selecting software tools for creating three-dimensional drawings and found a couple of good applications, but despite all the ease of use, these programs do not solve an important practical issue well. The fact is that in the foreseeable historical future, students will still be armed with a ruler and a pencil, and even having a high-quality “machine” drawing, many will not be able to correctly transfer it onto checkered paper. Therefore, in the manual, special attention is paid to the technique of manual construction, and a significant part of the page illustrations is a handmade product.

How does this reference material differ from analogues?

Having decent practical experience, I know very well which surfaces we most often have to deal with in real problems of higher mathematics, and I hope that this article will help you quickly replenish your luggage with the relevant knowledge and applied skills, which account for 90-95% there should be enough cases.

What do you need to be able to do at the moment?

The most basic:

Firstly, you need to be able to build correctly spatial Cartesian coordinate system (see the beginning of the article Graphs and properties of functions) .

What will you gain after reading this article?

Bottle After mastering the lesson materials, you will learn to quickly determine the type of surface by its function and/or equation, imagine how it is located in space, and, of course, make drawings. It’s okay if you don’t get everything in your head after the first reading - you can always return to any paragraph later as needed.

Information is within the power of everyone - to master it you do not need any super knowledge, special artistic talent or spatial vision.

Begin!

In practice, the spatial surface is usually given function of two variables or an equation of the form (the constant on the right side is most often equal to zero or one). The first designation is more typical for mathematical analysis, the second - for analytical geometry. The equation is essentially implicitly given a function of 2 variables, which in typical cases can easily be reduced to the form . Let me remind you of the simplest example c:

–plane equation kind .

– plane function in explicitly .

Let's start with it:

Common equations of planes

Typical options for the arrangement of planes in a rectangular coordinate system are discussed in detail at the very beginning of the article. Plane equation. However, let us once again dwell on the equations that are of great importance for practice.

First of all, you must fully automatically recognize the equations of planes that are parallel to coordinate planes. Fragments of planes are standardly depicted as rectangles, which in the last two cases look like parallelograms. By default, you can choose any dimensions (within reasonable limits, of course), but it is desirable that the point at which the coordinate axis “pierces” the plane is the center of symmetry:


Strictly speaking, the coordinate axes should be depicted with dotted lines in some places, but in order to avoid confusion we will neglect this nuance.

– (left drawing) the inequality specifies the half-space farthest from us, excluding the plane itself;

– (middle drawing) the inequality specifies the right half-space, including the plane;

– (right drawing) the double inequality defines a “layer” located between the planes, including both planes.

For self-warm-up:

Example 1

Draw a body bounded by planes
Create a system of inequalities that define a given body.

An old acquaintance should emerge from under the lead of your pencil. cuboid. Do not forget that invisible edges and faces must be drawn with a dotted line. Finished drawing at the end of the lesson.

Please, DON'T NEGLECT learning tasks, even if they seem too simple. Otherwise, it may happen that you missed it once, missed it twice, and then spent a solid hour trying to figure out a three-dimensional drawing in some real example. In addition, mechanical work will help you learn the material much more effectively and develop your intelligence! It is no coincidence that in kindergarten and elementary school children are loaded with drawing, modeling, construction toys and other tasks for fine motor skills of the fingers. Sorry for the digression, but my two notebooks on developmental psychology should not go missing =)

We will conditionally call the next group of planes “direct proportionality” - these are planes passing through the coordinate axes:

2) an equation of the form specifies a plane passing through the axis ;

3) an equation of the form specifies a plane passing through the axis.

Although the formal sign is obvious (which variable is missing from the equation – the plane passes through that axis), it is always useful to understand the essence of the events taking place:

Example 2

Construct plane

What is the best way to build? I propose the following algorithm:

First, let’s rewrite the equation in the form , from which it is clearly seen that the “y” can take any meanings. Let us fix the value, that is, we will consider the coordinate plane. Equations set space line, lying in a given coordinate plane. Let's depict this line in the drawing. The straight line passes through the origin of coordinates, so to construct it it is enough to find one point. Let . Set aside a point and draw a straight line.

Now we return to the equation of the plane. Since the "Y" accepts any values, then the straight line constructed in the plane is continuously “replicated” to the left and to the right. This is exactly how our plane is formed, passing through the axis. To complete the drawing, we lay two parallel lines to the left and right of the straight line and “close” the symbolic parallelogram with transverse horizontal segments:

Since the condition did not impose additional restrictions, a fragment of the plane could be depicted in slightly smaller or slightly larger sizes.

Let us once again repeat the meaning of spatial linear inequality using the example. How to determine the half-space it defines? Let's take some point not belonging to plane, for example, a point from the half-space closest to us and substitute its coordinates into the inequality:

Received true inequality, which means that the inequality specifies the lower (relative to the plane) half-space, while the plane itself is not included in the solution.

Example 3

Construct planes
A) ;
b) .

These are tasks for self-construction; in case of difficulties, use similar reasoning. Brief instructions and drawings at the end of the lesson.

In practice, planes parallel to the axis are especially common. The special case when the plane passes through the axis was just discussed in paragraph “be”, and now we will analyze a more general problem:

Example 4

Construct plane

Solution: the variable “z” is not explicitly included in the equation, which means that the plane is parallel to the applicate axis. Let's use the same technique as in the previous examples.

Let us rewrite the equation of the plane in the form from which it is clear that “zet” can take any meanings. Let’s fix it and draw a regular “flat” straight line in the “native” plane. To construct it, it is convenient to take reference points.

Since "Z" accepts All values, then the constructed straight line continuously “multiplies” up and down, thereby forming the desired plane . We carefully draw up a parallelogram of a reasonable size:

Ready.

Equation of a plane in segments

The most important applied variety. If All odds general equation of the plane non-zero, then it can be represented in the form which is called equation of the plane in segments. It is obvious that the plane intersects the coordinate axes at points , and the great advantage of such an equation is the ease of constructing a drawing:

Example 5

Construct plane

Solution: First, let's create an equation of the plane in segments. Let's throw the free term to the right and divide both sides by 12:

No, there is no typo here and all things happen in space! We examine the proposed surface using the same method that was recently used for planes. Let's rewrite the equation in the form , from which it follows that “zet” takes any meanings. Let us fix and construct an ellipse in the plane. Since "zet" accepts All values, then the constructed ellipse is continuously “replicated” up and down. It is easy to understand that the surface infinite:

This surface is called elliptical cylinder. An ellipse (at any height) is called guide cylinder, and parallel lines passing through each point of the ellipse are called forming cylinder (which literally form it). The axis is axis of symmetry surface (but not part of it!).

The coordinates of any point belonging to a given surface necessarily satisfy the equation .

Spatial the inequality defines the “inside” of the infinite “pipe”, including the cylindrical surface itself, and, accordingly, the opposite inequality defines the set of points outside the cylinder.

In practical problems, the most popular special case is when guide cylinder is circle:

Example 8

Construct the surface given by the equation

It is impossible to depict an endless “pipe”, so art is usually limited to “trimming”.

First, it is convenient to construct a circle of radius in the plane, and then a couple more circles above and below. The resulting circles ( guides cylinder) carefully connect with four parallel straight lines ( forming cylinder):

Don't forget to use dotted lines for lines that are invisible to us.

The coordinates of any point belonging to a given cylinder satisfy the equation . The coordinates of any point lying strictly inside the “pipe” satisfy the inequality , and the inequality defines a set of points of the external part. For a better understanding, I recommend considering several specific points in space and seeing for yourself.

Example 9

Construct a surface and find its projection onto the plane

Let's rewrite the equation in the form from which it follows that "x" takes any meanings. Let us fix and depict in the plane circle– with center at the origin, unit radius. Since "x" continuously accepts All values, then the constructed circle generates a circular cylinder with an axis of symmetry. Draw another circle ( guide cylinder) and carefully connect them with straight lines ( forming cylinder). In some places there were overlaps, but what to do, such a slope:

This time I limited myself to a piece of a cylinder in the gap, and this is not accidental. In practice, it is often necessary to depict only a small fragment of the surface.

Here, by the way, there are 6 generatrices - two additional straight lines “cover” the surface from the upper left and lower right corners.

Now let's look at the projection of a cylinder onto a plane. Many readers understand what projection is, but, nevertheless, let’s conduct another five-minute physical exercise. Please stand and bow your head over the drawing so that the point of the axis points perpendicular to your forehead. What a cylinder appears to be from this angle is its projection onto a plane. But it seems to be an endless strip, enclosed between straight lines, including the straight lines themselves. This projection is exactly domain functions (upper “gutter” of the cylinder), (lower “gutter”).

By the way, let’s clarify the situation with projections onto other coordinate planes. Let the sun's rays shine on the cylinder from the tip and along the axis. The shadow (projection) of a cylinder onto a plane is a similar infinite strip - a part of the plane bounded by straight lines (- any), including the straight lines themselves.

But the projection onto the plane is somewhat different. If you look at the cylinder from the tip of the axis, then it will be projected into a circle of unit radius , with which we began the construction.

Example 10

Construct a surface and find its projections onto coordinate planes

This is a task for you to solve on your own. If the condition is not very clear, square both sides and analyze the result; find out which part of the cylinder is specified by the function. Use the construction technique repeatedly used above. A short solution, drawing and comments at the end of the lesson.

Elliptical and other cylindrical surfaces can be offset relative to the coordinate axes, for example:

(based on familiar motives of the article about 2nd order lines) – a cylinder of unit radius with a line of symmetry passing through a point parallel to the axis. However, in practice, such cylinders are encountered quite rarely, and it is absolutely incredible to encounter a cylindrical surface that is “oblique” relative to the coordinate axes.

Parabolic cylinders

As the name suggests, guide such a cylinder is parabola.

Example 11

Construct a surface and find its projections onto coordinate planes.

I couldn't resist this example =)

Solution: Let's go along the beaten path. Let's rewrite the equation in the form, from which it follows that “zet” can take any value. Let us fix and construct an ordinary parabola on the plane, having previously marked the trivial support points. Since "Z" accepts All values, then the constructed parabola is continuously “replicated” up and down to infinity. We lay the same parabola, say, at a height (in the plane) and carefully connect them with parallel straight lines ( forming the cylinder):

I remind you useful technique: if you are initially unsure of the quality of the drawing, then it is better to first draw the lines very thinly with a pencil. Then we evaluate the quality of the sketch, find out the areas where the surface is hidden from our eyes, and only then apply pressure to the stylus.

Projections.

1) The projection of a cylinder onto a plane is a parabola. It should be noted that in this case it is impossible to talk about domain of definition of a function of two variables– for the reason that the cylinder equation is not reducible to functional form.

2) The projection of a cylinder onto a plane is a half-plane, including the axis

3) And finally, the projection of the cylinder onto the plane is the entire plane.

Example 12

Construct parabolic cylinders:

a) limit yourself to a fragment of the surface in the near half-space;

b) in the interval

In case of difficulties, we do not rush and reason by analogy with previous examples; fortunately, the technology has been thoroughly developed. It is not critical if the surfaces turn out a little clumsy - it is important to correctly display the fundamental picture. I myself don’t really bother with the beauty of the lines; if I get a passable drawing with a C grade, I usually don’t redo it. By the way, the sample solution uses another technique to improve the quality of the drawing ;-)

Hyperbolic cylinders

Guides such cylinders are hyperbolas. This type of surface, according to my observations, is much less common than previous types, so I will limit myself to a single schematic drawing of a hyperbolic cylinder:

The principle of reasoning here is exactly the same - the usual school hyperbole from the plane continuously “multiplies” up and down to infinity.

The considered cylinders belong to the so-called 2nd order surfaces, and now we will continue to get acquainted with other representatives of this group:

Ellipsoid. Sphere and ball

The canonical equation of an ellipsoid in a rectangular coordinate system has the form , where are positive numbers ( axle shafts ellipsoid), which in the general case different. An ellipsoid is called surface, so body, limited by a given surface. The body, as many have guessed, is determined by inequality and the coordinates of any interior point (as well as any surface point) necessarily satisfy this inequality. The design is symmetrical with respect to coordinate axes and coordinate planes:

The origin of the term “ellipsoid” is also obvious: if the surface is “cut” by coordinate planes, then the sections will result in three different (in the general case)

A first-order equation with three unknowns has the form Ax + Ву + Cz + D = 0, and at least one of the coefficients A, B, C must be different from zero. It specifies in space in rectangular coordinate system Oxyz algebraic surface of the first order.

The properties of a first-order algebraic surface are in many ways similar to the properties of a straight line on a plane - geometric image of a first order equation with two unknowns.

Theorem 5.1. Any plane in space is a surface of the first order and any surface of the first order in space is a plane.

◄ Both the statement of the theorem and its proof are similar to Theorem 4.1. Indeed, let the plane π be defined by its point M 0 and non-zero vector n, which is perpendicular to it. Then the set of all points in space is divided into three subsets. The first consists of points belonging to the plane, and the other two - of points located on one and the other side of the plane. Which of these sets belongs to an arbitrary point M of space depends on the sign dot product nM 0 M . If point M belongs to the plane (Fig. 5.1, a), then the angle between vectors n and M 0 M is straight, and therefore, according to Theorem 2.7, their scalar product is equal to zero:

nM 0 M = 0

If the point M does not belong to the plane, then the angle between the vectors n and M 0 M is acute or obtuse, and therefore nM 0 M > 0 or nM 0 M

Let's denote coordinates of points M 0 , M and vector n through (x 0 ; y 0 ; z 0), (x; y; z) and (A; B; C), respectively. Since M 0 M = (x - x 0 0; y - y 0; z - z 0 ), then, writing the scalar product from (5.1) in coordinate form (2.14) as the sum of pairwise products of the same coordinates of the vectors n and M 0 M , we obtain the condition for the point M to belong to the plane under consideration in the form

A(x - x 0) + B(y - y 0) + C (z - z 0) = 0. (5.2)

Opening the parentheses gives the equation

Ax + Wu + Cz + D = 0, (5.3)

where D = - Ax 0 - Ву 0 - Cz 0 and at least one of the coefficients A, B, or C is different from zero, since the vector n = (A; B; C) is non-zero. This means that the plane is the geometric image of equation (5.3), i.e. algebraic surface of the first order.

Carrying out the above proof of the first statement of the theorem in reverse order, we will prove that the geometric image of the equation Ax + Ву + Cz + D = 0, A 2 + В 2 + C 2 = 0, is a plane. Let's choose three numbers (x = x 0, y = y 0, z = z 0) that satisfy this equation. Such numbers exist. For example, when A ≠ 0 we can put y 0 = 0, z 0 = 0 and then x 0 = - D/A. The selected numbers correspond to the point M 0 (x 0 ; y 0 ; z 0), which belongs to the geometric image of the given equation. From the equality Ax 0 + Ву 0 + Cz 0 + D = 0 it follows that D = - Ax 0 - Ву 0 - Cz 0 . Substituting this expression into the equation under consideration, we obtain Ax + Ву + Cz - Ax 0 - Ву 0 - Cz 0 = 0, which is equivalent to (5.2). Equality (5.2) can be considered as vector orthogonality criterion n = (A; B; C) and M 0 M, where point M has coordinates (x; y; z). This criterion is fulfilled for points of the plane passing through the point M 0 perpendicular to the vector n = (A; B; C), and is not satisfied for other points in space. This means that equation (5.2) is the equation of the indicated plane.

The equation Ax + Wu + Cz + D = 0 is called general plane equation. The coefficients A, B, C for unknowns in this equation have a clear geometric meaning: the vector n = (A; B; C) is perpendicular to the plane. He is called normal plane vector. It, like the general equation of the plane, is determined up to a (non-zero) numerical factor.

Using the known coordinates of a point belonging to a certain plane and a non-zero vector perpendicular to it, using (5.2), the equation of the plane is written without any calculations.

Example 5.1. Let us find the general equation of a plane perpendicular to radius vector point A(2; 5; 7) and passing through point M 0 (3; - 4; 1).

Since the non-zero vector OA = (2; 5; 7) is perpendicular to the desired plane, its equation of type (5.2) has the form 2(x - 3) + 5(y + 4) + 7(z- 1) = 0. Opening the brackets , we obtain the desired general equation of the plane 2x + 5y + 7z + 7 = 0.