What coordinates determine the profile projection of a point. I stage. motivation for learning activities. The main types of documents of the compass graphic system

Word form	Graphic form
1. Set aside on the X, Y, Ζ axes the corresponding coordinates of point A. We get points A x , A y , A z
2. Horizontal projection A 1 is located at the intersection of communication lines from points A x and A y drawn parallel to the X and Y axes
3. Frontal projection A 2 is located at the intersection of communication lines from points A x and A z, drawn parallel to the axes X and z
4. Profile projection A 3 is located at the intersection of communication lines from points A z and A y drawn parallel to the axes Ζ and Y

3.2. Point position relative to projection planes

The position of a point in space relative to the projection planes is determined by its coordinates. The X coordinate determines the distance of the point from the P 3 plane (projection to P 2 or P 1), the Y coordinate - the distance from the P 2 plane (projection to P 3 or P 1), the Z coordinate - the distance from the P 1 plane (projection to P 3 or P 2). Depending on the value of these coordinates, a point can occupy both a general and a particular position in space with respect to the projection planes (Fig. 3.1).

Rice. 3.1. Point classification

Tpointsgeneralprovisions. Point coordinates general position not equal to zero ( x≠0, y≠0, z≠0 ), and depending on the sign of the coordinate, the point can be located in one of eight octants (Table 2.1).

On fig. 3.2 drawings of points in general position are given. An analysis of their images allows us to conclude that they are located in the following octants of space: A(+X;+Y; +Z(  Ioctant;B(+X;+Y;-Z(  IVoctant;C(-X;+Y; +Z(  Voctant;D(+X;+Y; +Z(  IIoctant.

Private position points. One of the coordinates of the point of particular position is equal to zero, so the projection of the point lies on the corresponding field of projections, the other two - on the axes of the projections. On fig. 3.3 such points are points A, B, C, D, G.A  P 3, then the point X A \u003d 0; IN  P 3, then the point X B \u003d 0; FROM  P 2, then point Y C \u003d 0; D  P 1, then point Z D \u003d 0.

A point can belong to two projection planes at once, if it lies on the line of intersection of these planes - the projection axis. For such points, only the coordinate on this axis is not equal to zero. On fig. 3.3, such a point is the point G(G  OZ, then point X G =0, Y G =0).

3.3. Mutual position of points in space

Consider three options relative position points depending on the ratio of the coordinates that determine their position in space.

On fig. 3.4 points A and B have different coordinates.

Their relative position can be estimated by the distance to the projection planes: Y A >Y B, then point A is located farther from the plane P 2 and closer to the observer than point B; Z A >Z B, then point A is located farther from the plane P 1 and closer to the observer than point B; X A

On fig. 3.5 shows points A, B, C, D, in which one of the coordinates is the same, and the other two are different.

Their relative position can be estimated by their distance to the projection planes as follows:

Y A \u003d Y B \u003d Y D, then points A, B and D are equidistant from the plane P 2, and their horizontal and profile projections are located respectively on the lines [A 1 B 1 ]llOX and [A 3 B 3 ]llOZ. The locus of such points is a plane parallel to П 2 ;

Z A \u003d Z B \u003d Z C, then points A, B and C are equidistant from the plane P 1, and their frontal and profile projections are located respectively on the lines [A 2 B 2 ]llOX and [A 3 C 3 ]llOY. The locus of such points is a plane parallel to П 1 ;

X A \u003d X C \u003d X D, then points A, C and D are equidistant from the plane P 3 and their horizontal and frontal projections are located respectively on the lines [A 1 C 1 ]llOY and [A 2 D 2 ]llOZ . The locus of such points is a plane parallel to П 3 .

3. If the points have two coordinates of the same name, then they are called competing. Competing points are located on the same projecting line. On fig. 3.3 three pairs of such points are given, in which: X A \u003d X D; Y A = Y D ; Z D > Z A; X A = X C ; Z A = Z C ; Y C > Y A ; Y A = Y B ; Z A = Z B ; X B > X A .

There are horizontally competing points A and D located on the horizontally projecting line AD, frontally competing points A and C located on the frontally projecting line AC, profile competing points A and B located on the profile projecting line AB.

Conclusions on the topic

1. A point is a linear geometric image, one of the basic concepts of descriptive geometry. The position of a point in space can be determined by its coordinates. Each of three projections points are characterized by two coordinates, their name corresponds to the names of the axes that form the corresponding projection plane: horizontal - A 1 (XA; YA); frontal - A 2 (XA; ZA); profile - A 3 (YA; ZA). Translation of coordinates between projections is carried out using communication lines. From two projections, you can build projections of a point either using coordinates or graphically.

3. A point in relation to the projection planes can occupy both a general and a particular position in space.

4. A point in general position is a point that does not belong to any of the projection planes, i.e., lies in the space between the projection planes. The coordinates of a point in general position are not equal to zero (x≠0,y≠0,z≠0).

5. A point of private position is a point belonging to one or two projection planes. One of the coordinates of a point of particular position is equal to zero, so the projection of the point lies on the corresponding field of the projection plane, the other two - on the axes of the projections.

6. Competing points are points whose coordinates of the same name are the same. There are horizontally competing points, frontally competing points, and profile competing points.

Keywords

Point coordinates

General point

Private position point

Competing points

Methods of activity necessary for solving problems

– construction of a point according to the given coordinates in the system of three projection planes in space;

– construction of a point according to the given coordinates in the system of three projection planes on the complex drawing.

Questions for self-examination

1. How is the connection of the location of coordinates on the complex drawing in the system of three projection planes P 1 P 2 P 3 with the coordinates of the projections of points established?

2. What coordinates determine the distance of points to the horizontal, frontal, profile projection planes?

3. What coordinates and projections of the point will change if the point moves in the direction perpendicular to the profile plane of the projections П 3 ?

4. What coordinates and projections of a point will change if the point moves in a direction parallel to the OZ axis?

5. What coordinates determine the horizontal (frontal, profile) projection of a point?

7. In what case does the projection of a point coincide with the point in space itself, and where are the other two projections of this point located?

8. Can a point belong to three projection planes at the same time, and in what case?

9. What are the names of the points whose projections of the same name coincide?

10. How can you determine which of the two points is closer to the observer if their frontal projections coincide?

Tasks for independent solution

1. Give a visual image of points A, B, C, D relative to the projection planes P 1, P 2. The points are given by their projections (Fig. 3.6).

2. Construct projections of points A and B according to their coordinates on a visual image and a complex drawing: A (13.5; 20), B (6.5; -20). Construct a projection of point C, located symmetrically to point A relative to the frontal plane of projections П 2 .

3. Construct projections of points A, B, C according to their coordinates on a visual image and a complex drawing: A (-20; 0; 0), B (-30; -20; 10), C (-10, -15, 0 ). Construct point D, located symmetrically to point C with respect to the OX axis.

An example of solving a typical problem

Task 1. Given the coordinates X, Y, Z of points A, B, C, D, E, F (Table 3.3)

PROJECTION OF A POINT ON TWO PLANES OF PROJECTIONS

The formation of a straight line segment AA 1 can be represented as a result of moving point A in any plane H (Fig. 84, a), and the formation of a plane can be represented as a displacement of a straight line segment AB (Fig. 84, b).

A point is the main geometric element of a line and surface, so the study of the rectangular projection of an object begins with the construction of rectangular projections of a point.

In the space of the dihedral angle formed by two perpendicular planes - the frontal (vertical) plane of projections V and the horizontal plane of projections H, we place the point A (Fig. 85, a).

The line of intersection of the projection planes is a straight line, which is called the projection axis and is denoted by the letter x.

The V plane is shown here as a rectangle, and the H plane as a parallelogram. The inclined side of this parallelogram is usually drawn at an angle of 45° to its horizontal side. The length of the inclined side is taken equal to 0.5 of its actual length.

From point A, perpendiculars are lowered on the planes V and H. Points a "and a of the intersection of perpendiculars with the projection planes V and H are rectangular projections of point A. The figure Aaa x a" in space is a rectangle. The side aax of this rectangle in the visual image is reduced by 2 times.

Let us align the H plane with the V plane by rotating V around the line of intersection of the x planes. The result is a complex drawing of point A (Fig. 85, b)

To simplify the complex drawing, the boundaries of the projection planes V and H are not indicated (Fig. 85, c).

Perpendiculars drawn from point A to the projection planes are called projecting lines, and the bases of these projecting lines - points a and a "are called projections of point A: a" - frontal projection of point A, a - horizontal projection points A.

Line a "a is called the vertical line of the projection connection.

The location of the projection of a point on a complex drawing depends on the position of this point in space.

If point A lies on the horizontal projection plane H (Fig. 86, a), then its horizontal projection a coincides with the given point, and the frontal projection a "is located on the axis. When point B is located on the frontal projection plane V, its frontal projection coincides with this point, and the horizontal projection lies on the x-axis. Horizontal and frontal projections given point C lying on the x-axis coincide with this point. A complex drawing of points A, B and C is shown in fig. 86b.

PROJECTION OF A POINT ON THREE PLANES OF PROJECTIONS

In cases where it is impossible to imagine the shape of an object from two projections, it is projected onto three projection planes. In this case, the profile plane of projections W is introduced, which is perpendicular to the planes V and H. A visual representation of the system of three projection planes is given in fig. 87 a.

The edges of a trihedral angle (the intersection of projection planes) are called projection axes and are denoted by x, y and z. The intersection of the projection axes is called the beginning of the projection axes and is denoted by the letter O. Let us drop the perpendicular from point A to the projection plane W and, marking the base of the perpendicular with the letter a, we will obtain the profile projection of point A.

To obtain a complex drawing, points A of the H and W planes are aligned with the V plane, rotating them around the Ox and Oz axes. A complex drawing of point A is shown in fig. 87b and c.

The segments of the projecting lines from point A to the projection planes are called the coordinates of point A and are denoted: x A, y A and z A.

For example, the coordinate z A of point A, equal to the segment a "a x (Fig. 88, a and b), is the distance from point A to the horizontal projection plane H. The coordinate at point A, equal to the segment aa x, is the distance from point A to the frontal plane of projections V. The x A coordinate equal to the segment aa y is the distance from point A to the profile plane of projections W.

Thus, the distance between the projection of a point and the projection axis determines the coordinates of the point and is the key to reading its complex drawing. By two projections of a point, all three coordinates of a point can be determined.

If the coordinates of point A are given (for example, x A \u003d 20 mm, y A \u003d 22 mm and z A \u003d 25 mm), then three projections of this point can be built.

To do this, from the origin of coordinates O in the direction of the Oz axis, the coordinate z A is laid up and the coordinate y A is laid down. segments equal to the x coordinate A. The resulting points a "and a are the frontal and horizontal projections of the point A.

According to two projections a "and a point A, its profile projection can be constructed in three ways:

1) from the origin O, an auxiliary arc is drawn with a radius Oa y equal to the coordinate (Fig. 87, b and c), from the obtained point a y1 draw a straight line parallel to the Oz axis, and lay a segment equal to z A;

2) from the point a y, an auxiliary straight line is drawn at an angle of 45 ° to the axis Oy (Fig. 88, a), a point a y1 is obtained, etc.;

3) from the origin O, draw an auxiliary straight line at an angle of 45 ° to the axis Oy (Fig. 88, b), get a point a y1, etc.

Projection apparatus

The projection apparatus (Fig. 1) includes three projection planes:

π 1 - horizontal projection plane;

π 2 - frontal projection plane;

π 3– profile plane of projections .

The projection planes are mutually perpendicular ( π 1^ π 2^ π 3), and their intersection lines form axes:

Plane intersection π 1 And π 2 form an axis 0X (π 1∩ π 2 = 0X);

Plane intersection π 1 And π 3 form an axis 0Y (π 1∩ π 3 = 0Y);

Plane intersection π 2 And π 3 form an axis 0Z (π 2∩ π 3 = 0Z).

The point of intersection of the axes (ОХ∩OY∩OZ=0) is considered to be the reference point (point 0).

Since the planes and axes are mutually perpendicular, such an apparatus is similar to the Cartesian coordinate system.

The projection planes divide the entire space into eight octants (in Fig. 1 they are indicated by Roman numerals). Projection planes are considered opaque, and the viewer is always in I th octane.

Projection orthogonal with projection centers S1, S2 And S3 respectively for the horizontal, frontal and profile projection planes.

BUT.

From projection centers S1, S2 And S3 projecting beams come out l 1, l 2 And l 3 BUT

- A 1 BUT;

- A 2– frontal projection of the point BUT;

- A 3– profile projection of a point BUT.

A point in space is characterized by its coordinates A(x,y,z). points A x, A y And Az respectively on the axes 0X, 0Y And 0Z show coordinates x, y And z points BUT. On fig. 1 gives all the necessary designations and shows the relationship between the point BUT space, its projections and coordinates.

point diagram

To plot a point BUT(Fig. 2), in the projection apparatus (Fig. 1) the plane π 1 A 1 0X π 2. Then the plane π 3 with point projection A 3, rotate counterclockwise around the axis 0Z, until it coincides with the plane π 2. Direction of rotation of planes π 2 And π 3 shown in fig. 1 arrows. At the same time, direct A 1 A x And A 2 A x 0X perpendicular A 1 A 2, and straight lines A 2 A x And A 3 A x will be located in common to the axis 0Z perpendicular A 2 A 3. These lines will be referred to as vertical And horizontal connection lines.

It should be noted that when moving from the projection apparatus to the diagram, the projected object disappears, but all information about its shape, geometric dimensions and its position in space are preserved.

BUT(x A , y A , z Ax A , y A And z A in the following sequence (Fig. 2). This sequence is called the point plotting technique.

1. Axes are drawn orthogonally OX, OY And oz.

2. On the axis OX x A points BUT and get the position of the point A x.

3. Through the dot A x perpendicular to the axis OX

A x in the direction of the axis OY the numerical value of the coordinate is postponed y A points BUT A 1 on the plot.

A x in the direction of the axis oz the numerical value of the coordinate is postponed zA points BUT A 2 on the plot.

6. Through the dot A 2 parallel to axis OX a horizontal line is drawn. The intersection of this line and the axis oz will give the position of the point A z.

7. On a horizontal line from the point A z in the direction of the axis OY the numerical value of the coordinate is postponed y A points BUT and the position of the profile projection of the point is determined A 3 on the plot.

Point characteristic

All points of space are subdivided into points of private and general positions.

Private position points. Points belonging to the projection apparatus are called points of particular position. These include points belonging to the projection planes, axes, origin and projection centers. The characteristic features of points of private position are:

Metamathematical - one, two or all numerical values ​​of the coordinates are equal to zero and (or) infinity;

On the diagram - two or all projections of a point are located on the axes and (or) are located at infinity.

Points in general position. Points in general position include points that do not belong to the projection apparatus. For example, dot BUT in fig. 1 and 2.

In the general case, the numerical values ​​of the coordinates of a point characterize its distance from the projection plane: the coordinate X from the plane π 3; coordinate y from the plane π 2; coordinate z from the plane π 1. It should be noted that the signs at the numerical values ​​of the coordinates indicate the direction of removal of the point from the projection planes. Depending on the combination of signs for the numerical values ​​of the coordinates of the point, it depends in which of the octane it is located.

Two Image Method

In practice, in addition to the full projection method, the two-image method is used. It differs in that the third projection of the object is excluded in this method. To obtain the projection apparatus of the two-image method, the profile projection plane with its projection center is excluded from the full projection apparatus (Fig. 3). In addition, on the axis 0X the origin is assigned (point 0 ) and from it perpendicular to the axis 0X in projection planes π 1 And π 2 spend axis 0Y And 0Z respectively.

In this apparatus, the entire space is divided into four quadrants. On fig. 3 are marked with Roman numerals.

Projection planes are considered opaque, and the viewer is always in I th quadrant.

Consider the operation of the device using the example of projecting a point BUT.

From projection centers S1 And S2 projecting beams come out l 1 And l 2. These rays pass through the point BUT and intersecting with the projection planes form its projections:

- A 1- horizontal projection of a point BUT;

- A 2– frontal projection of the point BUT.

To plot a point BUT(Fig. 4), in the projection apparatus (Fig. 3) the plane π 1 with the resulting point projection A 1 rotate clockwise around an axis 0X, until it coincides with the plane π 2. Plane rotation direction π 1 shown in fig. 3 arrows. At the same time, only one point remains on the diagram of the point obtained by the two-image method. vertical communication line A 1 A 2.

In practice, plotting a point BUT(x A , y A , z A) is carried out according to the numerical values ​​of its coordinates x A , y A And z A in the following sequence (Fig. 4).

1. An axis is drawn OX and the origin is assigned (point 0 ).

2. On the axis OX the numerical value of the coordinate is postponed x A points BUT and get the position of the point A x.

3. Through the dot A x perpendicular to the axis OX a vertical line is drawn.

4. On the vertical line from the point A x in the direction of the axis OY the numerical value of the coordinate is postponed y A points BUT and the position of the horizontal projection of the point is determined A 1 OY is not plotted, but its positive values ​​are assumed to be below the axis OX, while the negative ones are higher.

5. On the vertical line from the point A x in the direction of the axis oz the numerical value of the coordinate is postponed zA points BUT and the position of the frontal projection of the point is determined A 2 on the plot. It should be noted that on the diagram the axis oz is not drawn, but it is assumed that its positive values ​​are located above the axis OX, while the negative ones are lower.

Competing points

Points on the same projecting ray are called competing points. They have a common projection in the direction of the projecting beam, i.e. their projections coincide identically. A characteristic feature of competing points on the diagram is the identical coincidence of their projections of the same name. The competition lies in the visibility of these projections relative to the observer. In other words, in space for the observer, one of the points is visible, the other is not. And, accordingly, in the drawing: one of the projections of the competing points is visible, and the projection of the other point is invisible.

On a spatial projection model (Fig. 5) from two competing points BUT And IN visible dot BUT on two mutually complementary grounds. According to the chain S 1 →A→B dot BUT closer to the observer than a point IN. And, accordingly, further from the projection plane π 1(those. zA > zA).

Rice. 5 Fig.6

If the point itself is visible A, then its projection is also visible A 1. In relation to the projection coinciding with it B1. For clarity and, if necessary, on the diagram, invisible projections of points are usually enclosed in brackets.

Remove points on the model BUT And IN. Their coinciding projections on the plane will remain π 1 and separate projections - on π 2. We conditionally leave the frontal projection of the observer (⇩), located in the center of the projection S1. Then along the chain of images ⇩ → A2 → B2 it will be possible to judge that zA > zB and that the point itself is visible BUT and its projection A 1.

Similarly, consider the competing points FROM And D apparently relative to the plane π 2 . Since the common projecting beam of these points l 2 parallel to axis 0Y, then the sign of visibility of competing points FROM And D is determined by the inequality yC > yD. Therefore, the point D closed by a dot FROM and, accordingly, the projection of the point D2 will be covered by the projection of the point From 2 on surface π 2.

Let's consider how the visibility of competing points is determined in a complex drawing (Fig. 6).

According to matching projections A 1≡IN 1 the points themselves BUT And IN are on the same projecting beam parallel to the axis 0Z. So coordinates are to be compared zA And zB these points. To do this, we use the frontal projection plane with separate point images. In this case zA > zB. It follows from this that the projection is visible A 1.

points C And D in the complex drawing under consideration (Fig. 6) are also on the same projecting beam, but only parallel to the axis 0Y. Therefore, from a comparison yC > yD we conclude that the projection C 2 is visible.

General rule . Visibility for coinciding projections of competing points is determined by comparing the coordinates of these points in the direction of the common projecting beam. Visible is the projection of the point for which this coordinate is greater. In this case, the comparison of coordinates is carried out on the plane of projections with separate images of points.

projection(lat. Projicio - I throw forward) - the process of obtaining an image of an object (spatial object) on any surface using light or visual rays (rays that conditionally connect the observer's eye with any point of a spatial object), which are called projecting.

There are two projection methods: central And parallel .

Centralprojection is to pass through each point ( A, B, C,…) of the depicted object and in a certain way selected projection center (S) straight line ( SA, SB, >… — projecting beam).

Figure 1.1 - Central projection

Let's introduce the following notation (Figure 1.1):

S– projection center (observer's eye);

π 1 - projection plane;

A, B, C

SA, SB- projecting straight lines (projecting rays).

Note: left mouse button can move the point in the horizontal plane, when you click on the point with the left mouse button, the direction of movement will change and you can move it vertically.

Central projection point the point of intersection of the projecting line passing through the projection center and the projection object (point) with the projection plane is called.

Property 1 . Each point in space corresponds to a single projection, but each point in the projection plane corresponds to a set of points in space that lie on the projecting line.

Let's prove this statement.

Figure 1.1: dot BUT 1 is the central projection of point A on the plane of projections π 1 . But all points lying on the projecting line can have the same projection. Take on the projecting line SA point FROM. Central projection point FROM(FROM 1) on the plane of projections π 1 coincides with the projection of the point BUT(BUT 1):

1. FROM∈ SA;
2. SC∩ π 1 = C 1 →C 1 ≡ A 1 .

The conclusion follows that by the projection of a point it is impossible to judge unambiguously about its position in space.

To eliminate this uncertainty, i.e. make a drawing reversible, we introduce one more projection plane (π 2) and one more projection center ( S 2) (Figure 1.2).

Figure 1.2 - Illustration of the 1st and 2nd properties

Let's construct projections of a point BUT on the plane of projections π 2 . Of all points in space, only a point BUT has its projections BUT 1 to the plane π 1 and BUT 2 to π 2 at the same time. All other points lying on the projecting rays will have at least one different projection from the projections of the point BUT(e.g. dot IN).

Property 2 . The projection of a straight line is a straight line.

Let's prove this property.

Connect the dots BUT And IN among themselves (Figure 1.2). We get a segment AB defining a straight line. triangle SAB defines a plane, denoted by σ. It is known that two planes intersect in a straight line: σ∩π 1 = BUT 1 IN 1 , where BUT 1 IN 1 - central projection of a straight line given by a segment AB.

The central projection method is a model of image perception by the eye, it is mainly used when making perspective images of building objects, interiors, as well as in film technology and optics. The method of central projection does not solve the main task facing the engineer - to accurately reflect the shape, dimensions of the object, the ratio of the sizes of various elements.

1.2. Parallel projection

Consider the method of parallel projection. We will impose three restrictions that will allow us, albeit to the detriment of the visibility of the image, to get a drawing more convenient for using it in practice:

1. Let's delete both projection centers to infinity. Thus, we will ensure that the projecting rays from each center become parallel, and, therefore, the ratio of the true length of any line segment and the length of its projection will depend only on the angle of inclination of this segment to the projection planes and do not depend on the position of the projection center;
2. Let's fix the projection direction relative to the projection planes;
3. Let's arrange the projection planes perpendicular to each other, which will make it easy to move from the image on the projection planes to the real object in space.

Thus, having imposed these restrictions on the central projection method, we have come to its special case - parallel projection method(Figure 1.3). Projection, in which the projecting rays passing through each point of the object are parallel to the selected projection direction P, is called parallel .

Figure 1.3 - Parallel projection method

Let's introduce the notation:

R– direction of projection;

π 1 - horizontal plane of projections;

A,B– projection objects – points;

BUT 1 and IN 1 - projections of points BUT And IN onto the projection plane π 1 .

Parallel point projection is the point of intersection of the projecting line parallel to the given direction of projection R, with the projection plane π 1 .

Pass through the dots BUT And IN projecting beams parallel to a given projection direction R. Projecting ray passing through a point BUT intersects the projection plane π 1 at the point BUT one . Similarly, a projecting ray through a point IN intersects the projection plane at a point IN one . By connecting the dots BUT 1 and IN 1 , we get a segment BUT 1 IN 1 is the projection of the segment AB onto the plane π 1 .

1.3. Orthographic projection. Monge method

If the projection direction R perpendicular to the plane of projections p 1 , then the projection is called rectangular (Figure 1.4), or orthogonal (gr. orthos- straight, gonia- angle) if R not perpendicular to π 1, then the projection is called oblique .

quadrilateral AA 1 IN 1 IN defines the plane γ, which is called the projecting plane, since it is perpendicular to the plane π 1 (γ⊥π 1). In what follows, we will use only rectangular projection.

Figure 1.4 - Orthographic projection Figure 1.5 - Monge, Gaspard (1746-1818)

The French scientist Gaspard Monge is considered the founder of orthogonal projection (Figure 1.5).

Before Monge, builders, artists and scientists possessed quite significant information about projection methods, and yet only Gaspard Monge is the creator of descriptive geometry as a science.

Gaspard Monge was born on May 9, 1746 in the small town of Beaune (Burgundy) in eastern France in the family of a local merchant. He was the eldest of five children, for whom his father, despite the low origin and relative poverty of the family, tried to provide the most better education from available at that time for people from the humble class. His second son, Louis, became a professor of mathematics and astronomy, the youngest, Jean, also a professor of mathematics, hydrography and navigation. Gaspard Monge received his initial education at the city school of the Oratory order. After graduating in 1762 as the best student, he entered the college of Lyon, also owned by the Oratorians. Soon Gaspard was entrusted with teaching physics there. In the summer of 1764, Monge drew up a plan of his native city of Beaune, remarkably accurate. The necessary methods and instruments for measuring angles and drawing lines were invented by the compiler himself.

While studying in Lyon, he received an offer to join the order and remain a college teacher, however, instead, having shown great abilities in mathematics, drafting and drawing, he managed to enter the Mézieres School of Military Engineers, but (due to origin) only as an auxiliary non-commissioned officer officer department and without a paycheck. Nevertheless, success in the exact sciences and an original solution to one of the important problems of fortification (the placement of fortifications depending on the location of enemy artillery) allowed him in 1769 to become an assistant (teaching assistant) in mathematics, and then in physics, and already with a decent salary at 1800 livres a year.

In 1770, at the age of 24, Monge held the position of professor at the same time in two departments - mathematics and physics, and, in addition, conducts classes in cutting stones. Starting with the task of accurately cutting stones according to given sketches in relation to architecture and fortification, Monge came to the creation of methods that he later generalized in new science- descriptive geometry, the creator of which he is rightfully considered. Given the possibility of using the methods of descriptive geometry for military purposes in the construction of fortifications, the leadership of the Mézières school did not allow open publication until 1799, the book was published under the title descriptive geometry (Geometry descriptive) (a verbatim record of these lectures was made in 1795). The approach to lecturing on this science and doing the exercises outlined in it has survived to this day. Another significant work of Monge - Application of analysis to geometry (L'application de l'analyse à la geometrie, 1795) - is a textbook of analytic geometry, in which special emphasis is placed on differential relations.

In 1780 he was elected a member of the Paris Academy of Sciences, in 1794 he became director of the Polytechnic School. For eight months he served as minister of the sea in the government of Napoleon, was in charge of the gunpowder and cannon factories of the republic, and accompanied Napoleon on his expedition to Egypt (1798–1801). Napoleon granted him the title of count, honored him with many other distinctions.

The method of depicting objects according to Monge consists of two main points:

1. The position of a geometric object in space, in this example a point BUT, is considered relative to two mutually perpendicular planes π 1 and π 2(Figure 1.6).

They conditionally divide the space into four quadrants. Dot BUT located in the first quadrant. The Cartesian coordinate system served as the basis for the Monge projections. Monge replaced the concept of projection axes with the line of intersection of projection planes (coordinate axes) and proposed to combine the coordinate planes into one by rotating them around the coordinate axes.

Figure 1.6 - Model for constructing point projections

π 1 - horizontal (first) projection plane

π 2 - frontal (second) projection plane

π 1 ∩ π 2 is the axis of projections (we denote π 2 / π 1)

Consider an example of projecting a point BUT onto two mutually perpendicular projection planes π 1 and π 2 .

Drop from point BUT perpendiculars (projecting rays) on the planes π 1 and π 2 and mark their bases, that is, the points of intersection of these perpendiculars (projecting rays) with the projection planes. BUT 1 - horizontal (first) projection of the point BUT;BUT 2 - frontal (second) projection of the point BUT;AA 1 and AA 2 - projecting lines. Arrows show the direction of projection on the plane of projections π 1 and π 2 . Such a system allows you to uniquely determine the position of a point relative to the projection planes π 1 and π 2:

AA 1 ⊥π 1

BUT 2 BUT 0 ⊥π 2 /π 1 AA 1 = BUT 2 BUT 0 - distance from point A to plane π 1

AA 2 ⊥π 2

BUT 1 BUT 0 ⊥π 2 /π 1 AA 2 \u003d A 1 A 0 - the distance from point A to the plane π 2

2. Let's combine the rotation around the axis of projections π 2 / π 1 of the projection plane into one plane(π 1 with π 2), but so that the images do not overlap, (in the α direction, Figure 1.6), we get an image called a rectangular drawing (Figure 1.7):

Figure 1.7 - Orthogonal drawing

Rectangular or orthogonal is called Monge diagram .

Straight BUT 2 BUT 1 called projection link , which connects opposite projections of the point ( BUT 2 - frontal and BUT 1 - horizontal) is always perpendicular to the projection axis (coordinate axis) BUT 2 BUT 1 ⊥π 2 /π 1 . On the diagram, the segments indicated by curly brackets are:

• BUT 0 BUT 1 - distance from the point BUT to the plane π 2 corresponding to the coordinate y A;
• BUT 0 BUT 2 - distance from the point BUT to the plane π 1 corresponding to the coordinate z A.

1.4. Rectangular point projections. Orthographic drawing properties

1. Two rectangular projections of a point lie on the same projection connection line perpendicular to the projection axis.

2. Two rectangular projections of a point uniquely determine its position in space relative to the projection planes.

Let's verify the validity of the last statement, for which we turn the plane π 1 to its original position (when π 1 ⊥ π 2). To build a point BUT needed from points BUT 1 and BUT 2 to restore the projecting rays, and in fact - the perpendiculars to the planes π 1 and π 2 , respectively. The intersection point of these perpendiculars fixes the desired point in space BUT. Consider an orthogonal drawing of a point BUT(Figure 1.8).

Figure 1.8 - Plotting a point

Let us introduce the third (profile) plane of projections π 3 perpendicular to π 1 and π 2 (given by the axis of projections π 2 /π 3).

Distance from the profile projection of a point to the vertical axis of the projections BUT‘ 0 A 3 allows you to determine the distance from the point BUT to the frontal projection plane π 2 . It is known that the position of a point in space can be fixed relative to the Cartesian coordinate system using three numbers (coordinates) A(X A ; Y A ; Z A) or relative to the projection planes using its two orthogonal projections ( A 1 =(X A ; Y A); A 2 =(X A ; Z A)). On an orthogonal drawing, using two projections of a point, you can determine its three coordinates and, conversely, using three coordinates of a point, build its projections (Figure 1.9, a and b).

Figure 1.9 - Plotting a point according to its coordinates

By the location on the projection diagram of a point, one can judge its location in space:

• BUT — BUT 1 lies under the coordinate axis X, and the front BUT 2 - above the axis X, then we can say that the point BUT belongs to the 1st quadrant;
• if on the plot the horizontal projection of the point BUT — BUT 1 lies above the coordinate axis X, and the front BUT 2 - under the axle X, then the point BUT belongs to the 3rd quadrant;
• BUT — BUT 1 and BUT 2 lie above the axis X, then the point BUT belongs to the 2nd quadrant;
• if on the diagram there are horizontal and frontal projections of the point BUT — BUT 1 and BUT 2 lie under the axle X, then the point BUT belongs to the 4th quadrant;
• if on the diagram the projection of a point coincides with the point itself, then it means that the point belongs to the plane of projections;
• a point belonging to the projection plane or projection axis (coordinate axes) is called private point.

To determine in which quadrant of space a point is located, it is enough to determine the sign of the coordinates of the point.

Dependences of the quadrant of the position of the point and the signs of the coordinates

	X	Y	Z
I	+	+	+
II	+	—	+
III	+	—	—
IV	+	+	—

The exercise

Build orthogonal projections points with coordinates BUT(60, 20, 40) and determine in which quadrant the point is located.

Problem solution: along the axis OX set aside the value of the coordinate XA=60, then through this point on the axis OX restore the projection connection line perpendicular to OX, along which to set aside the value of the coordinate ZA=40, and down - the value of the coordinate YA=20(Figure 1.10). All coordinates are positive, which means that the point is located in the I quadrant.

Figure 1.10 - Solution of the problem

1.5. Tasks for independent solution

1. Based on the diagram, determine the position of the point relative to the projection planes (Figure 1.11).

Figure 1.11

2. Complete the missing orthogonal projections of points BUT, IN, FROM on the projection plane π 1 , π 2 , π 3 (Figure 1.12).

Figure 1.12

3. Build point projections:

• E, symmetric point BUT relative to the projection plane π 1 ;
• F, symmetric point IN relative to the plane of projections π 2 ;
• G, symmetric point FROM relative to the projection axis π 2 /π 1 ;
• H, symmetric point D relative to the bisector plane of the second and fourth quadrants.

4. Construct orthogonal projections of the point TO, located in the second quadrant and remote from the projection planes π 1 by 40 mm, from π 2 - by 15 mm.

In this article, we will find answers to questions about how to create a projection of a point onto a plane and how to determine the coordinates of this projection. In the theoretical part, we will rely on the concept of projection. We will give definitions of terms, accompany the information with illustrations. Let's consolidate the acquired knowledge by solving examples.

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Projection, types of projection

For convenience of consideration of spatial figures, drawings depicting these figures are used.

Definition 1

Projection of a figure onto a plane- a drawing of a spatial figure.

Obviously, there are a number of rules used to construct a projection.

Definition 2

projection- the process of constructing a drawing of a spatial figure on a plane using construction rules.

Projection plane is the plane in which the image is built.

The use of certain rules determines the type of projection: central or parallel.

A special case of parallel projection is perpendicular projection or orthogonal projection: in geometry, it is mainly used. For this reason, the adjective “perpendicular” itself is often omitted in speech: in geometry they simply say “projection of a figure” and mean by this the construction of a projection by the method of perpendicular projection. In special cases, of course, otherwise can be stipulated.

We note the fact that the projection of a figure onto a plane is, in fact, the projection of all points of this figure. Therefore, in order to be able to study a spatial figure in a drawing, it is necessary to acquire the basic skill of projecting a point onto a plane. What we will talk about below.

Recall that most often in geometry, speaking of projection onto a plane, they mean the use of perpendicular projection.

We will make constructions that will enable us to obtain the definition of the projection of a point onto a plane.

Suppose a three-dimensional space is given, and in it - a plane α and a point M 1 that does not belong to the plane α. Draw a straight line through a given point M 1 but perpendicular to the given plane α. The point of intersection of the line a and the plane α will be denoted as H 1 , by construction it will serve as the base of the perpendicular dropped from the point M 1 to the plane α .

If a point M 2 is given, belonging to a given plane α, then M 2 will serve as a projection of itself onto the plane α.

Definition 3

is either the point itself (if it belongs to a given plane), or the base of the perpendicular dropped from a given point to a given plane.

Finding the coordinates of the projection of a point on a plane, examples

Let in three-dimensional space given: rectangular coordinate system O x y z, plane α, point M 1 (x 1, y 1, z 1) . It is necessary to find the coordinates of the projection of the point M 1 onto a given plane.

The solution obviously follows from the above definition of the projection of a point onto a plane.

We denote the projection of the point M 1 onto the plane α as H 1 . According to the definition, H 1 is the point of intersection of the given plane α and the line a through the point M 1 (perpendicular to the plane). Those. the coordinates of the projection of the point M 1 we need are the coordinates of the point of intersection of the line a and the plane α.

Thus, to find the coordinates of the projection of a point onto a plane, it is necessary:

Get the equation of the plane α (in case it is not set). An article about the types of plane equations will help you here;

Determine the equation of a straight line apassing through the point M 1 and perpendicular to the plane α (study the topic of the equation of a straight line passing through a given point perpendicular to a given plane);

Find the coordinates of the point of intersection of the line a and the plane α (article - finding the coordinates of the point of intersection of the plane and the line). The data obtained will be the coordinates of the projection of the point M 1 onto the plane α that we need.

Let's consider the theory on practical examples.

Example 1

Determine the coordinates of the projection of the point M 1 (- 2, 4, 4) onto the plane 2 x - 3 y + z - 2 \u003d 0.

Solution

As we can see, the equation of the plane is given to us, i.e. there is no need to compose it.

Let's write the canonical equations of the straight line a passing through the point M 1 and perpendicular to the given plane. For these purposes, we determine the coordinates of the directing vector of the straight line a. Since the line a is perpendicular to the given plane, then the directing vector of the line a is the normal vector of the plane 2 x - 3 y + z - 2 = 0. In this way, a → = (2 , - 3 , 1) – direction vector of the line a .

Now we compose the canonical equations of a straight line in space passing through the point M 1 (- 2, 4, 4) and having a direction vector a → = (2 , - 3 , 1) :

x + 2 2 = y - 4 - 3 = z - 4 1

To find the desired coordinates, the next step is to determine the coordinates of the point of intersection of the line x + 2 2 = y - 4 - 3 = z - 4 1 and the plane 2 x - 3 y + z - 2 = 0 . To this end, we pass from the canonical equations to the equations of two intersecting planes:

x + 2 2 = y - 4 - 3 = z - 4 1 ⇔ - 3 (x + 2) = 2 (y - 4) 1 (x + 2) = 2 (z - 4) 1 ( y - 4) = - 3 (z + 4) ⇔ 3 x + 2 y - 2 = 0 x - 2 z + 10 = 0

Let's make a system of equations:

3 x + 2 y - 2 = 0 x - 2 z + 10 = 0 2 x - 3 y + z - 2 = 0 ⇔ 3 x + 2 y = 2 x - 2 z = - 10 2 x - 3 y + z = 2

And solve it using Cramer's method:

∆ = 3 2 0 1 0 - 2 2 - 3 1 = - 28 ∆ x = 2 2 0 - 10 0 - 2 2 - 3 1 = 0 ⇒ x = ∆ x ∆ = 0 - 28 = 0 ∆ y = 3 2 0 1 - 10 - 2 2 2 1 = - 28 ⇒ y = ∆ y ∆ = - 28 - 28 = 1 ∆ z = 3 2 2 1 0 - 10 2 - 3 2 = - 140 ⇒ z = ∆ z ∆ = - 140 - 28 = 5

Thus, the desired coordinates of a given point M 1 on a given plane α will be: (0, 1, 5) .

Answer: (0 , 1 , 5) .

Example 2

Points А (0 , 0 , 2) are given in a rectangular coordinate system O x y z of three-dimensional space; In (2, - 1, 0) ; C (4, 1, 1) and M 1 (-1, -2, 5). It is necessary to find the coordinates of the projection M 1 onto the plane A B C

Solution

First of all, we write the equation of a plane passing through three given points:

x - 0 y - 0 z - 0 2 - 0 - 1 - 0 0 - 2 4 - 0 1 - 0 1 - 2 = 0 ⇔ xyz - 2 2 - 1 - 2 4 1 - 1 = 0 ⇔ ⇔ 3 x - 6y + 6z - 12 = 0 ⇔ x - 2y + 2z - 4 = 0

Let's write down parametric equations straight line a, which will pass through the point M 1 perpendicular to the plane A B C. The plane x - 2 y + 2 z - 4 \u003d 0 has a normal vector with coordinates (1, - 2, 2), i.e. vector a → = (1 , - 2 , 2) – direction vector of the line a .

Now, having the coordinates of the point of the line M 1 and the coordinates of the directing vector of this line, we write the parametric equations of the line in space:

Then we determine the coordinates of the point of intersection of the plane x - 2 y + 2 z - 4 = 0 and the line

x = - 1 + λ y = - 2 - 2 λ z = 5 + 2 λ

To do this, we substitute into the equation of the plane:

x = - 1 + λ , y = - 2 - 2 λ , z = 5 + 2 λ

Now, using the parametric equations x = - 1 + λ y = - 2 - 2 λ z = 5 + 2 λ, we find the values ​​of the variables x, y and z at λ = - 1: x = - 1 + (- 1) y = - 2 - 2 (- 1) z = 5 + 2 (- 1) ⇔ x = - 2 y = 0 z = 3

Thus, the projection of the point M 1 onto the plane A B C will have coordinates (- 2, 0, 3) .

Answer: (- 2 , 0 , 3) .

Let us dwell separately on the question of finding the coordinates of the projection of a point on the coordinate planes and planes that are parallel to the coordinate planes.

Let points M 1 (x 1, y 1, z 1) and coordinate planes O x y , O x z and O y z be given. The projection coordinates of this point on these planes will be respectively: (x 1 , y 1 , 0) , (x 1 , 0 , z 1) and (0 , y 1 , z 1) . Consider also the planes parallel to the given coordinate planes:

C z + D = 0 ⇔ z = - D C , B y + D = 0 ⇔ y = - D B

And the projections of the given point M 1 on these planes will be points with coordinates x 1 , y 1 , - D C , x 1 , - D B , z 1 and - D A , y 1 , z 1 .

Let us demonstrate how this result was obtained.

As an example, let's define the projection of the point M 1 (x 1, y 1, z 1) onto the plane A x + D = 0. The rest of the cases are similar.

The given plane is parallel to the coordinate plane O y z and i → = (1 , 0 , 0) is its normal vector. The same vector serves as the directing vector of the straight line perpendicular to the plane O y z . Then the parametric equations of a straight line drawn through the point M 1 and perpendicular to a given plane will look like:

x = x 1 + λ y = y 1 z = z 1

Find the coordinates of the point of intersection of this line and the given plane. We first substitute into the equation A x + D = 0 the equalities: x = x 1 + λ, y = y 1, z = z 1 and get: A (x 1 + λ) + D = 0 ⇒ λ = - DA - x one

Then we calculate the desired coordinates using the parametric equations of the straight line for λ = - D A - x 1:

x = x 1 + - D A - x 1 y = y 1 z = z 1 ⇔ x = - D A y = y 1 z = z 1

That is, the projection of the point M 1 (x 1, y 1, z 1) onto the plane will be a point with coordinates - D A , y 1 , z 1 .

Example 2

It is necessary to determine the coordinates of the projection of the point M 1 (- 6 , 0 , 1 2) onto the coordinate plane O x y and onto the plane 2 y - 3 = 0 .

Solution

The coordinate plane O x y will correspond to an incomplete general equation plane z = 0 . The projection of the point M 1 onto the plane z \u003d 0 will have coordinates (- 6, 0, 0) .

The plane equation 2 y - 3 = 0 can be written as y = 3 2 2 . Now just write the coordinates of the projection of the point M 1 (- 6 , 0 , 1 2) onto the plane y = 3 2 2:

6 , 3 2 2 , 1 2

Answer:(- 6 , 0 , 0) and - 6 , 3 2 2 , 1 2

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