The projection of a point on three planes of projections of the coordinate angle begins with obtaining its image on the plane H - the horizontal plane of projections. To do this, through point A (Fig. 4.12, a) a projecting beam is drawn perpendicular to the plane H.

In the figure, the perpendicular to the H plane is parallel to the Oz axis. The point of intersection of the beam with the plane H (point a) is chosen arbitrarily. The segment Aa determines how far point A is from the plane H, thus indicating unambiguously the position of point A in the figure with respect to the projection planes. Point a is a rectangular projection of point A onto the plane H and is called the horizontal projection of point A (Fig. 4.12, a).

To obtain an image of point A on the plane V (Fig. 4.12, b), a projecting beam is drawn through point A perpendicular to the frontal projection plane V. In the figure, the perpendicular to the plane V is parallel to the Oy axis. On the H plane, the distance from point A to plane V will be represented by a segment aa x, parallel to the Oy axis and perpendicular to the Ox axis. If we imagine that the projecting beam and its image are carried out simultaneously in the direction of the plane V, then when the image of the beam intersects the Ox axis at the point a x, the beam intersects the plane V at the point a. Drawing from the point a x in the V plane perpendicular to the Ox axis , which is the image of the projecting beam Aa on the plane V, the point a is obtained at the intersection with the projecting beam. Point a "is the frontal projection of point A, i.e. its image on the plane V.

The image of point A on the profile plane of projections (Fig. 4.12, c) is built using a projecting beam perpendicular to the W plane. In the figure, the perpendicular to the W plane is parallel to the Ox axis. The projecting beam from point A to plane W on the plane H will be represented by a segment aa y, parallel to the Ox axis and perpendicular to the Oy axis. From the point Oy parallel to the Oz axis and perpendicular to the Oy axis, an image of the projecting beam aA is built and, at the intersection with the projecting beam, the point a is obtained. Point a is the profile projection of the point A, i.e., the image of the point A on the plane W.

The point a "can be constructed by drawing from the point a" the segment a "a z (the image of the projecting beam Aa" on the plane V) parallel to the Ox axis, and from the point a z - the segment a "a z parallel to the Oy axis until it intersects with the projecting beam.

Having received three projections of point A on the projection planes, the coordinate angle is deployed into one plane, as shown in Fig. 4.11, b, together with the projections of the point A and the projecting rays, and the point A and the projecting rays Aa, Aa "and Aa" are removed. The edges of the combined projection planes are not carried out, but only the projection axes Oz, Oy and Ox, Oy 1 (Fig. 4.13) are carried out.

An analysis of the orthogonal drawing of a point shows that three distances - Aa", Aa and Aa" (Fig. 4.12, c), characterizing the position of point A in space, can be determined by discarding the projection object itself - point A, on a coordinate angle deployed in one plane (Fig. 4.13). The segments a "a z, aa y and Oa x are equal to Aa" as opposite sides of the corresponding rectangles (Fig. 4.12, c and 4.13). They determine the distance at which point A is located from the profile plane of projections. Segments a "a x, a" a y1 and Oa y are equal to segment Aa, determine the distance from point A to the horizontal plane of projections, segments aa x, a "a z and Oa y 1 are equal to segment Aa", which determines the distance from point A to frontal projection plane.

The segments Oa x, Oa y and Oa z located on the projection axes are a graphic expression of the sizes of the X, Y and Z coordinates of point A. The point coordinates are denoted with the index of the corresponding letter. By measuring the size of these segments, you can determine the position of the point in space, i.e., set the coordinates of the point.

On the diagram, the segments a "a x and aa x are arranged as one line perpendicular to the Ox axis, and the segments a" a z and a "a z - to the Oz axis. These lines are called projection connection lines. They intersect the projection axes at points a x and and z, respectively.The line of the projection connection connecting the horizontal projection of point A with the profile one turned out to be “cut” at the point a y.

Two projections of the same point are always located on the same projection connection line perpendicular to the projection axis.

To represent the position of a point in space, two of its projections and a given origin (point O) are sufficient. 4.14, b, two projections of a point completely determine its position in space. Using these two projections, you can build a profile projection of point A. Therefore, in the future, if there is no need for a profile projection, diagrams will be built on two projection planes: V and H.

Rice. 4.14. Rice. 4.15.

Let's consider several examples of building and reading a drawing of a point.

Example 1 Determination of the coordinates of the point J given on the diagram by two projections (Fig. 4.14). Three segments are measured: segment Ov X (X coordinate), segment b X b (Y coordinate) and segment b X b "(Z coordinate). Coordinates are written in the following order: X, Y and Z, after the letter designation of the point, for example , B20; 30; 15.

Example 2. Construction of a point according to the given coordinates. Point C is given by coordinates C30; ten; 40. On the Ox axis (Fig. 4.15) find a point with x, at which the line of the projection connection intersects the projection axis. To do this, the X coordinate (size 30) is plotted along the Ox axis from the origin (point O) and a point with x is obtained. Through this point, perpendicular to the Ox axis, a projection connection line is drawn and the Y coordinate is laid down from the point (size 10), the point c is obtained - the horizontal projection of the point C. The coordinate Z (size 40) is plotted upwards from the point c x along the projection connection line (size 40), the point is obtained c" - frontal projection of point C.

Example 3. Construction of a profile projection of a point according to the given projections. The projections of the point D - d and d are set. Through the point O, the projection axes Oz, Oy and Oy 1 are drawn (Fig. 4.16, a). it to the right behind the Oz axis. The profile projection of the point D will be located on this line. It will be located at the same distance from the Oz axis as the horizontal projection of the point d is located: from the Ox axis, i.e. at a distance dd x. The segments d z d "and dd x are the same, since they determine the same distance - the distance from point D to the frontal projection plane. This distance is the Y coordinate of point D.

Graphically, the segment d z d "is built by transferring the segment dd x from the horizontal plane of projections to the profile one. To do this, draw a line of projection connection parallel to the Ox axis, get a point d y on the Oy axis (Fig. 4.16, b). Then transfer the size of the segment Od y to the Oy 1 axis , drawing from the point O an arc with a radius equal to the segment Od y, until it intersects with the axis Oy 1 (Fig. 4.16, b), get the point dy 1. This point can also be constructed, as shown in Fig. 4.16, c, by drawing a straight line at an angle 45 ° to the Oy axis from the point d y. From the point d y1 draw a projection connection line parallel to the Oz axis and lay a segment on it equal to the segment d "d x, get the point d".

Transferring the value of the segment d x d to the profile plane of the projections can be done using a constant straight line drawing (Fig. 4.16, d). In this case, the projection connection line dd y is drawn through the horizontal projection of the point parallel to the Oy 1 axis until it intersects with a constant straight line, and then parallel to the Oy axis until it intersects with the continuation of the projection connection line d "d z.

Particular cases of the location of points relative to projection planes

The position of a point relative to the projection plane is determined by the corresponding coordinate, i.e., the value of the segment of the projection connection line from the Ox axis to the corresponding projection. On fig. 4.17 the Y coordinate of point A is determined by the segment aa x - the distance from point A to plane V. The Z coordinate of point A is determined by the segment a "a x - the distance from point A to plane H. If one of the coordinates is zero, then the point is located on the projection plane Fig. 4.17 shows examples of different locations of points relative to the projection planes.The Z coordinate of point B is zero, the point is in plane H. Its frontal projection is on the Ox axis and coincides with point b x. The Y coordinate of point C is zero, the point is located on the plane V, its horizontal projection c is on the x-axis and coincides with the point c x.

Therefore, if a point is on the projection plane, then one of the projections of this point lies on the projection axis.

On fig. 4.17, the Z and Y coordinates of point D are zero, therefore, point D is on the projection axis Ox and its two projections coincide.

A point in space is defined by any two of its projections. If it is necessary to build a third projection according to two given ones, it is necessary to use the correspondence of the segments of the projection connection lines obtained when determining the distances from a point to the projection plane (see Fig. 2.27 and Fig. 2.28).

Examples of problem solving in the I octant

Given A 1 ; A 2	Build A 3
Given A 2 ; A 3	Build A 1
Given A 1 ; A 3	Build A 2

Consider the algorithm for constructing point A (Table 2.5)

Table 2.5

Algorithm for constructing point A
according to the given coordinates A ( x = 5, y = 20, z = -9)

In the following chapters we will consider images: straight lines and planes only in the first quarter. Although all the methods under consideration can be applied in any quarter.

findings

Thus, based on the theory of G. Monge, it is possible to transform the spatial image of the image (point) into a planar one.

This theory is based on the following points:

1. The whole space is divided into 4 quarters with the help of two mutually perpendicular planes p 1 and p 2, or into 8 octants by adding a third mutually perpendicular plane p 3 .

2. The image of a spatial image on these planes is obtained using a rectangular (orthogonal) projection.

3. To convert a spatial image into a planar image, it is considered that the p 2 plane is stationary, and the p 1 plane rotates around the axis x so that the positive half-plane p 1 coincides with the negative half-plane p 2 , the negative part of p 1 coincides with the positive part p 2 .

4. The p 3 plane rotates around the axis z(lines of intersection of the planes) until aligned with the plane p 2 (see Fig. 2.31).

Images obtained on the planes p 1 , p 2 and p 3 with a rectangular projection of images are called projections.

Planes p 1 , p 2 and p 3 together with the projections depicted on them form a planar complex drawing or diagrams.

Lines connecting the projections of the image ^ to the axes x, y, z, are called projection lines.

For a more accurate definition of images in space, a system of three mutually perpendicular planes p 1 , p 2 , p 3 can be applied.

Depending on the condition of the problem, you can choose for the image either the system p 1 , p 2 or p 1 , p 2 , p 3 .

The system of planes p 1 , p 2 , p 3 can be connected to the system of Cartesian coordinates, which makes it possible to specify objects not only graphically or (verbally) but also analytically (using numbers).

This way of depicting images, in particular points, makes it possible to solve such positional problems as:

• the location of the point relative to the projection planes (general position, belonging to the plane, axis);
• position of the point in quarters (in which quarter the point is located);
• the position of the points relative to each other (higher, lower, closer, farther relative to the planes of projections and the viewer);
• the position of the point projections relative to the projection planes (equal distance, closer, farther).

Metric tasks:

• equidistance of the projection from the projection planes;
• the ratio of projection removal from the projection planes (2–3 times, more, less);
• determination of the distance of a point from the projection planes (when introducing a coordinate system).

Questions for introspection

1. The line of intersection of which planes is the axis z?

2. The line of intersection of which planes is the axis y?

3. How is the line of projection connection of the frontal and profile projection of the point located? Show.

4. What coordinates determine the position of the point projection: horizontal, frontal, profile?

5. In what quarter is point F (10; -40; -20)? From which projection plane is point F farthest away?

6. The distance from which projection to which axis determines the distance of the point from the plane p 1 ? What is the coordinate of the point is this distance?

Consider the projections of points onto two planes, for which we take two perpendicular planes (Fig. 4), which we will call the horizontal frontal and planes. The line of intersection of these planes is called the projection axis. We project one point A onto the considered planes using a flat projection. To do this, it is necessary to lower the perpendiculars Aa and A from the given point onto the considered planes.

Projection onto a horizontal plane is called plan view points BUT, and the projection a? on the frontal plane is called front projection.

Points that are to be projected in descriptive geometry are usually denoted using capital Latin letters. A, B, C. Small letters are used to designate horizontal projections of points. a, b, c... Frontal projections are indicated in small letters with a stroke at the top a?, b?, c?…

The designation of points with Roman numerals I, II, ... is also used, and for their projections - with Arabic numerals 1, 2 ... and 1?, 2? ...

When the horizontal plane is rotated by 90°, a drawing can be obtained in which both planes are in the same plane (Fig. 5). This picture is called point plot.

Through perpendicular lines Ah and ah? draw a plane (Fig. 4). The resulting plane is perpendicular to the frontal and horizontal planes because it contains perpendiculars to these planes. Therefore, this plane is perpendicular to the line of intersection of the planes. The resulting straight line intersects the horizontal plane in a straight line aa x, and the frontal plane - in a straight line huh? X. Straight aah and huh? x are perpendicular to the axis of intersection of the planes. I.e Aaah? is a rectangle.

When combining the horizontal and frontal projection planes a and a? will lie on one perpendicular to the axis of intersection of the planes, since when the horizontal plane rotates, the perpendicularity of the segments aa x and huh? x is not broken.

We get that on the projection diagram a and a? some point BUT always lie on the same perpendicular to the axis of intersection of the planes.

Two projections a and a? of some point A can uniquely determine its position in space (Fig. 4). This is confirmed by the fact that when constructing a perpendicular from the projection a to the horizontal plane, it will pass through point A. Similarly, the perpendicular from the projection a? to the frontal plane will pass through the point BUT, i.e. point BUT lies on two definite lines at the same time. Point A is their intersection point, i.e. it is definite.

Consider a rectangle Aaa X a?(Fig. 5), for which the following statements are true:

1) Point distance BUT from the frontal plane is equal to the distance of its horizontal projection a from the axis of intersection of the planes, i.e.

ah? = aa X;

2) point distance BUT from the horizontal plane of projections is equal to the distance of its frontal projection a? from the axis of intersection of the planes, i.e.

Ah = huh? X.

In other words, even without the point itself on the plot, using only its two projections, you can find out at what distance from each of the projection planes this point is located.

The intersection of two projection planes divides the space into four parts, which are called quarters(Fig. 6).

The axis of intersection of the planes divides the horizontal plane into two quarters - the front and rear, and the frontal plane into the upper and lower quarters. The upper part of the frontal plane and the anterior part of the horizontal plane are considered as the boundaries of the first quarter.

Upon receipt of the diagram, the horizontal plane rotates and coincides with the frontal plane (Fig. 7). In this case, the front of the horizontal plane will coincide with the bottom of the frontal plane, and the back of the horizontal plane with the top of the frontal plane.

Figures 8-11 show points A, B, C, D, located in different quarters of space. Point A is in the first quarter, point B is in the second, point C is in the third, and point D is in the fourth.

When the points are located in the first or fourth quarters of their horizontal projections located on the front of the horizontal plane, and on the diagram they will lie below the axis of intersection of the planes. When a point is located in the second or third quarter, its horizontal projection will lie on the back of the horizontal plane, and on the plot it will be above the axis of intersection of the planes.

Front projections points that are located in the first or second quarters will lie on the upper part of the frontal plane, and on the diagram they will be located above the axis of intersection of the planes. When a point is located in the third or fourth quarter, its frontal projection is below the axis of intersection of the planes.

Most often, in real constructions, the figure is placed in the first quarter of the space.

In some particular cases, the point ( E) may lie on a horizontal plane (Fig. 12). In this case, its horizontal projection e and the point itself will coincide. The frontal projection of such a point will be on the axis of the intersection of the planes.

In the case where the point To lies on the frontal plane (Fig. 13), its horizontal projection k lies on the axis of intersection of the planes, and the frontal k? shows the actual location of that point.

For such points, the sign that it lies on one of the projection planes is that one of its projections is on the axis of intersection of the planes.

If a point lies on the intersection axis of the projection planes, it and both of its projections coincide.

When a point does not lie on the projection planes, it is called point of general position. In what follows, if there are no special marks, the point under consideration is a point in general position.

2. Lack of projection axis

To explain how to obtain on the model projections of a point onto perpendicular projection planes (Fig. 4), it is necessary to take a piece of thick paper in the form of an elongated rectangle. It needs to be bent between projections. The fold line will depict the axis of the intersection of the planes. If after that the bent piece of paper is straightened again, we get a diagram similar to the one shown in the figure.

Combining two projection planes with the drawing plane, you can not show the fold line, i.e., do not draw the axis of intersection of the planes on the diagram.

When constructing on a diagram, you should always place projections a and a? point A on one vertical line (Fig. 14), which is perpendicular to the axis of intersection of the planes. Therefore, even if the position of the axis of intersection of the planes remains undefined, but its direction is determined, the axis of the intersection of the planes can only be perpendicular to the straight line on the diagram ah?.

If there is no projection axis on the point diagram, as in the first figure 14 a, you can imagine the position of this point in space. To do this, draw in any place perpendicular to the line ah? projection axis, as in the second figure (Fig. 14) and bend the drawing along this axis. If we restore the perpendiculars at the points a and a? before they intersect, you can get a point BUT. When changing the position of the projection axis, different positions of the point relative to the projection planes are obtained, but the uncertainty of the position of the projection axis does not affect the relative position of several points or figures in space.

3. Projections of a point onto three projection planes

Consider the profile plane of projections. Projections on two perpendicular planes usually determine the position of the figure and make it possible to find out its real dimensions and shape. But there are times when two projections are not enough. Then apply the construction of the third projection.

The third projection plane is carried out so that it is perpendicular to both projection planes at the same time (Fig. 15). The third plane is called profile.

In such constructions, the common line of the horizontal and frontal planes is called axis X , the common line of the horizontal and profile planes - axis at , and the common straight line of the frontal and profile planes - axis z . Dot O, which belongs to all three planes, is called the point of origin.

Figure 15a shows the point BUT and three of its projections. Projection onto the profile plane ( a??) are called profile projection and denote a??.

To obtain a diagram of point A, which consists of three projections a, a a, it is necessary to cut the trihedron formed by all planes along the y axis (Fig. 15b) and combine all these planes with the plane of the frontal projection. The horizontal plane must be rotated about the axis X, and the profile plane is near the axis z in the direction indicated by the arrow in Figure 15.

Figure 16 shows the position of the projections ah, huh? and a?? points BUT, obtained as a result of combining all three planes with the drawing plane.

As a result of the cut, the y-axis occurs on the diagram in two different places. On a horizontal plane (Fig. 16), it takes a vertical position (perpendicular to the axis X), and on the profile plane - horizontal (perpendicular to the axis z).

Figure 16 shows three projections ah, huh? and a?? points A have a strictly defined position on the diagram and are subject to unambiguous conditions:

a and a? must always be located on one vertical straight line perpendicular to the axis X;

a? and a?? must always be located on the same horizontal line perpendicular to the axis z;

3) when drawn through a horizontal projection and a horizontal line, but through a profile projection a??- a vertical straight line, the constructed lines will necessarily intersect on the bisector of the angle between the projection axes, since the figure Oa at a 0 a n is a square.

When constructing three projections of a point, it is necessary to check the fulfillment of all three conditions for each point.

4. Point coordinates

The position of a point in space can be determined using three numbers called its coordinates. Each coordinate corresponds to the distance of a point from some projection plane.

Point distance BUT to the profile plane is the coordinate X, wherein X = huh?(Fig. 15), the distance to the frontal plane - by the coordinate y, and y = huh?, and the distance to the horizontal plane is the coordinate z, wherein z = aA.

In Figure 15, point A occupies the width of a rectangular box, and the measurements of this box correspond to the coordinates of this point, i.e., each of the coordinates is presented in Figure 15 four times, i.e.:

x \u003d a? A \u003d Oa x \u003d a y a \u003d a z a?;

y \u003d a? A \u003d Oa y \u003d a x a \u003d a z a?;

z = aA = Oa z = a x a? = a y a?.

On the diagram (Fig. 16), the x and z coordinates occur three times:

x \u003d a z a? \u003d Oa x \u003d a y a,

z = a x a? = Oa z = a y a?.

All segments that correspond to the coordinate X(or z) are parallel to each other. Coordinate at represented twice by the vertical axis:

y \u003d Oa y \u003d a x a

and twice - located horizontally:

y \u003d Oa y \u003d a z a?.

This difference appeared due to the fact that the y-axis is present on the diagram in two different positions.

It should be noted that the position of each projection is determined on the diagram by only two coordinates, namely:

1) horizontal - coordinates X and at,

2) frontal - coordinates x and z,

3) profile - coordinates at and z.

Using coordinates x, y and z, you can build projections of a point on the diagram.

If point A is given by coordinates, their record is defined as follows: A ( X; y; z).

When constructing point projections BUT the following conditions must be checked:

1) horizontal and frontal projections a and a? X X;

2) frontal and profile projections a? and a? should be located on the same perpendicular to the axis z, since they have a common coordinate z;

3) horizontal projection and also removed from the axis X, like the profile projection a away from axis z, since the projection ah? and huh? have a common coordinate at.

If the point lies in any of the projection planes, then one of its coordinates is equal to zero.

When a point lies on the projection axis, its two coordinates are zero.

If a point lies at the origin, all three of its coordinates are zero.

The surfaces of polyhedra are known to be limited to flat figures. Therefore, points given on the surface of a polyhedron by at least one projection are, in the general case, definite points. The same applies to the surfaces of other geometric bodies: cylinder, cone, ball and torus, bounded by curved surfaces.

Let us agree to depict visible points lying on the surface of the body as circles, invisible points as blackened circles (dots); Visible lines will be shown as solid lines, and invisible lines as dashed lines.

Let the horizontal projection A 1 of point A lying on the surface of a straight triangular prism be given (Fig. 162, a).

TBegin-->TEnd-->

As can be seen from the drawing, the front and rear bases of the prism are parallel to the frontal projection plane P 2 and are projected onto it without distortion, the lower side face of the prism is parallel to the horizontal projection plane P 1 and is also projected without distortion. The lateral edges of the prism are frontally projecting straight lines, therefore they are projected onto the frontal projection plane P 2 in the form of points.

Since the projection A 1 . is depicted by a light circle, then point A is visible and, therefore, is located on the right side face of the prism. This face is a frontal projection plane, and the frontal projection A2 of the point must coincide with the frontal projection of the plane represented by a straight line.

Having drawn a constant straight line k 123, we find the third projection A 3 of point A. When projected onto the profile plane of projections, point A will be invisible, therefore point A 3 is shown as a black circle. Specifying a point by frontal projection B 2 is undefined, since it does not determine the distance of point B from the front base of the prism.

Let's build an isometric projection of the prism and point A (Fig. 162, b). It is convenient to start construction from the front base of the prism. We build a triangle of the base according to the dimensions taken from the complex drawing; along the y-axis "we set aside the size of the edge of the prism. We build the axonometric image A" of point A using the coordinate polyline circled in both drawings with a double thin line.

Let the frontal projection C 2 of point C be given, lying on the surface of a regular quadrangular pyramid, given by two main projections (Fig. 163, a). It is required to build three projections of point C.

From the frontal projection, it can be seen that the top of the pyramid is higher than the square base of the pyramid. Under this condition, all four side faces will be visible when projected onto the horizontal projection plane П 1 . When projecting onto the frontal projection plane P 2, only the front face of the pyramid will be visible. Since the projection C 2 is shown in the drawing as a light circle, the point C is visible and belongs to the front face of the pyramid. To build a horizontal projection C 1, we draw an auxiliary line D 2 E 2 through the point C 2, parallel to the line of the base of the pyramid. We find its horizontal projection D 1 E 1 and point C 1 on it. If there is a third projection of the pyramid, we find the horizontal projection of point C 1 more simply: having found the profile projection C 3, we build the third one using two projections using horizontal and horizontal-vertical communication lines. The construction progress is shown in the drawing by arrows.

TBegin-->
Tend-->

Let's build a dimetric projection of the pyramid and point C (Fig. 163, b). We build the base of the pyramid; for this, through the point O "taken on the r axis", we draw the x" and y" axes; on the x-axis "we set aside the actual dimensions of the base, and on the y-axis" - halved. Through the obtained points we draw straight lines parallel to the axes x "and y". On the z-axis, we plot the height of the pyramid; we connect the resulting point with the base points, taking into account the visibility of the edges. To construct point C, we use the coordinate polyline circled in the drawings by a double thin line. To check the accuracy of the solution, we draw a straight line D "E" through the found point C, parallel x axis". Its length must be equal to the length of the straight line D 2 E 2 (or D 1 E 1).

The position of a point in space can be specified by its two orthogonal projections, for example, horizontal and frontal, frontal and profile. The combination of any two orthogonal projections allows you to find out the value of all coordinates of a point, build a third projection, determine the octant in which it is located. Let's consider some typical tasks from the course of descriptive geometry.

According to the given complex drawing of points A and B, it is necessary:

Let us first determine the coordinates of point A, which can be written in the form A (x, y, z). The horizontal projection of point A is point A ", having coordinates x, y. Draw from point A" perpendiculars to the x, y axes and find, respectively, A x, A y. The x-coordinate for point A is equal to the length of the segment A x O with a plus sign, since A x lies in the region of positive x-axis values. Taking into account the scale of the drawing, we find x \u003d 10. The y coordinate is equal to the length of the segment A y O with a minus sign, since t. A y lies in the region of negative y-axis values. Given the scale of the drawing, y = -30. The frontal projection of point A - point A"" has x and z coordinates. Let's drop the perpendicular from A"" to the z-axis and find A z . The z-coordinate of point A is equal to the length of the segment A z O with a minus sign, since A z lies in the region of negative values ​​of the z-axis. Given the scale of the drawing, z = -10. Thus, the coordinates of point A are (10, -30, -10).

The coordinates of point B can be written as B (x, y, z). Consider the horizontal projection of point B - point B. "Since it lies on the x axis, then B x \u003d B" and the coordinate B y \u003d 0. The abscissa x of point B is equal to the length of the segment B x O with a plus sign. Taking into account the scale of the drawing, x = 30. The frontal projection of the point B - point B˝ has the coordinates x, z. Draw a perpendicular from B"" to the z-axis, thus finding B z . The applicate z of point B is equal to the length of the segment B z O with a minus sign, since B z lies in the region of negative values ​​of the z-axis. Taking into account the scale of the drawing, we determine the value z = -20. So the B coordinates are (30, 0, -20). All necessary constructions are shown in the figure below.

Construction of projections of points

Points A and B in the P 3 plane have the following coordinates: A""" (y, z); B""" (y, z). In this case, A"" and A""" lie on the same perpendicular to the z-axis, since they have a common z-coordinate. In the same way, B"" and B""" lie on a common perpendicular to the z-axis. To find the profile projection of t. A, we set aside along the y-axis the value of the corresponding coordinate found earlier. In the figure, this is done using an arc of a circle of radius A y O. After that, we draw a perpendicular from A y to the intersection with the perpendicular restored from the point A "" to the z axis. The intersection point of these two perpendiculars determines the position of A""".

Point B""" lies on the z-axis, since the y-ordinate of this point is zero. To find the profile projection of point B in this problem, it is only necessary to draw a perpendicular from B"" to the z-axis. The point of intersection of this perpendicular with the z-axis is B """.

Determining the position of points in space

Visually imagining a spatial layout composed of projection planes P 1, P 2 and P 3, the location of octants, as well as the order of transformation of the layout into diagrams, you can directly determine that t. A is located in octant III, and t. B lies in the plane P 2 .

Another option for solving this problem is the method of exceptions. For example, the coordinates of point A are (10, -30, -10). The positive abscissa x makes it possible to judge that the point is located in the first four octants. A negative y-ordinate indicates that the point is in the second or third octant. Finally, the negative applicate of z indicates that point A is in the third octant. The given reasoning is clearly illustrated by the following table.

Octants	Coordinate signs
	x	y	z
1	+	+	+
2	+	–	+
3	+	–	–
4	+	+	–
5	–	+	+
6	–	–	+
7	–	–	–
8	–	+	–

Point B coordinates (30, 0, -20). Since the ordinate of t. B is equal to zero, this point is located in the projection plane П 2 . The positive abscissa and the negative applicate of point B indicate that it is located on the border of the third and fourth octants.

Construction of a visual image of points in the system of planes P 1, P 2, P 3

Using the frontal isometric projection, we built a spatial layout of the third octant. It is a rectangular trihedron, whose faces are the planes P 1, P 2, P 3, and the angle (-y0x) is 45 º. In this system, segments along the x, y, z axes will be plotted in full size without distortion.

The construction of a visual image of point A (10, -30, -10) will begin with its horizontal projection A ". Having set aside the corresponding coordinates along the abscissa and ordinates, we find the points A x and A y. The intersection of perpendiculars restored from A x and A y respectively to the x and y axes determines the position of point A". Putting from A" parallel to the z axis towards its negative values ​​the segment AA", whose length is equal to 10, we find the position of point A.

A visual image of point B (30, 0, -20) is constructed in a similar way - in the P 2 plane, the corresponding coordinates must be set aside along the x and z axes. The intersection of the perpendiculars reconstructed from B x and B z will determine the position of point B.