The construction of a plane p perpendicular to plane a can be done in two ways: I) plane p is drawn through a straight line perpendicular to plane a; 2) plane p is drawn perpendicular to a line lying in plane a or parallel to this plane. To obtain a unique solution, additional conditions are required. Figure 148 shows the construction of a plane perpendicular to the plane defined by the triangle CDE. An additional condition here is that the desired plane must pass through the straight line AB. Consequently, the desired plane is determined by the straight line AB and the perpendicular to the plane of the triangle. To draw this perpendicular to the CDE plane, the fronts CN and the horizontal CM are taken in it: if B"F" ± C"N" and B"G 1 CM\ then BFX of the CDF plane. The plane formed by intersecting straight lines AB and BF is perpendicular to the CDE plane, How does it pass through the perpendicular to this plane? Can the perpendicularity of the same-name traces of the planes serve as a sign of the perpendicularity of the planes themselves? Obvious cases where this is true also include the mutual perpendicularity of two horizontally projecting planes, in which the horizontal traces are mutually perpendicular. with the frontal traces of the frontally projecting planes being mutually perpendicular; these planes are mutually perpendicular. Consider (Figure 149) a horizontally projecting plane p perpendicular to the plane of general position a. If the plane p is perpendicular to the plane i and to the plane a, then p 1 as k. the line of intersection of the plane a and the plane i,. Hence h"0a 1р and, therefore, h"0u 1 р", as to one of the straight lines in the plane р. So, the perpendicularity of the horizontal traces of the general position plane and the horizontally projecting plane corresponds to the mutual perpendicularity of these planes. Obviously, the perpendicularity of the frontal traces of the frontally projecting plane and the general position plane also corresponds to the mutual perpendicularity of these planes. But if the traces of the same name of two planes in general position are mutually perpendicular, then the planes themselves are not perpendicular to each other, since none of the conditions stated at the beginning of this section are met. Questions for self-test 1. How is the plane defined in the drawing? 2. What is the trace of a plane on a projection plane? 3. Where are the frontal projection of the horizontal trace and the horizontal projection of the frontal trace of the plane located? L. How is it determined in the drawing whether a straight line belongs to a given plane? 5. How to construct a point on a drawing that belongs to a given plane? 6. How is nt located in the system? and 713 general position plane? 7. What are frontal projection, horizontal projection and profile projection planes? 8. How is the frotal-projecting plane drawn through a straight line drawn in general position shown in the drawing? 9. What relative position can two planes occupy? 10. What is the sign of parallelism of two planes? 11. How are the traces of the same name of two planes parallel to each other mutually located? 12. How to establish the relative position of a straight line and a plane? 13. What is the general method of constructing the line of intersection of two planes? 14. What is the general method for constructing the point of intersection of a line with a plane? 15. How to determine “visibility” when a line intersects a plane? 16. What determines the mutual parallelism of two planes? 17. How to draw a plane parallel to a given plane through a point? 18. How is the projection of the perpendicular to the plane located? 19. How to construct mutually perpendicular planes?
The construction of mutually perpendicular lines and planes is an important graphic operation in solving metric problems.
The construction of a perpendicular to a line or plane is based on the property of a right angle, which is formulated as follows: if one of the sides of the right angle is parallel to the projection plane and the other is not perpendicular to it, then the angle is projected in full size onto this plane.
Figure 28
Side BC of right angle ABC, shown in Figure 28, is parallel to plane P 1. Consequently, the projection of angle ABC onto this plane will represent a right angle A 1 B 1 C 1 =90.
A line is perpendicular to a plane if it is perpendicular to two intersecting lines lying in this plane. When constructing a perpendicular from a set of straight lines belonging to the plane, choose level straight lines - horizontal and frontal. In this case, the horizontal projection of the perpendicular is carried out perpendicular to the horizontal, and the frontal projection is perpendicular to the front. The example shown in Figure 29 shows the construction of a perpendicular to the plane defined by triangle ABC from point K. To do this, first draw the horizontal and frontal lines in the plane. Then from the frontal projection of point K we draw a perpendicular to the frontal projection of the frontal, and from the horizontal projection of the point - a perpendicular to the horizontal projection of the horizontal. Then we construct the point of intersection of this perpendicular with the plane using the auxiliary cutting plane Σ. The required point is F. Thus, the resulting segment KF is perpendicular to the plane ABC.
Figure 29
Figure 29 shows the construction of a perpendicular KF to the ABC plane.
Two planes are perpendicular if a line lying in one plane is perpendicular to two intersecting lines of the other plane. The construction of a plane perpendicular to this plane ABC is shown in Figure 30. A straight line MN is drawn through point M, perpendicular to the plane ABC. The horizontal projection of this line is perpendicular to AC, since AC is horizontal, and the frontal projection is perpendicular to AB, since AB is frontal. Then an arbitrary straight line EF is drawn through point M. Thus, the plane is perpendicular to ABC and is defined by two intersecting lines EF and MN.
Figure 30
This method is used to determine the natural values of segments in general position, as well as their angles of inclination to projection planes. In order to determine the natural size of a segment using this method, it is necessary to complete a right triangle to one of the projections of the segment. The other leg will be the difference in heights or depths of the end points of the segment, and the hypotenuse will be the natural value.
Let's consider an example: Figure 31 shows a segment AB in general position. It is required to determine its natural size and the angles of its inclination to the frontal and horizontal planes of projections.
We draw a perpendicular to one of the ends of the segment on a horizontal plane. We plot the height difference (ZA-ZB) of the ends of the segment on it and complete the construction of a right triangle. Its hypotenuse is the natural value of the segment, and the angle between the natural value and the projection of the segment is the natural value of the angle of inclination of the segment to the plane P 1. The order of construction on the frontal plane is the same. Along the perpendicular we plot the difference in the depths of the ends of the segment (YA-YB). The resulting angle between the natural size of the segment and its frontal projection is the angle of inclination of the segment to the P 2 plane.
Figure 31
1. State a theorem about the property of right angles.
2. In what case is a straight line perpendicular to a plane?
3. How many straight lines and how many planes perpendicular to a given plane can be drawn through a point in space?
4. What is the right triangle method used for?
5. How to use this method to determine the angle of inclination of a segment in general position to the horizontal plane of projections?
| Verbal form | Graphic form |
| 1. It is known that to construct a straight line perpendicular to a plane, it is necessary to construct a horizontal and a frontal line in the plane. a) Note that the construction of a perpendicular is simplified, since the sides of the plane Q(D ABC) are level straight lines: AB (A 1 B 1; A 2 B 2) – front AC (A 1 C 1; A 2 C 2) – horizontal . b) Take on a straight line l arbitrary point K | |
| 2. Through point K, which belongs to the line l, we conduct a direct n^Q, i.e. n 1 ^ A 1 C 1 and n 2 ^ A 2 B 2 . The desired plane will be determined by two intersecting lines, one of which is given - l, and the other - n is perpendicular to the given plane: P( l n)^Q (D ABC) |  |
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Descriptive geometry - T.V. Khrustaleva
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DESCRIPTIVE GEOMETRY
Recommended by the Far Eastern Regional Educational and Methodological Center as a textbook for students of specialty 210700 “Automation, telemechanics and railway communications”
Geometric images
1. Projection plane: p – arbitrary; p1 – horizontal; p2 – frontal; p3 – profile; S – center projection
Set-theoretic notation
The essence of the projection method is that the projection Ap of some geometric image
Projection central
Central is a projection in which all projecting rays emerge from one point S, called the projection center. In Fig. 1.3 gives an example of central projection, where p is flat
Parallel projection
Parallel projection is a projection in which all projecting rays are parallel to each other. Parallel projections can be oblique (Fig. 1.7) and rectangular (Fig. 1.8).
Properties of orthogonal projections
1. The projection of a point is a point (Fig. 1.9). Rice. 1.9 2. Projection of a line in general
Reversibility of the drawing. Monge method
The method of projecting projections onto one plane, discussed in § 2 and § 3, makes it possible to solve the direct problem (having an object, you can find its projection), but does not allow solving the inverse problem (having
System of two mutually perpendicular planes
The reversibility of the drawing, as mentioned earlier, i.e., the unambiguous determination of the position of a point in space from its projections, can be ensured by projecting onto two mutually perpendicular
System of three mutually perpendicular planes
In practice, research and imaging, a system of two mutually perpendicular planes does not always provide the possibility of an unambiguous solution. So, for example, if you move point A along the axis
Complex drawing and visual representation of a point in octants I–IV
Let's consider an example of constructing points A, B, C, D in various octants (Table 2.4). Table 2.4 Octant Visual representation
General provisions
A line is a one-dimensional geometric image that has a length; the set of all successive positions of a moving point. According to Euclid’s definition: “A line is length without width.” Floor
Direct levels
Definition Visual representation Complex drawing A horizontal line is any line parallel to a horizontal one
Projecting straight lines
Definition Visual representation Complex drawing A horizontally projecting line is a straight line perpendicular to
Construction of the third projection of a segment based on two given
In our example, we will consider the construction of a general line in the first quarter (Table 3.3). Table 3.3 Verbal form
Right triangle method. Determination of the natural size of a straight line segment and the angles of inclination of the straight line to the projection planes
Constructing projections of a straight line segment in general and particular position makes it possible to solve not only positional problems (location relative to the projection planes), but also metric ones - determining the length from
Determination of the natural value of a line segment in general position
To determine the natural value of a segment of a straight line in general position from its projections, the right-angled triangle method is used. Let us consider the sequence of this position (Table.
General provisions
Two straight lines in space can have different locations: intersect (lie in the same plane). A special case of intersection is at a right angle; may be parallel
Determining the visibility of lines relative to projection planes
Competing points are used to determine the visibility of lines relative to projection planes. Let's consider a complex drawing of intersecting straight lines a and b (Fig. 4.1 and Fig. 4.2). Let's determine which
Algorithm for constructing intersecting lines
Verbal form Graphic form 1. Through point K draw a straight line h|| p1 and intersecting line a
Projecting planes
Definition Visual image Complex drawing A horizontally projecting plane is a plane perpendicular
Level planes
Characteristics Visual representation Diagram The frontal plane is a plane parallel to the p2 plane. This
Straight lines of special position in the plane
The lines of special position in the plane are the horizontal h, the frontal f and the lines of greatest inclination to the projection planes. Let's look at the graphical representation of these lines (Table 5.6). Ta
Frontal construction algorithm
Verbal form Graphic form Given a plane a (a|| b), therefore a1 || b1; a2
Algorithm for constructing the second projection of point K
Verbal form Graphic form Plane a – defined by a flat figure a (D ABC), K2 – frontal projection of point K
Algorithm for constructing a plane parallel to a given one
Verbal form Graphic form 1. To solve the problem in a given plane P(D ABC), any intersecting lines are taken. For example, AB
Intersecting planes
Two planes intersect in a straight line. To construct a line of their intersection, it is necessary to find two points belonging to this line. The problem is simplified if one of the intersecting planes is occupied
Algorithm for constructing a straight line parallel to a plane
Verbal form Graphic form 1. Let us construct in the plane P(D ABC) a straight line A1, which belongs to the plane P
Algorithm for the intersection of a straight line with a generic plane
Verbal form Graphic form 1. To construct the point of intersection of a straight line l with a plane
Algorithm for constructing a perpendicular to a plane
Verbal form Graphic form 1. In order to construct a perpendicular to the plane P(D ABC) through point D, you must first
To Chapter 3
1. Construct the projection of line AB (Fig. 3) if it: a) is parallel to p1; b) parallel to p2; c) parallel to OX; d) perpendicular to p1
To Chapter 5
In the plane defined by two parallel straight lines, construct a frontal at a distance of 15 mm from p1 (Fig. 9):
To Chapter 6
1. Given a plane P(a|| b) and a frontal projection m2 of a line m passing through point D. Construct a horizontal projection of line m1 so that line m is parallel to the plane
Tests for Chapter 3
Select the correspondence between the designation of the segment AB and its image (Fig. 6): 1. AB || p 1 2. AB || p 2 3. AB ^ p 1 4.
Tests for Chapter 6
1. In which of the drawings (Fig. 12) is the plane S (D ABC) parallel to the plane P(m C n).
Recommended bibliography
1. GOST 2.001-70. General provisions // In collection. Unified system of design documentation. Basic provisions. – M.: Publishing House of Standards, 1984. – P. 3–5. 2. GOST 2.104-68. Main inscriptions // B
It would not be an exaggeration to say that the construction of mutually perpendicular lines and planes, along with determining the distance between two points, are the main graphic operations when solving metric problems.
The theoretical prerequisite for constructing projections of lines and planes perpendicular to each other in space on the Monge diagram is the property noted earlier (see § 6)
projections of a right angle, one of the sides of which is parallel to any projection plane:
1. Mutually perpendicular lines.
In order to be able to use the noted property to construct two straight lines intersecting at an angle of 90° on a Monge diagram, it is necessary that one of them be parallel to some projection plane. Let us explain what has been said with examples.
EXAMPLE 1. Through point A, draw a straight line l intersecting the horizontal line h at a right angle (Fig. 249).
Since one of the sides h of the right angle is parallel to the plane π 1, the right angle will be projected onto this plane without distortion. Therefore, through A" we draw a horizontal projection l" ⊥ h". We mark the point M" = l" ∩ h". We find M" (M" ∈ h"). Points A" and M" determine l" (see Fig. 249, a).
If instead of a horizontal line a frontal f is given, then the geometric constructions for drawing the straight line l ⊥ f are similar to those just considered with the only difference being that the construction of an undistorted projection of a right angle should begin with a frontal projection (see Fig. 249, b).
EXAMPLE 2. Through point A, draw a straight line l intersecting straight line a, defined by the segment [BC], at an angle of 90° (Fig. 250).
Since this segment occupies an arbitrary position in relation to the projection planes, we cannot, as in the previous example, use the property about the special case of projecting a right angle, so first we need to transfer [BC] to a position parallel to some projection plane.
In Fig. 250 [BC] is moved to a position parallel to the π 3 plane. This is done using the method of replacing projection planes by replacing the plane π 1 → π 3 || [Sun].
As a result of such a replacement in the new system, x 1 π 2 /π 3 [BC] defines a horizontal line, therefore all further constructions are carried out in the same way as was done in the previous example: after the point M "1 was found, it was translated onto the original projection planes at the positions M" and M", these points together with A" and A" determine the projections of straight line l.
EXAMPLE 3. Conduct a horizontal projection of the side [BC] of the right angle ABC, if its frontal projection ∠A"B"C" and the horizontal projection of the side [A"B"] are known (Fig. 251).
1. Move the side of the angle [BA] to position || π 3 by moving from the system of projection planes xπ 2 /π 1 to the new x 1 π 3 /π 2
2. Define a new frontal projection.
From B" 1 we construct a perpendicular to [B" 1 A" 1]. On this perpendicular we define point C" 1 (C" 1 is removed from the x 1 axis by a distance |C x 1 C" 1 | = |C x C"| ).
4. Horizontal projection C" is defined as the point of intersection of the lines (C" 1 C x 1) ∩ (C"C x) = C".
2. Mutually perpendicular straight line and plane.
From the stereometry course we know that a straight line is perpendicular to a plane if it is perpendicular to at least two intersecting lines belonging to this plane.
If we take not arbitrary intersecting lines in the plane, but its horizontal and frontal lines, then it becomes possible to use the property of right angle projection, as was done in Example 1, Fig. 249.
Consider the following example; Let us assume that from a point A ∈ α we need to restore a perpendicular to the plane α (Fig. 252).
Through point A we draw the horizontal line h and the frontal line f of the plane α. Then, by definition (AB), perpendicular to the plane α, must be perpendicular to the straight lines h and f, i.e. . But the AM side ∠ YOU || π 1 , therefore ∠VAM is projected onto the plane π 1 , without distortion, i.e. 
. AK side ∠ VAK || π 2 and, therefore, onto the plane π 2 this angle is also projected without distortion, i.e. 
. The above reasoning can be formulated in the form of the following theorem: In order for a straight line in space to be perpendicular to a plane, it is necessary and sufficient that on the diagram the horizontal projection of the straight line is perpendicular to the horizontal projection of the horizontal of the plane, and the frontal projection to the frontal projection of the frontal of this plane.
If the plane is given by traces, then the theorem can be formulated differently: In order for a line in space to be perpendicular to a plane, it is necessary and sufficient that the projections of this line be perpendicular to the traces of the same name on the plane.
The relationships established by the theorem between a line in space perpendicular to the plane and the projections of this line to the projections of level lines (traces) of the plane underlie the graphical algorithm for solving the problem of drawing a line perpendicular to the plane, as well as constructing a plane perpendicular to a given line.
EXAMPLE 1. Restore the perpendicular AD to the plane ΔАВС at vertex A (Fig. 253).
In order to determine the direction of the projections of the perpendicular, we carry out projections of the horizontal h and frontal f of the ΔABC plane. After this, from point A" we restore a perpendicular to h", and from A" - to f".
EXAMPLE 2. From point A, belonging to the plane α (m || n), construct a perpendicular to this plane (Fig. 254).
SOLUTION. To determine the direction of the perpendicular projections l" and l", as in the previous example, draw a horizontal line h(h", h") through point A (A", A"), belonging to the plane α. Knowing the direction h", we construct a horizontal projection of the perpendicular l" (l" ⊥ h"). To determine the direction of the frontal projection of the perpendicular through point A (A", A"), draw the frontal f (f", f") of the α plane. Due to the parallelism of f to the frontal projection plane, the right angle between l and f is projected onto π 2 without distortion, so we draw l" ⊥ f".
In Fig. 255 the same problem was solved for the case when the plane α is given by traces. To determine the directions of perpendicular projections, there is no need to draw horizontal and front lines.
waist, since their functions are performed by the traces of the plane h 0α and f 0α . As can be seen from the drawing, the solution boils down to drawing projections l" ⊥ h 0α and l" ⊥ f 0α through points A" and A".
EXAMPLE 3. Construct a plane γ perpendicular to a given line l and passing through a given point A (Fig. 256).
SOLUTION. Through point A we draw a horizontal line h and a front line f. These two intersecting lines define a plane; for it to be perpendicular to straight line l, it is necessary that straight lines h and f make an angle of 90° with straight l. To do this, we draw h" ⊥ l" and f" ⊥ l". Frontal projection h" and horizontal projection f" are parallel to the x-axis.
The considered case allows us to solve the problem given in example 3 in a different way (p. 175 Fig. 251). Side [BC] ∠ABC must belong to the plane γ ⊥ [AB] and pass through point B (Fig. 257).
This condition determines the course of solving the problem, which is as follows: we enclose point B in the plane γ ⊥ [AB], for this, through point B we draw the horizontal and frontal of the plane γ so that h" ⊥ A"B" and f" ⊥ A "B".
Point C ∈ (BC), belonging to the plane γ, therefore, to find its horizontal projection, we draw an arbitrary straight line 1"2" through C" belonging to the plane γ; determine the horizontal projection of this line 1"2" and mark point C on it" (C "is determined by the intersection of the connection line - the perpendicular dropped from C" with the horizontal projection of straight line 1"2"). C" together with B" determine the horizontal projection (BC) ⊥ (AB).
3. Mutually perpendicular planes..
Two planes are perpendicular if one of them contains a line perpendicular to the other plane.
Based on the definition of perpendicularity of planes, we solve the problem of constructing a plane β perpendicular to the plane α in the following way: draw a straight line l perpendicular to the plane α; we enclose the line l in the plane β. The plane β ⊥ α, since β ⊃ l ⊥ α.
Many planes can be drawn through the line l, so the problem has many solutions. To make the answer more specific, additional conditions must be specified.
EXAMPLE 1. Through a given straight line a draw a plane β perpendicular to the plane α (Fig. 258).
SOLUTION. We determine the direction of the projections of the perpendicular to the plane α, for this we find the horizontal projection of the horizontal (h") and the frontal projection of the frontal (f"); From the projections of an arbitrary point A ∈ α we draw the projections of the perpendicular l" ⊥ h" and l" ⊥ f". The plane β ⊥ α, since β ⊃ l ⊥ α.
EXAMPLE 2. Through a given point A, draw a horizontally projecting plane γ, perpendicular to the plane α, specified by the tracks (Fig. 259, a).
The required plane γ must contain a line perpendicular to the plane α, or be perpendicular to a line belonging to the plane α. Since the plane γ must be horizontally projecting, then the straight line perpendicular to it must be parallel to the plane π 1, i.e., be a horizontal of the plane α or (which is the same) a horizontal trace of this plane - h 0α Therefore, through a horizontal projection point A" draw a horizontal trace h 0γ ⊥ h 0α frontal trace f 0γ ⊥ x axis.
In Fig. 259, b shows the frontally projecting plane γ, passing through point B and perpendicular to the plane π 2.
It is clear from the drawing that a distinctive feature of the diagram, on which two mutually perpendicular planes are specified, one of which is frontally projecting, is the perpendicularity of their frontal traces f 0γ ⊥ f 0α , the horizontal trace of the frontally projecting plane is perpendicular to the x axis.
The sign of perpendicularity of a line and a plane allows us to construct mutually perpendicular lines and planes, that is, to prove the existence of such lines and planes. Let's start by constructing a plane perpendicular to a given line and passing through a given point. Let us solve two construction problems corresponding to two possibilities in the location of a given point and a given line.
Problem 1. Through a given point A on a given line a, draw a plane perpendicular to this line.
Let us draw any two planes through straight line a and in each of these planes through point A we draw along a straight line perpendicular to straight line a, denoting them b and c (Fig. 2.17). Plane a passing through straight lines bis contains point A and is perpendicular to straight line a (based on the perpendicularity of the straight line and the plane). Therefore, plane a is the desired one. The problem is solved.
The problem has only one (i.e., unique) solution. Indeed, let's assume the opposite. Then, in addition to plane a, another plane P passes through point A, perpendicular to straight line a (Fig. 2.18). Let us take in the plane P any line passing through point A and not lying in the plane a. Let us draw the plane y through the intersecting lines a and . The y plane intersects the a plane along the straight line q. The line q does not coincide with the line , since q lies in and does not lie in a. Both of these lines lie in the y plane, pass through point A and are perpendicular to the line a since and similarly since and. But this contradicts the well-known theorem of planimetry, according to which in a plane only one straight line passes through each point, perpendicular to a given straight line.
So, assuming that two planes perpendicular to line a pass through point A, we have arrived at a contradiction. Therefore, the problem has a unique solution.
Problem 2. Through a given point A, which does not lie on a given line a, draw a plane perpendicular to this line.
Through point A we draw a line b perpendicular to line a. Let B be the intersection point of a and b. Through point B we also draw a straight line c, perpendicular to straight line a (Fig. 2.19). A plane passing through both drawn lines will be perpendicular to a according to the perpendicularity criterion (Theorem 2).
As in Problem 1, the constructed plane is unique. Indeed, let us take any plane passing through point A perpendicular to straight line a. Such a plane contains a line perpendicular to line a and passing through point A. But there is only one such line. This is a line b that passes through point B. This means that the plane passing through A and perpendicular to line a must contain point B, and only one plane passes through point B, perpendicular to line a (problem 1). So, having solved these construction problems and proved the uniqueness of their solutions, we have proven the following important theorem.
Theorem 3 (about a plane perpendicular to a line). Through each point there passes a plane perpendicular to a given line, and, moreover, only one.
Corollary (about the plane of perpendiculars). Lines perpendicular to a given line at a given point lie in the same plane and cover it.
Let a be a given line and A be any point on it. A plane passes through it. By definition of perpendicularity of a line and a plane, it is covered
covered by straight lines perpendicular to straight line a at point A, i.e. through each point of the plane a there passes a line perpendicular to the line a.
Let us assume that a straight line passes through point A and does not lie in the plane a. Let us draw a plane P through it and straight line a. Plane P will intersect a along a certain straight line c (Fig. 2.20). And since it turns out that through point A in the plane P there pass two straight lines b and c, perpendicular to straight line a. This is impossible. This means that there are no lines perpendicular to line a at point A and not lying in the plane a. They all lie in this plane.
An example of the corollary of Theorem 3 is given by the spokes in a wheel, perpendicular to its axis: when rotating, they draw a plane (more precisely, a circle), taking all positions perpendicular to the axis of rotation.
Theorems 2 and 3 help provide a simple solution to the following problem.
Problem 3. Through a point on a given plane, draw a line perpendicular to this plane.
Let a plane a and a point A in the plane a be given. Let us draw some line a in the plane a through point A. Through point A we draw a plane perpendicular to line a (problem 1). The plane will intersect plane a along some straight line b (Fig. 2.21). Let us draw a line c in the plane P through point A, perpendicular to line b. Since (since c lies in the plane
And), then by Theorem 2. The uniqueness of its solution is established in section 2.1.
Comment. About constructions in space. Recall that in Chapter 1 we study “structural geometry”. And at this point we solved three problems on construction in space. What is meant in stereometry by the terms “construct”, “draw”, “inscribe”, etc.? First, let’s remember about constructions on a plane. By indicating, for example, how to construct a circle circumscribed about a triangle, we thereby prove its existence. In general, when solving a construction problem, we prove a theorem for the existence of a figure with given properties. This solution boils down to drawing up a certain algorithm for constructing the desired figure, i.e., indicating the sequence of performing the simplest operations leading to the required result. The simplest operations are drawing segments. circles and finding their intersection points. Then, using drawing tools, the figure is directly constructed on paper or on a board.
So, in planimetry, the solution to the construction problem has, as it were, two sides: theoretical - the construction algorithm - and practical - the implementation of this algorithm, for example, with a compass and a ruler.
The stereometric construction task has only one side left - the theoretical, since there are no tools for construction in space, similar to a compass and a ruler.
The basic constructions in space are taken to be those provided by axioms and theorems on the existence of straight lines and planes. This is drawing a line through two points, drawing a plane (propositions of clause 1.1 and axiom 1 of clause 1.4), as well as constructing a line of intersection of any two constructed planes (axiom 2 of clause 1.4). In addition, we will naturally assume that it is possible to carry out planimetric constructions in already constructed planes.
Solving a construction problem in space means indicating the sequence of basic constructions that result in the desired figure. Usually, not all basic constructions are explicitly indicated, but references are made to already solved construction problems, i.e. on already proven propositions and theorems about the possibility of such constructions.
In addition to constructions - existence theorems in stereometry, two more types of problems related to constructions are possible.
First, the tasks are in the picture or drawing. These are problems for cutting polyhedra or other bodies. We are not actually constructing the section itself, but only depicting it on
drawing or drawing that we already have. Such constructions are carried out as planimetric ones, taking into account the axioms and theorems of stereometry and image rules. Problems of this type are constantly solved in drawing and design practice.
Secondly, tasks on constructing bodies on surfaces. The task: “Construct points on the surface of a cube that are distant from a given vertex at a given distance” - can be solved using a compass (how?). The task: “Construct points on the surface of a ball that are distant from a given point at a given distance” - can also be solved using a compass (how?). Problems of this type are not solved in geometry lessons - they are constantly solved by the marker, of course, with the accuracy that his tools allow him to achieve. But when solving such problems, he relies on geometry.
