Three perpendicular planes. System of three mutually perpendicular planes

There are many parts whose shape information cannot be conveyed by two drawing projections. In order for information about the complex shape of a part to be presented sufficiently fully, projection is used on three mutually perpendicular projection planes: frontal - V, horizontal - H and profile - W (read “double ve”).

Complex drawing A drawing presented in three views or projections, in most cases gives a complete picture of the shape and design of the part (item and object) and is also called a complex drawing. main drawing. If a drawing is constructed with coordinate axes, it is called an axis drawing. axisless If the drawing is constructed without coordinate axes, it is called axisless profile If the plane W is perpendicular to the frontal and horizontal planes of projections, then it is called profile

An object is placed in a trihedral corner so that its formative edge and base are parallel to the frontal and horizontal projection planes, respectively. Then, projection rays are passed through all points of the object, perpendicular to all three projection planes, on which frontal, horizontal and profile projections of the object are obtained. After projection, the object is removed from the trihedral angle, and then the horizontal and profile projection planes are rotated 90°, respectively, around the Ox and Oz axes until aligned with the frontal projection plane and a drawing of the part containing three projections is obtained.

The three projections of the drawing are interconnected with each other. Frontal and horizontal projections preserve the projection connection of images, i.e. projection connections are established between frontal and horizontal, frontal and profile, as well as horizontal and profile projections. Projection lines define the location of each projection on the drawing field. The shape of most objects is a combination of various geometric bodies or their parts. Therefore, to read and execute drawings you need to know how geometric bodies are depicted in the system of three projections in production

1. Faces parallel to the projection planes are projected onto it without distortion, in natural size. 2. Faces perpendicular to the projection plane are projected in a segment of straight lines. 3. Faces located obliquely to the projection planes, images on it with distortion (reduced)

& 3. pg questions in writing task 4.1. pp pp, & 5, pp. 37-45, written assignment questions

The position of the plane in space is determined:

three points that do not lie on the same line;
a straight line and a point taken outside the straight line;
two intersecting lines;
two parallel lines;
flat figure.

In accordance with this, the plane can be specified on the diagram:

projections of three points that do not lie on the same line (Figure 3.1, a);
projections of a point and a line (Figure 3.1, b);
projections of two intersecting lines (Figure 3.1c);
projections of two parallel lines (Figure 3.1d);
flat figure (Figure 3.1, d);
traces of a plane;
line of the greatest slope of the plane.

Figure 3.1 – Methods for defining planes

General plane is a plane that is neither parallel nor perpendicular to any of the projection planes.

Following the plane is a straight line obtained as a result of the intersection of a given plane with one of the projection planes.

A generic plane can have three traces: horizontal – απ 1, frontal – απ 2 and profile – απ 3, which it forms when intersecting with known projection planes: horizontal π 1, frontal π 2 and profile π 3 (Figure 3.2).

Figure 3.2 – Traces of a general plane

3.2. Partial planes

Partial plane– a plane perpendicular or parallel to the plane of projections.

The plane perpendicular to the projection plane is called projecting and onto this projection plane it will be projected as a straight line.

Property of the projection plane: all points, lines, flat figures belonging to the projecting plane have projections on the inclined trace of the plane(Figure 3.3).

Figure 3.3 – Frontally projecting plane, which includes: points A, IN, WITH; lines AC, AB, Sun; triangle plane ABC

Front projection plane – plane perpendicular to the frontal plane of projections(Figure 3.4, a).

Horizontal projection plane – plane perpendicular to the horizontal plane of projections(Figure 3.4, b).

Profile-projecting plane – plane perpendicular to the profile plane of projections.

Planes parallel to projection planes are called level planes or double projecting planes.

Front level plane – plane parallel to the frontal plane of projections(Figure 3.4, c).

Horizontal level plane – plane parallel to the horizontal plane of projections(Figure 3.4, d).

Profile plane of the level – plane parallel to the profile plane of projections(Figure 3.4, e).

Figure 3.4 – Diagrams of planes of particular position

3.3. A point and a straight line in a plane. Belonging of a point and a straight plane

A point belongs to a plane if it belongs to any line lying in this plane(Figure 3.5).

A straight line belongs to a plane if it has at least two common points with the plane(Figure 3.6).

Figure 3.5 – Belonging of a point to a plane

α = m // n

D∈ n⇒ D∈ α

Figure 3.6 – Belonging to a straight plane

Exercise

Given a plane defined by a quadrilateral (Figure 3.7, a). It is necessary to complete the horizontal projection of the top WITH.


A	b

Figure 3.7 – Solution to the problem

Solution :

ABCD– a flat quadrilateral defining a plane.
Let's draw diagonals in it A.C. And BD(Figure 3.7, b), which are intersecting straight lines, also defining the same plane.
According to the criterion of intersecting lines, we will construct a horizontal projection of the point of intersection of these lines - K according to its known frontal projection: A 2 C 2 ∩ B 2 D 2 =K 2 .
Let us restore the projection connection line until it intersects with the horizontal projection of the straight line BD: on the diagonal projection B 1 D 1 we are building TO 1 .
Through A 1 TO 1 we carry out a diagonal projection A 1 WITH 1 .
Full stop WITH 1 is obtained through the projection connection line until it intersects with the horizontal projection of the extended diagonal A 1 TO 1 .

3.4. Main plane lines

An infinite number of straight lines can be constructed in a plane, but there are special straight lines lying in the plane, called main lines of the plane (Figure 3.8 – 3.11).

Straight level or parallel to the plane is a straight line lying in a given plane and parallel to one of the projection planes.

Horizontal or horizontal level line h(first parallel) is a straight line lying in a given plane and parallel to the horizontal plane of projections (π 1)(Figure 3.8, a; 3.9).

Front or front level straight f(second parallel) is a straight line lying in a given plane and parallel to the frontal plane of projections (π 2)(Figure 3.8, b; 3.10).

Level profile line p(third parallel) is a straight line lying in a given plane and parallel to the profile plane of projections (π 3)(Figure 3.8, c; 3.11).

Figure 3.8 a – Horizontal straight line of the level in the plane defined by the triangle

Figure 3.8 b – Frontal straight line of the level in the plane defined by the triangle

Figure 3.8 c – Level profile line in the plane defined by the triangle

Figure 3.9 – Horizontal straight line of the level in the plane defined by the tracks

Figure 3.10 – Frontal straight line of the level in the plane defined by the tracks

Figure 3.11 – Level profile line in the plane defined by the tracks

3.5. Mutual position of straight line and plane

A straight line with respect to a given plane can be parallel and can have a common point with it, that is, intersect.

3.5.1. Parallelism of a straight plane

Sign of parallelism of a straight plane: a line is parallel to a plane if it is parallel to any line belonging to this plane(Figure 3.12).

Figure 3.12 – Parallelism of a straight plane

3.5.2. Intersection of a line with a plane

To construct the point of intersection of a straight line with a general plane (Figure 3.13), you must:

Conclude direct A to the auxiliary plane β (planes of particular position should be selected as the auxiliary plane);
Find the line of intersection of the auxiliary plane β with the given plane α;
Find the intersection point of a given line A with the line of intersection of planes MN.

Figure 3.13 – Construction of the meeting point of a straight line with a plane

Exercise

Given: straight AB general position, plane σ⊥π 1. (Figure 3.14). Construct the intersection point of a line AB with plane σ.

Solution :

The plane σ is horizontally projecting, therefore, the horizontal projection of the plane σ is the straight line σ 1 (horizontal trace of the plane);
Dot TO must belong to the line AB ⇒ TO 1 ∈A 1 IN 1 and a given plane σ ⇒ TO 1 ∈σ 1 , therefore, TO 1 is located at the intersection point of the projections A 1 IN 1 and σ 1 ;
Frontal projection of the point TO we find through the projection communication line: TO 2 ∈A 2 IN 2 .

Figure 3.14 – Intersection of a general line with a particular plane

Exercise

Given: plane σ = Δ ABC– general position, straight E.F.(Figure 3.15).

It is required to construct the point of intersection of a line E.F. with plane σ.


A	b

Figure 3.15 – Intersection of a straight line and a plane

Let's conclude a straight line E.F. into an auxiliary plane, for which we will use the horizontally projecting plane α (Figure 3.15, a);
If α⊥π 1, then onto the projection plane π 1 the plane α is projected into a straight line (horizontal trace of the plane απ 1 or α 1), coinciding with E 1 F 1 ;
Let's find the line of intersection (1-2) of the projecting plane α with the plane σ (the solution to a similar problem will be considered);
Straight line (1-2) and specified straight line E.F. lie in the same plane α and intersect at the point K.

Algorithm for solving the problem (Figure 3.15, b):

Through E.F. Let's draw an auxiliary plane α:

3.6. Visibility determination using the competing point method

When assessing the position of a given line, it is necessary to determine which point of the line is located closer (further) to us, as observers, when looking at the projection plane π 1 or π 2.

Points that belong to different objects, and on one of the projection planes their projections coincide (that is, two points are projected into one), are called competing on this projection plane.

It is necessary to separately determine visibility on each projection plane.

Visibility at π 2 (Fig. 3.15)

Let us choose points competing on π 2 – points 3 and 4. Let point 3∈ VS∈σ, point 4∈ E.F..

To determine the visibility of points on the projection plane π 2, it is necessary to determine the location of these points on the horizontal projection plane when looking at π 2.

The direction of view towards π 2 is shown by the arrow.

From the horizontal projections of points 3 and 4, when looking at π 2, it is clear that point 4 1 is located closer to the observer than 3 1.

4 1 ∈E 1 F 1 ⇒ 4∈E.F.⇒ on π 2 point 4 will be visible, lying on the straight line E.F., therefore, straight E.F. in the area of the competing points under consideration is located in front of the σ plane and will be visible up to the point K

Visibility at π 1

To determine visibility, we select points that compete on π 1 - points 2 and 5.

To determine the visibility of points on the projection plane π 1, it is necessary to determine the location of these points on the frontal projection plane when looking at π 1.

The direction of view towards π 1 is shown by the arrow.

From the frontal projections of points 2 and 5, when looking at π 1, it is clear that point 2 2 is located closer to the observer than 5 2.

2 1 ∈A 2 IN 2 ⇒ 2∈AB⇒ on π 1 point 2 will be visible, lying on the straight line AB, therefore, straight E.F. in the area of the competing points under consideration is located under the plane σ and will be invisible until the point K– points of intersection of the straight line with the plane σ.

The visible one of the two competing points will be the one whose “Z” and/or “Y” coordinates are greater.

3.7. Perpendicularity to a straight plane

Sign of perpendicularity of a straight plane: a line is perpendicular to a plane if it is perpendicular to two intersecting lines lying in a given plane.


A	b

Figure 3.16 – Defining a straight line perpendicular to the plane

Theorem. If the straight line is perpendicular to the plane, then 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 of the straight line is perpendicular to the frontal projection of the frontal (Figure 3.16, b)

The theorem is proven through the theorem on the projection of a right angle in a special case.

If the plane is defined by traces, then the projections of a straight line perpendicular to the plane are perpendicular to the corresponding traces of the plane (Figure 3.16, a).

Let it be straight p perpendicular to the plane σ=Δ ABC and passes through the point K.

Let's construct the horizontal and frontal lines in the plane σ=Δ ABC : A-1∈σ; A-1//π 1 ; S-2∈σ; S-2//π 2 .
Let's restore from the point K perpendicular to a given plane: p 1⊥h 1 And p2⊥f 2, or p 1⊥απ 1 And p2⊥απ 2

3.8. Relative position of two planes

3.8.1. Parallelism of planes

Two planes can be parallel and intersecting.

Sign of parallelism of two planes: two planes are mutually parallel if two intersecting lines of one plane are correspondingly parallel to two intersecting lines of another plane.

Exercise

The general position plane is given α=Δ ABC and period F∉α (Figure 3.17).

Through the point F draw plane β parallel to plane α.

Figure 3.17 – Construction of a plane parallel to a given one

Solution :

As intersecting lines of the plane α, let us take, for example, the sides of the triangle AB and BC.

Through the point F we conduct a direct m, parallel, for example, AB.
Through the point F, or through any point belonging to m, we draw a straight line n, parallel, for example, Sun, and m∩n=F.
β = m∩n and β//α by definition.

3.8.2. Intersection of planes

The result of the intersection of 2 planes is a straight line. Any straight line on a plane or in space can be uniquely defined by two points. Therefore, in order to construct a line of intersection of two planes, you should find two points common to both planes, and then connect them.

Let's consider examples of the intersection of two planes with different ways of defining them: by traces; three points that do not lie on the same line; parallel lines; intersecting lines, etc.

Exercise

Two planes α and β are defined by traces (Figure 3.18). Construct a line of intersection of planes.

Figure 3.18 – Intersection of general planes defined by traces

The procedure for constructing the line of intersection of planes:

Find the point of intersection of horizontal traces - this is the point M(her projections M 1 And M 2, while M 1 =M, because M – private point belonging to the plane π 1).
Find the point of intersection of the frontal tracks - this is the point N(her projections N 1 and N 2, while N 2 = N, because N – private point belonging to the plane π 2).
Construct a line of intersection of planes by connecting the projections of the resulting points of the same name: M 1 N 1 and M 2 N 2 .

MN– line of intersection of planes.

Exercise

Given plane σ = Δ ABC, plane α – horizontally projecting (α⊥π 1) ⇒α 1 – horizontal trace of the plane (Figure 3.19).

Construct the line of intersection of these planes.

Solution :

Since the plane α intersects the sides AB And AC triangle ABC, then the points of intersection K And L these sides with the plane α are common to both given planes, which will allow, by connecting them, to find the desired intersection line.

Points can be found as the points of intersection of straight lines with the projecting plane: we find horizontal projections of points K And L, that is K 1 and L 1, at the intersection of the horizontal trace (α 1) of a given plane α with horizontal projections of the sides Δ ABC: A 1 IN 1 and A 1 C 1 . Then, using projection communication lines, we find the frontal projections of these points K2 And L 2 on frontal projections of straight lines AB And AC. Let's connect the projections of the same name: K 1 and L 1 ; K2 And L 2. The intersection line of the given planes is drawn.

Algorithm for solving the problem:

KL– intersection line Δ ABC and σ (α∩σ = KL).

Figure 3.19 – Intersection of general and particular planes

Exercise

Given planes α = m//n and plane β = Δ ABC(Figure 3.20).

Construct a line of intersection of the given planes.

Solution :

To find points common to both given planes and defining the intersection line of planes α and β, it is necessary to use auxiliary planes of particular position.
As such planes, we will choose two auxiliary planes of particular position, for example: σ // τ; σ⊥π 2 ; τ⊥π 2 .
The newly introduced planes intersect with each of the given planes α and β along straight lines parallel to each other, since σ // τ:

— the result of the intersection of planes α, σ and τ are straight lines (4-5) and (6-7);

— the result of the intersection of planes β, σ and τ are straight lines (3-2) and (1-8).

Lines (4-5) and (3-2) lie in the σ plane; their point of intersection M simultaneously lies in the planes α and β, that is, on the straight line of intersection of these planes;
Similarly, we find the point N, common to the α and β planes.
Connecting the dots M And N, let's construct the straight line of intersection of the planes α and β.

Figure 3.20 – Intersection of two planes in general position (general case)

Algorithm for solving the problem:

Exercise

Given planes α = Δ ABC and β = a//b. Construct a line of intersection of the given planes (Figure 3.21).

Figure 3.21 Solving the plane intersection problem

Solution :

Let us use auxiliary secant planes of particular position. Let us introduce them in such a way as to reduce the number of constructions. For example, let’s introduce the plane σ⊥π 2 by enclosing the straight line a into the auxiliary plane σ (σ∈ a). The plane σ intersects the plane α along a straight line (1-2), and σ∩β= A. Therefore (1-2)∩ A=K.

Dot TO belongs to both planes α and β.

Therefore, the point K, is one of the required points through which the intersection line of the given planes α and β passes.

To find the second point belonging to the line of intersection of α and β, we conclude the line b into the auxiliary plane τ⊥π 2 (τ∈ b).

Connecting the dots K And L, we obtain the straight line of intersection of the planes α and β.

3.8.3. Mutually perpendicular planes

Planes are mutually perpendicular if one of them passes through the perpendicular to the other.

Exercise

Given a plane σ⊥π 2 and a line in general position – DE(Figure 3.22)

Required to build through DE plane τ⊥σ.

Solution .

Let's draw a perpendicular CD to the plane σ – C 2 D 2 ⊥σ 2 (based on ).

Figure 3.22 – Construction of a plane perpendicular to a given plane

By the right angle projection theorem C 1 D 1 must be parallel to the projection axis. Intersecting lines CD∩DE define the plane τ. So, τ⊥σ.

Similar reasoning in the case of a general plane.

Exercise

Given plane α = Δ ABC and period K outside the α plane.

It is required to construct a plane β⊥α passing through the point K.

Solution algorithm(Figure 3.23):

Let's build a horizontal line h and front f in a given plane α = Δ ABC;
Through the point K let's draw a perpendicular b to the plane α (along perpendicular to the plane theorem: if a straight line is perpendicular to a plane, then its projections are perpendicular to the inclined projections of the horizontal and frontal lines lying in the plane:b 2⊥f 2; b 1⊥h 1;
We define the plane β in any way, for example, β = a∩b, thus, a plane perpendicular to the given one is constructed: α⊥β.

Figure 3.23 – Construction of a plane perpendicular to a given Δ ABC

3.9. Problems to solve independently

1. Given plane α = m//n(Figure 3.24). It is known that K∈α.

Construct a frontal projection of a point TO.

Figure 3.24

2. Construct traces of a line given by a segment C.B., and identify the quadrants through which it passes (Figure 3.25).

Figure 3.25

3. Construct the projections of a square belonging to the plane α⊥π 2 if its diagonal MN//π 2 (Figure 3.26).

Figure 3.26

4. Construct a rectangle ABCD with the larger side Sun on a straight line m, based on the condition that the ratio of its sides is 2 (Figure 3.27).

Figure 3.27

5. Given plane α= a//b(Figure 3.28). Construct a plane β parallel to the plane α and distant from it at a distance of 20 mm.

Figure 3.28

6. Given plane α=∆ ABC and period D D plane β⊥α and β⊥π 1 .

7. Given plane α=∆ ABC and period D out of plane. Build through point D direct DE//α and DE//π 1 .

System of three mutually perpendicular planes

Formation of a complex drawing (diagram)

For the convenience of using the resulting images from the spatial system of planes, let’s move on to the planar one.

For this:

1. Let us apply the method of rotating the plane p 1 around the X axis until it aligns with the plane p 2 (Fig. 1)

2. Combine planes p 1 and p 2 into one drawing plane (Fig. 2)


Picture 1	Figure 2

Projections A 1 and A 2 are located on the same connection line perpendicular to the X axis. This line is usually called the projection connection line (Fig. 3).

Figure 3

Since the projection plane is considered infinite in space, the boundaries of the plane p 1, p 2 need not be depicted (Fig. 4).

Figure 4

As a result of combining the planes p 1 and p 2, a complex drawing or diagram is obtained (from the French epure drawing), ᴛ.ᴇ. drawing in the system p 1 and p 2 or in the system of two projection planes. Having replaced the visual image with a diagram, we have lost the spatial picture of the location of projection planes and points. But the diagrams provide accuracy and easy-to-measure images with significant simplicity of construction.

A point defined in space can have different positions relative to projection planes.

Constructing point images can be done in various ways:

words (verbal);
graphically (drawings);
visual image (volumetric);
planar (complex drawing).

Table 1

An example of an image of points belonging to the planes p 1 and p 2

Point position	Visual representation	Complex drawing	Characteristic signs
Point A belongs to the plane p 1			A 1 – below the X axis, A 2 – on the X axis
Point B belongs to plane p 1			B 1 – above the X axis, B 2 – on the X axis
Point C belongs to the plane p 2			C 2 – above the X axis, C 1 – on the X axis
Point D belongs to the plane p 2			D 1 – on the X axis, D 2 – below the X axis
Point E belongs to the X axis			E 1 coincides with E 2 and belongs to the X axis

Picture 1

Consider three mutually perpendicular planes p 1 , p2 , p 3 ( rice. 1). The vertical plane p 3 is called I profile projection plane. Intersecting each other, planes 1 , p2 , p 3 form the projection axes, while the space is divided into 8 octants.

p 1 p 2 = x; -x

p 1 p 3 = y; -y

p 2 p 3 = z; -z

0 – point of intersection of the projection axes.

The projection planes, intersecting in pairs, define three axes x, y, z, which can be considered as a system of Cartesian coordinates: axis X usually called the abscissa axis, the axis y– ordinate axis, axis Z– applicate axis, the point of intersection of the axes, denoted by the letter ABOUT, is the origin of coordinates.

To obtain a complex drawing, we apply the method of rotating the planes p 1 and p 3 until they align with the plane p 2. The final view of all planes in the first octant is shown in Fig. 2.

Figure 2

Here are the axes Oh And Oz, lying in the fixed plane p 2, are depicted only once, the axis Oh shown twice. This is explained by the fact that, rotating with the plane p 1, the axis y on the diagram it is combined with the axis Oz, and rotating with the plane p 3, this same axis coincides with the axis Oh.

Any point in space is specified by coordinates. By the signs of the coordinates, you can determine the octant in which a given point is located. To do this, we will use the table. 1, in which the signs of coordinates in octants 1–4 are considered (octants 5–8 are not presented, they have a negative value X, A y And z are repeated).

Table 1

x	y	z	Octant
+	+	+	I
+	_	+	II
+	_	_	III
+	+	_	IV

10.1 Dihedral angle. Angle between planes

Two intersecting lines form two pairs of vertical angles. Just as two intersecting lines on a plane form a pair of vertical angles (Fig. 89, a), so two intersecting planes in space form two pairs of vertical dihedral angles (Fig. 89, b).

Rice. 89

A dihedral angle is a figure that consists of two half-planes that have a common boundary straight line and do not lie in the same plane (Fig. 90). The half-planes themselves are called the faces of a dihedral angle, and their common boundary straight line is called its edge.

Rice. 90

Dihedral angles are measured as follows.

Let us take point O on edge p of a dihedral angle with faces α and β. Draw rays a and b from point O at its faces, perpendicular to edge p: a - in face α and b - in face β (Fig. 91, a).

Rice. 91

An angle with sides a, b is called a linear dihedral angle.

The magnitude of the linear angle does not depend on the choice of its vertex on the edge of the dihedral angle.

Indeed, let’s take another point O 1 of the edge p and draw the rays a 1 ⊥ p and b 1 ⊥ p in the faces α and β (Fig. 91, b).

Let us plot on ray a the segment OA, on ray a 1 the segment O 1 A 1, equal to the segment OA, on ray b the segment OB, and on ray b 1 the segment O 1 B 1, equal to the segment OB (Fig. 91, c).

In rectangles OAA 1 O 1 and 0BB 1 0 1, the sides AA 1 and BB 1 are equal to their common side OO 1 and parallel to it. Therefore AA 1 = BB 1 and AA 1 || BB 1.

Consequently, the quadrilateral ABV 1 A 1 is a parallelogram (Fig. 91, d), which means AB = A 1 B 1. Therefore, triangles ABO and A 1 B 1 O 1 are equal (on three sides) and angle ab is equal to angle a 1 b 1.

Now we can give the following definition: the magnitude of a dihedral angle is the magnitude of its linear angle.

The angle between intersecting planes is the size of the smaller of the dihedral angles formed by them. If this angle is 90°, then the planes are called mutually perpendicular. The angle between parallel planes is assumed to be 0°.

The angle between the planes α and β, as well as the value of the dihedral angle with faces α and β, is denoted ∠αβ.

The angle between the faces of a polyhedron that have a common edge is the value of the dihedral angle corresponding to these faces.

10.2 Properties of mutually perpendicular planes

Property 1. A straight line lying in one of two mutually perpendicular planes and perpendicular to their common straight line is perpendicular to the other plane.

Proof. Let the planes α and β be mutually perpendicular and intersect along a straight line c. Let straight line a lie in the plane α and a ⊥ с (Fig. 92). Line a intersects c at some point O. Let us draw a line b in the plane β through point O, perpendicular to line c. Since α ⊥ β, then a ⊥ b. Since a ⊥ b and a ⊥ c, then α ⊥ β based on the perpendicularity of the line and the plane.

Rice. 92

The second property is the converse of the first property.

Property 2. A straight line that has a common point with one of two mutually perpendicular planes and is perpendicular to the other plane lies in the first of them.

Proof. Let the planes α and β be mutually perpendicular and intersect along a straight line c, the straight line a ⊥ β and a have a common point A with a (Fig. 93). Through point A we draw a straight line p in the plane α, perpendicular to the straight line c. According to property 1 p ⊥ β. Lines a and p pass through point A and are perpendicular to the plane β. Therefore, they coincide, since only one straight line passes through a point, perpendicular to a certain plane. Since the straight line p lies in the α plane, then the straight line a lies in the α plane.

Rice. 93

A consequence of property 2 is the following sign of perpendicularity of a line and a plane: if two planes perpendicular to a third plane intersect, then the line of their intersection is perpendicular to the third plane.

Proof. Let two planes α and β, intersecting along a straight line a, be perpendicular to the plane γ (Fig. 94). Then through any point of line a we draw a line perpendicular to the plane γ. According to property 2, this line lies both in the plane α and in the plane β, i.e., it coincides with line a. So, a ⊥ γ.

Rice. 94

10.3 Sign of perpendicularity of planes

Let's start with practical examples. The plane of a door hung on a jamb perpendicular to the floor is perpendicular to the plane of the floor in any position of the door (Fig. 95). When they want to check whether a flat surface (wall, fence, etc.) is installed vertically, they do this using a plumb line - a rope with a load. The plumb line is always directed vertically, and the wall stands vertically if the plumb line, located along it, does not deviate. These examples tell us the following simple sign of the perpendicularity of planes: if a plane passes through a perpendicular to another plane, then these planes are mutually perpendicular.

Rice. 95

Proof. Let the plane α contain a line a perpendicular to the plane β (see Fig. 92). Then straight line a intersects plane β at some point O. Point O lies on line c along which planes α and β intersect. Let us draw a line b in the β plane through point O, perpendicular to line c. Since a ⊥ β, then a ⊥ b and a ⊥ c. This means that the linear angles of the dihedral angles formed by intersecting planes α and β are straight. Therefore, planes α and β are mutually perpendicular.

Note that each two of the three straight lines a, b and c, considered now (see Fig. 92), are mutually perpendicular. If we build another line passing through point O and perpendicular to two of these three lines, then it will coincide with the third line. This fact speaks about the three-dimensionality of the space around us: there is no fourth line perpendicular to each of the lines a, b and c.

Questions for self-control

How is the dihedral angle calculated?
How to calculate the angle between planes?
What planes are called mutually perpendicular?
What properties of mutually perpendicular planes do you know?
What sign of perpendicularity of planes do you know?

Task No. 4.

Task No. 3.

Task No. 2.

Task No. 1.

Formation of a complex drawing (diagram)

For the convenience of using the resulting images from the spatial system of planes, let’s move on to the planar one.

For this:

1. Apply the method of rotating the plane p 1 around the X axis until it aligns with the plane p 2 (Fig. 2.7)

2. Combine planes p 1 and p 2 into one drawing plane (Fig. 2.8)


Rice. 2.7	Rice. 2.8

Projections A 1 and A 2 are located on the same connection line perpendicular to the X axis. This line is called the projection connection line (Fig. 2.9).

Since the projection plane is considered infinite in space, the boundaries of the plane p 1, p 2 need not be depicted (Fig. 2.10).

As a result of combining the planes p 1 and p 2, a complex drawing or diagram is obtained (from the French epure drawing), i.e. drawing in the system p 1 and p 2 or in the system of two projection planes. Having replaced the visual image with a diagram, we have lost the spatial picture of the location of projection planes and points. But the diagrams provide accuracy and easy-to-measure images with significant simplicity of construction. To imagine a spatial picture from a diagram requires the work of imagination: for example, according to Fig. 2.11 you need to imagine the picture shown in Fig. 2.12.

If there is a projection axis in the complex drawing along projections A 1 and A 2, you can establish the position of point A relative to p 1 and p 2 (see Fig. 2.5 and 2.6). Comparing Fig. 2.11 and 2.12 it is easy to establish that the segment A 2 A X is the distance from point A to the plane p 1, and the segment A 1 A X is the distance from point A to p 2. The location of A 2 above the projection axis means that point A is located above the plane p 1. If A 1 on the diagram is located below the projection axis, then point A is in front of the plane p 2. Thus, the horizontal projection of the geometric image determines its position relative to the frontal plane of projections p 2 , and the frontal projection of the geometric image - relative to the horizontal plane of projections p 1 .


Rice. 2.11	Rice. 2.12

§ 4. Characteristics of the position of a point in the system p 1 and p 2

A point defined in space can have different positions relative to the projection planes (Fig. 2.13).

Let's consider possible options for the location of a point in the space of the first quarter:

1. A point is located in the space of the first quarter at any distance from the X axis and planes p 1 p 2, for example, points A, B (such points are called points of general position) (Fig. 2.14 and Fig. 2.15).

3. Point K belongs simultaneously to both the plane p 1 and p 2, that is, it belongs to the X axis (Fig. 2.18):

Based on the above, we can draw the following conclusion:

1. If a point is located in the space of the first quarter, then its projection A 2 is located above the X axis, and A 1 is below the X axis; A 2 A 1 – lie on the same perpendicular (connection line) to the X axis (Fig. 2.14).

2. If a point belongs to the plane p 2, then its projection C 2 C (coincides with the point C itself) and the projection C 1 X (belongs to the X axis) and coincides with C X: C 1 C X.

3. If a point belongs to the plane p 1, then its projection D 1 onto this plane coincides with the point D D 1 itself, and the projection D 2 belongs to the X axis and coincides with D X: D 2 D X.

4. If a point belongs to the X axis, then all its projections coincide and belong to the X axis: K K 1 K 2 K X.

Exercise:

1. Characterize the position of points in the space of the first quarter (Fig. 2.19).

2. Construct a visual image and a comprehensive drawing of the point according to the description:

a) point C is located in the first quarter, and is equidistant from the planes p 1 and p 2.

b) point M belongs to the plane p 2.

c) point K is located in the first quarter, and its distance to p 1 is twice as large as to the plane p 2.

d) point L belongs to the X axis.

3. Construct a complex drawing of a point according to the description:

a) point P is located in the first quarter, and its distance from the plane p 2 is greater than from the plane p 1.

b) point A is located in the first quarter and its distance to the plane p 1 is 3 times greater than to the plane p 2.

c) point B is located in the first quarter, and its distance to the plane is p 1 =0.

4. Compare the position of the points relative to the projection planes p 1 and p 2 and with each other. The comparison is made based on characteristics or features. For points, these characteristics are the distance to the planes p 1; p 2 (Fig. 2.20).

The application of the above theory when constructing images of a point can be carried out in various ways:

words (verbal);
graphically (drawings);
visual image (volumetric);
planar (complex drawing).

The ability to translate information from one method to another contributes to the development of spatial thinking, i.e. from verbal to visual (volumetric), and then to planar, and vice versa.

Let's look at this with examples (Table 2.1 and Table 2.2).

Table 2.1

Example of dot image
in a system of two projection planes

Quarter space	Visual representation	Complex drawing	Characteristic signs
I			Frontal projection of point A above the X axis, horizontal projection of point A below the X axis
II			Frontal and horizontal projections of point B above the X axis
III			Frontal projection of point C below the X axis, horizontal projection of point C above the X axis
IV			Frontal and horizontal projections of point D below the X axis

Table 2.2

An example of an image of points belonging to the planes p 1 and p 2

Point position	Visual representation	Complex drawing	Characteristic signs
Point A belongs to the plane p 1			A 1 – below the X axis, A 2 – on the X axis
Point B belongs to plane p 1			B 1 – above the X axis, B 2 – on the X axis
Point C belongs to the plane p 2			C 2 – above the X axis, C 1 – on the X axis
Point D belongs to the plane p 2			D 1 – on the X axis, D 2 – below the X axis
Point E belongs to the X axis			E 1 coincides with E 2 and belongs to the X axis

Construct a complex drawing of point A if:

1. The point is located in the II quarter and is equidistant from the planes p 1 and p 2.

2. The point is located in the third quarter, and its distance to the plane p 1 is twice as large as to the plane p 2.

3. The point is located in the IV quarter, and its distance to the p1 plane is greater than to the p2 plane.

Determine in which quarters the points are located (Fig. 2.21).

1. Construct a visual image of the points in the quarters:

a) A – general position in the third quarter;

b) B – general position in the IV quarter;

c) C – in the second quarter, if its distance from p 1 is 0;

d) D – in the first quarter, if its distance from p 2 is 0.

Construct a complex drawing of points A, B, C, D (see task 3).

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 X axis, its image will not change.

The position of the point in space (Fig. 2.22) has changed (Fig. 2.24), but the images in the complex drawing remain unchanged (Fig. 2.23 and Fig. 2.25).


Rice. 2.22	Rice. 2.23

Rice. 2.24	Rice. 2.25

To solve this problem, a system of three mutually perpendicular planes is introduced, since when drawing up drawings, for example, machines and their parts, not two, but more images are required. On this basis, in some constructions when solving problems, it is necessary to introduce p 1, p 2 and other projection planes into the system.

These planes divide the entire space into VIII parts, which are called octants (from the Latin okto eight). The planes have no thickness, are opaque and infinite. The observer is located in the first quarter (for systems p 1, p 2) or the first octant (for systems p 1, p 2, p 3) at an infinite distance from the projection planes.

§ 6. Point in the system p 1, p 2, p 3

The construction of projections of a certain point A, located in the first octant, onto three mutually perpendicular planes p 1, p 2, p 3 is shown in Fig. 2.27. Using the combination of projection planes with the p 2 plane and using the method of rotating the planes, we obtain a complex drawing of point A (Fig. 2.28):

AA 1 ^ p 1 ; AA 2 ^ p 2 ; AA 3 ^ p 3,

where A 3 – profile projection of point A; А Х, А y, А Z – axial projections of point A.

Projections A 1, A 2, A 3 are called, respectively, the frontal, horizontal and profile projection of point A.


Rice. 2.27	Rice. 2.28

The projection planes, intersecting in pairs, define three axes x, y, z, which can be considered as a system of Cartesian coordinates: axis X called the abscissa axis, axis y– ordinate axis, axis Z– applicate axis, the point of intersection of the axes, denoted by the letter ABOUT, is the origin of coordinates.

Thus, the viewer looking at the object is in the first octant.

To obtain a complex drawing, we apply the method of rotating the planes p 1 and p 3 (as shown in Fig. 2.27) until aligned with the plane p 2. The final view of all planes in the first octant is shown in Fig. 2.29.

Let's look at Fig. 2.30, where is the point in space A, given by coordinates (5,4,6). These coordinates are positive, and she herself is in the first octant. The construction of an image of the point itself and its projections on a spatial model is carried out using a coordinate rectangular parallelogram. To do this, we plot segments on the coordinate axes, corresponding to the length segments: Oah = 5, OAy = 4, OAz= 6. On these segments ( ОАx, ОАy, ОАz), as on the edges, we build a rectangular parallelepiped. One of its vertices will define a given point A.