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How to construct a straight line in a drawing that lies in a given plane? This construction is based on two provisions known from geometry.

1. A straight line belongs to a plane if it passes through two points belonging to this plane.
2. A straight line belongs to a plane if it passes through a point belonging to a given plane and is parallel to a line located in or parallel to this plane.

Let us assume that pl.α (Fig. 106) is defined by two intersecting straight lines AB and CB, and pl. β - two parallel ones - DE and FG. According to the first point

The line that intersects the lines defining the plane is located in the given plane.

It follows from this that if the plane is defined by traces, then a line belongs to a plane if the traces of the line are on traces of the same name as the plane(Fig. 107).

Let's assume that pl. γ (Fig. 106) is determined by point A and straight line BC. According to the second position, a straight line drawn through point A parallel to straight line BC belongs to the square. γ. From here a straight line belongs to a plane if it is parallel to one of the traces of this plane and has a common point with the other trace(Fig. 108).

Examples of constructions in Fig. 107 and 108 should not be understood in such a way that in order to construct a straight line in a plane, one must first construct traces of this plane. This is not required.

For example, in Fig. 109, the construction of straight line AM was completed in the plane specified by point A and the straight line passing through point L. Let us assume that straight line AM should be parallel to the square. π 1. The construction began with the projection А "М" perpendicular to the communication line А "А". Using point M" point M" was found and then projection A"M" was carried out. The straight line AM meets the condition: it is parallel to the square. π 1 And lies in this plane, since it passes through two points (A and M) that obviously belong to this plane.

How to construct a point in a drawing that lies in a given plane? In order to do this, first construct a straight line lying in a given plane, and take a point on this straight line.

For example, it is required to find the frontal projection of point D if its horizontal projection D" is given and it is known that point D must lie in the plane defined by triangle ABC (Fig. 110).

First, a horizontal projection of a certain line is constructed so that point D could be on this line, and the latter would be located in a given plane. To do this, draw a straight line through points A" and D" and mark point M", at which straight line A"D" intersects segment B"C". By constructing a frontal projection of M" onto B"C", a straight line AM is obtained, located in this plane : this line passes through points A and M, of which the first obviously belongs to a given plane, and the second is built in it.

The required frontal projection D" of point D must be on the frontal projection of straight line AM.

Another example is given in Fig. 111. In pl. β, given by parallel lines AB and CD, there must be a point K, for which only a horizontal projection is given - point K

Through point K" a certain straight line is drawn, which is accepted as a horizontal projection of a line in a given plane. Using points E" and F" we construct E" on A"B" and F" on C"D". The constructed straight line EF belongs to the area β, since it passes through points E and F, which obviously belong to the plane. If we take point K" on E"F", then point K will be in the area β.

Among the straight lines that occupy a special position in the plane we include horizontal, frontal 1) and lines of greatest inclination to projection planes. The line of greatest inclination to the square. π 1, we will call plane slope line 2).

Horizontals of a plane are straight lines lying in a line and parallel to the horizontal plane of projections.

Let's construct a horizontal plane defined by triangle ABC. It is required to draw a horizontal line through vertex A (Fig. 112).

Since the horizontal of the plane is a straight line parallel to the plane π 1, we obtain the frontal projection of this straight line by drawing A"K"⊥A"A". To construct a horizontal projection of this horizontal line, we construct point K" and draw a straight line through points A" and K".

The constructed straight line AK is indeed a horizontal line of this plane: this straight line lies in the plane, since it passes through two points that obviously belong to it, and is parallel to the projection plane π 1.

Now let's consider the construction of the horizontal plane defined by the traces.

The horizontal trace of a plane is one of its horizontals (the “zero” horizontal). Therefore, the construction of any of the horizontal planes is reduced to

to draw a straight line in this plane parallel to the horizontal trace of the plane (Fig. 108, left). The horizontal projection of the horizontal is parallel to the horizontal trace of the plane; the frontal projection of the horizontal is parallel to the axis of the projections.

The fronts of a plane are lines lying in it and parallel to the projection planeπ 2.

An example of constructing a frontal in a plane is given in Fig. 113. The construction is carried out similarly to the construction of a horizontal line (see Fig. 112).

Let the front pass through point A (Fig. 113). We begin the construction by drawing a horizontal projection of the frontal - straight line A"K", since the direction of this projection is known: A K"⊥A"A". Then we build a frontal projection of the frontal - straight line A"K".

1) Along with the horizontals and fronts of a plane, one can also consider its profile straight lines - straight lines lying in a given plane and parallel to the square. π 3. For horizontals, fronts and profile lines there is a common name - level line. However, this name corresponds to the usual idea of ​​only horizontality.

2) For the slope line of a plane, the name “line of the greatest slope” is common, but the concept of “slope” in relation to a plane does not require the addition of “largest”.

The constructed straight line is indeed the frontal of the given plane: this straight line lies in the plane, since it passes through two points that obviously belong to it and is parallel to the plane, π 2.

Let us now construct the frontal of the plane defined by the tracks. Looking at Fig. 108, on the right, which shows the square. β and straight line MV, we establish that this straight line is the frontal of the plane. Indeed, it is parallel to the frontal trace (“zero” frontal) of the plane. The horizontal projection of the frontal is parallel to the x-axis, the frontal projection of the frontal is parallel to the frontal trace of the plane.

The lines of the greatest inclination of the plane to the planes π 1, π 2 and π 3 are the straight lines lying in it and perpendicular either to the horizontals of the plane, or to its fronts, or to its profile straight lines. In the first case, the slope to square π 1 is determined, in the second - to square. π 2, in the third - to pl. π 3. To draw the lines of the greatest inclination of the plane, you can, of course, take its traces accordingly.

As mentioned above, the line of greatest inclination of the plane to the square. to π 1, is called slope line of the plane.

According to the rules for projecting right angles (see, § 15), the horizontal projection of the slope line of a plane is perpendicular to the horizontal projection of the horizontal of this plane or to its horizontal trace. The frontal projection of the slope line is constructed after the horizontal one and can occupy different positions depending on the specification of the plane. Figure 114 shows the slope line Pl. α: ВК⊥h" 0α. Since В"К is also perpendicular to h" 0α, then ∠ВКВ" is a linear angle

dihedral, formed by planes α and π 1 Therefore, the slope line of the plane can be used to determine the angle of inclination of this plane to the projection planeπ 1.

Similarly, the line of the greatest inclination of the plane to the plane, π 2, serves to determine the angle between this plane and the plane, π 2, and the line of the greatest inclination to the plane, π 3, to determine the angle with the plane. π 3.

In Fig. 115, slope lines are plotted in given planes. The angle pl, α with pl.π 1 is expressed by projections - frontal in the form of angle B "K" B" and horizontal in the form of segment K "B". The value of this angle can be determined by constructing a right triangle along legs equal to K "B" and B"B".

Obviously, the line of greatest inclination of the plane determines the position of this plane. For example, if (Fig. 115) the slope line KV is given, then by drawing a horizontal line AN perpendicular to it or by specifying the x axis of projections and drawing h" 0α ⊥ K"B", we completely determine the plane for which KV is the slope line.

The straight lines of special position in the plane that we have considered, mainly horizontal and frontal, are very often used in various constructions and in solving problems. This is explained by the significant simplicity of constructing these straight lines; Therefore, it is convenient to use them as auxiliary ones.

In Fig. 116, the horizontal projection K" of point K was given. It was necessary to find the frontal projection K" if point K should be in the plane defined by two parallel straight lines drawn from points A and B.

First, a certain straight line was drawn passing through point K and lying in a given plane. The frontal MN was chosen as such a straight line: its horizontal projection is drawn through this projection K." Then the points M" and N" were constructed, defining the frontal projection of the frontal.

The required projection K" must be on the straight line M"N".

In Fig. 117 on the left, based on the given frontal projection A" of point A, belonging to square α, its horizontal projection (A") was found; the construction was made using the horizontal line EK. In Fig. 117 on the right, a similar problem was solved using the MN frontal.

Another example of constructing a missing projection of a point belonging to a certain plane is given in Fig. 118. On the left is shown the task: the slope line of the plane (AB) and the horizontal projection of the point (K"). On the right in Fig. 118 the construction is shown; through the point K" a horizontal projection of the horizontal line on which should lie is drawn (perpendicular to A"B") point K, at point L" the frontal projection of this horizontal line was found and on it the desired projection K".

In Fig. 119 gives an example of constructing the second projection of a certain plane curve if one projection (horizontal) and pl. α, in which this curve is located. Taking a number of points on the horizontal projection of the curve, we find the points for constructing the frontal projection of the curve using horizontal lines.

The arrows show the progress of constructing the frontal projection A" along the horizontal projection A".

Questions for §§ 16-18

1. How is a plane defined in a 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?
4. How is it determined in a 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. What is the frontal, horizontal and slope line of the plane?
7. Can the slope line of a plane serve to determine the angle of inclination of this plane to the plane of projections π 1?
8. Does a straight line define the plane for which this straight line is a slope line?

A point belongs to a plane if it belongs to any line lying in this plane.

Constructing a point in a plane comes down to two operations: constructing an auxiliary line in the plane and constructing a point on this line.

Task: Plane S defined by intersecting lines A And b(Fig. 2-3). Dot M(M 2) belongs to the plane.

Find M 1.

Brief description of the problem conditions: S(aÇ b), M(M 2)Î S; M 1 = ?

Solution: Through the point M 2(Fig. 2-4) draw an auxiliary straight line

kÌ S: k 2Ç a 2 =1 2 ; k 2Ç b 2 =2 2 ;

then we find the horizontal projections of the points 1 And 2 according to the condition of belonging to direct A And b respectively; through two points 1 1 And 2 1 we conduct a direct k 1 and on it, using the communication line, we find a point M 1. And you can draw as many such lines as you like, that is, there are countless possible solutions.

A straight line belongs to a plane if it:

1. Passes through two points of the plane;

2. Passes through one point of the plane and is parallel to some line lying in this plane.

In the previous example, we looked at how to construct a straight line in a plane using two points. For the second case, the plane G let's define it as a triangle ABC .

Task: Plane G given DABC(Fig. 2-5).

Dot M(M 1) belongs G. Find M 2.

M(M 1)О Г(АВС). M 2 =?

Solution:

Through the point M 1(Fig. 2-6) let's draw a straight line k, parallel to the side of the triangle AB. She'll cross the side AC at the point 1 : k 1|| A 1 B 1 ; k 1 A 1Ç C 1 =1 1; using the communication line we will find 1 2 , let's conduct k 2 parallel A 2 B 2 let's find a point M 2:

Algorithmic record of the solution:

1 1 Î A 1C 1Þ 1 2Î A 2C2; 12Î k2,k 2|| A 2B2;M 2Î k2.

How do you think?

How many solutions does this problem have?

Constructing a point in a plane comes down to two operations: constructing an auxiliary line in the plane and constructing a point on this line.

Task: Plane S defined by intersecting lines A And b(Fig. 2-3). Dot M(M 2) belongs to the plane.

Find M 1.

Brief description of the problem conditions: S(a Ç b), M(M 2)Î S; M 1 = ?

Solution: Through the point M 2(Fig. 2-4) draw an auxiliary straight line

kÌ S: k 2 Ç a 2 =1 2 ; k 2 Ç b 2 =2 2 ;

then we find the horizontal projections of the points 1 And 2 according to the condition of belonging to direct A And b respectively; through two points 1 1 And 2 1 we conduct a direct k 1 and on it, using the communication line, we find a point M 1. And you can draw as many such lines as you like, that is, there are countless possible solutions.

A straight line belongs to a plane if it:

1. Passes through two points of the plane;

Passes through one point of the plane and is parallel to some line lying in this plane.

In the previous example, we looked at how to construct a straight line in a plane using two points. For the second case, the plane G let's define it as a triangle ABC .

Task: Plane G given DABC(Fig. 2-5).

Dot M(M 1) belongs G. Find M 2.

М(М 1)О Г(АВС). M 2 =?

Solution:

Through the point M 1(Fig. 2-6) let's draw a straight line k, parallel to the side of the triangle AB. She'll cross the side AC at the point 1 : k 1 || A 1 B 1 ; k 1 A 1 Ç C 1 =1 1; using the communication line we will find 1 2 , let's conduct k 2 parallel A 2 B 2 let's find a point M 2:

Algorithmic record of the solution:

1 1 Î A 1 C 1 Þ 1 2 Î A 2 C 2 ; 1 2 О k 2 , k 2 || A 2 B 2 ; M 2 О k 2 .

How do you think?

How many solutions does this problem have?

Partial planes

Planes parallel or perpendicular to one of the projection planes are called planes of particular position.

There are two groups of such planes:

1. Projection planes
2. Level planes

Projection planes

If a plane is perpendicular to only one projection plane, then it is called projecting.

One of its projections degenerates into a straight line called main projection and having collective properties.

Horizontal projection plane

This is a plane perpendicular to the horizontal plane of projections: G^^ P 1

(Fig. 2-7a, 2-7b).

Graphic sign:

Horizontal projection G 1 horizontally to the projecting plane is a straight line, neither parallel nor perpendicular to the communication lines. This home projection.

For example:

G ^^ P 1- horizontally projecting plane.

Г^ П 1 Þ Г 1- straight line, main projection.

Ðb- plane inclination angle G to P 2.

Spatial drawing

One of the problems for which level lines are used is the problem of constructing projections of a point belonging to a plane. Let there be a frontal projection D 2 of a point D belonging to the plane defined by the traces k X l (Fig. 111, a). It is required to find the horizontal projection D 1 of point D.

A point belongs to a plane if it belongs to a line belonging to the plane. We solve the problem using the horizontal h of the k X l plane. Through point D 2 we draw a frontal projection h 2 of this horizontal line, which, as is known, should be parallel to the x 12 axis (Fig. 111 b). It will intersect the frontal projection k 2 of the frontal trace k to point N 2 ; Having drawn a vertical connection line, we find on the x 12 projection axis the horizontal projection of the frontal trace N of the horizontal (see Fig. 108).

TBegin-->TEnd-->

The horizontal projection h 1 of the horizontal must be parallel to l 1. We will find the horizontal projection D 1 of point D on the horizontal projection h 1 of the horizontal at the point of its intersection with the vertical connection line drawn through point D 2.

This problem could also be solved using the frontal. In this case, it would be necessary to draw a frontal projection f 2 ||k 2 through point D 2. We advise students to complete the construction themselves. The result should be the same as the first construction.

Let us slightly change the conditions of the problem. Let the horizontal projection E 1 of point E and the plane ABC, defined by the projections of the triangle (Fig. 112, a), be given. In this problem, you cannot use the horizontal of the plane, since there is no frontal projection of point E. We use the frontal f; through point E 1 we draw a horizontal projection (x frontal), find its frontal projection l2 and point E 1 on it.

A point in a plane can be constructed not only using the horizontal and frontal, but also using a straight line in general position. In some cases it is even more convenient.

TBegin-->
TEnd-->

The construction of a general line belonging to a general plane is not fundamentally different from the construction of horizontals and fronts belonging to the plane. The construction is based on a position known from geometry: a straight line belongs to a plane if it has two common points with this plane. Thus, if we intersect one of the projections of the plane with an arbitrary line and use two points of intersection of this line with lines belonging to the plane to construct a second projection of the line, then we can solve the problem. For example, let's solve the previous problem using a straight line in general position (Fig. 112, b). Through point E 1 we draw a straight line D 1 F 1 of any slope; we find the frontal projection D 2 F 2 of the DF line using the intersection points of D 1 and F 1. At the intersection of the frontal projection D 2 F 2 with the vertical communication line, we find the frontal projection E 1 of point E.

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3. Flatness

3.1. Methods for defining a plane in orthogonal drawings

 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.
  

• 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 general position plane can have three traces: horizontal απ1, frontal απ2 and profile απ3, which it forms when intersecting with known projection planes: horizontal π1, frontal π2 and profile π3 (Figure 3.2).

Figure 3.2 - Traces of a general plane

3.2. Partial planes

Partial plane - a plane perpendicular or parallel to the projection plane.

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 - Frontal projection plane, to which belong: points A, B, C , lines,
AC, AB, BC triangle plane

ABC Horizontal projection plane - plane perpendicular to the horizontal plane of projections

(Figure 3.4, b). Front projection plane- plane perpendicular to the frontal plane of projections

(Figure 3.4, a). Profile-projecting plane

- a plane perpendicular to the profile plane of projections. Planes parallel to projection planes are called level planes or .

double projecting planes - plane parallel to the horizontal plane of projections(Figure 3.4, d).

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

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. Point and line in a plane

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



α = Figure 3.5. Belonging of a point to a plane // m
n ∈ m ⇒ n ∈ α

D

α = Figure 3.5. Belonging of a point to a plane // m
n ∈ α
Figure 3.6. Belonging to a straight plane ∈ α ⇒ WITH ∈ α

CD

 Exercise Figure 3.6. Belonging to a straight plane. 

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

Figure 3.7 - Condition (a) and solution (b) of the problem

1. Solution :ABCD
   
2. - a flat quadrilateral defining a plane.Let's draw diagonals in it A.C. AndBD
   
3. (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 linesKaccording to its known frontal projection: 2 A 2 ∩ C 2 n 2 B 2 .
   
4. =KAndLet us restore the projection connection line until it intersects with the horizontal projection of the straight lineC 1 n: on the diagonal projection 1 we are building 1 .
   
5. TO Through 1 1 we are building 1 AThrough 1 Figure 3.6. Belonging to a straight plane 1 .
   
6. carry out a diagonal projection Figure 3.6. Belonging to a straight plane 1 Full stopThrough 1 1 we are building 1 .

(Figure 3.8, a; 3.9).

Figure 3.8.a. Horizontal straight line of a level in a plane defined by a triangle 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 (π2)

 (Figure 3.8, b; 3.10).

Figure 3.8.b. Front line of the level in the plane defined by the triangle Level profile line (p third parallel) is a straight line lying in a given plane and parallel to the profile plane of projections (π3)


 (Figure 3.8, c; 3.11).

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 3.19).



Figure 3.19. Parallelism of a straight plane

3.5.2. Intersection of a line with a plane

1. To construct the line of intersection of a straight line with a plane, it is necessary (Figure 3.20):AConclude direct
   
2. to the auxiliary plane β (planes of particular position should be selected as the auxiliary plane);
   
3. Find the line of intersection of the auxiliary plane β with the given plane α;Find the intersection point of a given lineawith the line of intersection of planes.

MN

CD

Figure 3.20. Constructing the meeting point of a straight line with a plane AB Given: straight

Figure 3.7 - Condition (a) and solution (b) of the problem

1. general position, plane σ ⊥ π1 (Figure 3.21). Construct the point of intersection of straight line AB with plane σ. The plane σ is horizontally projecting, therefore, the horizontal trace is σπ 1 (or σ 1
   
2. ) is straight; 1 we are buildingDotAB ⇒ 1 we are building 1 ∈ Through 1 must belong to the line 1 IN1 we are building and a given plane σ ⇒ 1 we are building 1 1 ∈ σ 1, therefore,according to its known frontal projection: 1 C is located at the point of intersection of projections
   
3. 1 and σ 1 ;1 we are buildingFrontal projection of the pointAccording to the criterion of intersecting lines, we will construct a horizontal projection of the point of intersection of these lines 2 ∈ according to its known frontal projection: 2 C 2 .

we find through the projection communication line:

CD

Figure 3.21. The intersection of a general line with a particular plane triangle plane Given: plane σ = Δ - general position, straight E.F.
(Figure 3.22). - general position, straight It is required to construct the point of intersection of a line

with plane σ.
A                     b

Figure 3.7 - Condition (a) and solution (b) of the problem

1. Figure 3.22. The intersection of a straight line with a plane (a - model, b - drawing) - general position, straightLet's conclude a straight line
   
2. into an auxiliary plane, for which we will use the horizontally projecting plane α (Figure 3.22, a); If α ⊥ π 1 1 , then onto the projection plane π plane α is projected onto a straight line (horizontal trace of plane απ 1 or α 1 ), coinciding with 1 E 1 ;
   
3. F
   
4. Let's find the line of intersection (1-2) of the projecting plane α with the plane σ (the solution to a similar problem was considered earlier);- general position, straightStraight line (1-2) and specified straight lineAccording to the criterion of intersecting lines, we will construct a horizontal projection of the point of intersection of these lines.
   

(Figure 3.22, b):

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 π1 or π2.

Points that in space 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 the visibility on each projection plane!

Visibility at π2

Let's choose points that compete on π2 - points 3 and 4 (Figure 3.23). Let point 3 ∈ Sun∈ σ, point 4 ∈ - general position, straight.

To determine the visibility of points on the projection plane π2, it is necessary to determine the location of these points on the horizontal projection plane when looking at π2.

The direction of view towards π2 is shown by the arrow.

From the horizontal projections of points 3 and 4, when looking at π2, it is clear that point 41 is located closer to the observer than 31.

41 ∈ 1 or α 1 ), coinciding with 1 E 1 → 4 ∈ - general position, straight⇒ onπ 2 point 4 will be visible, lying on the line - general position, straight, therefore, straight - general position, straight 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 According to the criterion of intersecting lines, we will construct a horizontal projection of the point of intersection of these lines

Visibility at π1

To determine visibility, we select points that compete on π1 - points 2 and 5.

To determine the visibility of points on the projection plane π1, it is necessary to determine the location of these points on the frontal projection plane when looking at π1.

The direction of view towards π1 is shown by the arrow.

According to the frontal projections of points 2 and 5, when looking at π1, point 22 is located closer to the observer than 52.

22 ∈ Through 2 must belong to the line 2 → 2 ∈ AB⇒ on π1 point 2 will be visible, lying on the straight line AB, therefore, straight - general position, straight in the area of ​​the competing points under consideration is located under the plane σ and will be invisible until the point According to the criterion of intersecting lines, we will construct a horizontal projection of the point of intersection of these lines- intersection of a straight line with the plane σ.

The visible one of the two competing points will be the one whose coordinate is " Z» or(s) « Y" more.

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.

Figure 3.24. Specifying a straight line perpendicular to the plane

If the straight line is perpendicular to the plane, then on the diagram: the projections of the straight line are perpendicular to the inclined projections of the horizontal and frontal lines lying in the plane, or traces of the plane (Figure 3.24).

1. Let it be straight Level profile lineperpendicular to the plane σ = Δtriangle planeand passes through the pointAccording to the criterion of intersecting lines, we will construct a horizontal projection of the point of intersection of these lines.
2. Let's construct the horizontal and frontal lines in the plane σ = Δtriangle plane :
   according to its known frontal projection:-1 ∈ σ; according to its known frontal projection:-1 // π 1 ; Figure 3.6. Belonging to a straight plane-2 ∈ σ; Figure 3.6. Belonging to a straight plane-2 // π 2 .
3. Let's restore from the pointAccording to the criterion of intersecting lines, we will construct a horizontal projection of the point of intersection of these linesperpendicular to a given plane:
   Level profile line 1 ⊥ horizontal level line 1 and Level profile line 2 ⊥ front level straight 2 .