Determination of perpendicular lines and planes in space. Perpendicular line and plane, sign and conditions of perpendicularity of line and plane

Outline of a geometry lesson in grade 10 on the topic “Perpendicularity of a line and a plane”

Lesson objectives:

educational

introduction of the sign of perpendicularity of a line and a plane;

to form students’ ideas about the perpendicularity of a straight line and a plane, their properties;

to develop students’ ability to solve typical problems on a topic, the ability to prove statements;

developing

develop independence and cognitive activity;

develop the ability to analyze, draw conclusions, systematize the information received,

develop logical thinking;

develop spatial imagination.

educational

nurturing students’ speech culture and perseverance;

instill in students an interest in the subject.

Lesson type: Lesson of studying and primary consolidation of knowledge.

Forms of student work: frontal survey.

Equipment: computer, projector, screen.

Literature:"Geometry 10-11", Textbook. Atanasyan L.S. etc.

(2009, 255 pp.)

Lesson plan:

Organizational moment (1 minute);

Updating knowledge (5 minutes);

Learning new material (15 minutes);

Primary consolidation of the studied material (20 minutes);

Summing up (2 minutes);

Homework (2 minutes).

Progress of the lesson.

Organizational moment (1 minutes)

Greeting students. Checking students' readiness for the lesson: checking the availability of notebooks and textbooks. Checking absences from class.

Updating knowledge (5 minutes)

Teacher. Which line is called perpendicular to the plane?

Student. A line perpendicular to any line lying in this plane is called a line perpendicular to this plane.

Teacher. What is the lemma about two parallel lines perpendicular to a third?

Student. If one of two parallel lines is perpendicular to the third line, then the other line is perpendicular to this line.

Teacher. Theorem on the perpendicularity of two parallel lines to a plane.

Student. If one of two parallel lines is perpendicular to a plane, then the second line is perpendicular to this plane.

Teacher. What is the converse of this theorem?

Student. If two lines are perpendicular to the same plane, then they are parallel.

Checking homework

Homework is checked if students have difficulty solving it.

Learning new material (15 minutes)

Teacher. You and I know that if a line is perpendicular to a plane, then it will be perpendicular to any line lying in this plane, but in the definition, the perpendicularity of a line to a plane is given as a fact. In practice, it is often necessary to determine whether a straight line will be perpendicular to the plane or not. Such examples can be given from life: during the construction of buildings, piles are driven perpendicular to the surface of the earth, otherwise the structure may collapse. In this case, it is impossible to use the definition of a straight perpendicular plane. Why? How many straight lines can be drawn in a plane?

Student. An infinite number of straight lines can be drawn in a plane.

Teacher. Right. And it is impossible to check the perpendicularity of a straight line to each individual plane, since this will take an infinitely long time. In order to understand whether a line is perpendicular to a plane, we introduce the sign of perpendicularity of a line and a plane. Write it down in your notebook. If a line is perpendicular to two intersecting lines lying in a plane, then it is perpendicular to this plane.

Writing in a notebook. If a line is perpendicular to two intersecting lines lying in a plane, then it is perpendicular to this plane.

Teacher. Thus, we do not need to check the perpendicularity of a straight line for each straight plane; it is enough to check the perpendicularity only for two straight lines of this plane.

Teacher. Let's prove this sign.

Given: p And q– straight, p ∩ q = O, a⊥ p, a⊥ q, p ϵ α, q ϵ α.

Prove: a⊥ α.

Teacher. And yet, to prove it, we will use the definition of a straight line perpendicular to a plane, how does it sound?

Student. If a line is perpendicular to a plane, then it is perpendicular to any line lying in this plane.

Teacher. Right. Let us draw any straight line m in the α plane. Let's draw a straight line l ║ m through point O. On line a, mark points A and B so that point O is the midpoint of segment AB. Let's draw a straight line z in such a way that it intersects the lines p, q, l; we denote the intersection points of these lines as P, Q, L, respectively. Let's connect the ends of the segment AB with points P,Q and L.

Teacher. What can we say about triangles ∆APQ and ∆BPQ?

Student. These triangles will be equal (according to the 3rd sign of equality of triangles).

Teacher. Why?

Student. Because lines p and q are perpendicular bisectors, then AP = BP, AQ = BQ, and side PQ is common.

Teacher. Right. What can we say about triangles ∆APL and ∆BPL?

Student. These triangles will also be equal (according to 1 sign of equality of triangles).

Teacher. Why?

Student. AP = B.P., P.L.– general side, APL =  BPL(from the equality ∆ APQ and ∆ B.P.Q.)

Teacher. Right. This means AL = BL. So what will ∆ALB be?

Student. This means that ∆ALB will be isosceles.

Teacher. LO is the median in ∆ALB, so what will it be in this triangle?

Student. This means that LO will also be the height.

Teacher. Therefore straightlwill be perpendicular to the linea. And since it’s straightlis any straight line belonging to the plane α, then by definition a straight linea⊥ α. Q.E.D.

Proved by presentation

Teacher. What to do if line a does not intersect point O, but remains perpendicular to lines p and q? What if straight line a intersects any other point of the given plane?

Student. You can construct a straight line 1 , which will be parallel to line a, will intersect point O, and using the lemma about two parallel lines perpendicular to the third, it can be proven thata 1 ⊥ p, a 1 ⊥ q.

Teacher. Right.

Primary consolidation of the studied material (20 minutes)

Teacher. In order to consolidate the material we have studied, we will solve number 126. Read the task.

Student. The straight line MB is perpendicular to sides AB and BC of triangle ABC. Determine the type of triangle МВD, where D is an arbitrary point of the line AC.

Drawing.

Given: ∆ ABC, M.B.⊥ B.A., M.B.⊥ B.C., D ϵ A.C..

Find: ∆ MBD.

Solution.

Teacher. Is it possible to draw a plane through the vertices of a triangle?

Student. Yes, you can. The plane can be drawn along three points.

Teacher. How will straight lines BA and NE be located relative to this plane?

Student. These lines will lie in this plane.

Teacher. It turns out that we have a plane, and in it there are two intersecting lines. How does the direct MV relate to these direct lines?

Student. Direct MV⊥ VA, MV ⊥ VS.

Write on the board and in notebooks. Because MV⊥ VA, MV ⊥ VS

Teacher. If a line is perpendicular to two intersecting lines lying in a plane, will the line be related to this plane?

Student. The straight line MV will be perpendicular to the ABC plane.

⊥ ABC.

Teacher. Point D is an arbitrary point on the segment AC, so how will straight line BD relate to plane ABC?

Student. This means that BD belongs to the ABC plane.

Write on the board and in notebooks. Because BD ϵ ABC

Teacher. What will the direct MV and BD be relative to each other?

Student. These lines will be perpendicular by definition of a line perpendicular to the plane.

Write on the board and in notebooks. ↔ MV⊥ BD

Teacher. If MB is perpendicular to BD, then what will be the triangle MBD?

Student. Triangle MBD will be rectangular.

Write on the board and in notebooks. ↔ ∆MBD – rectangular.

Teacher. Right. Let's solve number 127. Read the task.

Student. In a triangleABC sum of angles A And Bequal to 90°. StraightBDperpendicular to the planeABC. Prove that CD⊥ AC.

The student goes to the board. Draws a drawing.

Write on the board and in your notebook.

Given: ∆ ABC,  A +  B= 90°, BD⊥ ABC.

Prove: CD⊥ A.C..

Proof:

Teacher. What is the sum of the angles of a triangle?

Student. The sum of the angles in a triangle is 180°.

Teacher. What will be the angle C in triangle ABC?

Student. Angle C in triangle ABC will be equal to 90°.

Write on the board and in notebooks. C = 180° - A- B= 90°

Teacher. If angle C is 90°, then how will straight lines AC and BC be positioned relative to each other?

Student. So AC⊥ Sun.

Write on the board and in notebooks. ↔ AC⊥ Sun

Teacher. Line BD is perpendicular to plane ABC. What follows from this?

Student. So BD is perpendicular to any line from ABC.

BD⊥ ABC ↔ BDperpendicular to any straight lineABC(by definition)

Teacher. According to this, how will direct BD and AC relate?

Student. This means that these lines will be perpendicular.

BD⊥ A.C.

Teacher. AC is perpendicular to two intersecting lines lying in the DBC plane, but AC does not pass through the intersection point. How to fix this?

Student. Through point B we draw a straight line a parallel to AC. Since AC is perpendicular to BC and BD, then a will be perpendicular to BC and BD by lemma.

Write on the board and in notebooks. Through point B we draw a straight line a ║AC ↔ a⊥ B.C., and ⊥ BD

Teacher. If straight line a is perpendicular to BC and BD, then what can be said about the relative position of straight line a and plane BDC?

Student. This means that straight line a will be perpendicular to the plane BDC, and therefore straight line AC will be perpendicular to BDC.

Write on the board and in notebooks. ↔ a⊥ BDC↔ AC ⊥ BDC.

Teacher. If AC is perpendicular to BDC, then how will the lines AC and DC be positioned relative to each other?

Student. AC and DC will be perpendicular by definition of a line perpendicular to the plane.

Write on the board and in notebooks. Because AC⊥ BDC↔ AC ⊥ DC

Teacher. Well done. Let's solve number 129. Read the assignment.

Student. StraightA.M.perpendicular to the plane of the squareABCD, whose diagonals intersect at point O. Prove that: a) straight lineBDperpendicular to the planeAMO; b)M.O.⊥ BD.

A student comes to the board. Draws a drawing.

Write on the board and in your notebook.

Given:ABCD- square,A.M.⊥ ABCD, A.C. ∩ BD = O

Prove:BD⊥ AMO, MO⊥ BD

Proof:

Teacher. We need to prove that the straight lineBD⊥ AMO. What conditions must be met for this to happen?

Student. It needs to be straight BD was perpendicular to at least two intersecting straight lines from the plane AMO.

Teacher. The condition says that BD perpendicular to two intersecting lines of AMO?

Student. No.

Teacher. But we know that A.M. perpendicular ABCD . What conclusion can be drawn from this?

Student. This means that A.M. perpendicular to any straight line from this plane, that is A.M. perpendicular B.D.

A.M.⊥ ABCD ↔ A.M.⊥ BD(by definition).

Teacher. One line is perpendicular BD There is. Pay attention to the square, how the straight lines will be located relative to each other AC and BD?

Student. A.C. will be perpendicular BD by the property of the diagonals of a square.

Write on the board and in your notebook. BecauseABCD- square, thenA.C.⊥ BD(by the property of the diagonals of a square)

Teacher. We found two intersecting lines lying in the plane AMO perpendicular to a straight line BD . What follows from this?

Student. This means that BD perpendicular to the plane AMO.

Write on the board and in notebooks. BecauseA.C.⊥ BDAndA.M.⊥ BD ↔ BD⊥ AMO(by attribute)

Teacher. Which line is called a line perpendicular to a plane?

Student. A line is called perpendicular to a plane if it is perpendicular to any line from this plane.

Teacher. This means how the lines are interconnected BD and OM?

Student. So BD perpendicular OM . Q.E.D.

Write on the board and in notebooks. ↔BD⊥ M.O.(by definition). Q.E.D.

Summing up (2 minutes)

Teacher. Today we studied the sign of perpendicularity of a line and a plane. What does it sound like?

Student. If a line is perpendicular to two intersecting lines lying in a plane, then this line is perpendicular to this plane.

Teacher. Right. We learned to use this feature when solving problems. Well done to those who answered at the board and helped from the spot.

Homework (2 minutes)

Teacher. Paragraph 1, paragraphs 15-17, teach: lemma, definition and all theorems. No. 130, 131.

In this article we will talk about the perpendicularity of a line and a plane. First, the definition of a line perpendicular to a plane is given, a graphic illustration and example are given, and the designation of a line perpendicular to a plane is shown. After this, the sign of perpendicularity of a straight line and a plane is formulated. Next, conditions are obtained that make it possible to prove the perpendicularity of a straight line and a plane, when the straight line and the plane are specified by certain equations in a rectangular coordinate system in three-dimensional space. In conclusion, detailed solutions to typical examples and problems are shown.

Page navigation.

Perpendicular straight line and plane - basic information.

We recommend that you first repeat the definition of perpendicular lines, since the definition of a line perpendicular to a plane is given through the perpendicularity of the lines.

Definition.

They say that line is perpendicular to the plane, if it is perpendicular to any line lying in this plane.

We can also say that a plane is perpendicular to a line, or a line and a plane are perpendicular.

To indicate perpendicularity, use an icon like “”. That is, if line c is perpendicular to the plane, then we can briefly write .

An example of a line perpendicular to a plane is the line along which two adjacent walls of a room intersect. This line is perpendicular to the plane and to the plane of the ceiling. A rope in a gym can also be considered as a straight line segment perpendicular to the plane of the floor.

In conclusion of this paragraph of the article, we note that if a straight line is perpendicular to a plane, then the angle between the straight line and the plane is considered equal to ninety degrees.

Perpendicularity of a straight line and a plane - a sign and conditions of perpendicularity.

In practice, the question often arises: “Are the given straight line and plane perpendicular?” To answer this there is sufficient condition for perpendicularity of a line and a plane, that is, such a condition, the fulfillment of which guarantees the perpendicularity of the straight line and the plane. This sufficient condition is called the sign of perpendicularity of a line and a plane. Let us formulate it in the form of a theorem.

Theorem.

For a given line and plane to be perpendicular, it is sufficient that the line be perpendicular to two intersecting lines lying in this plane.

You can look at the proof of the sign of perpendicularity of a line and a plane in a geometry textbook for grades 10-11.

When solving problems of establishing the perpendicularity of a line and a plane, the following theorem is also often used.

Theorem.

If one of two parallel lines is perpendicular to a plane, then the second line is also perpendicular to the plane.

At school, many problems are considered, for the solution of which the sign of perpendicularity of a line and a plane is used, as well as the last theorem. We will not dwell on them here. In this section of the article we will focus on the application of the following necessary and sufficient condition for the perpendicularity of a line and a plane.

This condition can be rewritten in the following form.

Let is the direction vector of line a, and is the normal vector of the plane. For straight line a and plane to be perpendicular, it is necessary and sufficient that And : , where t is some real number.

The proof of this necessary and sufficient condition for the perpendicularity of a line and a plane is based on the definitions of the direction vector of a line and the normal vector of a plane.

Obviously, this condition is convenient to use to prove the perpendicularity of a line and a plane, when the coordinates of the directing vector of the line and the coordinates of the normal vector of the plane in a fixed three-dimensional space can be easily found. This is true for cases when the coordinates of the points through which the plane and the line pass are given, as well as for cases when the line is determined by some equations of a line in space, and the plane is given by an equation of a plane of some type.

Let's look at solutions to several examples.

Example.

Prove the perpendicularity of the line and planes.

Solution.

We know that the numbers in the denominators of the canonical equations of a line in space are the corresponding coordinates of the direction vector of this line. Thus, - direct vector .

The coefficients of the variables x, y and z in the general equation of a plane are the coordinates of the normal vector of this plane, that is, is the normal vector of the plane.

Let us check the fulfillment of the necessary and sufficient condition for the perpendicularity of a line and a plane.

Because , then the vectors and are related by the relation , that is, they are collinear. Therefore, straight perpendicular to the plane.

Example.

Are the lines perpendicular? and plane.

Solution.

Let us find the direction vector of a given straight line and the normal vector of the plane in order to check whether the necessary and sufficient condition for the perpendicularity of the line and the plane is satisfied.

The directing vector is straight is

In order for a straight line in space to be a plane, it is necessary and sufficient that on the diagram the horizontal projection of the line is a horizontal projection of the horizontal, and the frontal projection is to the frontal projection of the front of this plane.

Determining the distance from a point to a plane(Fig. 19)

1. From a point, lower a perpendicular to the plane (to do this in the plane

hold h,f);

2. Find the point of intersection of the straight line with the plane (see Fig. 18);

3. Find n.v. perpendicular segment (see Fig. 7).

Second section Method of replacing projection planes

(for tasks 5, 6,7)

This geometric figure is left motionless in the system of projection planes. New projection planes are installed so that the projections obtained on them provide a rational solution to the problem under consideration. In this case, each new system of projection planes must be an orthogonal system. After projecting objects onto planes, they are combined into one by rotating them around common straight lines (projection axes) of each pair of mutually perpendicular planes.

For example, let point A be specified in a system of two planes P 1 and P 2. Let us supplement the system with another plane P 4 (Fig. 20), P 1 P 4. It has a common line X 14 with plane P 1. We build a projection of A 4 onto P 4.

AA 1 =A 2 A 12 =A 4 A 14.

In Fig. 21, where planes P 1, P 2 and P 4 are aligned, this fact is determined by the result A 1 A 4 X 14, and A 14 A 4 A 2 A 12.

The distance of the new projection of the point to the new projection axis (A 4 A 14) is equal to the distance from the replaced projection of the point to the replaced axis (A 2 A 12).

A large number of metric problems of descriptive geometry are solved based on the following four problems:

1. Transformation of a general position straight line into a level straight line (Fig. 22):

a) P 4 || AB (axis X 14 || A 1 B 1);

b) A 1 A 4 X 14; B 1 B 4 X 14 ;

c) A 4 A 14 = A 12 A 2;

V 4 V 14 = V 12 V 2;

A 4 B 4 - n.v.

2. Converting a general line into a projecting line (Fig. 23):

a) P 4 || AB (X 14 || A 1 B 1);

A 1 A 4 X 14;

B 1 B 4 X 14 ;

A 14 A 4 = A 12 A 2;

V 14 V 4 = V 12 V 2;

A 4 B 4 - present;

b) P 5 AB (X 45 A 4 B 4);

A 4 A 5 X 45;

B 4 B 5 X 45;

A 45 A 5 =B 45 V 5 =A 14 A 1 =B 14 V 1;

3. Converting the general position plane to the projecting position (Fig. 24):

The plane can be brought into a projecting position if one straight line of the plane is made projecting. In the ABC plane we draw a horizontal line (h 2 ,h 1), which can be made projecting in one transformation. Let's draw plane P 4 perpendicular to the horizontal; onto this plane it will be projected as a point, and the plane of the triangle as a straight line.

4. Transformation of the general position plane into the level plane (Fig. 25).

Make the plane a level plane using two transformations. First, the plane must be made projecting (see Fig. 25), and then draw P 5 || A 4 B 4 C 4, we get A 5 B 5 C 5 - n.v.

Problem #5

Determine the distance from point C to a straight line in general position (Fig. 26).

The solution comes down to the 2nd main problem. Then the distance in the diagram is defined as the distance between two points

A 5 B 5 D 5 and C 5.

Projection C 4 D 4 || X 45.

Problem #6

Determine the distance from ()D to the plane specified by points A, B, C (Fig. 27).

The problem is solved using the 2nd main problem. The distance (E 4 D 4), from ()D 4 to the straight line A 4 C 4 B 4, into which the ABC plane was projected, is the natural value of the segment ED.

Projection D 1 E 1 || X 14;

E 2 E X12 = E 4 E X14.

Build it yourself D 1 E 1.

Build it yourself D 2 E 2.

Problem No. 7

Determine the actual size of triangle ABC (see solution to the 4th main problem) (Fig. 25)

Construction a straight line perpendicular to a given plane from a point taken outside this plane.
Let a- plane, A – the point from which the perpendicular must be lowered. Let's draw a straight line in the plane A. Through point A and straight line A let's draw a plane b(a straight line and a point define a plane, and only one). In plane b from point A we drop to a straight line A perpendicular AB. From point B to the plane a Let us restore the perpendicular and designate the straight line on which this perpendicular lies beyond With. Through segment AB and straight line With let's draw a plane g(two intersecting lines define a plane, and only one). In plane g from point A we drop to a straight line With perpendicular to AC. Let us prove that the segment AC is perpendicular to the plane b. Proof. Straight A perpendicular to straight lines With and AB (by construction), which means it is perpendicular to the plane itself g, in which these two intersecting lines lie (based on the perpendicularity of the line and the plane). And since it is perpendicular to this plane, then it is perpendicular to any straight line in this plane, which means it is a straight line A perpendicular to AC. Line AC is perpendicular to two lines lying in the plane α: With(by construction) and A(according to what has been proven), it means that it is perpendicular to the plane α (based on the perpendicularity of the line and the plane)

Theorem 1 . If two intersecting lines are parallel to two perpendicular lines, then they are also perpendicular.
Proof. Let A And b- perpendicular lines, A 1 and b 1 - intersecting lines parallel to them. Let us prove that the straight lines A 1 and b 1 are perpendicular.
If straight A, b, A 1 and b 1 lie in the same plane, then they have the property specified in the theorem, as is known from planimetry.
Let us now assume that our lines do not lie in the same plane. Then straight A And b lie in some plane α, and the straight lines A 1 and b 1 - in some plane β. Based on the parallelism of planes, planes α and β are parallel. Let C be the point of intersection of the lines A And b, and C 1 - intersections of lines A 1 and b 1. Let us draw in the plane of parallel lines A And A A And A 1 at points A and A 1. In the plane of parallel lines b And b 1 line parallel to straight line CC 1. She will cross the lines b And b 1 at points B and B 1.
Quadrilaterals CAA 1 C 1 and SVV 1 C 1 are parallelograms, since their opposite sides are parallel. Quadrilateral ABC 1 A 1 is also a parallelogram. Its sides AA 1 and BB 1 are parallel, because each of them is parallel to the line CC 1. Thus, the quadrilateral lies in the plane passing through the parallel lines AA 1 and BB 1. And it intersects parallel planes α and β along parallel straight lines AB and A 1 B 1.
Since the opposite sides of a parallelogram are equal, then AB = A 1 B 1, AC = A 1 C 1, BC = B 1 C 1. According to the third sign of equality, triangles ABC and A 1 B 1 C 1 are equal. So, angle A 1 C 1 B 1, equal to angle ACB, is straight, i.e. straight A 1 and b 1 are perpendicular. Etc.

Properties perpendicular to a straight line and a plane.
Theorem 2 . If a plane is perpendicular to one of two parallel lines, then it is also perpendicular to the other.
Proof. Let A 1 and A 2 - two parallel lines and α - a plane perpendicular to the line A 1. Let us prove that this plane is perpendicular to the straight line A 2 .
Let's draw 2 intersections of a line through point A A 2 with plane α an arbitrary straight line With 2 in the α plane. Let us draw in the plane α through point A 1 the intersection of the line A 1 with plane α straight With 1 parallel to the line With 2. Since it's straight A 1 is perpendicular to the plane α, then straight lines A 1 and With 1 are perpendicular. And according to Theorem 1, the intersecting lines parallel to them A 2 and With 2 are also perpendicular. Thus, straight A 2 is perpendicular to any line With 2 in the α plane. And this means that straight A 2 is perpendicular to the plane α. The theorem is proven.

Theorem 3 . Two lines perpendicular to the same plane are parallel to each other.
We have a plane α and two lines perpendicular to it A And b. Let's prove that A || b.
Through the points of intersection of the straight lines of the plane, draw a straight line With. Based on the characteristic we get A ^ c And b ^ c. Through straight lines A And b Let's draw a plane (two parallel lines define a plane, and only one). In this plane we have two parallel lines A And b and secant With. If the sum of internal one-sided angles is 180°, then the lines are parallel. We have just such a case - two right angles. That's why A || b.

The construction of mutually perpendicular lines and planes is an important graphic operation in solving metric problems.

The construction of a perpendicular to a straight 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?