In this article, we will find answers to questions about how to create a projection of a point onto a plane and how to determine the coordinates of this projection. In the theoretical part, we will rely on the concept of projection. We will give definitions of terms, accompany the information with illustrations. Let's consolidate the acquired knowledge by solving examples.

Projection, types of projection

For convenience of consideration of spatial figures, drawings depicting these figures are used.

Definition 1

Projection of a figure onto a plane- a drawing of a spatial figure.

Obviously, there are a number of rules used to construct a projection.

Definition 2

projection- the process of constructing a drawing of a spatial figure on a plane using construction rules.

Projection plane is the plane in which the image is built.

The use of certain rules determines the type of projection: central or parallel.

A special case of parallel projection is perpendicular projection or orthogonal projection: in geometry, it is mainly used. For this reason, the adjective “perpendicular” itself is often omitted in speech: in geometry they simply say “projection of a figure” and mean by this the construction of a projection by the method of perpendicular projection. In special cases, of course, otherwise can be stipulated.

We note the fact that the projection of a figure onto a plane is, in fact, the projection of all points of this figure. Therefore, in order to be able to study a spatial figure in a drawing, it is necessary to acquire the basic skill of projecting a point onto a plane. What we will talk about below.

Recall that most often in geometry, speaking of projection onto a plane, they mean the use of perpendicular projection.

We will make constructions that will enable us to obtain the definition of the projection of a point onto a plane.

Suppose a three-dimensional space is given, and in it - a plane α and a point M 1 that does not belong to the plane α. Draw a straight line through a given point M 1 A perpendicular to the given plane α. The point of intersection of the line a and the plane α will be denoted as H 1 , by construction it will serve as the base of the perpendicular dropped from the point M 1 to the plane α .

If a point M 2 is given, belonging to a given plane α, then M 2 will serve as a projection of itself onto the plane α.

Definition 3

is either the point itself (if it belongs to a given plane), or the base of the perpendicular dropped from a given point to a given plane.

Finding the coordinates of the projection of a point on a plane, examples

Let in three-dimensional space given: rectangular coordinate system O x y z, plane α, point M 1 (x 1, y 1, z 1) . It is necessary to find the coordinates of the projection of the point M 1 onto a given plane.

The solution obviously follows from the above definition of the projection of a point onto a plane.

We denote the projection of the point M 1 onto the plane α as H 1 . According to the definition, H 1 is the point of intersection of the given plane α and the line a through the point M 1 (perpendicular to the plane). Those. the coordinates of the projection of the point M 1 we need are the coordinates of the point of intersection of the line a and the plane α.

Thus, to find the coordinates of the projection of a point onto a plane, it is necessary:

Get the equation of the plane α (in case it is not set). An article about the types of plane equations will help you here;

Determine the equation of the line a passing through the point M 1 and perpendicular to the plane α (study the topic of the equation of the straight line passing through a given point perpendicular to a given plane);

Find the coordinates of the point of intersection of the line a and the plane α (article - finding the coordinates of the point of intersection of the plane and the line). The data obtained will be the coordinates of the projection of the point M 1 onto the plane α that we need.

Let's consider the theory on practical examples.

Example 1

Determine the coordinates of the projection of the point M 1 (- 2, 4, 4) onto the plane 2 x - 3 y + z - 2 \u003d 0.

Solution

As we can see, the equation of the plane is given to us, i.e. there is no need to compose it.

Let's write the canonical equations of the straight line a passing through the point M 1 and perpendicular to the given plane. For these purposes, we determine the coordinates of the directing vector of the straight line a. Since the line a is perpendicular to the given plane, then the directing vector of the line a is normal vector plane 2 x - 3 y + z - 2 \u003d 0. Thus, a → = (2 , - 3 , 1) – direction vector of the line a .

Now we compose the canonical equations of a straight line in space passing through the point M 1 (- 2, 4, 4) and having a direction vector a → = (2 , - 3 , 1) :

x + 2 2 = y - 4 - 3 = z - 4 1

To find the desired coordinates, the next step is to determine the coordinates of the point of intersection of the line x + 2 2 = y - 4 - 3 = z - 4 1 and the plane 2 x - 3 y + z - 2 = 0 . For this purpose, we move from canonical equations to the equations of two intersecting planes:

x + 2 2 = y - 4 - 3 = z - 4 1 ⇔ - 3 (x + 2) = 2 (y - 4) 1 (x + 2) = 2 (z - 4) 1 ( y - 4) = - 3 (z + 4) ⇔ 3 x + 2 y - 2 = 0 x - 2 z + 10 = 0

Let's make a system of equations:

3 x + 2 y - 2 = 0 x - 2 z + 10 = 0 2 x - 3 y + z - 2 = 0 ⇔ 3 x + 2 y = 2 x - 2 z = - 10 2 x - 3 y + z = 2

And solve it using Cramer's method:

∆ = 3 2 0 1 0 - 2 2 - 3 1 = - 28 ∆ x = 2 2 0 - 10 0 - 2 2 - 3 1 = 0 ⇒ x = ∆ x ∆ = 0 - 28 = 0 ∆ y = 3 2 0 1 - 10 - 2 2 2 1 = - 28 ⇒ y = ∆ y ∆ = - 28 - 28 = 1 ∆ z = 3 2 2 1 0 - 10 2 - 3 2 = - 140 ⇒ z = ∆ z ∆ = - 140 - 28 = 5

Thus, the desired coordinates of a given point M 1 on a given plane α will be: (0, 1, 5) .

Answer: (0 , 1 , 5) .

Example 2

Points А (0 , 0 , 2) are given in a rectangular coordinate system O x y z of three-dimensional space; In (2, - 1, 0) ; C (4, 1, 1) and M 1 (-1, -2, 5). It is necessary to find the coordinates of the projection M 1 onto the plane A B C

Solution

First of all, we write the equation of a plane passing through three given points:

x - 0 y - 0 z - 0 2 - 0 - 1 - 0 0 - 2 4 - 0 1 - 0 1 - 2 = 0 ⇔ x y z - 2 2 - 1 - 2 4 1 - 1 = 0 ⇔ ⇔ 3 x - 6y + 6z - 12 = 0 ⇔ x - 2y + 2z - 4 = 0

Let's write the parametric equations of the straight line a, which will pass through the point M 1 perpendicular to the plane A B C. The plane x - 2 y + 2 z - 4 \u003d 0 has a normal vector with coordinates (1, - 2, 2), i.e. vector a → = (1 , - 2 , 2) – direction vector of the line a .

Now, having the coordinates of the point of the line M 1 and the coordinates of the directing vector of this line, we write the parametric equations of the line in space:

Then we determine the coordinates of the point of intersection of the plane x - 2 y + 2 z - 4 = 0 and the line

x = - 1 + λ y = - 2 - 2 λ z = 5 + 2 λ

To do this, we substitute into the equation of the plane:

x = - 1 + λ , y = - 2 - 2 λ , z = 5 + 2 λ

Now, using the parametric equations x = - 1 + λ y = - 2 - 2 λ z = 5 + 2 λ, we find the values ​​of the variables x, y and z at λ = - 1: x = - 1 + (- 1) y = - 2 - 2 (- 1) z = 5 + 2 (- 1) ⇔ x = - 2 y = 0 z = 3

Thus, the projection of the point M 1 onto the plane A B C will have coordinates (- 2, 0, 3) .

Answer: (- 2 , 0 , 3) .

Let us dwell separately on the question of finding the coordinates of the projection of a point on the coordinate planes and planes that are parallel to the coordinate planes.

Let points M 1 (x 1, y 1, z 1) and coordinate planes O x y , O x z and O y z be given. The projection coordinates of this point on these planes will be respectively: (x 1 , y 1 , 0) , (x 1 , 0 , z 1) and (0 , y 1 , z 1) . Consider also the planes parallel to the given coordinate planes:

C z + D = 0 ⇔ z = - D C , B y + D = 0 ⇔ y = - D B

And the projections of the given point M 1 on these planes will be points with coordinates x 1 , y 1 , - D C , x 1 , - D B , z 1 and - D A , y 1 , z 1 .

Let us demonstrate how this result was obtained.

As an example, let's define the projection of the point M 1 (x 1, y 1, z 1) onto the plane A x + D = 0. The rest of the cases are similar.

The given plane is parallel to the coordinate plane O y z and i → = (1 , 0 , 0) is its normal vector. The same vector serves as the directing vector of the straight line perpendicular to the plane O y z . Then the parametric equations of a straight line drawn through the point M 1 and perpendicular to a given plane will look like:

x = x 1 + λ y = y 1 z = z 1

Find the coordinates of the point of intersection of this line and the given plane. We first substitute into the equation A x + D = 0 equalities: x = x 1 + λ, y = y 1, z = z 1 and get: A (x 1 + λ) + D = 0 ⇒ λ = - D A - x 1

Then we calculate the desired coordinates using the parametric equations of the straight line for λ = - D A - x 1:

x = x 1 + - D A - x 1 y = y 1 z = z 1 ⇔ x = - D A y = y 1 z = z 1

That is, the projection of the point M 1 (x 1, y 1, z 1) onto the plane will be a point with coordinates - D A , y 1 , z 1 .

Example 2

It is necessary to determine the coordinates of the projection of the point M 1 (- 6 , 0 , 1 2) onto the coordinate plane O x y and onto the plane 2 y - 3 = 0 .

Solution

The coordinate plane O x y will correspond to an incomplete general equation plane z = 0 . The projection of the point M 1 onto the plane z \u003d 0 will have coordinates (- 6, 0, 0) .

The plane equation 2 y - 3 = 0 can be written as y = 3 2 2 . Now just write the coordinates of the projection of the point M 1 (- 6 , 0 , 1 2) onto the plane y = 3 2 2:

6 , 3 2 2 , 1 2

Answer:(- 6 , 0 , 0) and - 6 , 3 2 2 , 1 2

If you notice a mistake in the text, please highlight it and press Ctrl+Enter

This article is the answer to two questions: "What is" and "How to find coordinates of the projection of a point on a plane"? First, the necessary information about projection and its types is given. Next, the definition of the projection of a point onto a plane is given and a graphic illustration is given. After that, a method was obtained for finding the coordinates of the projection of a point onto a plane. In conclusion, solutions of examples are analyzed in which the coordinates of the projection of a given point onto a given plane are calculated.

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Projection, types of projection - necessary information.

When studying spatial figures, it is convenient to use their images in the drawing. The drawing of a spatial figure is a so-called projection this figure to the plane. The process of constructing an image of a spatial figure on a plane occurs according to certain rules. So the process of constructing an image of a spatial figure on a plane, together with a set of rules by which this process is carried out, is called projection figures on this plane. The plane in which the image is built is called projection plane.

Depending on the rules by which the projection is carried out, there are central And parallel projection. We will not go into details, as this is beyond the scope of this article.

In geometry, a special case of parallel projection is mainly used - perpendicular projection, which is also called orthogonal. In the name of this type of projection, the adjective "perpendicular" is often omitted. That is, when in geometry they talk about the projection of a figure onto a plane, they usually mean that this projection was obtained using perpendicular projection (unless, of course, otherwise is specified).

It should be noted that the projection of a figure onto a plane is a set of projections of all points of this figure onto the projection plane. In other words, in order to get the projection of a certain figure, it is necessary to be able to find the projections of the points of this figure onto a plane. The next paragraph of the article just shows how to find the projection of a point onto a plane.

Projection of a point onto a plane - definition and illustration.

We emphasize once again that we will talk about the perpendicular projection of a point onto a plane.

Let's make constructions that will help us define the projection of a point onto a plane.

Let in three-dimensional space we are given a point M 1 and a plane. Let's draw a straight line a through the point M 1, perpendicular to the plane. If the point M 1 does not lie in the plane, then we denote the intersection point of the line a and the plane as H 1. Thus, by construction, the point H 1 is the base of the perpendicular dropped from the point M 1 to the plane.

Definition.

Projection of point M 1 onto a plane is the point M 1 itself, if , or the point H 1, if .

This definition projection of a point onto a plane is equivalent to the following definition.

Definition.

Projection of a point onto a plane- this is either the point itself, if it lies in a given plane, or the base of the perpendicular dropped from this point to a given plane.

In the drawing below, the point H 1 is the projection of the point M 1 onto the plane; point M 2 lies in the plane, therefore M 2 is the projection of the point M 2 itself onto the plane.

Finding the coordinates of the projection of a point on a plane - solving examples.

Let Oxyz be introduced in three-dimensional space, a point and plane. Let's set ourselves the task: to determine the coordinates of the projection of the point M 1 onto the plane.

The solution of the problem follows logically from the definition of the projection of a point onto a plane.

Denote the projection of the point M 1 onto the plane as H 1 . By definition, the projection of a point onto a plane, H 1 is the intersection point of a given plane and a straight line a passing through the point M 1 perpendicular to the plane. Thus, the desired coordinates of the projection of the point M 1 onto the plane are the coordinates of the point of intersection of the line a and the plane.

Hence, to find the projection coordinates of a point on the plane you need:

Let's consider examples.

Example.

Find the projection coordinates of a point to the plane .

Solution.

In the condition of the problem, we are given a general equation of the plane of the form , so it does not need to be compiled.

Let's write the canonical equations of the straight line a, which passes through the point M 1 perpendicular to the given plane. To do this, we obtain the coordinates of the directing vector of the straight line a. Since the line a is perpendicular to the given plane, the direction vector of the line a is the normal vector of the plane . That is, - directing vector of straight line a . Now we can write the canonical equations of a straight line in space that passes through the point and has a direction vector :
.

To obtain the required coordinates of the projection of a point onto a plane, it remains to determine the coordinates of the point of intersection of the line and plane . To do this, from the canonical equations of the straight line, we pass to the equations of two intersecting planes, we compose a system of equations and find its solution. We use:

So the projection of the point to the plane has coordinates.

Answer:

Example.

In a rectangular coordinate system Oxyz in three-dimensional space, points and . Determine the coordinates of the projection of the point M 1 onto the plane ABC.

Solution.

Let us first write the equation of a plane passing through three given points:

But let's look at an alternative approach.

Let's get the parametric equations of the straight line a , which passes through the point and perpendicular to the plane ABC. The normal vector of the plane has coordinates , therefore, the vector is the direction vector of the line a . Now we can write the parametric equations of a straight line in space, since we know the coordinates of a point on a straight line ( ) and the coordinates of its direction vector ( ):

It remains to determine the coordinates of the point of intersection of the line and planes. To do this, we substitute into the equation of the plane:
.

Now by parametric equations calculate the values ​​of the variables x , y and z at :
.

Thus, the projection of the point M 1 onto the plane ABC has coordinates.

Answer:

In conclusion, let's discuss finding the coordinates of the projection of some point on the coordinate planes and planes parallel to the coordinate planes.

point projections to the coordinate planes Oxy , Oxz and Oyz are the points with coordinates and correspondingly. And the projections of the point on the plane and , which are parallel to the coordinate planes Oxy , Oxz and Oyz respectively, are points with coordinates And .

Let us show how these results were obtained.

For example, let's find the projection of a point onto the plane (other cases are similar to this).

This plane is parallel to the coordinate plane Oyz and is its normal vector. The vector is the direction vector of the line perpendicular to the Oyz plane. Then the parametric equations of the straight line passing through the point M 1 perpendicular to the given plane have the form .

Find the coordinates of the point of intersection of the line and the plane. To do this, first we substitute into the equation of equality: , and the projection of the point

Bugrov Ya.S., Nikolsky S.M. Higher Mathematics. Volume One: Elements of Linear Algebra and Analytic Geometry.

Ilyin V.A., Poznyak E.G. Analytic geometry.

Goals:

• Studying the rules for constructing projections of points on the surface of an object and reading drawings.
• Develop spatial thinking, the ability to analyze the geometric shape of an object.
• To cultivate industriousness, the ability to cooperate when working in groups, interest in the subject.

DURING THE CLASSES

I STAGE. MOTIVATION OF LEARNING ACTIVITIES.

II STAGE. FORMATION OF KNOWLEDGE, SKILLS AND SKILLS.

HEALTH-SAVING PAUSE. REFLECTION (MOOD)

STAGE III. INDIVIDUAL WORK.

I STAGE. MOTIVATION OF LEARNING ACTIVITIES

1) Teacher: Check your workplace Is everything in place? Is everyone ready to go?

BREATHED DEEPLY, HOLD THE BREATH ON EXHAUST, EXHALED.

Determine your mood at the beginning of the lesson according to the scheme (such a scheme is on the table for everyone)

I WISH YOU GOOD LUCK.

2)Teacher: Practical work on this topic " Projections of Vertices, Edges, Faces” showed that there are guys who make mistakes when projecting. They get confused which of the two matching points in the drawing is the visible vertex and which is the invisible one; when the edge is parallel to the plane, and when it is perpendicular. Same thing with edges.

To avoid repeating mistakes, complete the necessary tasks using the consulting card and correct mistakes in practical work (by hand). And as you work, remember:

“EVERYBODY CAN MAKE MISTAKES, STAY AT HIS MISTAKE - ONLY THE CRAZY”.

And those who have mastered the topic well will work in groups with creative tasks (see. Annex 1 ).

II STAGE. FORMATION OF KNOWLEDGE, SKILLS AND SKILLS

1)Teacher: In production, there are many parts that are attached to each other in a certain way.
For example:
The desktop cover is attached to the vertical posts. Pay attention to the table at which you are, how and with what the lid and racks are attached to each other?

Answer: Bolt.

Teacher: What is required for a bolt?

Answer: Hole.

Teacher: Really. And in order to make a hole, you need to know its location on the product. When making a table, the carpenter cannot contact the customer every time. So, what is the need to provide a carpenter?

Answer: Drawing.

Teacher: Drawing!? What do we call a drawing?

Answer: A drawing is an image of an object by rectangular projections in a projection connection. According to the drawing, you can represent the geometric shape and design of the product.

Teacher: We have completed rectangular projections, and then? Will we be able to determine the location of the holes from one projection? What else do we need to know? What to learn?

Answer: Build points. Find projections of these points in all views.

Teacher: Well done! This is the purpose of our lesson, and the topic: Construction of projections of points on the surface of an object. Write the topic of the lesson in your notebook.
You and I know that any point or segment on the image of an object is a projection of a vertex, edge, face, i.e. each view is an image not from one side (ch. view, top view, left view), but the whole object.
In order to correctly find the projections of individual points lying on the faces, you must first find the projections of this face, and then use the connection lines to find the projections of the points.

(We look at the drawing on the board, we work in a notebook where 3 projections of the same part are made at home).

- Opened a notebook with a completed drawing (An explanation of the construction of points on the surface of an object with leading questions on the board, and students fix it in a notebook.)

Teacher: Consider a point IN. What plane is the face with this point parallel to?

Answer: The face is parallel to the frontal plane.

Teacher: We set the projection of a point b' in frontal projection. Draw down from the point b' vertical line of communication to the horizontal projection. Where will the horizontal projection of the point be? IN?

Answer: At the intersection with the horizontal projection of the face that was projected into the edge. And is at the bottom of the projection (view).

Teacher: Point profile projection b'' where will it be located? How will we find it?

Answer: At the intersection of the horizontal line of communication from b' with a vertical edge on the right. This edge is the projection of the face with a point IN.

THOSE WANTING TO CONSTRUCT THE NEXT PROJECTION OF THE POINT ARE CALLED TO THE BOARD.

Teacher: Point projections A are also located using communication lines. Which plane is parallel to the edge with a point A?

Answer: The face is parallel to the profile plane. We set a point on the profile projection A'' .

Teacher: On what projection is the face projected into the edge?

Answer: On the front and horizontal. Let's draw a horizontal connection line to the intersection with a vertical edge on the left on the frontal projection, we get a point A' .

Teacher: How to find the projection of a point A on a horizontal projection? After all, communication lines from the projection of points A' And A'' do not intersect the projection of the face (edge) on the horizontal projection on the left. What can help us?

Answer: You can use a constant straight line (it determines the position of the view on the left) from A'' draw a vertical line of communication until it intersects with a constant straight line. From the intersection point, a horizontal line of communication is drawn, until it intersects with a vertical edge on the left. (This is the face with point A) and denotes the projection with a point A .

2) Teacher: Everyone has a task card on the table, with a tracing paper attached. Consider the drawing, now try on your own, without redrawing the projections, to find the given projections of points on the drawing.

– Find in the textbook p. 76 fig. 93. Test yourself. Who performed correctly - score "5" "; one mistake - ''4''; two - ''3''.

(The grades are set by the students themselves in the self-control sheet).

- Collect cards for testing.

3)Group work: Time limited: 4min. + 2 min. checks. (Two desks with students are combined, and a leader is selected within the group).

For each group, tasks are distributed in 3 levels. Students choose tasks by levels, (as they wish). Solve problems on the construction of points. Discuss the construction under the supervision of the leader. Then the correct answer is displayed on the board with the help of a codoscope. Everyone checks that the points are projected correctly. With the help of the group leader, grades are given on assignments and in self-control sheets (see. Annex 2 And Annex 3 ).

HEALTH-SAVING PAUSE. REFLECTION

"Pharaoh's Pose"- sit on the edge of a chair, straighten your back, bend your arms at the elbows, cross your legs and put on your toes. Inhale, tighten all the muscles of the body while holding the breath, exhale. Do 2-3 times. Close your eyes tightly, to the stars, open. Mark your mood.

STAGE III. PRACTICAL PART. (Individual tasks)

There are task cards to choose from with different levels. Students choose their own option. Find projections of points on the surface of an object. Works are handed over and evaluated for the next lesson. (Cm. Appendix 4 , Annex 5 , Appendix 6 ).

STAGE IV. FINAL

1) Homework assignment. (Instruction). Performed by levels:

B - understanding, on "3". Exercise 1 fig. 94a p. 77 - according to the assignment in the textbook: complete the missing projections of points on these projections.

B - application, on "4". Exercise 1 Fig. 94 a, b. complete the missing projections and mark the vertices on the visual image in 94a and 94b.

A - analysis, on "5". (Increased difficulty.) Ex. 4 fig.97 - construct missing projections of points and designate them with letters. There is no visual image.

2)Reflective analysis.

1. Determine the mood at the end of the lesson, mark it on the self-control sheet with any sign.
2. What new did you learn at the lesson today?
3. What form of work is most effective for you: group, individual and would you like it to be repeated in the next lesson?
4. Collect checklists.

3)"Wrong Teacher"

Teacher: You have learned how to build projections of vertices, edges, faces and points on the surface of an object, following all the construction rules. But here you were given a drawing, where there are errors. Now try yourself as a teacher. Find the mistakes yourself, if you find all 8–6 mistakes, then the score is “5”, respectively; 5–4 errors - “4”, 3 errors - “3”.

Answers:

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

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

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

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

Construction of projections of points

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

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

Determining the position of points in space

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

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

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

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

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

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

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

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

To construct images of a number of details, it is necessary to be able to find the projections of individual points. For example, it is difficult to draw a top view of the part shown in Fig. 139 without building horizontal projections of points A, B, C, D, E, F, etc.

The problem of finding the projections of points by one given on the surface of the object is solved as follows. First, the projections of the surface on which the point is located are found. Then, drawing a connection line to the projection, where the surface is represented by a line, the second projection of the point is found. The third projection lies at the intersection of communication lines.

Consider an example.

Three projections of the part are given (Fig. 140, a). The horizontal projection a of the point A lying on the visible surface is given. We need to find the other projections of this point.

First of all, you need to draw an auxiliary line. If two views are given, then the place of the auxiliary line in the drawing is chosen arbitrarily, to the right of the top view, so that the view on the left is at the required distance from the main view (Fig. 141).

If three views have already been built (Fig. 142, a), then the place of the auxiliary line cannot be arbitrarily chosen; you need to find the point through which it will pass. To do this, just continue to mutual intersection horizontal and profile projections of the axis of symmetry and through the resulting point k (Fig. 142, b) draw a straight line segment at an angle of 45 °, which will be an auxiliary straight line.

If there are no axes of symmetry, then continue until the intersection at point k 1 horizontal and profile projections of any face projected in the form of straight line segments (Fig. 142, b).

Having drawn an auxiliary straight line, they begin to build the projections of the point (see Fig. 140, b).

Frontal a" and profile a" projections of point A must be located on the corresponding projections of the surface to which point A belongs. These projections are found. On fig. 140, b they are highlighted in color. Draw communication lines as indicated by the arrows. At the intersections of the communication lines with the projections of the surface, the desired projections a" and a" are found.

The construction of projections of points B, C, D is shown in fig. 140, in lines of communication with arrows. The given projections of points are colored. Communication lines are drawn to the projection on which the surface is depicted as a line, and not as a figure. Therefore, the frontal projection from the point C is first found. The profile projection from the point C is determined by the intersection of the communication lines.

If the surface is not depicted by a line on any projection, then an auxiliary plane must be used to construct the projections of points. For example, a frontal projection d of point A is given, lying on the surface of a cone (Fig. 143, a). An auxiliary plane is drawn through a point parallel to the base, which will intersect the cone in a circle; its frontal projection is a straight line segment, and its horizontal projection is a circle with a diameter equal to the length of this segment (Fig. 143, b). By drawing a communication line to this circle from point a, a horizontal projection of point A is obtained.

The profile projection a" of point A is found in the usual way at the intersection of communication lines.

In the same way, one can find the projections of a point lying, for example, on the surface of a pyramid or a ball. When a pyramid is intersected by a plane parallel to the base and passing through a given point, a figure similar to the base is formed. The projections of the given point lie on the projections of this figure.

Answer the questions

1. At what angle is the auxiliary line drawn?

2. Where is the auxiliary line drawn if front and top views are given, but you need to build a view from the left?

3. How to determine the place of the auxiliary line in the presence of three types?

4. What is the method of constructing projections of a point according to one given one, if one of the surfaces of the object is represented by a line?

5. For what geometric bodies and in what cases are the projections of a point given on their surface found using an auxiliary plane?

Assignments to § 20

Exercise 68

Write to workbook, which projections of the points indicated by numbers in the views correspond to the points indicated by letters in the visual image in the example indicated to you by the teacher (Fig. 144, a-d).

Exercise 69

On fig. 145, a-b letters indicated by only one projection of some of the vertices. Find in the example given to you by the teacher, the remaining projections of these vertices and designate them with letters. Construct in one of the examples the missing projections of points given on the edges of the object (Fig. 145, d and e). Highlight with color the projections of the edges on which the points are located. Complete the task on transparent paper, overlaying it on the page of the textbook. There is no need to redraw Fig. 145.

Exercise 70

Find the missing projections of points given by one projection on the visible surfaces of the object (Fig. 146). Label them with letters. Highlight the given projections of points with color. A visual image will help you solve the problem. The task can be completed both in a workbook and on transparent paper, overlaying it on the page of the textbook. In the latter case, redraw Fig. 146 is not necessary.

Exercise 71

In the example given to you by the teacher, draw three types (Fig. 147). Construct the missing projections of the points given on the visible surfaces of the object. Highlight the given projections of points with color. Label all point projections. To build projections of points, use an auxiliary straight line. Make a technical drawing and mark the given points on it.