Frontal projection of a point. An example of constructing the third projection of a point according to two given

The projection of a point on three planes of projections of the coordinate angle begins with obtaining its image on the plane H - the horizontal plane of projections. To do this, through point A (Fig. 4.12, a) a projecting beam is drawn perpendicular to the plane H.

In the figure, the perpendicular to the H plane is parallel to the Oz axis. The point of intersection of the beam with the plane H (point a) is chosen arbitrarily. The segment Aa determines how far point A is from the plane H, thus indicating unambiguously the position of point A in the figure with respect to the projection planes. Point a is a rectangular projection of point A onto the plane H and is called the horizontal projection of point A (Fig. 4.12, a).

To obtain an image of point A on the plane V (Fig. 4.12, b), a projecting beam is drawn through point A perpendicular to the frontal projection plane V. In the figure, the perpendicular to the plane V is parallel to the Oy axis. On the H plane, the distance from point A to plane V will be represented by a segment aa x, parallel to the Oy axis and perpendicular to the Ox axis. If we imagine that the projecting beam and its image are carried out simultaneously in the direction of the plane V, then when the image of the beam intersects the Ox axis at the point a x, the beam intersects the plane V at the point a. Drawing from the point a x in the V plane perpendicular to the Ox axis , which is the image of the projecting beam Aa on the plane V, the point a is obtained at the intersection with the projecting beam. Point a "is the frontal projection of point A, i.e. its image on the plane V.

The image of point A on the profile plane of projections (Fig. 4.12, c) is built using a projecting beam perpendicular to the W plane. In the figure, the perpendicular to the W plane is parallel to the Ox axis. The projecting beam from point A to plane W on the plane H will be represented by a segment aa y, parallel to the Ox axis and perpendicular to the Oy axis. From the point Oy parallel to the Oz axis and perpendicular to the Oy axis, an image of the projecting beam aA is built and, at the intersection with the projecting beam, the point a is obtained. Point a is the profile projection of the point A, i.e., the image of the point A on the plane W.

The point a "can be constructed by drawing from the point a" the segment a "a z (the image of the projecting beam Aa" on the plane V) parallel to the Ox axis, and from the point a z - the segment a "a z parallel to the Oy axis until it intersects with the projecting beam.

Having received three projections of point A on the projection planes, the coordinate angle is deployed into one plane, as shown in Fig. 4.11, b, together with the projections of the point A and the projecting rays, and the point A and the projecting rays Aa, Aa "and Aa" are removed. The edges of the combined projection planes are not carried out, but only the projection axes Oz, Oy and Ox, Oy 1 (Fig. 4.13) are carried out.

An analysis of the orthogonal drawing of a point shows that three distances - Aa", Aa and Aa" (Fig. 4.12, c), characterizing the position of point A in space, can be determined by discarding the projection object itself - point A, on a coordinate angle deployed in one plane (Fig. 4.13). The segments a "a z, aa y and Oa x are equal to Aa" as opposite sides of the corresponding rectangles (Fig. 4.12, c and 4.13). They determine the distance at which point A is located from the profile plane of projections. Segments a "a x, a" a y1 and Oa y are equal to segment Aa, determine the distance from point A to the horizontal projection plane, segments aa x, a "a z and Oa y 1 are equal to segment Aa", which determines the distance from point A to frontal projection plane.

The segments Oa x, Oa y and Oa z located on the projection axes are a graphic expression of the sizes of the X, Y and Z coordinates of point A. The point coordinates are denoted with the index of the corresponding letter. By measuring the size of these segments, you can determine the position of the point in space, i.e., set the coordinates of the point.

On the diagram, the segments a "a x and aa x are arranged as one line perpendicular to the Ox axis, and the segments a" a z and a "az - to the Oz axis. These lines are called projection connection lines. They intersect the projection axes at points a x and and z, respectively.The line of the projection connection connecting the horizontal projection of point A with the profile one turned out to be “cut” at the point a y.

Two projections of the same point are always located on the same projection connection line perpendicular to the projection axis.

To represent the position of a point in space, two of its projections and a given origin (point O) are sufficient. 4.14, b, two projections of a point completely determine its position in space. Using these two projections, you can build a profile projection of point A. Therefore, in the future, if there is no need for a profile projection, diagrams will be built on two projection planes: V and H.

Rice. 4.14. Rice. 4.15.

Let's consider several examples of building and reading a drawing of a point.

Example 1 Determination of the coordinates of the point J given on the diagram by two projections (Fig. 4.14). Three segments are measured: segment Ov X (X coordinate), segment b X b (Y coordinate) and segment b X b "(Z coordinate). Coordinates are written in the following order: X, Y and Z, after the letter designation of the point, for example , B20; 30; 15.

Example 2. Construction of a point according to the given coordinates. Point C is given by coordinates C30; 10; 40. On the Ox axis (Fig. 4.15) find a point with x, at which the line of the projection connection intersects the projection axis. To do this, the X coordinate (size 30) is plotted along the Ox axis from the origin (point O) and a point with x is obtained. Through this point, perpendicular to the Ox axis, a projection connection line is drawn and the Y coordinate is laid down from the point (size 10), the point c is obtained - the horizontal projection of the point C. The coordinate Z (size 40) is plotted upwards from the point c x along the projection connection line (size 40), the point is obtained c" - frontal projection of point C.

Example 3. Construction of a profile projection of a point according to the given projections. The projections of the point D - d and d are set. Through the point O, the projection axes Oz, Oy and Oy 1 are drawn (Fig. 4.16, a). it to the right behind the Oz axis. The profile projection of the point D will be located on this line. It will be located at the same distance from the Oz axis as the horizontal projection of the point d is located: from the Ox axis, i.e. at a distance dd x. The segments d z d "and dd x are the same, since they determine the same distance - the distance from point D to the frontal projection plane. This distance is the Y coordinate of point D.

Graphically, the segment dzd "is built by transferring the segment dd x from the horizontal projection plane to the profile one. To do this, draw a projection connection line parallel to the Ox axis, get a point dy on the Oy axis (Fig. 4.16, b). Then transfer the size of the segment Od y to the Oy 1 axis , drawing from point O an arc with a radius equal to the segment Od y, until it intersects with the axis Oy 1 (Fig. 4.16, b), get the point dy 1. This point can be constructed and, as shown in Fig. 4.16, c, drawing a straight line at an angle 45 ° to the Oy axis from the point dy... From the point d y1, draw a projection connection line parallel to the Oz axis and lay a segment on it equal to the segment d "dx, get the point d".

Transferring the value of the segment d x d to profile plane projections can be carried out using a constant straight line drawing (Fig. 4.16, d). In this case, the projection connection line dd y is drawn through the horizontal projection of the point parallel to the Oy 1 axis until it intersects with a constant straight line, and then parallel to the Oy axis until it intersects with the continuation of the projection connection line d "d z.

Particular cases of the location of points relative to projection planes

The position of a point relative to the projection plane is determined by the corresponding coordinate, i.e., the value of the segment of the projection connection line from the Ox axis to the corresponding projection. On fig. 4.17 the Y coordinate of point A is determined by the segment aa x - the distance from point A to plane V. The Z coordinate of point A is determined by the segment a "a x - the distance from point A to plane H. If one of the coordinates is zero, then the point is located on the projection plane Fig. 4.17 shows examples of different locations of points relative to the projection planes.The Z coordinate of point B is zero, the point is in the H plane. front projection is on the x-axis and coincides with point b x. The Y coordinate of point C is zero, the point is located on the V plane, its horizontal projection c is on the Ox axis and coincides with the point with x.

Therefore, if a point is on the projection plane, then one of the projections of this point lies on the projection axis.

On fig. 4.17, the Z and Y coordinates of point D are zero, therefore, point D is on the projection axis Ox and its two projections coincide.

In some cases, for the convenience of solving problems, it is necessary to use additional projection planes perpendicular to the existing projection planes.

If the horizontal and frontal projections of a point are given, then the profile projection is determined by the following algorithm.

We draw a line of projection connection perpendicular to the axis Oz.

On this line of the projection connection, we postpone the segment A 1 A X =A Z A 3 .

Using this rule, it is possible to build projections of points onto additional projection planes (method of plane replacements).

Let a point be given A(A 2 ,A 1 ) and a new additional projection plane P 4 P 1 . Build A 4 – point projection A on the P 4 .

Solution

a) We build a line of intersection of planes P 1 and P 4 = x 1,4 ;

b) Through a point A we draw a line of projection communication x 1,4 .

c) Building a projection A 4 , I use line segments A 2 A X =A 4 A X .

Two point projections A 1 and A 4 lie on the same line of projection connection perpendicular to the axis X 1,4 .

Distance from the “new” point projection A 4 to the “new” axis x 1,4 is equal to the distance from the “old” point projection A 2 to the “old” axis x 1,2 .

Competing points

competing points call a pair of points lying on the same projecting ray.

Of the two competing points, the visible point is the point that is further from the projection plane.

points A and V called horizontally competitive.

points WITH and D are called frontal competitors.

Enter an additional plane so that the points A and V became competitive.

Solution plan:

1 Building an axis x 1,4 A 1 , B 1 ;

2 We build a line of projection connection x 1,4 ;

3 On the line of the projection connection, lay off the segments A x A 2 = A / x A 4 , B x B 2 = B / x B 4 .

Self-study material Modeling 2d graphics objects in the compass graphics system Starting the compass system and shutting down

The KOMPAS-3D-V8 system is launched similarly to other programs. To start the system, select the menu \ Start\ All Pprograms\ ASCON \KOMPAS-3D- V8 and run COMPASS. You can select the program shortcut with the mouse pointer on the desktop field and double-click the left mouse button. To open a document, click the button Open on the panel Standard . To start a new document, click the button Create on the panel Standard or run the command File > Create and in the dialog box that opens, select the type of document to be created and click OK.

Select menu to finish. File\Exit, the Alt-F4 key combination, or click the Close button.

The main types of documents of the compass graphic system

The type of document created in the KOMPAS system depends on the type of information stored in this document. Each document type has a file name extension and its own icon.

1 Drawing- the main type of graphic document in KOMPAS. The drawing contains a graphic image of the product in one or more views, a title block, a frame. A KOMPAS drawing always contains one sheet of a user-defined format. The drawing file has the extension .cdw.

2 Fragment- auxiliary type of graphic document in KOMPAS. A fragment differs from a drawing by the absence of a frame, title block, and other design objects of the design document. Fragments store created standard solutions for later use in other documents. The snippet file has the extension .frw.

3 Text Document(file extension . kdw);

4 Specification(file extension . spw);

5 Assembly(file extension . a3 d);

6 Detail- 3D modeling (file extension . m3 d);

Word form	Graphic form
1. Set aside on the X, Y, Ζ axes the corresponding coordinates of point A. We get points A x , A y , A z
2. Horizontal projection A 1 is located at the intersection of communication lines from points A x and A y drawn parallel to the X and Y axes
3. Frontal projection A 2 is located at the intersection of communication lines from points A x and A z, drawn parallel to the axes X and z
4. Profile projection A 3 is located at the intersection of communication lines from points A z and A y drawn parallel to the axes Ζ and Y

3.2. Point position relative to projection planes

The position of a point in space relative to the projection planes is determined by its coordinates. The X coordinate determines the distance of the point from the P 3 plane (projection to P 2 or P 1), the Y coordinate - the distance from the P 2 plane (projection to P 3 or P 1), the Z coordinate - the distance from the P 1 plane (projection to P 3 or P 2). Depending on the value of these coordinates, a point can occupy both a general and a particular position in space with respect to the projection planes (Fig. 3.1).

Rice. 3.1. Point classification

Tpointsgeneralprovisions. Point coordinates general position not equal to zero ( x≠0, y≠0, z≠0 ), and depending on the sign of the coordinate, the point can be located in one of eight octants (Table 2.1).

On fig. 3.2 drawings of points in general position are given. An analysis of their images allows us to conclude that they are located in the following octants of space: A(+X;+Y; +Z(  Ioctant;B(+X;+Y;-Z(  IVoctant;C(-X;+Y; +Z(  Voctant;D(+X;+Y; +Z(  IIoctant.

Private position points. One of the coordinates of a particular position point is equal to zero, so the projection of the point lies on the corresponding projection field, the other two lie on the projection axes. On fig. 3.3 such points are points A, B, C, D, G.A  P 3, then the point X A \u003d 0; V  P 3, then the point X B \u003d 0; WITH  P 2, then point Y C \u003d 0; D  P 1, then point Z D \u003d 0.

A point can belong to two projection planes at once, if it lies on the line of intersection of these planes - the projection axis. For such points, only the coordinate on this axis is not equal to zero. On fig. 3.3, such a point is the point G(G  OZ, then point X G =0, Y G =0).

3.3. Mutual position of points in space

Let us consider three options for the mutual arrangement of points depending on the ratio of the coordinates that determine their position in space.

On fig. 3.4 points A and B have different coordinates.

Their relative position can be estimated by the distance to the projection planes: Y A >Y B, then point A is located farther from the plane P 2 and closer to the observer than point B; Z A >Z B, then point A is located farther from the plane P 1 and closer to the observer than point B; X A

On fig. 3.5 shows points A, B, C, D, in which one of the coordinates is the same, and the other two are different.

Their relative position can be estimated by their distance to the projection planes as follows:

Y A \u003d Y B \u003d Y D, then points A, B and D are equidistant from the plane P 2, and their horizontal and profile projections are located respectively on the lines [A 1 B 1 ]llOX and [A 3 B 3 ]llOZ. The locus of such points is a plane parallel to П 2 ;

Z A \u003d Z B \u003d Z C, then points A, B and C are equidistant from the plane P 1, and their frontal and profile projections are located respectively on the lines [A 2 B 2 ]llOX and [A 3 C 3 ]llOY. The locus of such points is a plane parallel to П 1 ;

X A \u003d X C \u003d X D, then points A, C and D are equidistant from the plane P 3 and their horizontal and frontal projections are located respectively on the lines [A 1 C 1 ]llOY and [A 2 D 2 ]llOZ . The locus of such points is a plane parallel to П 3 .

3. If the points have two coordinates of the same name, then they are called competing. Competing points are located on the same projecting line. On fig. 3.3 three pairs of such points are given, in which: X A \u003d X D; Y A = Y D ; Z D > Z A; X A = X C ; Z A = Z C ; Y C > Y A ; Y A = Y B ; Z A = Z B ; X B > X A .

There are horizontally competing points A and D located on the horizontally projecting line AD, frontally competing points A and C located on the frontally projecting line AC, profile competing points A and B located on the profile projecting line AB.

Conclusions on the topic

1. A point is a linear geometric image, one of the basic concepts of descriptive geometry. The position of a point in space can be determined by its coordinates. Each of the three projections of a point is characterized by two coordinates, their name corresponds to the names of the axes that form the corresponding projection plane: horizontal - A 1 (XA; YA); frontal - A 2 (XA; ZA); profile - A 3 (YA; ZA). Translation of coordinates between projections is carried out using communication lines. From two projections, you can build projections of a point either using coordinates or graphically.

3. A point in relation to the projection planes can occupy both a general and a particular position in space.

4. A point in general position is a point that does not belong to any of the projection planes, i.e., lies in the space between the projection planes. The coordinates of a point in general position are not equal to zero (x≠0,y≠0,z≠0).

5. A point of private position is a point belonging to one or two projection planes. One of the coordinates of a point of particular position is equal to zero, so the projection of the point lies on the corresponding field of the projection plane, the other two - on the axes of the projections.

6. Competing points are points whose coordinates of the same name are the same. There are horizontally competing points, frontally competing points, and profile competing points.

Keywords

Point coordinates

General point

Private position point

Competing points

Methods of activity necessary for solving problems

– construction of a point according to the given coordinates in the system of three projection planes in space;

– construction of a point according to the given coordinates in the system of three projection planes on the complex drawing.

Questions for self-examination

1. How is the connection of the location of coordinates on the complex drawing in the system of three projection planes P 1 P 2 P 3 with the coordinates of the projections of points established?

2. What coordinates determine the distance of points to the horizontal, frontal, profile projection planes?

3. What coordinates and projections of the point will change if the point moves in the direction perpendicular to the profile plane of the projections П 3 ?

4. What coordinates and projections of a point will change if the point moves in a direction parallel to the OZ axis?

5. What coordinates determine the horizontal (frontal, profile) projection of a point?

7. In what case does the projection of a point coincide with the point in space itself, and where are the other two projections of this point located?

8. Can a point belong to three projection planes at the same time, and in what case?

9. What are the names of the points whose projections of the same name coincide?

10. How can you determine which of the two points is closer to the observer if their frontal projections coincide?

Tasks for independent solution

1. Give a visual image of points A, B, C, D relative to the projection planes P 1, P 2. The points are given by their projections (Fig. 3.6).

2. Construct projections of points A and B according to their coordinates on a visual image and a complex drawing: A (13.5; 20), B (6.5; -20). Construct a projection of point C, located symmetrically to point A relative to the frontal plane of projections П 2 .

3. Build projections of points A, B, C according to their coordinates on a visual image and a complex drawing: A (-20; 0; 0), B (-30; -20; 10), C (-10, -15, 0 ). Construct point D, located symmetrically to point C with respect to the OX axis.

An example of solving a typical problem

Task 1. Given the coordinates X, Y, Z of points A, B, C, D, E, F (Table 3.3)

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.

Yandex.RTB R-A-339285-1

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 the normal vector of the plane 2 x - 3 y + z - 2 = 0. In this way, 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

In a rectangular coordinate system O x y z three-dimensional space given points A (0, 0, 2); 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 ⇔ xyz - 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 the equalities: x = x 1 + λ, y = y 1, z = z 1 and get: A (x 1 + λ) + D = 0 ⇒ λ = - DA - x one

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

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, it is enough to continue until the mutual intersection of the 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.