Horizontal projection of points a and b. A short course in descriptive geometry. Method of rotation around an axis perpendicular to the projection plane

POINT PROJECTIONS.

ORTHOGONAL SYSTEM OF TWO PLANES OF PROJECTIONS.

The essence of the orthogonal projection method lies in the fact that the object is projected onto two mutually perpendicular planes by rays orthogonal (perpendicular) to these planes.

One of the projection planes H is placed horizontally, and the other V is placed vertically. Plane H is called the horizontal plane of projections, V - frontal. The planes H and V are infinite and opaque. The line of intersection of the projection planes is called the coordinate axis and is denoted OX. Projection planes divide space into four dihedral angles - quarters.

Considering orthogonal projections, assume that the observer is in the first quadrant at an infinite distance from the projection planes. Since these planes are opaque, only those points, lines and figures that are located within the same first quarter will be visible to the observer.

When constructing projections, it is necessary to remember that point orthogonal projectionon a plane is called the base of the perpendicular dropped from a given pointto this plane.

The figure shows the dot BUT and its orthogonal projections a 1 and a 2 .

Point a 1 called plan view points BUT, point a 2- her front projection. Each of them is the base of the perpendicular dropped from the point BUT respectively on the plane H and V.

It can be proved that point projectionalways located on straight lines, perpendicularcular axisOH and crossing this axisat the same point. Indeed, projecting rays BUTa 1 and BUTa 2 define a plane perpendicular to the planes of projections and the lines of their intersection - axes OH. This plane intersects H and V in straight lines a 1 ax and a 1 ax, which form with the axis OX and with each other right angles with vertex at a point ax.

The opposite is also true, i.e. if points are given on the projection planesa 1 and a 2 , located on straight lines intersecting axis OXat this point at a right angle,then they are projections of somepoints A. This point is determined by the intersection of the perpendiculars constructed from the points a 1 and a 2 to planes H and V.

Note that the position of the projection planes in space may be different. For example, both planes, being mutually perpendicular, can be vertical. But in this case, the above assumption about the orientation of opposite projections of points relative to the axis remains valid.

To get a flat drawing consisting of the above projections, the plane H aligned by rotation around an axis OX with plane V as shown by the arrows in the figure. As a result, the front half-plane H will be aligned with the lower half-plane V, and the rear half-plane H- with upper half-plane V.

A projection drawing, in which the projection planes with everything that is depicted on them, are combined in a certain way with one another, is called diagram(from French epure - drawing). The figure shows a diagram of a point BUT.

With this method of combining planes H and V projections a 1 and a 2 will be located on the same perpendicular to the axis OX. At the same time, the distance a 1 a x — from the horizontal projection of the point to the axis OX BUT up to the plane V, and the distance a 2 a x — from the frontal projection of the point to the axis OX equal to the distance from the point BUT up to the plane H.

Straight lines connecting opposite projections of a point on the diagram, we agree to call projection communication lines.

The position of the projections of points on the diagram depends on the quarter in which the given point is located. So if the point AT is located in the second quarter, then after the alignment of the planes, both projections will lie above the axis OX.

If point With is in the third quarter, then its horizontal projection after the alignment of the planes will be above the axis, and the frontal projection will be below the axis OX. Finally, if the point D located in the fourth quarter, then both its projections will be under the axis OX. The figure shows the points M and N lying on the projection planes. In this position, the point coincides with one of its projections, while its other projection turns out to be lying on the axis OX. This feature is also reflected in the designation: near the projection with which the point itself coincides, a capital letter without an index is written.

It should also be noted that the case when both projections of the point coincide. This will happen if the point is in the second or fourth quarter at the same distance from the projection planes. Both projections are combined with the point itself, if the latter is located on the axis OX.

ORTHOGONAL SYSTEM OF THREE PLANES OF PROJECTIONS.

It was shown above that two projections of a point determine its position in space. Since each figure or body is a collection of points, it can be argued that two orthogonal projections of an object (in the presence of letter designations) completely determine its shape.

However, in the practice of depicting building structures, machines and various engineering structures, it becomes necessary to create additional projections. They do this for the sole purpose of making the projection drawing clearer, more readable.

The model of three projection planes is shown in the figure. The third plane, perpendicular and H and V, denoted by the letter W and called profile.

The projections of points on this plane will also be called profile, and they are denoted by capital letters or numbers with index 3 (ah,bh,ch, ...1h, 2h, 3 3 ...).

Projection planes, intersecting in pairs, define three axes: OX, OY and OZ, which can be considered as a system of rectangular Cartesian coordinates in space with origin at point O. The sign system shown in the figure corresponds to the "right system" of coordinates.

Three projection planes divide space into eight trihedral angles - these are the so-called octants. The numbering of octants is given in the figure.

To get a plot of a plane H and W rotate as shown in the figure until aligned with the plane V. As a result of rotation, the front half-plane H turns out to be aligned with the lower half-plane V, and the rear half-plane H- with upper half-plane V. When rotated 90° around the axis OZ front half-plane W coincides with the right half-plane V, and the rear half-plane W- with the left half-plane V.

The final view of all combined projection planes is given in the figure. In this drawing, the axes OX and OZ, lying in a fixed plane V, are shown only once, and the axis OY shown twice. This is explained by the fact that, rotating with the plane H, axis OY on the diagram is aligned with the axis OZ, while rotating with the plane W, the same axis is aligned with the axis OX.

In the future, when designating the axes on the diagram, the negative semiaxes (- OX, — OY, — OZ) will not be indicated.

THREE COORDINATES AND THREE PROJECTIONS OF A POINT AND ITS RADIUS-VECTOR.

Coordinates are numbers thatput in correspondence with a point to determineniya of its position in space or onsurfaces.

In three-dimensional space, the position of a point is set using rectangular Cartesian coordinates x, y and z.

Coordinate X called abscissa, at— ordinate and z — applique. Abscissa X defines the distance from a given point to a plane W, ordinate y - up to the plane V and applique z - up to the plane H. Having adopted the system shown in the figure for counting the coordinates of a point, we will compile a table of signs of coordinates in all eight octants. Any point in space BUT, given by coordinates, will be denoted as follows: A(x, y,z).

If x = 5, y = 4 and z = 6, then the entry will take the following form BUT(5, 4, 6). This point BUT, all coordinates of which are positive, is in the first octant

Point coordinates BUT are, at the same time, the coordinates of its radius-vector

OA with respect to the origin of coordinates. If a i, j, k are unit vectors directed respectively along the coordinate axes x, y,z(picture), then

OA =OA x i+OAyj + OAzk , where OA X, OA U, OA g - vector coordinates OA

It is recommended to build an image of the point itself and its projections on a spatial model (figure) using a coordinate rectangular parallelepiped. First of all, on the coordinate axes from the point O put off segments, respectively equal 5, 4 and 6 units of length. On these segments (Oa x , Oa y , Oa z ), as on the edges, build a rectangular parallelepiped. Its vertex, opposite the origin, will determine the given point BUT. It is easy to see that in order to determine the point BUT it is enough to construct only three edges of the parallelepiped, for example Oa x , a x a 1 and a 1 BUT or Oa y , a y a 1 and a 1 A and so on. These edges form a coordinate polyline, the length of each link of which is determined by the corresponding coordinate of the point.

However, the construction of a parallelepiped allows us to determine not only the point BUT, but also all three of its orthogonal projections.

Rays projecting a point on a plane H, V, W are the three edges of the parallelepiped that intersect at the point BUT.

Each of the orthogonal projections of the point BUT, being located on a plane, is determined by only two coordinates.

Yes, the horizontal projection a 1 determined by coordinates X and y, front projection a 2 - coordinates x andz, profile projection a 3 — coordinates at and z. But any two projections are determined by three coordinates. That is why specifying a point with two projections is equivalent to specifying a point with three coordinates.

On the diagram (figure), where all the projection planes are combined, the projections a 1 and a 2 will be on the same perpendicular to the axis OX, and projections a 2 and a 3 — one perpendicular to the axis oz.

As for projections a 1 and a 3 , then they are connected by straight lines a 1 a y and a 3 a y , perpendicular to the axis OY. But since this axis occupies two positions on the diagram, the segment a 1 a y cannot be a continuation of a segment a 3 a y .

Construction of point projections A (5, 4, 6) on the diagram at the given coordinates, they are performed in the following sequence: first of all, on the abscissa axis from the origin, a segment is laid Oa x = x(in our case x =5), then through the dot a x draw perpendicular to the axis OX, on which, taking into account the signs, we postpone the segments a x a 1 = y(we get a 1 ) and a x a 2 = z(we get a 2 ). It remains to construct the profile projection of the point a 3 . Since the profile and frontal projections of the point must be located on the same perpendicular to the axis oz , then through a 3 direct a 2 a z ^ oz.

Finally, the last question arises: at what distance from the axis OZ should be a 3 ?

Considering the coordinate box (see figure), the edges of which a z a 3 =O a y = a x a 1 = y we conclude that the desired distance a z a 3 equals y. Line segment a z a 3 set aside to the right of the OZ axis if y>0, and to the left if y

Let's see what changes will occur on the diagram when the point starts to change its position in space.

Let, for example, a point A (5, 4, 6) will move in a straight line perpendicular to the plane V. With such a movement, only one coordinate will change y, showing the distance from a point to a plane V. The coordinates will remain constant. x andz , and the projection of the point defined by these coordinates, i.e. a 2 will not change his position.

As for projections a 1 and a 3 , then the first will begin to approach the axis OX, the second - to the axis OZ. In the figures, the new position of the point corresponds to the designations a 1 (a 1 1 a 2 1 a 3 1 ). When the point is on the plane V(y = 0), two of the three projections ( a 1 2 and a 3 2 ) will lie on the axes.

Having moved from I octant in II, the point will start moving away from the plane V, coordinate at becomes negative, its absolute value will increase. The horizontal projection of this point, being located on the back half-plane H, on the plot will be above the axis OX, and the profile projection, being on the back half-plane W, on the diagram will be to the left of the axis OZ. As always, cut a za 3 3 = y.

In subsequent diagrams, we will not denote by letters the points of intersection of the coordinate axes with the lines of the projection connection. This will simplify the drawing to some extent.

In the future, there will be diagrams without coordinate axes. This is done in practice when depicting objects, when only the image itself is essentialobject, not its position relative toprojection planes.

The projection planes in this case are determined with an accuracy only up to parallel translation (figure). They are usually moved parallel to themselves in such a way that all points of the object are above the plane. H and in front of the plane V. Since the position of the X 12 axis turns out to be indefinite, the formation of the diagram in this case does not need to be associated with the rotation of the planes around the coordinate axis. When switching to a plane plot H and V are combined so that opposite projections of points are located on vertical lines.

Axisless plot of points A and B(picture) notdetermines their position in space,but allows us to judge their relative orientation. So, the segment △x characterizes the displacement of the point BUT in relation to the point AT in the direction, parallel to planes H and V. In other words, △x indicates how much the point BUT located to the left of the point AT. The relative offset of the point in the direction perpendicular to the V plane is determined by the segment △y, i.e. the point And in in our example, closer to the observer than the point AT, a distance equal to △y.

Finally, the segment △z shows the excess of the point BUT over the dot AT.

Supporters of axisless study of the course descriptive geometry rightly point out that in solving many problems it is possible to do without coordinate axes. However, a complete rejection of them cannot be considered expedient. Descriptive geometry is designed to prepare the future engineer not only for the competent execution of drawings, but also for solving various technical problems, among which the problems of spatial statics and mechanics occupy not the last place. And for this it is necessary to cultivate the ability to orient this or that object relative to the Cartesian coordinate axes. These skills will also be necessary when studying such sections of descriptive geometry as perspective and axonometry. Therefore, on a number of diagrams in this book, we save images of the coordinate axes. Such drawings determine not only the shape of the object, but also its location relative to the projection planes.

A point, as a mathematical concept, has no dimensions. Obviously, if the object of projection is a zero-dimensional object, then it is meaningless to talk about its projection.

Fig.9 Fig.10

In geometry under a point, it is advisable to take a physical object that has linear dimensions. Conventionally, a ball with an infinitely small radius can be taken as a point. With this interpretation of the concept of a point, we can talk about its projections.

When constructing orthogonal projections of a point, one should be guided by the first invariant property of orthogonal projection: the orthogonal projection of a point is a point.

The position of a point in space is determined by three coordinates: X, Y, Z, showing the distances at which the point is removed from the projection planes. To determine these distances, it is enough to determine the meeting points of these lines with the projection planes and measure the corresponding values, which will indicate the values ​​of the abscissa, respectively. X, ordinates Y and appliques Z points (Fig. 10).

The projection of a point is the base of the perpendicular dropped from the point to the corresponding projection plane. Horizontal projection points a call the rectangular projection of a point on the horizontal plane of projections, frontal projection a /- respectively on the frontal plane of projections and profile a // – on the profile projection plane.

Direct Aa, Aa / and Aa // are called projecting lines. At the same time, direct Ah, projecting point BUT on the horizontal plane of projections, called horizontally projecting line, Аa / and Aa //- respectively: frontally and profile-projecting straight lines.

Two projecting lines passing through a point BUT define the plane, which is called projecting.

When converting the spatial layout, the frontal projection of the point A - a / remains in place as belonging to a plane that does not change its position under the considered transformation. Horizontal projection - a together with the horizontal projection plane will turn in the direction of clockwise movement and will be located on one perpendicular to the axis X with front projection. Profile projection - a // will rotate together with the profile plane and by the end of the transformation will take the position indicated in Figure 10. At the same time - a // will be perpendicular to the axis Z drawn from the point a / and will be removed from the axis Z the same distance as the horizontal projection a away from axis X. Therefore, the connection between the horizontal and profile projections of a point can be established using two orthogonal segments aa y and a y a // and a conjugating arc of a circle centered at the point of intersection of the axes ( O- origin). The marked connection is used to find the missing projection (for two given ones). The position of the profile (horizontal) projection according to the given horizontal (profile) and frontal projections can be found using a straight line drawn at an angle of 45 0 from the origin to the axis Y(this bisector is called a straight line) k is the Monge constant). The first of these methods is preferable, as it is more accurate.

Therefore:

1. Point in space removed:

from the horizontal plane H Z,

from the frontal plane V by the value of the given coordinate Y,

from profile plane W by the value of the coordinate. x.

2. Two projections of any point belong to the same perpendicular (one connection line):

horizontal and frontal - perpendicular to the axis x,

horizontal and profile - perpendicular to the Y axis,

frontal and profile - perpendicular to the Z axis.

3. The position of a point in space is completely determined by the position of its two orthogonal projections. Therefore - from any two given orthogonal projections of a point, it is always possible to construct its missing third projection.

If a point has three definite coordinates, then such a point is called point in general position. If a point has one or two coordinates equal to zero, then such a point is called private position point.

Rice. 11 Fig. 12

Figure 11 shows a spatial drawing of points of particular position, Figure 12 shows a complex drawing (diagrams) of these points. Dot BUT belongs to the frontal projection plane, the point AT– horizontal plane of projections, point With– profile plane of projections and point D– abscissa axis ( X).

projection(lat. Projicio - I throw forward) - the process of obtaining an image of an object (spatial object) on any surface using light or visual rays (rays that conditionally connect the observer's eye with any point of a spatial object), which are called projecting.

There are two projection methods: central and parallel .

Centralprojection is to pass through each point ( A, B, C,…) of the depicted object and in a certain way selected projection center (S) straight line ( SA, SB, >… — projecting beam).

Figure 1.1 - Central projection

Let's introduce the following notation (Figure 1.1):

S– projection center (observer's eye);

π 1 - projection plane;

A, B, C

SA, SB- projecting straight lines (projecting rays).

Note: left mouse button can move the point in the horizontal plane, when you click on the point with the left mouse button, the direction of movement will change and you can move it vertically.

Central projection point the point of intersection of the projecting line passing through the projection center and the projection object (point) with the projection plane is called.

Property 1 . Each point in space corresponds to a single projection, but each point in the projection plane corresponds to a set of points in space that lie on the projecting line.

Let's prove this statement.

Figure 1.1: dot BUT 1 is the central projection of point A on the plane of projections π 1 . But all points lying on the projecting line can have the same projection. Take on the projecting line SA point With. Central projection point With(With 1) on the plane of projections π 1 coincides with the projection of the point BUT(BUT 1):

1. With∈ SA;
2. SC∩ π 1 = C 1 →C 1 ≡ A 1 .

The conclusion follows that by the projection of a point it is impossible to judge unambiguously about its position in space.

To eliminate this uncertainty, i.e. make a drawing reversible, we introduce one more projection plane (π 2) and one more projection center ( S 2) (Figure 1.2).

Figure 1.2 - Illustration of the 1st and 2nd properties

Let's construct projections of a point BUT on the plane of projections π 2 . Of all points in space, only a point BUT has its projections BUT 1 to the plane π 1 and BUT 2 to π 2 at the same time. All other points lying on the projecting rays will have at least one different projection from the projections of the point BUT(e.g. dot AT).

Property 2 . The projection of a straight line is a straight line.

Let's prove this property.

Connect the dots BUT and AT among themselves (Figure 1.2). We get a segment AB defining a straight line. triangle SAB defines a plane, denoted by σ. It is known that two planes intersect in a straight line: σ∩π 1 = BUT 1 AT 1 , where BUT 1 AT 1 - central projection of a straight line given by a segment AB.

The central projection method is a model of image perception by the eye, it is mainly used when making perspective images of building objects, interiors, as well as in film technology and optics. The method of central projection does not solve the main task facing the engineer - to accurately reflect the shape, dimensions of the object, the ratio of the sizes of various elements.

1.2. Parallel projection

Consider the method of parallel projection. We will impose three restrictions that will allow us, albeit to the detriment of the visibility of the image, to get a drawing more convenient for using it in practice:

1. Let's delete both projection centers to infinity. Thus, we will ensure that the projecting rays from each center become parallel, and, therefore, the ratio of the true length of any line segment and the length of its projection will depend only on the angle of inclination of this segment to the projection planes and do not depend on the position of the projection center;
2. Let's fix the projection direction relative to the projection planes;
3. Let's place the projection planes perpendicular to each other, which will make it easy to move from the image on the projection planes to the real object in space.

Thus, having imposed these restrictions on the central projection method, we have come to its special case - parallel projection method(Figure 1.3). Projection, in which the projecting rays passing through each point of the object are parallel to the selected projection direction P, is called parallel .

Figure 1.3 - Parallel projection method

Let's introduce the notation:

R– direction of projection;

π 1 - horizontal plane of projections;

A,B– projection objects – points;

BUT 1 and AT 1 - projections of points BUT and AT onto the projection plane π 1 .

Parallel point projection is the point of intersection of the projecting line parallel to the given direction of projection R, with the projection plane π 1 .

Pass through the dots BUT and AT projecting beams parallel to a given projection direction R. Projecting ray passing through a point BUT intersects the projection plane π 1 at the point BUT one . Similarly, a projecting ray through a point AT intersects the projection plane at a point AT one . By connecting the dots BUT 1 and AT 1 , we get a segment BUT 1 AT 1 is the projection of the segment AB onto the plane π 1 .

1.3. Orthographic projection. Monge method

If the projection direction R perpendicular to the plane of projections p 1 , then the projection is called rectangular (Figure 1.4), or orthogonal (gr. orthos- straight, gonia- angle) if R not perpendicular to π 1, then the projection is called oblique .

quadrilateral AA 1 AT 1 AT defines the plane γ, which is called the projecting plane, since it is perpendicular to the plane π 1 (γ⊥π 1). In what follows, we will use only rectangular projection.

Figure 1.4 - Orthographic projection Figure 1.5 - Monge, Gaspard (1746-1818)

The French scientist Gaspard Monge is considered the founder of orthogonal projection (Figure 1.5).

Before Monge, builders, artists and scientists possessed quite significant information about projection methods, and yet only Gaspard Monge is the creator of descriptive geometry as a science.

Gaspard Monge was born on May 9, 1746 in the small town of Beaune (Burgundy) in eastern France in the family of a local merchant. He was the eldest of five children, for whom his father, despite the low origin and relative poverty of the family, tried to provide the most better education from available at that time for people from the humble class. His second son, Louis, became a professor of mathematics and astronomy, the youngest, Jean, also a professor of mathematics, hydrography and navigation. Gaspard Monge received his initial education at the city school of the Oratory order. After graduating in 1762 as the best student, he entered the college of Lyon, also owned by the Oratorians. Soon Gaspard was entrusted with teaching physics there. In the summer of 1764, Monge drew up a plan of his native city of Beaune, remarkably accurate. The necessary methods and instruments for measuring angles and drawing lines were invented by the compiler himself.

While studying in Lyon, he received an offer to join the order and remain a college teacher, however, instead, having shown great abilities in mathematics, drafting and drawing, he managed to enter the Mézieres School of Military Engineers, but (due to origin) only as an auxiliary non-commissioned officer officer department and without a paycheck. Nevertheless, success in the exact sciences and an original solution to one of the important problems of fortification (the placement of fortifications depending on the location of enemy artillery) allowed him in 1769 to become an assistant (teaching assistant) in mathematics, and then in physics, and already with a decent salary at 1800 livres a year.

In 1770, at the age of 24, Monge held the position of professor at the same time in two departments - mathematics and physics, and, in addition, conducts classes in cutting stones. Starting with the task of accurately cutting stones according to given sketches in relation to architecture and fortification, Monge came to the creation of methods that he later generalized in new science- descriptive geometry, the creator of which he is rightfully considered. Given the possibility of using the methods of descriptive geometry for military purposes in the construction of fortifications, the leadership of the Mézières school did not allow open publication until 1799, the book was published under the title descriptive geometry (Geometry descriptive) (a verbatim record of these lectures was made in 1795). The approach to lecturing on this science and doing the exercises outlined in it has survived to this day. Another significant work of Monge - Application of analysis to geometry (L'application de l'analyse à la geometrie, 1795) - is a textbook of analytic geometry, in which special emphasis is placed on differential relations.

In 1780 he was elected a member of the Paris Academy of Sciences, in 1794 he became director of the Polytechnic School. For eight months he served as minister of the sea in the government of Napoleon, was in charge of the gunpowder and cannon factories of the republic, and accompanied Napoleon on his expedition to Egypt (1798–1801). Napoleon granted him the title of count, honored him with many other distinctions.

The method of depicting objects according to Monge consists of two main points:

1. The position of a geometric object in space, in this example a point BUT, is considered relative to two mutually perpendicular planes π 1 and π 2(Figure 1.6).

They conditionally divide the space into four quadrants. Dot BUT located in the first quadrant. The Cartesian coordinate system served as the basis for the Monge projections. Monge replaced the concept of projection axes with the line of intersection of projection planes (coordinate axes) and proposed to combine the coordinate planes into one by rotating them around the coordinate axes.

Figure 1.6 - Model for constructing point projections

π 1 - horizontal (first) projection plane

π 2 - frontal (second) projection plane

π 1 ∩ π 2 is the axis of projections (we denote π 2 / π 1)

Consider an example of projecting a point BUT onto two mutually perpendicular projection planes π 1 and π 2 .

Drop from point BUT perpendiculars (projecting rays) on the planes π 1 and π 2 and mark their bases, that is, the points of intersection of these perpendiculars (projecting rays) with the projection planes. BUT 1 - horizontal (first) projection of the point BUT;BUT 2 - frontal (second) projection of the point BUT;AA 1 and AA 2 - projecting lines. Arrows show the direction of projection on the plane of projections π 1 and π 2 . Such a system allows you to uniquely determine the position of a point relative to the projection planes π 1 and π 2:

AA 1 ⊥π 1

BUT 2 BUT 0 ⊥π 2 /π 1 AA 1 = BUT 2 BUT 0 - distance from point A to plane π 1

AA 2 ⊥π 2

BUT 1 BUT 0 ⊥π 2 /π 1 AA 2 \u003d A 1 A 0 - the distance from point A to the plane π 2

2. Let's combine the rotation around the axis of projections π 2 / π 1 of the projection plane into one plane(π 1 with π 2), but so that the images do not overlap each other (in the α direction, Figure 1.6), we get an image called a rectangular drawing (Figure 1.7):

Figure 1.7 - Orthogonal drawing

Rectangular or orthogonal is called Monge diagram .

Straight BUT 2 BUT 1 called projection link , which connects opposite projections of the point ( BUT 2 - frontal and BUT 1 - horizontal) is always perpendicular to the projection axis (coordinate axis) BUT 2 BUT 1 ⊥π 2 /π 1 . On the diagram, the segments indicated by curly brackets are:

• BUT 0 BUT 1 - distance from the point BUT to the plane π 2 corresponding to the coordinate y A;
• BUT 0 BUT 2 - distance from the point BUT to the plane π 1 corresponding to the coordinate z A.

1.4. Rectangular point projections. Orthographic drawing properties

1. Two rectangular projections of a point lie on the same projection connection line perpendicular to the projection axis.

2. Two rectangular projections of a point uniquely determine its position in space relative to the projection planes.

Let's verify the validity of the last statement, for which we turn the plane π 1 to its original position (when π 1 ⊥ π 2). To build a point BUT needed from points BUT 1 and BUT 2 to restore the projecting rays, and in fact - the perpendiculars to the planes π 1 and π 2 , respectively. The intersection point of these perpendiculars fixes the desired point in space BUT. Consider an orthogonal drawing of a point BUT(Figure 1.8).

Figure 1.8 - Plotting a point

Let us introduce the third (profile) plane of projections π 3 perpendicular to π 1 and π 2 (given by the axis of projections π 2 /π 3).

Distance from profile projection points to the vertical projection axis BUT‘ 0 A 3 allows you to determine the distance from the point BUT to the frontal projection plane π 2 . It is known that the position of a point in space can be fixed relative to the Cartesian coordinate system using three numbers (coordinates) A(X A ; Y A ; Z A) or relative to the projection planes using its two orthogonal projections ( A 1 =(X A ; Y A); A 2 =(X A ; Z A)). On an orthogonal drawing, using two projections of a point, you can determine its three coordinates and, conversely, using three coordinates of a point, build its projections (Figure 1.9, a and b).

Figure 1.9 - Plotting a point according to its coordinates

By the location on the projection diagram of a point, one can judge its location in space:

• BUT — BUT 1 lies under the coordinate axis X, and the front BUT 2 - above the axis X, then we can say that the point BUT belongs to the 1st quadrant;
• if on the plot the horizontal projection of the point BUT — BUT 1 lies above the coordinate axis X, and the front BUT 2 - under the axle X, then the point BUT belongs to the 3rd quadrant;
• BUT — BUT 1 and BUT 2 lie above the axis X, then the point BUT belongs to the 2nd quadrant;
• if on the diagram there are horizontal and frontal projections of the point BUT — BUT 1 and BUT 2 lie under the axle X, then the point BUT belongs to the 4th quadrant;
• if on the diagram the projection of a point coincides with the point itself, then it means that the point belongs to the plane of projections;
• a point belonging to the projection plane or projection axis (coordinate axes) is called private point.

To determine in which quadrant of space a point is located, it is enough to determine the sign of the coordinates of the point.

Dependences of the quadrant of the position of the point and the signs of the coordinates

	X	Y	Z
I	+	+	+
II	+	—	+
III	+	—	—
IV	+	+	—

An exercise

Construct orthogonal projections of a point with coordinates BUT(60, 20, 40) and determine in which quadrant the point is located.

Problem solution: along the axis OX set aside the value of the coordinate XA=60, then through this point on the axis OX restore the projection connection line perpendicular to OX, along which to set aside the value of the coordinate ZA=40, and down - the value of the coordinate YA=20(Figure 1.10). All coordinates are positive, which means that the point is located in the I quadrant.

Figure 1.10 - Solution of the problem

1.5. Tasks for independent solution

1. Based on the diagram, determine the position of the point relative to the projection planes (Figure 1.11).

Figure 1.11

2. Complete the missing orthogonal projections of points BUT, AT, With on the projection plane π 1 , π 2 , π 3 (Figure 1.12).

Figure 1.12

3. Build point projections:

• E, symmetric point BUT relative to the projection plane π 1 ;
• F, symmetric point AT relative to the plane of projections π 2 ;
• G, symmetric point With relative to the projection axis π 2 /π 1 ;
• H, symmetric point D relative to the bisector plane of the second and fourth quadrants.

4. Construct orthogonal projections of the point To, located in the second quadrant and remote from the projection planes π 1 by 40 mm, from π 2 - by 15 mm.

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 "a z - 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; ten; 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 such a distance from the Oz axis, at which the horizontal projection of the point d is located: from the Ox axis, that is, 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 d z d "is built by transferring the segment dd x from the horizontal plane of projections to the profile one. To do this, draw a line of projection connection parallel to the Ox axis, get a point d y 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 d y. 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 "d x, get the point d".

Transferring the value of the segment d x d to the profile plane of the projections can be done 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 the 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 plane H. Its frontal projection is on the Ox axis and coincides with point b x. The Y coordinate of point C is zero, the point is located on the plane V, its horizontal projection c is on the x-axis and coincides with the point c 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 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.

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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 a straight line apassing through the point M 1 and perpendicular to the plane α (study the topic of the equation of a 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.

Decision

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. 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 . To this end, we pass from the 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

Decision

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 down parametric equations 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 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 .

Decision

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

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