Find the projection of the point on the plane given by the equation. Finding the coordinates of the projection of a point on a plane, examples

Projection apparatus

The projection apparatus (Fig. 1) includes three projection planes:

π 1 - horizontal projection plane;

π 2 - frontal projection plane;

π 3– profile plane of projections .

The projection planes are mutually perpendicular ( π 1^ π 2^ π 3), and their intersection lines form axes:

Plane intersection π 1 and π 2 form an axis 0X (π 1∩ π 2 = 0X);

Plane intersection π 1 and π 3 form an axis 0Y (π 1∩ π 3 = 0Y);

Plane intersection π 2 and π 3 form an axis 0Z (π 2∩ π 3 = 0Z).

The point of intersection of the axes (ОХ∩OY∩OZ=0) is considered to be the reference point (point 0).

Since the planes and axes are mutually perpendicular, such an apparatus is similar Cartesian system coordinates.

The projection planes divide the entire space into eight octants (in Fig. 1 they are indicated by Roman numerals). Projection planes are considered opaque, and the viewer is always in I th octane.

Projection orthogonal with projection centers S1, S2 and S3 respectively for the horizontal, frontal and profile projection planes.

BUT.

From projection centers S1, S2 and S3 projecting beams come out l 1, l 2 and l 3 BUT

- A 1 BUT;

- A 2– frontal projection of the point BUT;

- A 3– profile projection of a point BUT.

A point in space is characterized by its coordinates A(x,y,z). points A x, A y and Az respectively on the axes 0X, 0Y and 0Z show coordinates x, y and z points BUT. On fig. 1 gives all the necessary designations and shows the relationship between the point BUT space, its projections and coordinates.

point diagram

To plot a point BUT(Fig. 2), in the projection apparatus (Fig. 1) the plane π 1 A 1 0X π 2. Then the plane π 3 with point projection A 3, rotate counterclockwise around the axis 0Z, until it coincides with the plane π 2. Direction of rotation of planes π 2 and π 3 shown in fig. 1 arrows. At the same time, direct A 1 A x and A 2 A x 0X perpendicular A 1 A 2, and straight lines A 2 A x and A 3 A x will be located in common to the axis 0Z perpendicular A 2 A 3. These lines will be referred to as vertical and horizontal connection lines.

It should be noted that when moving from the projection apparatus to the diagram, the projected object disappears, but all information about its shape, geometric dimensions and the place of its position in space are preserved.

BUT(x A , y A , z Ax A , y A and z A in the following sequence (Fig. 2). This sequence is called the point plotting technique.

1. Axes are drawn orthogonally OX, OY and oz.

2. On the axis OX x A points BUT and get the position of the point A x.

3. Through the dot A x perpendicular to the axis OX

A x in the direction of the axis OY the numerical value of the coordinate is postponed y A points BUT A 1 on the plot.

A x in the direction of the axis oz the numerical value of the coordinate is postponed z A points BUT A 2 on the plot.

6. Through the dot A 2 parallel to axis OX a horizontal line is drawn. The intersection of this line and the axis oz will give the position of the point A z.

7. On a horizontal line from the point A z in the direction of the axis OY the numerical value of the coordinate is postponed y A points BUT and the position of the profile projection of the point is determined A 3 on the plot.

Point characteristic

All points of space are subdivided into points of private and general positions.

Private position points. Points belonging to the projection apparatus are called points of particular position. These include points belonging to the projection planes, axes, origin and projection centers. The characteristic features of points of private position are:

Metamathematical - one, two or all numerical values ​​of the coordinates are equal to zero and (or) infinity;

On the diagram - two or all projections of a point are located on the axes and (or) are located at infinity.

points general position. Points in general position include points that do not belong to the projection apparatus. For example, dot BUT in fig. 1 and 2.

AT general case the numerical values ​​of the coordinates of a point characterize its distance from the projection plane: the coordinate X from the plane π 3; coordinate y from the plane π 2; coordinate z from the plane π 1. It should be noted that the signs at the numerical values ​​of the coordinates indicate the direction of removal of the point from the projection planes. Depending on the combination of signs for the numerical values ​​of the coordinates of the point, it depends in which of the octane it is located.

Two Image Method

In practice, in addition to the full projection method, the two-image method is used. It differs in that the third projection of the object is excluded in this method. To obtain the projection apparatus of the two-image method, the profile projection plane with its projection center is excluded from the full projection apparatus (Fig. 3). In addition, on the axis 0X the origin is assigned (point 0 ) and from it perpendicular to the axis 0X in projection planes π 1 and π 2 spend axis 0Y and 0Z respectively.

In this apparatus, the entire space is divided into four quadrants. On fig. 3 are marked with Roman numerals.

Projection planes are considered opaque, and the viewer is always in I th quadrant.

Consider the operation of the device using the example of projecting a point BUT.

From projection centers S1 and S2 projecting beams come out l 1 and l 2. These rays pass through the point BUT and intersecting with the projection planes form its projections:

- A 1- horizontal projection of a point BUT;

- A 2– frontal projection of the point BUT.

To plot a point BUT(Fig. 4), in the projection apparatus (Fig. 3) the plane π 1 with the resulting point projection A 1 rotate clockwise around an axis 0X, until it coincides with the plane π 2. Plane rotation direction π 1 shown in fig. 3 arrows. At the same time, only one point remains on the diagram of the point obtained by the two-image method. vertical communication line A 1 A 2.

In practice, plotting a point BUT(x A , y A , z A) is carried out according to the numerical values ​​of its coordinates x A , y A and z A in the following sequence (Fig. 4).

1. An axis is drawn OX and the origin is assigned (point 0 ).

2. On the axis OX the numerical value of the coordinate is postponed x A points BUT and get the position of the point A x.

3. Through the dot A x perpendicular to the axis OX a vertical line is drawn.

4. On the vertical line from the point A x in the direction of the axis OY the numerical value of the coordinate is postponed y A points BUT and the position of the horizontal projection of the point is determined A 1 OY not drawn, but assumed to be positive values located below the axis OX, while the negative ones are higher.

5. On the vertical line from the point A x in the direction of the axis oz the numerical value of the coordinate is postponed z A points BUT and the position of the frontal projection of the point is determined A 2 on the plot. It should be noted that on the diagram the axis oz is not drawn, but it is assumed that its positive values ​​are located above the axis OX, while the negative ones are lower.

Competing points

Points on the same projecting ray are called competing points. They have a common projection in the direction of the projecting beam, i.e. their projections coincide identically. characteristic feature competing points on the diagram is the identical coincidence of their projections of the same name. The competition lies in the visibility of these projections relative to the observer. In other words, in space for the observer, one of the points is visible, the other is not. And, accordingly, in the drawing: one of the projections of the competing points is visible, and the projection of the other point is invisible.

On a spatial projection model (Fig. 5) from two competing points BUT and AT visible dot BUT on two mutually complementary grounds. According to the chain S 1 →A→B dot BUT closer to the observer than a point AT. And, accordingly, further from the projection plane π 1(those. z A > z A).

Rice. 5 Fig.6

If the point itself is visible A, then its projection is also visible A 1. In relation to the projection coinciding with it B1. For clarity and, if necessary, on the diagram, invisible projections of points are usually enclosed in brackets.

Remove points on the model BUT and AT. Their coinciding projections on the plane will remain π 1 and separate projections - on π 2. We conditionally leave the frontal projection of the observer (⇩), located in the center of the projection S1. Then along the chain of images ⇩ → A2 → B2 it will be possible to judge that z A > zB and that the point itself is visible BUT and its projection A 1.

Similarly, consider the competing points With and D apparently relative to the plane π 2 . Since the common projecting beam of these points l 2 parallel to axis 0Y, then the sign of visibility of competing points With and D is determined by the inequality yC > yD. Therefore, the point D closed by a dot With and, accordingly, the projection of the point D2 will be covered by the projection of the point From 2 on surface π 2.

Let's consider how the visibility of competing points is determined in a complex drawing (Fig. 6).

According to matching projections A 1≡IN 1 the points themselves BUT and AT are on the same projecting beam parallel to the axis 0Z. So coordinates are to be compared z A and zB these points. To do this, we use the frontal projection plane with separate point images. AT this case z A > zB. It follows from this that the projection is visible A 1.

points C and D in the complex drawing under consideration (Fig. 6) are also on the same projecting beam, but only parallel to the axis 0Y. Therefore, from a comparison yC > yD we conclude that the projection C 2 is visible.

General rule . Visibility for coinciding projections of competing points is determined by comparing the coordinates of these points in the direction of the common projecting beam. Visible is the projection of the point for which this coordinate is greater. In this case, the comparison of coordinates is carried out on the plane of projections with separate images of points.

The study of the properties of figures in space and on a plane is impossible without knowing the distances between a point and such geometric objects as a straight line and a plane. In this article, we will show how to find these distances by considering the projection of a point onto a plane and onto a line.

Equation of a straight line for two-dimensional and three-dimensional spaces

Calculation of distances of a point to a straight line and a plane is carried out using its projection onto these objects. To be able to find these projections, one should know in what form the equations for lines and planes are given. Let's start with the first.

A straight line is a collection of points, each of which can be obtained from the previous one by transferring to vectors parallel to each other. For example, there is a point M and N. The vector MN¯ connecting them takes M to N. There is also a third point P. If the vector MP¯ or NP¯ is parallel to MN¯, then all three points lie on the same line and form it.

Depending on the dimension of the space, the equation that defines the straight line can change its form. Yes, everyone knows linear dependence y-coordinates from x in space describe a plane that is parallel to the third z-axis. In this regard, in this article we will consider only the vector equation for a straight line. It has the same look for plane and three-dimensional spaces a.

In space, a straight line can be defined following expression:

(x; y; z) = (x 0 ; y 0 ; z 0) + α*(a; b; c)

Here, the values ​​of coordinates with zero indices correspond to some point belonging to the line, u¯(a; b; c) are the coordinates of the direction vector that lies on the given line, α is an arbitrary real number, changing which you can get all the points of the line. This equation is called vector.

Often the above equation is written in expanded form:

Similarly, you can write an equation for a straight line that is in a plane, that is, in two-dimensional space:

(x; y) = (x 0 ; y 0) + α*(a; b);

Plane equation

To be able to find the distance from a point to projection planes, you need to know how a plane is specified. Just like a straight line, it can be represented in several ways. Here we consider only one: the general equation.

Suppose that the point M(x 0 ; y 0 ; z 0) belongs to the plane, and the vector n¯(A; B; C) is perpendicular to it, then for all points (x; y; z) of the plane the equality will be valid:

A*x + B*y + C*z + D = 0 where D = -1*(A*x 0 + B*y 0 + C*z 0)

It should be remembered that in this general equation of the plane, the coefficients A, B and C are the coordinates of the vector normal to the plane.

Calculation of distances by coordinates

Before proceeding to the consideration of projections onto the plane of a point and onto a straight line, it should be recalled how the distance between two known points should be calculated.

Let there be two spatial points:

A 1 (x 1 ; y 1 ; z 1) and A 2 (x 2 ; y 2 ​​; z 2)

Then the distance between them is calculated by the formula:

A 1 A 2 \u003d √ ((x 2 -x 1) 2 + (y 2 -y 1) 2 + (z 2 -z 1) 2)

Using this expression, the length of the vector A 1 A 2 ¯ is also determined.

For the case on the plane, when two points are given by just a pair of coordinates, we can write a similar equality without the presence of a term with z in it:

A 1 A 2 \u003d √ ((x 2 -x 1) 2 + (y 2 -y 1) 2)

Now we consider various cases of projection on a plane of a point onto a straight line and onto a plane in space.

Point, line and distance between them

Suppose there is some point and a line:

P 2 (x 1 ; y 1);

(x; y) = (x 0 ; y 0) + α*(a; b)

The distance between these geometric objects will correspond to the length of the vector, the beginning of which lies at the point P 2 , and the end is located at a point P on the specified line, for which the vector P 2 P ¯ is perpendicular to this line. The point P is called the projection of the point P 2 onto the line under consideration.

The figure below shows the point P 2 , its distance d to the straight line, as well as the guide vector v 1 ¯. Also on the line is selected arbitrary point P 1 and from it to P 2 a vector is drawn. Point P here coincides with the place where the perpendicular intersects the line.

It can be seen that the orange and red arrows form a parallelogram, the sides of which are the vectors P 1 P 2 ¯ and v 1 ¯, and the height is d. It is known from geometry that to find the height of a parallelogram, its area should be divided by the length of the base, on which the perpendicular is lowered. Since the area of ​​a parallelogram is calculated as the vector product of its sides, we get the formula for calculating d:

d = ||/|v 1 ¯|

All vectors and point coordinates in this expression are known, so you can use it without performing any transformations.

This problem could have been solved differently. For this, two equations should be written:

• the scalar product of P 2 P ¯ and v 1 ¯ must be equal to zero, since these vectors are mutually perpendicular;
• the coordinates of point P must satisfy the equation of a straight line.

These equations are enough to find the coordinates P and then the length d using the formula given in the previous paragraph.

Finding the distance between a line and a point

Let's show you how to use the data theoretical information to solve a specific problem. Suppose the following point and line are known:

(x; y) = (3; 1) - α*(0; 2)

It is necessary to find the projection points on the line on the plane, as well as the distance from M to the line.

Denote the projection to be found by the point M 1 (x 1 ; y 1). We solve this problem in two ways, described in the previous paragraph.

Method 1. Direction vector v 1 ¯ coordinates has (0; 2). To construct a parallelogram, we select some point belonging to the line. For example, a point with coordinates (3; 1). Then the vector of the second side of the parallelogram will have coordinates:

(5; -3) - (3; 1) = (2; -4)

Now you should calculate the product of the vectors that define the sides of the parallelogram:

We substitute this value into the formula, we get the distance d from M to the straight line:

Method 2. Now let's find in another way not only the distance, but also the coordinates of the projection of M onto the straight line, as required by the condition of the problem. As mentioned above, to solve the problem, it is necessary to compose a system of equations. It will take the form:

(x 1 -5)*0+(y 1 +3)*2 = 0;

(x 1 ; y 1) = (3; 1)-α*(0; 2)

Let's solve this system:

The projection of the original point of the coordinate has M 1 (3; -3). Then the desired distance is:

d = |MM 1 ¯| = √(4+0) = 2

As you can see, both methods of solving gave the same result, which indicates the correctness of the performed mathematical operations.

Projection of a point onto a plane

Now consider what is the projection of a point given in space onto a certain plane. It is easy to guess that this projection is also a point, which, together with the original one, forms a vector perpendicular to the plane.

Suppose that the projection onto the plane of the point M has the following coordinates:

The plane itself is described by the equation:

A*x + B*y + C*z + D = 0

Based on these data, we can formulate the equation of a straight line intersecting the plane at a right angle and passing through M and M 1:

(x; y; z) = (x 0 ; y 0 ; z 0) + α*(A; B; C)

Here, the variables with zero indices are the coordinates of the point M. The position on the plane of the point M 1 can be calculated based on the fact that its coordinates must satisfy both written equations. If these equations are not enough when solving the problem, then the condition of parallelism of MM 1 ¯ and the guide vector for a given plane can be used.

Obviously, the projection of a point belonging to the plane coincides with itself, and the corresponding distance is zero.

Problem with point and plane

Let a point M(1; -1; 3) and a plane be given, which is described by the following general equation:

You should calculate the coordinates of the projection onto the plane of the point and calculate the distance between these geometric objects.

To begin with, we construct the equation of a straight line passing through M and perpendicular to the indicated plane. It looks like:

(x; y; z) = (1; -1; 3) + α*(-1; 3; -2)

Let's denote the point where this line intersects the plane, M 1 . Equalities for a plane and a straight line must be satisfied if the coordinates M 1 are substituted into them. Writing explicitly the equation of a straight line, we obtain the following four equalities:

X 1 + 3*y 1 -2*z 1 + 4 = 0;

y 1 \u003d -1 + 3 * α;

From the last equality we get the parameter α, then we substitute it into the penultimate and into the second expression, we get:

y 1 \u003d -1 + 3 * (3-z 1) / 2 \u003d -3 / 2 * z 1 + 3.5;

x 1 \u003d 1 - (3-z 1) / 2 \u003d 1/2 * z 1 - 1/2

We substitute the expression for y 1 and x 1 into the equation for the plane, we have:

1*(1/2*z 1 - 1/2) + 3*(-3/2*z 1 + 3.5) -2*z 1 + 4 = 0

Where do we get:

y 1 \u003d -3 / 2 * 15/7 + 3.5 \u003d 2/7;

x 1 = 1/2*15/7 - 1/2 = 4/7

We have determined that the projection of the point M onto given plane corresponds to coordinates (4/7; 2/7; 15/7).

Now let's calculate the distance |MM 1 ¯|. The coordinates of the corresponding vector are:

MM 1 ¯(-3/7; 9/7; -6/7)

The required distance is:

d = |MM 1 ¯| = √126/7 ≈ 1.6

Three projection points

During the preparation of drawings, it is often necessary to obtain projections of sections on mutually perpendicular three planes. Therefore, it is useful to consider what the projections of some point M with coordinates (x 0 ; y 0 ; z 0) on three coordinate planes.

It is not difficult to show that the xy plane is described by the equation z = 0, the xz plane corresponds to the expression y = 0, and the remaining yz plane is denoted by the equality x = 0. It is easy to guess that the projections of a point on 3 planes will be equal:

for x = 0: (0; y 0 ; z 0);

for y = 0: (x 0 ; 0 ; z 0);

for z = 0: (x 0 ; y 0 ; 0)

Where is it important to know the projections of a point and its distances to planes?

Determining the position of the projection of points on a given plane is important when finding such quantities as surface area and volume for inclined prisms and pyramids. For example, the distance from the top of the pyramid to the plane of the base is the height. The latter is included in the formula for the volume of this figure.

The considered formulas and methods for determining projections and distances from a point to a straight line and a plane are quite simple. It is only important to remember corresponding forms equations of a plane and a straight line, and also have a good spatial imagination to apply them successfully.

The projection of a point onto a plane is a special case common task finding the projection of a point onto a surface. Due to the simplicity of calculating the projection of a point onto a plane tangent to the surface, it is used as a zero approximation in solving the general problem.

Consider the problem of projecting a point onto a plane given by the radius vector

We will assume that the vectors are not collinear. Assume that in the general case the vectors are not orthogonal and have non-unit length. The plane passes through the point where the parameters are equal to zero, and the vectors define the parametric directions. The given point has a unique projection onto the plane (4.6.1). Let's construct a unit normal to the plane

Rice. 4.6.1. Projection of a point onto the plane s(u, v)

Let us calculate the radius vector of the point projection onto the plane as the difference between the radius vector of the projected point and the component of the vector parallel to the normal to the plane,

(4.6.4)

On fig. 4.6.1 shows the vectors of the plane, its starting point and projection given point.

The parameters and lengths of the projections are related by the equations

where the cosine of the angle between the vectors is determined by the formula (1.7.13).

From the system of these equations, we find the parameters of the projection of a point onto a plane

(4.6.6)

where are the coefficients of the first main quadratic form planes (1.7.8), which are also covariant components of the metric surface tensor, are contravariant components of the metric surface tensor. If the vectors are orthogonal, then formulas (4.6.6) and (4.6.7) take the form

The distance from a point to its projection onto a plane is generally calculated as the length of a vector. The distance from a point to its projection onto a plane can be determined without calculating the projection of the point, but by calculating the projection of the vector onto the normal to the plane

(4.6.8)

Special cases.

The projections of a point onto some analytic surfaces can be found without involving numerical methods. For example, to find the projection of a point onto the surface of a circular cylinder, cone, sphere, or torus, you need to translate the projected point into local system surface coordinates, where it is easy to find projection parameters. Similarly, projections on extrusion and rotation surfaces can be found. In some particular cases, the positions of the projected point of its projection can be easily found on other surfaces as well.

General case.

Consider the problem of projecting a point onto a surface in the general case. Let it be required to find all projections of a point on a surface . Each desired point surface satisfies the system of two equations

The system of equations (4.6.9) contains two unknown quantities - the parameters u and v. This problem is solved in the same way as the problem of finding the projections of a given point onto a curve.

At the first stage, we determine the zero approximations of the surface parameters for the projections of a point, and at the second stage, we find the exact values ​​of the parameters that determine the projections of a given point onto the surface

Let's go over the surface with steps calculated by formulas (4.2.4) and (4.2.5), described above by the way of moving along the parametric region. Let's denote the parameters of the points through which we will pass through . At each point, we will calculate the scalar products of vectors

(4.6.10)

If the desired solution lies near a point with parameters , then we will have different signs, as well as and will have different signs. Change of signs scalar products indicates that the desired solution is nearby. For the zero approximation of the parameters, we take the values ​​Starting from the zero approximation of the parameters, one of the methods for solving nonlinear equations find a solution to the problem with a given accuracy. For example, in Newton's method, at iterations, the increments of the projection parameters can be found from the system of linear equations

where are the partial derivatives of the radius vector with respect to the parameters. next approximation point projection parameters are equal to . The process of refining the parameters will be completed when the inequalities are fulfilled at the next iteration , where is the specified error. In the same way, we find all other roots of the system of equations (4.6.9).

If you need to find only the nearest projection of a given point onto the surface, then you can go through the same points of a geometric object and select the one closest to the given point. The parameters of the nearest point and should be chosen as the zero approximation of the problem solution.

Projection of a point onto a surface in a given direction.

In certain cases, the problem arises of determining the projection of a point onto a surface not along the normal to it, but along a given direction. Let the direction of projection be given by a vector of unit length q. Let's build a straight line

(4.6.12)

passing through a given point and having a direction given vector. Projections of a point onto a surface in given direction we define as the points of intersection of the surface with the line (4.6.12) passing through the given point in the given direction.

PROJECTION OF A POINT ON TWO PLANES OF PROJECTIONS

The formation of a straight line segment AA 1 can be represented as a result of moving point A in any plane H (Fig. 84, a), and the formation of a plane can be represented as a displacement of a straight line segment AB (Fig. 84, b).

Dot - main geometric element lines and surfaces, so the study of the rectangular projection of an object begins with the construction of rectangular projections of a point.

Into space dihedral angle, formed by two perpendicular planes - the frontal (vertical) plane of projections V and the horizontal plane of projections H, we place point A (Fig. 85, a).

The line of intersection of the projection planes is a straight line, which is called the projection axis and is denoted by the letter x.

The V plane is shown here as a rectangle, and the H plane as a parallelogram. The inclined side of this parallelogram is usually drawn at an angle of 45° to its horizontal side. The length of the inclined side is taken equal to 0.5 of its actual length.

From point A, perpendiculars are lowered on the planes V and H. Points a "and a of the intersection of perpendiculars with the projection planes V and H are rectangular projections points A. The figure Aaa x a "in space is a rectangle. The side aa of this rectangle in the visual image is reduced by 2 times.

Let us align the H plane with the V plane by rotating V around the line of intersection of the x planes. The result is a complex drawing of point A (Fig. 85, b)

To simplify the complex drawing, the boundaries of the projection planes V and H are not indicated (Fig. 85, c).

Perpendiculars drawn from point A to the projection planes are called projecting lines, and the bases of these projecting lines - points a and a "are called projections of point A: a" is the frontal projection of point A, a is the horizontal projection of point A.

Line a "a is called the vertical line of the projection connection.

The location of the projection of a point on a complex drawing depends on the position of this point in space.

If point A lies on the horizontal projection plane H (Fig. 86, a), then its horizontal projection a coincides with the given point, and the frontal projection a "is located on the axis. When point B is located on the frontal projection plane V, its frontal projection coincides with this point, and the horizontal projection lies on the x-axis. The horizontal and front projection a given point C lying on the x-axis coincide with this point. A complex drawing of points A, B and C is shown in fig. 86b.

PROJECTION OF A POINT ON THREE PLANES OF PROJECTIONS

In cases where it is impossible to imagine the shape of an object from two projections, it is projected onto three projection planes. In this case, the profile plane of projections W is introduced, perpendicular to planes V and H. A visual representation of a system of three projection planes is given in fig. 87 a.

The edges of a trihedral angle (the intersection of projection planes) are called projection axes and are denoted by x, y and z. The intersection of the projection axes is called the beginning of the projection axes and is denoted by the letter O. Let us drop the perpendicular from the point A to the projection plane W and, marking the base of the perpendicular with the letter a, we get profile projection points A.

To obtain a complex drawing, points A of the H and W planes are aligned with the V plane, rotating them around the Ox and Oz axes. A complex drawing of point A is shown in fig. 87b and c.

The segments of the projecting lines from point A to the projection planes are called the coordinates of point A and are denoted: x A, y A and z A.

For example, the coordinate z A of point A, equal to the segment a "a x (Fig. 88, a and b), is the distance from point A to the horizontal projection plane H. The coordinate at point A, equal to the segment aa x, is the distance from point A to the frontal plane of projections V. The x A coordinate equal to the segment aa y is the distance from point A to the profile plane of projections W.

Thus, the distance between the projection of a point and the projection axis determines the coordinates of the point and is the key to reading its complex drawing. By two projections of a point, all three coordinates of a point can be determined.

If the coordinates of point A are given (for example, x A \u003d 20 mm, y A \u003d 22 mm and z A \u003d 25 mm), then three projections of this point can be built.

To do this, from the origin of coordinates O in the direction of the Oz axis, the coordinate z A is laid up and the coordinate y A is laid down. segments equal to the x coordinate A. The resulting points a "and a - frontal and horizontal projection points A.

According to two projections a "and a point A, its profile projection can be constructed in three ways:

1) from the origin O, an auxiliary arc is drawn with a radius Oa y equal to the coordinate (Fig. 87, b and c), from the obtained point a y1 draw a straight line parallel to the Oz axis, and lay a segment equal to z A;

2) from the point a y, an auxiliary straight line is drawn at an angle of 45 ° to the axis Oy (Fig. 88, a), a point a y1 is obtained, etc.;

3) from the origin O, draw an auxiliary straight line at an angle of 45 ° to the axis Oy (Fig. 88, b), get a point a y1, etc.