Intersection of two planes online. Construction of a line of intersection of planes defined in various ways

A straight line obtained at the mutual intersection of two planes is completely determined by two points, each of which belongs to both planes. So, the straight line K 1 K 2 (Fig. 163), along which the plane given by the triangle ABC and pl. β given by lines DE and DF passes through points K 1 and K 2 ; but at these points the lines AB and AC of the first plane intersect pl. β i.e. points K 1 and K 2 belong to both planes.

Consequently, in the general case, to construct a line of intersection of two planes, one must find any two points, each of which belongs to both planes; these points define the line of intersection of the planes.

To find each of these two points, one usually has to perform special constructions. But if at least one of the intersecting planes is perpendicular to the projection plane, then the construction of projections of the intersection line is simplified. Let's start with such a case.

On fig. 164 shows the intersection of two planes, of which one (given by the triangle DEF) is located perpendicular to the square. π 2 . Since the triangle DEF is projected onto the square π 2 in the form of a straight line (D "F"), the frontal projection of the straight line segment along which both triangles intersect is the segment K "1 K" 2 on the projection D "F". Further construction is clear from the drawing.

Another example is given in Fig. 165. The horizontally projecting plane α intersects the plane of the triangle ABC. The horizontal projection of the line of intersection of these planes - the segment M "N" - is determined on the trace α ".

Now consider general case of constructing a line of intersection of two planes. Let one of the planes, β, be given by two intersecting lines, and the other, γ, by two parallel lines. The construction is shown in fig. 166. As a result of the mutual intersection of the planes β and γ, a straight line K 1 K 2 is obtained. Let's express it by writing: β × γ = К 1 K 2 .

To determine the position of the points K 1 and K 2, we take two auxiliary front-projecting planes (α 1 and α 2) intersecting each of the planes β and γ. At the intersection of the planes β and γ by the plane α 1 . we get straight lines with projections 1"2", 1"2" and 3"4", 3"4". These lines, located in pl. α 1 , in its intersection define the first point, K 1 , the line of intersection of the planes β and γ.

Having received the projections K" 1 and K" 2, we find on the traces and α" 1 and α" 2 the projections K" 1 and K" 2. This determines the projections K "1 K" 2 and K "1 K" 2 of the desired line of intersection of the planes β and γ (the projections are drawn by a dash-dotted line).

When constructing, you can keep in mind the following: since the auxiliary secant planes α 1 and α 2 are mutually parallel, then, having built the projections 1 "2" and 3 "4", one should take one point for the projections 5 "6" and 7 "8" , at least 5 and 8, since 5"6"||1"2" and 7"8"||3"4".

In the considered construction, two frontal-projecting planes were taken as auxiliary. Of course, other planes could also be taken, for example, two horizontal or one horizontal, another frontal, etc. The essence of the constructions does not change from this. However, such a case may occur. Let us assume that two horizontal planes were taken as auxiliary and obtained when they intersect

planes β and γ, the horizontals turned out to be mutually parallel. But fig. 167 shows that β and γ intersect, although their horizontals are parallel. Therefore, having received mutually parallel horizontal projections of the horizontals AB and CD and knowing that the planes are not necessarily parallel, but can intersect (along a common horizontal), it is necessary to test the planes β and γ using at least a horizontally projecting plane (see . Fig. 167); if the lines along which this auxiliary plane σ intersects β and γ would also turn out to be parallel to one another, then the planes β and γ do not intersect, but are parallel to one another. On fig. 167 these lines intersect at the point K, through which the line of intersection of the planes β and γ passes parallel to the lines BA and CD.

If the planes are given by their traces on the projection planes, then it is natural to look for points that define the line of intersection of the planes at the points of intersection of the traces of the planes of the same name (Fig. 168): the straight line passing through these points is common to both planes, i.e. their line intersections.

The scheme for constructing a line of intersection of two planes (see Fig. 166) can, of course, be extended to the case of specifying planes by their traces. Here, the role of auxiliary cutting planes is played by the projection planes themselves:

α × π 1 \u003d h "0α; β × π 1 \u003d h" 0β; h" 0α × h" 0β =M;

α × π 2 \u003d f "0α; β × π 2 \u003d f" 0β; f" 0α × f" 0β =N.

The intersection points of the same-name traces of planes are the traces of the line of intersection of these planes. Therefore, to construct projections of the line of intersection of the planes α and β (Fig. 168), it is necessary: ​​1) find the point M "at the intersection of traces h" 0α and h "0β

and point N" at the intersection of f" 0α and f" 0β, and along them - the projections M" and N"; 2) draw straight lines M"N" and M"N",

On fig. 169-171 show cases where the direction of the intersection line is known. Therefore, it is sufficient to have only one point from the intersection of traces and then draw a straight line through this point, based on the position of the planes and their traces.

Questions to §§ 22-24

1. What is the relative position of the two planes?
2. What is the sign of parallelism of two planes?
3. How are the frontal traces of two parallel front-projecting planes mutually arranged?
4. How are the horizontal traces of two parallel horizontally projecting planes mutually arranged?
5. How are the traces of the same name mutually arranged in two parallel planes?
6. Does the intersection of at least one pair of their traces of the same name serve as a sign of the mutual intersection of two planes?
7. How to set the relative position of a line and a plane?
8. How is the point of intersection of a straight line with a plane perpendicular to one or two projection planes constructed?
9. Which point among those located on a common perpendicular to a) pl. π 1 b) pl. π 2 is considered visible on π 1, respectively, on π 2 ?
10. How is the line of intersection of two planes, of which at least one is perpendicular to the square, constructed? π 1 or to square. p2?
11. What is the general way to draw a line of intersection of two planes?

Two planes intersect each other in a straight line. To construct it, it is necessary to determine two points that simultaneously belong to each of the given planes. Let's see how this is done with the following examples.

Let's find the line of intersection of planes in general position α and β for the case when pl. α is given by the projections of triangle ABC, and pl. β - parallel lines d and e. The solution of this problem is carried out by constructing points L 1 and L 2 belonging to the line of intersection.

Solution

1. We introduce an auxiliary horizontal plane γ 1 . It intersects α and β in straight lines. The frontal projections of these lines, 1""C"" and 2""3"", coincide with the frontal trace of the square. γ 1 . It is designated in the figure as f 0 γ 1 and is located parallel to the x axis.
2. We define horizontal projections 1"C" and 2"3" along the communication lines.
3. We find the horizontal projection of the point L 1 at the intersection of lines 1"C" and 2"3". The frontal projection of the point L 1 lies on the frontal trace of the plane γ.
4. We introduce an auxiliary horizontal plane γ 2 . Using constructions similar to those described in paragraphs 1, 2, 3, we find the projections of the point L 2 .
5. Through L 1 and L 2 we draw the required line l.

It should be noted that as γ it is convenient to use both level planes and projecting planes.

Let us find the line of intersection of the planes α and β given by traces. This task is much easier than the previous one. It does not require the introduction of auxiliary planes. Their role is played by the projection planes P 1 and P 2.

Construction algorithm

1. We find the point L" 1 located at the intersection of the horizontal traces h 0 α and h 0 β. The point L"" 1 lies on the x axis. Its position is determined using a connection line drawn from L" 1.
2. We find the point L "" 2 at the intersection of the frontal traces of pl. α and β. Point L" 2 lies on the x-axis. Its position is determined by the connection line drawn from L"" 2 .
3. We draw straight lines l" and l"" through the corresponding projections of the points L 1 and L 2, as shown in the figure.

Thus, the line l passing through the points of intersection of the traces of the planes is the desired one.

Intersection of planes of triangles

Let's consider the construction of the line of intersection of the planes given by the triangles ABC and DEF, and the determination of their visibility by the method of competing points.

Construction algorithm

1. We draw a front-projecting plane σ through the straight line DE: its trace f 0σ is indicated on the drawing. The plane σ intersects the triangle ABC along the straight line 35. Having marked the points 3""=A""B""∩f 0σ and 5""=A""С""∩f 0σ , we determine the position (∙)3" and (∙) 5" over communication lines to ΔA"B"C".
2. We find the horizontal projection N"=D"E"∩3"5" of the point N of intersection of the lines DE and 35, which lie in the auxiliary plane σ. The projection N"" is located on the frontal wake f 0σ on the same line of communication with N".

We draw a front-projecting plane τ through the straight line BC: its trace f 0τ is indicated on the drawing. Using constructions similar to those described in paragraphs 1 and 2 of the algorithm, we find the projections of the point K.

3. Through N and K we draw the desired line NK - the line of intersection of ΔABC and ΔDEF.

Visibility definition

Frontally competing points 4 and 5, belonging to ΔDEF and ΔABC, respectively, are on the same frontally projecting straight line, but are located at different distances from the projection plane π 2 . Since (∙)5" is closer to the observer than (∙)4", then the section ΔABC with (∙)5 belonging to it is visible in the projection on the square. π 2 . On the opposite side of the N""K"" line, the visibility of the triangles changes.

Horizontally competing points 6 and 7, belonging to ΔABC and ΔDEF, respectively, are on the same horizontally projecting straight line, but are located at different distances from the projection plane π 1 . Since (∙)6"" is higher than (∙)7"", then the section ΔABC with (∙)6 belonging to it is visible in the projection on the square. pi 1 . On the opposite side of the N"K" line, the visibility of the triangles changes.

Two planes in space can be parallel or intersecting, a special case of intersecting planes are mutually perpendicular planes.

The construction of the line of intersection of planes is one of the main tasks of descriptive geometry, which are of great practical importance. It belongs to the so-called positional tasks.

positional called tasks to determine the common elements of various conjugated geometric shapes. These include tasks for belonging geometric elements and to the intersection geometric objects, for example, the intersection of a line and a plane with a surface, the intersection of two surfaces and, in particular, the problem of the intersection of two planes.

The line of intersection of two planes is a straight line that simultaneously belongs to both intersecting planes. Therefore, to construct a line of intersection of planes, it is necessary to determine two points of this line or one point and the direction of the line of intersection.

Consider special case intersection of planes when one of them is projecting. On fig. 3.6 shows a plane in general position - given by the triangle ABC and horizontally projecting P. Two common points belonging to both planes are points D and E, which determine the line of intersection.

To determine these points, the points of intersection of the sides AB and BC with the projecting plane were found. The construction of points D and E both on a spatial drawing (Fig. 3.6, a) and on a diagram (Fig. 3.6, b) does not cause difficulties, because is based on the collective property of projecting traces of planes discussed above.

By connecting the projections of the same name of the points D and E, we obtain the projections of the line of intersection of the plane of the triangle ABC and the plane P. Thus, the horizontal projection D 1 E 1 of the line of intersection of the given planes coincides with the horizontal projection of the projecting plane P - with its horizontal trace.

Consider general case intersections when both planes are in common position. On fig. 3.7. two planes in general position are shown, given by a triangle and two parallel lines. To determine the two common points of the line of intersection of the planes, we draw two auxiliary (horizontal) level planes R and T. The auxiliary plane R intersects the given planes along two horizontals h and h 1, which in their intersection define point 1, common to the planes P and Q, so as they simultaneously belong to the auxiliary secant plane R. The second plane - mediator T also intersects each of the given planes along the horizontals h 2 and h 3, which are parallel to the first two horizontals. At the intersection of the contour lines, we get the second common point of 2 given planes. Combining on the diagram (Fig. 3.8, b) the projections of the same name of these points, we obtain the projections of the line of intersection of the planes.

On fig. 3.8 shows two planes defined by traces. The common points of the planes are the intersection points of M and N traces of the same name. Connecting the projections of the same name of these points with a straight line, I received the projections of the line of intersection of the planes.

If the intersection points of the traces of the same name are outside the drawing field (see example 5), and also in cases where the planes are specified not by traces, but by other geometric elements, then to determine the line of intersection of the planes, you should use auxiliary level planes- horizontal or frontal. It should be noted that when constructing the line of intersection of the planes specified by traces, the role of auxiliary secant planes is performed by the projection planes P 1 and P 2 .

On fig. 3.9 shows the case of the intersection of two planes, when the direction of the intersection line is known, since the plane P is the level plane (P||P 1). Therefore, it is sufficient to have only one point of intersection of traces and then draw a straight line through this point, based on the position of the planes and their traces. In our case, the line of intersection is the common horizontal of the NA planes P and T.

A straight line in space can be defined as a line of intersection of two non-parallel planes and, that is, as a set of points that satisfy a system of two linear equations

(V.5)

The converse statement is also true: a system of two independent linear equations of the form (V.5) defines a straight line as a line of intersection of planes (if they are not parallel). The equations of system (V.5) are called general equation straight in space
.

ExampleV.12 . Compose the canonical equation of the straight line given by the general equations of the planes

Solution. To write the canonical equation of a line or, which is the same, the equation of a line passing through two given points, you need to find the coordinates of any two points on the line. They can be the points of intersection of a straight line with any two coordinate planes, for example Oyz and Oxz.

Point of intersection of a line with a plane Oyz has an abscissa
. Therefore, assuming in this system of equations
, we get a system with two variables:

Her decision
,
together with
defines a point
desired straight line. Assuming in this system of equations
, we get the system

whose solution
,
together with
defines a point
intersection of a line with a plane Oxz.

Now we write the equations of a straight line passing through the points
and
:
or
, where
will be the direction vector of this straight line.

ExampleV.13. The straight line is given by the canonical equation
. Write the general equation for this line.

Solution. The canonical equation of a straight line can be written as a system of two independent equations:



We have obtained the general equation of a straight line, which is now given by the intersection of two planes, one of which
parallel to axis Oz (
), and the other
– axes OU (
).

This line can be represented as a line of intersection of two other planes by writing its canonical equation as another pair of independent equations:



Comment . The same straight line can be given by different systems of two linear equations (that is, by the intersection of different planes, since countless planes can be drawn through one straight line), as well as by different canonical equations (depending on the choice of a point on the line and its direction vector) .

A non-zero vector parallel to a straight line, we will call it guide vector .

Let in three-dimensional space given straight line l passing through the point
, and its direction vector
.

Any vector
, where
, lying on a straight line, is collinear with the vector , so their coordinates are proportional, that is

. (V.6)

This equation is called the canonical equation of the line. In the particular case when ﻉ is a plane, we obtain the equation of a straight line on a plane

. (V.7)

ExampleV.14. Find the equation of a straight line passing through two points
,
.

,

where
,
,
.

It is convenient to write equation (V.6) in parametric form. Since the coordinates of the direction vectors of the parallel lines are proportional, then, assuming

,

where t - parameter,
.

Distance from point to line

Consider a two-dimensional Euclidean space ﻉ with a Cartesian coordinate system. Let the point
ﻉ and lﻉ. Find the distance from this point to the line. Let's put
, and a straight line l is given by the equation
(Fig. V.8).

Distance
, vector
, where
is the normal line vector l,
and are collinear, so their coordinates are proportional, i.e.
, Consequently,
,
.

From here
or multiplying these equations by A and B respectively, and adding them together, we find
, hence

.

(V.8)

defines the distance from a point
to straight
.

ExampleV.15. Find the equation of a straight line passing through a point
perpendicular to the line l:
and find the distance from
to straight l.

From fig. V.8 we have
, and the normal vector is a straight line l
. From the perpendicularity condition, we have

Because
, then

. (V.9)

This is the equation of the line passing through the point
, perpendicular to the line
.

Let we have the equation of the straight line (V.9) passing through the point
, perpendicular to the line l:
. Find the distance from the point
to straight l, using formula (V.8).

To find the desired distance, it is enough to find the equation of a straight line passing through two points
and point
lying on the line at the base of the perpendicular. Let
, then

Because
, and the vector
, then

. (V.11)

Since the point
lies on a straight line l, then we have another equality
or

Let us bring the system to a form convenient for applying Cramer's method

Its solution looks like

,

. (V.12)

Substituting (V.12) into (V.10), we obtain the original distance.

ExampleV.16. A point is given in two-dimensional space
and direct
. Find distance from point
to a straight line; write the equation of a line passing through a point
perpendicular to a given line and find the distance from the point
to the base of the perpendicular to the original line.

By formula (V.8) we have

The equation of a straight line containing a perpendicular can be found as a straight line passing through two points
and
, using formula (V.11). Because
, then, taking into account that
, a
, we have

.

To find coordinates
we have a system taking into account the fact that the point
lies on the original line

Consequently,
,
, from here.

Consider a three-dimensional Euclidean space ﻉ. Let the point
ﻉ and plane ﻉ. Find the distance from this point
to the plane  given by the equation (Fig. V.9).

Similarly to the two-dimensional space, we have
and vector
ah, from here

. (V.13)

We write the equation of a straight line containing a perpendicular to the plane  as the equation of a straight line passing through two points
and
lying in the plane :

. (V.14)

To find the coordinates of a point
to any two equalities of formula (V.14) we add the equation

Solving the system of three equations (V.14), (V.15), we find ,,- point coordinates
. Then the perpendicular equation can be written as

.

To find the distance from a point
to the plane instead of formula (V.13) we use

According to the given coordinates of points A, B, C, D, E, F (Table 2), construct horizontal and frontal projections of triangles ∆ABC and ∆DEF, find the line of their intersection and determine the visibility of triangle elements.

2.2. An example of completing task number 2

The second task is a set of tasks on the topics:

1. Orthographic projection, Monge plot, point, line, plane: by known coordinates of six points A, B, C, D, E, F build horizontal and frontal projections of 2 planes given by ∆ ABC and ∆ DEF;

2. Planes of general and particular position, intersection of a line and a plane, intersection of planes, competing points: draw a line of intersection of given planes and determine the visibility of their elements.

Construct horizontal and frontal projections of given planes ∆ ABC and ∆ DEF(Figure 2.1).

To build the desired line of intersection of the given planes, you must:

1. Select one of the sides of the triangle and construct the point of intersection of this side with the plane of the other triangle: in Figure 2.1, a point is plotted M line intersection EF with plane ∆ ABC; for this direct EF enclosed in an auxiliary horizontally projecting plane δ;

2. Build a frontal projection 1 2 2 2 lines of intersection of the plane δ with the plane ∆ ABC;

3. Find the frontal projection M 2 search points M at the intersection frontal projection 1 2 2 2 with frontal projection E 2 F 2 straight EF;

4. Find the horizontal projection M 1 point M with the help of a projection communication line;

5. Construct the second point in the same way N belonging to the required line of intersection of the given planes: enclose a straight line in the front-projecting plane β sun; find line of intersection 34 plane with the plane ∆ DEF; at the intersection of the line 34 and direct sun find a point N;

6. Determine with the help of competing points, for each plane separately, the visible sections of the triangles.

Figure 2.1 - Construction of a line of intersection of two planes defined by triangles

Figure 2.2 - Example of task 2

Video example of completing task No. 2

2.3. Job options 2

Table 2 – Values ​​of point coordinates