With the help of this online calculator you can find the line of intersection of the planes. A detailed solution with explanations is given. To find the equation of the line of intersection of the planes, enter the coefficients in the equations of the planes and click on the "Solve" button. See the theoretical part and numerical examples below.

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Line of intersection of planes - theory, examples and solutions

Two planes in space may be parallel, may coincide or intersect. In this article, we will determine the relative position of two planes, and if these planes intersect, we will derive the equation for the line of intersection of the planes.

Let a Cartesian rectangular coordinate system be given Oxyz and let the planes be given in this coordinate system α 1 and α 2:

Since the vectors n 1 and n 2 are collinear, then there is a number λ ≠0, which satisfies the equality n 1 =λ n 2 , i.e. A 1 =λ A 2 , B 1 =λ B 2 , C 1 =λ C 2 .

Multiplying equation (2) by λ , we get:

If the equality D 1 =λ D 2 , then the plane α 1 and α 2 match if D 1 ≠λ D 2 then plane α 1 and α 2 are parallel, that is, they do not intersect.

2. Normal vectors n 1 and n 2 planes α 1 and α 2 are not collinear (Fig.2).

If the vectors n 1 and n 2 are not collinear, then we solve the system of linear equations (1) and (2). To do this, we translate the free terms to the right side of the equations and compose the corresponding matrix equation:

Where x 0 , y 0 , z 0 , m, p, l real numbers, and t− variable.

Equality (5) can be written in the following form:

Example 1. Find the line of intersection of the planes α 1 and α 2:

α 1: x+2y+z+54=0.

(7)

Let us solve the system of linear equations (9) with respect to x, y, z. To solve the system, we construct an augmented matrix:

Second phase. Reverse Gauss.

Exclude the elements of the 2nd column of the matrix above the element a 22. To do this, add row 1 with row 2, multiplied by −2/5:

We get a solution:

Obtained the equation of the line of intersection of the planes α 1 and α 2 in parametric form. We write it in canonical form.

Answer. Equation of line of intersection of planes α 1 and α 2 looks like:

(15)

α 1 has a normal vector n 1 ={A 1 , B 1 , C 1 )=(1, 2, 7). Plane α 2 has a normal vector n 2 ={A 2 , B 2 , C 2 }={2, 4, 14}.

n 1 and n 2 are collinear ( n 1 can be obtained by multiplying n 2 by the number 1/2), then the plane α 1 and α 2 are parallel or coincide.

α 2 multiplied by the number 1/2:

(18)

Solution. Let us first determine the relative position of these planes. Plane α 1 has a normal vector n 1 ={A 1 , B 1 , C 1 )=(5, −2, 3). Plane α 2 has a normal vector n 2 ={A 2 , B 2 , C 2 }={15, −6, 9}.

Since the direction vectors n 1 and n 2 are collinear ( n 1 can be obtained by multiplying n 2 by the number 1/3), then the plane α 1 and α 2 are parallel or coincide.

Multiplying an equation by a non-zero number does not change the equation. Let's transform the equation of the plane α 2 multiplied by the number 1/3:

(19)

Since the normal vectors of equations (17) and (19) coincide and the free terms are equal, the planes α 1 and α 2 match.

The problem of the intersection of planes, due to its importance, is called "positional problem No. 2" by a number of authors.

From stereometry it is known that the line of intersection of two planes is a straight line. In the previous preliminary problems, where we dealt with particular cases of intersection of planes, we proceeded from this definition.

As is known, in order to construct one or another line, in the simplest case, it is required to find two points belonging to this line. In the case of specifying a plane by traces, these two points are the points of intersection of traces of the same name of intersecting planes.

Examples for independent work

Exercise 5.1

Construct lines of intersection of planes given by traces (Fig. 72):

• a) horizontally projecting I and frontally projecting A;
• b) horizontally projecting Z and a plane in general position Q;
• c) two planes in general position I and 0.

Rice. 72

On fig. 73 shows the answers to this exercise.

For cases where planes are defined by local flat figures, it is appropriate to use at least two different solutions.

Rice. 73

The first solution is use of a three-stage algorithm for finding the meeting point of a line in general position with a plane in general position. To find the line of intersection of two triangles, one of the triangles is left unchanged, and the second is mentally divided into separate segments, representing them as straight lines in general position. First, find the point of intersection of one of the straight lines in general position with the plane of the triangle. Then they find one more missing point belonging to the desired line. This is done in a similar way, repeating the entire sequence of actions described.

Exercise 5.2

According to the given coordinates of the vertices of two triangles LAN And DEK construct a diagram of the latter and find the line of their intersection. Specify the visibility of the elements of both triangles on the diagram: A(0, 9, 2); ?(10, 1, 16); C (23, 14, 9); D(3, 17, 18); ?(22, 11, 17); ?(12.0, 2). To find the lines of intersection of triangles, it is recommended to first find the meeting point of the line KD with a triangle ABC, and then the meeting point of the straight line SW with a triangle EDK.

The general view of the resulting diagram is shown in fig. 74.

The second solution is use of two auxiliary cutting planes of the level.

The given intersecting flat figures should be crossed twice by auxiliary level planes (similar or opposite - it doesn't matter), for example, by two horizontal level planes.

It is easy to understand that a one-time dissection allows you to find two intersecting lines hl And AND 2 , giving one point A, belonging to the desired line of intersection (Fig. 75). Drawing another similar auxiliary plane at some distance

Rice. 74

Rice. 75

from the first, get a similar construction and one more point. By connecting the projections of the same name of the two obtained points, they find the desired line of intersection of the two planes.

Exercise 5.3

According to the given coordinates of the points of two triangular figures, construct a diagram of the latter, on which to construct, using auxiliary planes, the line of intersection of the triangles. Specify the visibility of the elements of both triangles on the diagram:

to ABC. A(16, 5, 17); Me (10, 19,

A DEF:D (24, 12, 14); ? (4, 18,

The general view of the solved problem is shown in fig. 76.

Exercise 5.4

To consolidate the skills of finding the line of intersection of two planes, a problem is given, the solution of which is given in the dynamics of constructions in accordance with the steps of the algorithm.

Find the line of intersection of two planes of common position R IS jq

lines defined by two triangles ABC And DEF, and determine the visibility of their interpenetration (Fig. 77).

The solution of the example is reduced to finding the points of intersection of the sides (straight lines) A ABC with a generic plane given by A DEF. The algorithm for solving this example is known.

We conclude the side (straight) AC LAN into the auxiliary frontally projecting plane t _1_ P 2 (Fig. 78).

The front trace of this auxiliary plane intersects the projections of the sides D 2 E 2 GLE 2 - 1 2 and D 2 F 2 pt 2 = 2 2 at points 1 2 and 2 2 . Projection communication lines allow you to determine the line of intersection on the horizontal projection plane (1 !~2 2) = n A D X E X F ( . Then the point K 1 and its projection K 2 determine the point of intersection of the line AC with A DEF.

We repeat the algorithm for finding the point of intersection of side A ABC straight sun with ADEF. We conclude the aircraft in an auxiliary frontally projecting plane p _L P 2 (Fig. 79).

We find the projections of points 3 and 4 and on the horizontal plane of the projections we determine the projection of the point of intersection of the line In 1 C [ with the line of intersection (3,-4,):

The projection line of communication allows you to find its frontal projection point M 2 .

We connect the found points Ki Mi find the line of intersection of two planes in general position A ABC nA DEF= AF (Fig. 80).

Side Visibility AABC relatively ADEF determined by competing points. First, we determine the visibility of geometric shapes on the plane of projections П 2 . To do this, through competing points 5 and 6 (5 2 = 6 2) we draw a projection line of communication perpendicular to the axis of projections x n(Fig. 81).

On horizontal projections 5 U And 6 { points 5 and 6, in which the line of the projection connection, respectively, intersects the intersecting straight lines AC 4 D.F., it turns out that point 6 is more distant from the projection plane P 2 than point 5. Therefore, point 6 and the line D.F., to which it belongs are visible relative to the plane of projections P 2 . It follows from this that the segment (K 2 -6 2) will be invisible. Similarly, we determine the visibility of sides A LAN and A DEF - Sun And D.F., those. the segment (W 2 -8 2) will be invisible.

Visibility AABC And ADEF relative to the plane of projections П j, is established similarly. To determine the visibility of intersecting lines AC * DF And BC ±DF relative to the plane of projections P] through competing points 9 1 = 10 1 and 11 1 = 12 1 we draw projection lines of communication perpendicularly x p. Based on the frontal projections of these competing points, we establish that the projections of points 10 2 and 12 2 are more distant from the projection plane P ( . Consequently, the segments (A^-YUD and (M g 2 1) will be invisible. Hence the visibility AABC And ADEF clearly shown in Fig. 82.

The task needs find the line of intersection of two planes and determine the actual size of one of them the method of plane-parallel movement.

To solve such a classical problem in descriptive geometry, you need to know the following theoretical material:

- drawing projections of points in space on a complex drawing according to given coordinates;

- methods for specifying a plane on a complex drawing, a plane of general and particular position;

- the main lines of the plane;

- determination of the point of intersection of a straight line with a plane (finding "meeting points");

- the method of plane-parallel movement to determine the natural size of a flat figure;

— definition of visibility on the drawing of straight lines and planes with the help of competing points.

Procedure for solving the Problem

1. According to the Assignment by point coordinates option, we put two planes on the complex drawing, specified in the form of triangles ABC(A’, B’, C’; A, B, C) and DKE(D', K', E'; D, K, E) ( fig.1.1).

Fig.1.1

2 . To find the line of intersection, we use projection plane method. Its essence is that one side (line) of the first plane (triangle) is taken and lies in the projecting plane. The point of intersection of this line with the plane of the second triangle is determined. Repeating this task again, but for the line of the second triangle and the plane of the first triangle, we determine the second point of intersection. Since the obtained points simultaneously belong to both planes, they must be on the line of intersection of these planes. By connecting these points with a straight line, we will have the desired line of intersection of the planes.

3. The problem is solved as follows:

A) enclosing in a projection plane F(F') side AB(A’ B’) of the first triangle in the frontal projection plane V. We mark the points of intersection of the projecting plane with the sides DK And DE second triangle, getting points 1(1') and 2(2'). We transfer them along the communication lines to the horizontal plane of projections H on the corresponding sides of the triangle, point 1 (1) on the side DE and dot 2(2) on the side DK.

Fig.1.2

b) by connecting the projections of points 1 and 2, we will have the projection of the projecting plane F. Then the point of intersection of the line AB with the plane of the triangle DKE is determined (according to the rule) together with the intersection of the projection of the projecting plane 1-2 and the projection of the same name AB. Thus, we got a horizontal projection of the first point of intersection of the planes - M, along which we determine (project along communication lines) its frontal projection - M’ on a straight line A’ B’ (fig.1.2.a);

V) we find the second point in the same way. We conclude in the projecting plane G(G) side of the second triangle DK(DK) . We mark the points of intersection of the projecting plane with the sides of the first triangle ACAndBC in a horizontal projection, getting projections of points 3 and 4. We project them onto the corresponding sides in the frontal plane, we get 3’ and 4'. Connecting them with a straight line, we have the projection of the projecting plane. Then the second point of intersection of the planes will be at the intersection of the line 3’-4’ with the side of a triangle D’ K’ , which was enclosed in a projecting plane. Thus, we got the frontal projection of the second intersection point - N’ , along the communication line we find the horizontal projection - N (fig.1.2.b).

G) by connecting the points MN(MN) And (M’ N’) on the horizontal and frontal planes, we have the desired line of intersection of the given planes.

4. With the help of competing points, we determine the visibility of the planes. Take a pair of competing points, for example, 1’=5’ in frontal projection. We project them onto the corresponding sides in the horizontal plane, we get 1 and 5. We see that the point 1 lying on the side DE has a large coordinate to the axis x than dot 5 lying on the side AIN. Therefore, according to the rule of greater coordinate, the point 1 and the side of the triangle D'E’ in the frontal plane will be visible. Thus, the visibility of each side of the triangle in the horizontal and frontal planes is determined. Visible lines in the drawings are drawn with a solid contour line, and non-visible lines are drawn with a dashed line. Recall that at the points of intersection of the planes ( M— N AndM’- N’ ) will change visibility.

Fig.1.3

RFig.1.4 .

The plot additionally shows the definition of visibility in the horizontal plane using competing points 3 And 6 on straight lines DK And AB.

5. Using the method of plane-parallel displacement, we determine the actual size of the plane of the triangle ABC, For what:

A) in the specified plane through a point C(C) conduct a frontal C— F(WITH-FAndC’- F’) ;

b) on the free field of the drawing in a horizontal projection, we take (mark) an arbitrary point From 1, assuming that this is one of the vertices of the triangle (specifically, the vertex C). From it we restore the perpendicular to the frontal plane (through x-axis);

Fig.1.5

V) by plane-parallel movement we translate the horizontal projection of the triangle ABC, to a new position A 1 B 1 C 1 in such a way that in the frontal projection it takes a projecting position (transformed into a straight line). To do this: on the perpendicular from the point From 1, postpone the frontal projection of the horizontal C 1 — F 1 (length lCF) we get a point F 1 . A solution of a compass from a point F1 size F-A we make an arc serif, and from a point C 1 - notch size CA, then at the intersection of arc lines we get a point A 1 (second vertex of the triangle);

- similarly we get a point B 1 (from point C 1 make a notch with the size C— B(57mm), and from the point F 1 magnitude F— B(90mm). Note that with the correct solution, three points A 1 F’ 1 And B’ 1 must lie on one straight line (the side of the triangle A 1 — B 1 ) the other two sides WITH 1 — A 1 And C 1 — B 1 are obtained by connecting their vertices;

G) it follows from the rotation method that when moving or rotating a point in some projection plane - on the conjugate plane, the projection of this point should move in a straight line, in our particular case, along a straight parallel axis X. Then we draw from the points A’ B’ C’ From the frontal projection, these are straight lines (they are called the planes of rotation of the points), and from the frontal projections of the displaced points A 1 IN 1C 1 restore perpendiculars (connection lines) ( fig.1.6).

Fig.1.6

The intersection of these lines with the corresponding perpendiculars gives new positions of the frontal projection of the triangle ABC, specifically A’ 1 IN 1C’ 1 which should become a projecting (straight line) since the horizontal h 1 we drew perpendicular to the frontal projection plane ( fig.1.6);

5) then, to obtain the natural size of the triangle, it is enough to expand its frontal projection to parallelism with the horizontal plane. The reversal is carried out using a compass through a point A' 1, considering it as a center of rotation, we put a triangle A’ 1 IN 1C’ 1 parallel to axis X, we get A’ 2 AT 2C’ 2 . As mentioned above, when the point rotates, on the conjugate (now horizontal) projection, they move along straight lines parallel to the axis X. Omitting perpendiculars (link lines) from frontal projections of points A’ 2 AT 2C’ 2 crossing them with the corresponding lines we find the horizontal projection of the triangle ABC (A 2 AT 2C 2 ) real size ( fig.1.7).

Rice. 1.7

I have all ready-made solutions to problems with such coordinates, you can buy

Price 55 rubles, drawings on descriptive geometry from Frolov's book, you can easily download immediately after payment or I will send you an email. They are in a ZIP archive in various formats:
*.jpg – the usual color drawing of the drawing on a scale of 1 to 1 in a good resolution of 300 dpi;
*.cdw – format of the program Compass 12 and higher or version LT;
*.dwg and .dxf — AUTOCAD, nanoCAD program format;

Section: Descriptive geometry /

Canonical equations of the straight line

Formulation of the problem. Find the canonical equations of a straight line defined as a line of intersection of two planes (general equations)

Solution plan. Canonical equations of a straight line with a direction vector passing through this point , have the form

. (1)

Therefore, in order to write the canonical equations of a straight line, it is necessary to find its directing vector and some point on the straight line.

1. Since the line belongs to both planes simultaneously, its direction vector is orthogonal to the normal vectors of both planes, i.e. according to the definition of a vector product, we have

. (2)

2. Choose some point on the line. Since the directing vector of the line is not parallel to at least one of the coordinate planes, the line intersects this coordinate plane. Therefore, as a point on a line, the point of its intersection with this coordinate plane can be taken.

3. We substitute the found coordinates of the directing vector and point into the canonical equations of the straight line (1).

Comment. If the vector product (2) is equal to zero, then the planes do not intersect (parallel) and it is not possible to write down the canonical equations of the straight line.

Task 12. Write the canonical equations of the line.

Canonical equations of a straight line:

,

Where are the coordinates of any point on the line, is its direction vector.

Find any point on the line . Let then

Hence, are the coordinates of a point belonging to the line.

The canonical equations of a straight line in space are equations that define a straight line passing through a given point collinearly to a direction vector.

Let a point and a direction vector be given. An arbitrary point lies on a line l only if the vectors and are collinear, i.e., they satisfy the condition:

.

The above equations are the canonical equations of the line.

Numbers m , n And p are projections of the direction vector onto the coordinate axes. Since the vector is non-zero, then all numbers m , n And p cannot be zero at the same time. But one or two of them may be zero. In analytical geometry, for example, the following notation is allowed:

,

which means that the projections of the vector on the axes Oy And Oz are equal to zero. Therefore, both the vector and the straight line given by the canonical equations are perpendicular to the axes Oy And Oz, i.e. planes yOz .

Example 1 Compose equations of a straight line in space perpendicular to a plane and passing through the point of intersection of this plane with the axis Oz .

Solution. Find the point of intersection of the given plane with the axis Oz. Since any point on the axis Oz, has coordinates , then, assuming in the given equation of the plane x=y= 0 , we get 4 z- 8 = 0 or z= 2 . Therefore, the point of intersection of the given plane with the axis Oz has coordinates (0; 0; 2) . Since the desired line is perpendicular to the plane, it is parallel to its normal vector. Therefore, the normal vector can serve as the directing vector of the straight line given plane.

Now we write the desired equations of the straight line passing through the point A= (0; 0; 2) in the direction of the vector :

Equations of a straight line passing through two given points

A straight line can be defined by two points lying on it And In this case, the directing vector of the straight line can be the vector . Then the canonical equations of the line take the form

.

The above equations define a straight line passing through two given points.

Example 2 Write the equation of a straight line in space passing through the points and .

Solution. We write the desired equations of the straight line in the form given above in the theoretical reference:

.

Since , then the desired line is perpendicular to the axis Oy .

Straight as a line of intersection of planes

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

The equations of the system are also called the general equations of a straight line in space.

Example 3 Compose canonical equations of a straight line in the space given by general equations

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

Point of intersection of a line with a plane yOz has an abscissa x= 0 . Therefore, assuming in this system of equations x= 0 , we get a system with two variables:

Her decision y = 2 , z= 6 together with x= 0 defines a point A(0; 2; 6) of the desired line. Assuming then in the given system of equations y= 0 , we get the system

Her decision x = -2 , z= 0 together with y= 0 defines a point B(-2; 0; 0) intersection of a line with a plane xOz .

Now we write the equations of a straight line passing through the points A(0; 2; 6) and B (-2; 0; 0) :

,

or after dividing the denominators by -2:

,