Scientific electronic library. Constructing the point of intersection of a plane with a line 3 points of intersection of a line with a plane

To determine the point of intersection of a line with a plane, we use the following algorithm: we enclose the line in an auxiliary plane, find the line of intersection of these two planes (given and auxiliary), and the line of intersection of the planes at the intersection with the given line will give the desired point. The last step in the construction is to determine the visibility of the line using competing points.

Example 1. The plane is defined by traces (Fig. 70)

1. To construct the point of intersection of a line l with a plane, it is necessary to draw an auxiliary plane of particular position through a straight line, for example, front-projecting β π 2, l"" f oβ , f oβ – collecting trace, h oβ x (Fig. 71).

2. Constructing an intersection line MN given and auxiliary plane M"=h oα ∩ h oβ, N""= f oβ ∩ f oα (Fig. 72).

3. Determine the intersection point TO given straight line l with the line of intersection MN. K"=M"N"∩l ", K""– at the intersection of the projection connection line drawn from K" and l".

4. Direct visibility l in the case of specifying a plane, we do not define it with traces.

Example 2. The intersection of a straight line with a projecting plane (Fig. 73).

When constructing the point of intersection of a straight line with a projecting plane, the task is simplified, because one of the projections of the desired point will lie on the collecting trace. Figure 73 shows a horizontally projecting plane. Searched point TO will simultaneously belong to the plane α and the line A.

Example 3 . The plane is defined by a flat figure (Fig. 74).

Via direct l we draw an auxiliary plane of particular position, for example, a horizontally projecting β π 1 . l"h oβ, h oβ – collecting trace, f oβ x (Fig. 75).

2. Constructing an intersection line MN given and auxiliary planes. M"=A"C"∩ hoβ M"" A""C"" and N"=B"C"∩ hoβ N"" B""C""(Fig. 76).

3. Constructing the intersection point TO given straight line l with the line of intersection MN. K""= M""N""∩l"". TO" is located at the intersection of the projection connection line drawn from K"" and M"N".

4. Determine the visibility of the line relative to Δ ABC using competing points.

Determining visibility relative to the plane π 2.Note the frontal projection 1"" coinciding with 2"" . Horizontal projection 2" note on A"C", A 1" on l". Horizontal projection 1" lies in front 2" 2"" relatively invisible π 2. Dot 1 lies on the straight line l, it is visible on π 2, therefore, frontal projection l" from 1"2"" to TO"" visible at the point TO"" visibility is reversed.

Let's determine the visibility of the line l relative to the plane π 1. Note the horizontal projection 3" , coinciding with the horizontal projection M. M"" A""C"" already marked 3""l"". Frontal projection M"" lies above the frontal projection 3"" , therefore, point M visible relative π 1. Dot 3 lies on l, therefore, from M"≡3" to TO", horizontal projection l" invisible. In horizontal projection TO" visibility is reversed. Beyond Δ ABC straight l visible everywhere.

This chapter talks about how to find the coordinates of the point of intersection of a line with a plane given the equations that define this plane. The concept of the point of intersection of a line with a plane and two ways of finding the coordinates of the point of intersection of a line with a plane will be considered.

For an in-depth study of the theory, it is necessary to begin consideration with the concept of a point, a straight line, a plane. The concept of a point and a straight line is considered both on the plane and in space. For a detailed consideration, it is necessary to turn to the topic of straight lines and planes in space.

There are several variations in the location of the line relative to the plane and space:

• a straight line lies in a plane;
• a straight line is parallel to a plane;
• a straight line intersects a plane.

If we consider the third case, we can clearly see that when a straight line and a plane intersect, they form a common point, which is called the point of intersection of the straight line and the plane. Let's look at this case using an example.

Finding the coordinates of the intersection point of a line and a plane

A rectangular coordinate system O x y z of three-dimensional space was introduced. Each straight line has its own equation, and each plane corresponds to its own given equation, each point has a certain number of real numbers - coordinates.

To understand in detail the topic of intersection coordinates, you need to know all types of straight line equations in space and plane equations. in this case, knowledge about the transition from one type of equation to another will be useful.

Consider a problem that is based on a given intersection of a line and a plane. it comes down to finding the coordinates of the intersections.

Example 1

Calculate whether point M 0 with coordinates - 2, 3, - 5 can be the point of intersection of the straight line x + 3 - 1 = y - 3 = z + 2 3 with the plane x - 2 y - z + 3 = 0.

Solution

When a point belongs to a certain line, the coordinates of the intersection point are the solution to both equations. From the definition we have that at intersection a common point is formed. To solve the problem, you need to substitute the coordinates of the point M 0 into both equations and calculate. If it is the intersection point, then both equations will correspond.

Let's imagine the coordinates of the point - 2, 3, - 5 and get:

2 + 3 - 1 = 3 - 3 = - 5 + 2 3 ⇔ - 1 = - 1 = - 1 - 2 - 2 3 - (- 5) + 3 = 0 ⇔ 0 = 0

Since we obtain the correct equalities, we conclude that point M 0 is the point of intersection of the given line with the plane.

Answer: the given point with coordinates is the intersection point.

If the coordinates of the intersection point are solutions to both equations, then they intersect.

The first method is to find the coordinates of the intersection of a line and a plane.

When a straight line a is specified with a plane α of a rectangular coordinate system, it is known that they intersect at the point M 0. First, let's look for the coordinates of a given intersection point for a given plane equation, which has the form A x + B y + C z + D = 0 with a straight line a, which is the intersection of the planes A 1 x + B 1 y + C 1 z + D 1 = 0 and A 2 x + B 2 y + C 2 z + D 2 = 0. This method of defining a line in space is discussed in the article equations of a line and equations of two intersecting planes.

The coordinates of the straight line a and the plane α we need must satisfy both equations. Thus, a system of linear equations is specified, having the form

A x + B y + C z + D = 0 A 1 x + B 1 y + C 1 z + D 1 = 0 A 2 x + B 2 y + C 2 z + D 2 = 0

Solving the system implies turning each identity into a true equality. It should be noted that with this solution we determine the coordinates of the intersection of 3 planes of the form A x + B y + C z + D = 0, A 1 x + B 1 y + C 1 z + D 1 = 0, A 2 x + B 2 y + C 2 z + D 2 = 0 . To consolidate the material, we will consider solving these problems.

Example 2

The straight line is defined by the equation of two intersecting planes x - y + 3 = 0 5 x + 2 z + 8 = 0, and intersects another one 3 x - z + 7 = 0. It is necessary to find the coordinates of the intersection point.

Solution

We obtain the necessary coordinates by compiling and solving a system that has the form x - y + 3 = 0 5 x + 2 z + 8 = 0 3 x - z + 7 = 0.

You should pay attention to the topic of solving systems of linear equations.

Let's take a system of equations of the form x - y = - 3 5 x + 2 z = - 8 3 x - z = - 7 and carry out calculations using the determinant of the main matrix of the system. We get that

∆ = 1 - 1 0 5 0 2 3 0 - 1 = 1 0 (- 1) + (- 1) 2 3 + 0 5 0 - 0 0 3 - 1 2 0 - (- 1) · 5 · (- 1) = - 11

Since the determinant of the matrix is ​​not equal to zero, the system has only one solution. To do this, we will use Cramer's method. It is considered very convenient and suitable for this occasion.

∆ x = - 3 - 1 0 - 8 0 2 - 7 0 - 1 = (- 3) 0 (- 1) + (- 1) 2 (- 7) + 0 (- 8) 0 - - 0 0 (- 7) - (- 3) 2 0 - (- 1) (- 8) (- 1) = 22 ⇒ x = ∆ x ∆ = 22 - 11 = - 2 ∆ y = 1 - 3 0 5 - 8 2 3 - 7 - 1 = 1 · (- 8) · (- 1) + (- 3) · 2 · 3 + 0 · 5 · (- 7) - - 0 · ( - 8) 3 - 1 2 (- 7) - (- 3) 5 (- 1) = - 11 ⇒ y = ∆ y ∆ = - 11 - 11 = 1 ∆ z = 1 - 1 - 3 5 0 - 8 3 0 - 7 = 1 0 (- 7) + (- 1) (- 8) 3 + (- 3) 5 0 - - (- 3) 0 3 - 1 · (- 8) · 0 - (- 1) · 5 · (- 7) = - 11 ⇒ z = ∆ z ∆ = - 11 - 11 = 1

It follows that the coordinates of the point of intersection of a given line and plane have the value (- 2, 1, 1).

Answer: (- 2 , 1 , 1) .

A system of equations of the form A x + B y + C z + D = 0 A 1 x + B 1 y + C 1 z + D 1 = 0 A 2 x + B 2 y + C 2 z + D 2 = 0 has only one solution. When line a is defined by equations such as A 1 x + B 1 y + C 1 z + D 1 = 0 A 2 x + B 2 y + C 2 z + D 2 = 0, and the plane α is given by A x + B y + C z + D = 0, then they intersect. When a straight line lies in a plane, the system produces an infinite number of solutions. If they are parallel, the equation has no solutions, since there are no common points of intersection.

Example 3

Find the intersection point of the straight line z - 1 = 0 2 x - y - 2 = 0 and the plane 2 x - y - 3 z + 1 = 0.

Solution

The given equations must be converted into the system z - 1 = 0 2 x - y - 2 = 0 2 x - y - 3 z + 1 = 0. When it has a unique solution, we will obtain the required intersection coordinates at the point. Provided that if there are no solutions, then they are parallel, or the straight line lies in the same plane.

We obtain that the main matrix of the system is A = 0 0 1 2 - 1 0 2 - 1 - 3, the extended matrix is ​​T = 0 0 1 1 2 - 1 0 2 2 - 1 - 3 - 1. We need to determine the rank of the matrix A and T using the Gaussian method:

1 = 1 ≠ 0 , 0 1 - 1 0 = 1 ≠ 0 , 0 0 1 2 - 1 0 2 - 1 - 3 = 0 , 0 1 1 - 1 0 2 - 1 - 3 - 1 = 0

Then we find that the rank of the main matrix is ​​equal to the rank of the extended one. Let us apply the Kronecker-Capelli theorem, which shows that the system has an infinite number of solutions. We obtain that the straight line z - 1 = 0 2 x - y - 2 = 0 belongs to the plane 2 x - y - 3 z + 1 = 0, which indicates the impossibility of their intersection and the presence of a common point.

Answer: there are no coordinates of the intersection point.

Example 4

Given the intersection of the straight line x + z + 1 = 0 2 x + y - 4 = 0 and the plane x + 4 y - 7 z + 2 = 0, find the coordinates of the intersection point.

Solution

It is necessary to assemble the given equations into a system of the form x + z + 1 = 0 2 x + y - 4 = 0 x + 4 y - 7 z + 2 = 0. To solve, we use the Gaussian method. With its help we will determine all available solutions in a short way. To do this, let's write

x + z + 1 = 0 2 x + y - 4 x + 4 y - 7 z + 2 = 0 ⇔ x + z = - 1 2 x + y = 4 x + 4 y - 7 z = - 2 ⇔ ⇔ x + z = - 1 y - 2 z = 6 4 y - 8 z = - 1 ⇔ x + z = - 1 y - 2 z = 6 0 = - 25

Having applied the Gauss method, it became clear that the equality is incorrect, since the system of equations has no solutions.

We conclude that the straight line x + z + 1 = 0 2 x + y - 4 = 0 with the plane x + 4 y - 7 z + 2 = 0 have no intersections. It follows that it is impossible to find the coordinates of the point, since they do not intersect.

Answer: there are no points of intersection, since the line is parallel to the plane.

When a straight line is given by a parametric or canonical equation, then from here you can find the equation of the intersecting planes that define the straight line a, and then look for the necessary coordinates of the intersection point. There is another method that is used to find the coordinates of the intersection point of a line and a plane.

The second method of finding a point begins with specifying a straight line a intersecting the plane α at the point M 0. It is necessary to find the coordinates of a given intersection point for a given plane equation A x + B y + C z + D = 0. We define straight line a by parametric equations of the form x = x 1 + a x · λ y = y 1 + a y · λ z = z 1 + a z · λ, λ ∈ R.

When substitution is made into the equation A x + B y + C z + D = 0 x = x 1 + a x · λ , y = y 1 + a y · λ , z = z 1 + a z · λ , the expression takes the form of an equation with an unknown λ. It is necessary to resolve it with respect to λ, then we obtain λ = λ 0, which corresponds to the coordinates of the point at which they intersect. The coordinates of the point are calculated from x = x 1 + a x · λ 0 y = y 1 + a y · λ 0 z = z 1 + a z · λ 0 .

This method will be discussed in more detail using the examples given below.

Example 5

Find the coordinates of the point of intersection of the line x = - 1 + 4 · λ y = 7 - 7 · λ z = 2 - 3 · λ, λ ∈ R with the plane x + 4 y + z - 2 = 0.

Solution

To solve the system, it is necessary to make a substitution. Then we get that

1 + 4 λ + 4 7 - 7 λ + 2 - 3 λ - 2 = 0 ⇔ - 27 λ + 27 = 0 ⇔ λ = 1

Let's find the coordinates of the point of intersection of the plane with the straight line using parametric equations with the value λ = 1.

x = - 1 + 4 1 y = 7 - 7 1 z = 2 - 3 1 ⇔ x = 3 y = 0 z = - 1

Answer: (3 , 0 , - 1) .

When a line of the form x = x 1 + a x · λ y = y 1 + a y · λ z = z 1 + a z · λ, λ ∈ R belongs to the plane A x + B y + C z + D = 0, then it is necessary to substitute there equation of the plane of expression x = x 1 + a x · λ, y = y 1 + a y · λ, z = z 1 + a z · λ, then we obtain an identity of this form 0 ≡ 0. If the plane and the line are parallel, we obtain an incorrect equality, since there are no points of intersection.

If a line is given by a canonical equation, having the form x - x 1 a x = y - y 1 a y = z - z 1 a z , then it is necessary to move from canonical to parametric when searching for the coordinates of the point of intersection of the line with the plane A x + B y + C z + D = 0, that is, we get x - x 1 a x = y - y 1 a y = z - z 1 a z ⇔ x = x 1 + a x · λ y = y 1 + a y · λ z = z 1 + a z · λ and We will apply the necessary method to find the coordinates of the point of intersection of a given line and a plane in space.

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If a line does not lie in a plane and is not parallel to it, it intersects the plane.
The task of determining the point of intersection of a line with a plane comes down to the following:
1) drawing an auxiliary plane ( It is recommended to choose an auxiliary plane that will give the simplest graphical solution to the problem) through this line;
2) finding the line of intersection of the auxiliary plane with the given plane;
3) determining the point of intersection of a given straight line with the line of intersection of planes, and therefore with a given plane.

Example 1. In (Fig. 250, a) the plane δ (δ 1 ) and the straight line AB (A 1 B 1 and A 2 B 2 ) are given; it is necessary to determine the point of their intersection.

In this case, there is no need to resort to an auxiliary plane, since this plane δ is horizontal - projecting. According to the property of projecting planes, the horizontal projection of the intersection point, lying in the plane δ, merges with the horizontal projection δ 1.
Therefore, the point K 1 of intersection of the horizontal projection A 1 B 1 of straight AB with the horizontal projection δ 1 is the horizontal projection of the point of intersection K; the front projection K 2 is determined by drawing a vertical communication line until it intersects with the front projection A 2 B 2.
Example 2. Figure 250b shows an example of the intersection of straight line AB with the frontally projecting plane δ.

Example 1. Given: a plane in general position a and a line in general position AB (A 1 B 1 A 2 B 2); you need to find the point of their intersection (Fig. 251, a).
We draw some auxiliary plane through straight line AB, for example horizontally - projecting plane δ (δ 1 ), as shown in (Fig. 251, b); it will intersect plane a along straight line NM (N 1 M 1, N 2 M 2), which, in turn, will intersect straight line AB (A 1 B 1 A 2 B 2) at point C (C 1 C 2), which can be seen on (Fig. 251, c). Point C is the point of intersection of line AB with plane a.

Example 2. (Fig. 252) shows an example of finding the projections of the point of intersection of straight line AB with a general plane using the horizontal h.
Example 3. Given: triangle ABC and line NM; it is necessary to determine the point of their intersection (Fig. 253, a).
Let us take the horizontal projection plane δ as an auxiliary plane, then the horizontal projection og will merge with the horizontal projection N 1 M 1 of the straight line NM and intersect the projections of the sides of the triangle at points E 1 and F 1 (Fig. 253, b). The segment E 1 F 1 will be the horizontal projection of the intersection line. Then we find the frontal projection of the intersection line: using vertical communication lines, we obtain points E 2 and F 2, draw a straight line E 2 F 2 through them, which will be the frontal projection of the intersection line.
Line E 2 F 2 intersects line N 2 M 2 at point K 2. Point K 2 will be the frontal projection of the point of intersection of straight line MN with straight line EF; the horizontal projection K 1 of this point is determined using a vertical communication line.
Point K (K 1, K 2) will be the point of intersection of this line MN with this triangle ABC, as it simultaneously belongs to them, because the straight line MN intersects in it with the line EF lying in the plane of the triangle ABC.

Exercise 1
Construct a complex drawing of triangle ABC using the given coordinates of the vertices. Find the actual size of the sides of the triangle and construct it in full size. Using the same coordinates, construct a visual image
Exercise 2
Based on the data from the frontal projection of the polygon and the horizontal projections of its two adjacent sides, complete the horizontal projection of the polygon.
Construct projections of an arbitrary triangle in the plane of the polygon. Construct a point outside the polygon, but lying in the same plane as it (

In this article we will answer the question: “How to find the coordinates of the point of intersection of a line and a plane if the equations defining the line and the plane are given”? Let's start with the concept of the point of intersection of a line and a plane. Next, we will show two ways to find the coordinates of the point of intersection of a line and a plane. To consolidate the material, consider detailed solutions to the examples.

Page navigation.

The point of intersection of a line and a plane - definition.

There are three possible options for the relative position of the straight line and the plane in space:

• a straight line lies in a plane;
• a straight line is parallel to a plane;
• a straight line intersects a plane.

We are interested in the third case. Let us recall what the phrase “a straight line and a plane intersect” means. A line and a plane are said to intersect if they have only one common point. This common point of intersecting line and plane is called the point of intersection of a line and a plane.

Let's give a graphic illustration.

Finding the coordinates of the intersection point of a line and a plane.

Let us introduce Oxyz in three-dimensional space. Now, each line corresponds to a straight line equation of some type (the article is devoted to them; types of equations of a line in space), each plane corresponds to an equation of a plane (you can read the article: types of equations of a plane), and each point corresponds to an ordered triple of numbers - the coordinates of the point. Further presentation implies knowledge of all types of equations of a line in space and all types of equations of a plane, as well as the ability to move from one type of equations to another. But don’t be alarmed, throughout the text we will provide links to the necessary theory.

Let's first analyze in detail the problem, the solution of which we can obtain based on determining the point of intersection of a straight line and a plane. This task will prepare us for finding the coordinates of the point of intersection of a line and a plane.

Example.

Is the point M 0 with coordinates the intersection point of the line and planes .

Solution.

We know that if a point belongs to a certain line, then the coordinates of the point satisfy the equations of the line. Similarly, if a point lies in a certain plane, then the coordinates of the point satisfy the equation of this plane. By definition, the intersection point of a line and a plane is a common point of the line and the plane, then the coordinates of the intersection point satisfy both the equations of the line and the equation of the plane.

Thus, to solve the problem, we should substitute the coordinates of the point M 0 into the given equations of the straight line and into the equation of the plane. If in this case all equations turn into correct equalities, then point M 0 is the point of intersection of the given line and plane, otherwise point M 0 is not the point of intersection of the line and plane.

Substitute the coordinates of the point :

All equations turned into correct equalities, therefore, the point M 0 simultaneously belongs to the straight line and planes , that is, M 0 is the intersection point of the indicated straight line and plane.

Answer:

Yes, period is the point of intersection of the line and planes .

So, the coordinates of the point of intersection of a line and a plane satisfy both the equations of the line and the equation of the plane. We will use this fact when finding the coordinates of the point of intersection of a line and a plane.

The first method is to find the coordinates of the intersection point of a line and a plane.

Let a straight line a and a plane be given in the rectangular coordinate system Oxyz, and it is known that straight a and the plane intersect at point M 0 .

The required coordinates of the point of intersection of the line a and the plane, as we have already said, satisfy both the equations of the line a and the equation of the plane, therefore, they can be found as a solution to a system of linear equations of the form . This is true, since solving a system of linear equations turns each equation of the system into an identity.

Note that with this formulation of the problem, we actually find the coordinates of the intersection point of three planes specified by the equations , and .

Let's solve an example to consolidate the material.

Example.

A straight line given by the equations of two intersecting planes as , intersects the plane . Find the coordinates of the point of intersection of the line and the plane.

Solution.

We obtain the required coordinates of the point of intersection of the line and the plane by solving a system of equations of the form . In this case, we will rely on the information in the article.

First, let's rewrite the system of equations in the form and calculate the determinant of the main matrix of the system (if necessary, refer to the article):

The determinant of the main matrix of the system is nonzero, so the system of equations has a unique solution. To find it, you can use any method. We use:

This is how we got the coordinates of the point of intersection of the line and the plane (-2, 1, 1).

Answer:

(-2, 1, 1) .

It should be noted that the system of equations has a unique solution if the line a defined by the equations , and the plane defined by the equation intersect. If straight line a lies in the plane, then the system has an infinite number of solutions. If straight line a is parallel to the plane, then the system of equations has no solutions.

Example.

Find the point of intersection of the line and planes , if possible.

Solution.

The “if possible” clause means that the line and the plane may not intersect.

. If this system of equations has a unique solution, then it will give us the desired coordinates of the point of intersection of the line and the plane. If this system has no solutions or has infinitely many solutions, then finding the coordinates of the intersection point is out of the question, since the straight line is either parallel to the plane or lies in this plane.

The main matrix of the system has the form , and the extended matrix is . Let's define A and the rank of matrix T:
. That is, the rank of the main matrix is ​​equal to the rank of the extended matrix of the system and is equal to two. Therefore, based on the Kronecker-Capelli theorem, it can be argued that the system of equations has an infinite number of solutions.

Thus, straight lies in a plane , and we cannot talk about finding the coordinates of the point of intersection of the line and the plane.

Answer:

It is impossible to find the coordinates of the intersection point of a line and a plane.

Example.

If straight intersects the plane, then find the coordinates of the point of their intersection.

Solution.

Let's create a system from the given equations . To find its solution we use . The Gauss method will allow us not only to determine whether the written system of equations has one solution, an infinite number of solutions, or does not have any solutions, but also to find solutions if they exist.

The last equation of the system after the direct passage of the Gauss method became an incorrect equality, therefore, the system of equations has no solutions. From this we conclude that the straight line and the plane do not have common points. Thus, we cannot talk about finding the coordinates of their intersection point.

Answer:

The line is parallel to the plane and they do not have an intersection point.

Note that if line a corresponds to parametric equations of a line in space or canonical equations of a line in space, then we can obtain the equations of two intersecting planes that define line a, and then find the coordinates of the point of intersection of line a and the plane in a parsed way. However, it is easier to use another method, which we now describe.

77*. Find the point of intersection of straight line AB with the plane defined by triangle CDE (Fig. 75, a).

Solution. As is known, to find the point of intersection of a straight line with a general plane, one should draw an auxiliary plane (R) through the straight line, construct a line of intersection of this plane with a given one (1-2) and find

the point of intersection (K) of the given and constructed lines. Point K is the desired point of intersection of the line with the plane (Fig. 75, b). A horizontal or frontal projection plane is usually used as an auxiliary plane.

In Fig. 75, in c, a frontally projecting plane R is drawn through the straight line AB, its trace R ϑ coincides with a "c". horizon. The trace of the plane is not needed in this problem and is therefore not shown.

We construct the line of intersection of the plane R and the plane defined by the triangle CDE (for an example of such a construction, see Problem 67). Having constructed line 1-2 (Fig. 75, c), we find the point of intersection of it with line AB - point K (k, k").

To determine the sections of line AB that will be covered by a triangle, you should use an analysis of the position of points on intersecting lines.

For example, points 1 and 3 are on the intersecting lines (respectively) ED and AB. The frontal projections of these points coincide, i.e. points 1 and 3 are equally distant from the square. N. But their distances from the square. V are different: point 3 is further from the square. V than point 1. Therefore, in relation to pl. V point 3 covers point 1 (the direction of view is indicated by the arrow S). Consequently, straight line AB passes in front of triangle CDE to point K. Starting from point K to the left, straight line AB is covered by a triangle, and therefore this section of the straight line is shown by a dashed line.

To identify an invisible area on the horizon. projections of straight line AB, consider points 4 and 5, lying respectively on straight lines AB and CD.

If we look at these points in the direction s 1, we first see point 5. Point 4 is covered by point 5. Consequently, line AB in this place is covered by triangle CDE, and the section of its projection from point k to point 4 should be shown by a dashed line. In this case, point K was inside the contour of triangle CDE.

If the relative position of the intersecting elements is different, it is possible that point K will be outside the triangle (Fig. 75, d). This means that line AB intersects the plane defined by triangle CDE outside the contour of this triangle. AB becomes invisible behind point K (to the left).

78. Find the points of intersection of straight line AB with the faces of the pyramid (Fig. 76). The faces of the pyramid should be considered as planes defined by triangles.

79. Find the points of intersection of straight line AB with the faces of the prism (Fig. 77). The faces of the prism should be considered as planes defined by parallel straight lines.

80*. Find the points of intersection of straight line AB with plane P (Fig. 78, a).

Solution. We draw the frontally projecting plane R through the straight line AB (Fig. 78, biv) (its trace R ϑ coincides with a "b") and construct the line MN of the intersection of both planes - given and drawn through AB (the construction is similar to that performed in problem 70). The required point K(k, k") of intersection of straight line AB with plane P is located at the point of intersection of MN with AB.

In this problem, the visibility of the straight section from point A to K is obvious; however, in more complex cases, the visible section of the straight line should be determined based on

analysis of the position of points. For example, taking point 1 (on line AB) and point N (on trace P ϑ). we see that point 1 is located further relative to the square. V than point N. Therefore, straight line AB to point K is visible. Beyond point K, the straight line is shown as a dashed line and is invisible. Visibility to the horizon is determined similarly. projections.

81. Find the point of intersection of straight line AB with plane P (Fig. 79).

82*. Find the point of intersection of straight line AB with plane P (Fig. 80, a).

Solution. Through straight line AB we draw a horizontally projecting plane R (the trace of R h coincides with ab) and construct the line of intersection of planes P and R,

using the points M and N of the intersection of their tracks of the same name (Fig. 80, b and c). The desired point (k", k) is located at the intersection point of MN with AB. In Fig. 80, d, point K is constructed using plot W. Since plot P is profile-projecting (Fig. 80, b).

then the profile projection k" lies at the point of intersection of the trace P ω with a"b". Knowing k", we construct k" on a"b" and k on ab. The visible sections of the straight line AB are determined in the same way as in problems 77 and 80.

83. Find the point of intersection of straight line AB with plane P (Fig. 81).

84*. Find the point of intersection of straight line AB with the plane defined by triangle CDE (Fig. 82, a).

Solution. Through straight line AB we draw (Fig. 82, b and c) square. R, parallel to the square W. It intersects the given plane along the straight line MN (points m", n", m and n lie at the intersection of the traces R ϑ and R h with the same projections of the corresponding sides

triangle CDE). Since the lines AB and MN are profile, to find the point (K) of their intersection we construct profile projections a"b" and m"n". The projection k" is located at the intersection of a"b" and m"m". Using k" we construct k" on a"b" and k on ab.

85. Find the point of intersection of straight line EF with the plane defined by the quadrilateral ABCD (Fig. 83).