The point of intersection of a straight line with the coordinate plane xoz. The intersection of a straight line with a plane. Determination of line visibility. Finding the coordinates of the intersection point of a line and a plane

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 the 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 given planeδ - 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 line 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
Build complex drawing triangle ABC given the 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 (

Constructing the point of intersection of a straight line with a projecting plane comes down to constructing a second projection of a point on a diagram, since one projection of a point always lies on the trace of the projecting plane, because everything that is in the projecting plane is projected onto one of the traces of the plane. In Fig. 224,a shows the construction of the point of intersection of straight line EF with the frontally projecting plane triangle ABC(perpendicular to plane V) On plane V, triangle ABC is projected onto segment a"c" of a straight line, and point k" will also lie on this straight line and be located at the point of intersection of e"f" with a"c". A horizontal projection is constructed using projection connection lines. The visibility of the straight line relative to the plane of the triangle ABC is determined by relative position projections of triangle ABC and straight line EF on plane V. Direction of view in Fig. 224, and indicated by an arrow. That section of the straight line, the frontal projection of which is above the projection of the triangle, will be visible. To the left of point k" the projection of the straight line is above the projection of the triangle, therefore, on the plane H this section is visible.

In Fig. 224, b straight line EF intersects the horizontal plane P. The frontal projection k" of point K - the point of intersection of the straight line EF with the plane P - will be located at the point of intersection of the projection e"f" with the trace of the plane Pv, since the horizontal plane is a front-projecting plane. The horizontal projection k of point K is found using the projection link line.

Constructing the line of intersection of two planes comes down to finding two points common to these two planes. To construct an intersection line, this is enough, since the intersection line is a straight line, and a straight line is defined by two points. When a projecting plane intersects a generic plane, one of the projections of the intersection line coincides with the trace of the plane located in the projection plane to which the projecting plane is perpendicular. In Fig. 225, and the frontal projection m"n" of the intersection line MN coincides with the trace Pv of the frontally projecting plane P, and in Fig. 225, b, the horizontal projection kl coincides with the trace of the horizontally projecting plane R. Other projections of the intersection line are constructed using projection connection lines.

Constructing the point of intersection of a line and a plane general position (Fig. 226, a) is performed using an auxiliary projection plane R, which is drawn through this straight line EF. The intersection line 12 of the auxiliary plane R with the given plane of the triangle ABC is constructed, two straight lines are obtained in the plane R: EF - the given straight line and 12 - the constructed intersection line, which intersect at point K.

Finding the projections of point K is shown in Fig. 226, b. Constructions are carried out in the following sequence.

An auxiliary horizontally projecting plane R is drawn through straight line EF. Its trace R H coincides with the horizontal projection ef of straight EF.

A frontal projection 1"2" of the intersection line 12 of the R plane with the given plane of the triangle ABC is constructed using projection communication lines, since the horizontal projection of the intersection line is known. It coincides with the horizontal trace R H of the R plane.

The frontal projection k" of the desired point K is determined, which is located at the intersection of the frontal projection of this straight line with the projection 1"2" of the intersection line. The horizontal projection of the point is constructed using a projection connection line.

The visibility of a line relative to the plane of triangle ABC is determined by the method of competing points. To determine the visibility of a straight line on the frontal plane of projections (Fig. 226, b), we compare the Y coordinates of points 3 and 4, whose frontal projections coincide. The Y coordinate of point 3, lying on line BC, is less than the Y coordinate of point 4, lying on line EF. Consequently, point 4 is closer to the observer (the direction of view is indicated by the arrow) and the projection of the straight line is depicted on the plane V visible. The straight line passes in front of the triangle. To the left of point K" the straight line is closed by the plane of triangle ABC.

Visibility on the horizontal projection plane is shown by comparing the Z coordinates of points 1 and 5. Since Z 1 > Z 5, point 1 is visible. Consequently, to the right of point 1 (up to point K) the straight line EF is invisible.

To construct the line of intersection of two general planes, auxiliary cutting planes are used. This is shown in Fig. 227, a. One plane is defined by triangle ABC, the other by parallel lines EF and MN. The given planes (Fig. 227, a) are intersected by the third auxiliary plane. For ease of construction, horizontal or frontal planes are taken as auxiliary planes. In this case, the auxiliary plane R is the horizontal plane. It intersects the given planes along straight lines 12 and 34, which at the intersection give a point K, belonging to all three planes, and therefore to two given ones, i.e., lying on the line of intersection of the given planes. The second point is found using the second auxiliary plane Q. The two points K and L found determine the line of intersection of the two planes.

In Fig. 227,b the auxiliary plane R is specified by the frontal trace. The frontal projections of the intersection lines 1"2" and 3"4 of the R plane with given planes coincide with the frontal trace Rv of the R plane, since the R plane is perpendicular to the V plane, and everything that is in it (including the intersection lines) is projected onto its frontal trace Rv. Horizontal projections of these lines are constructed using projection connection lines drawn from the frontal projections of points 1", 2", 3", 4" to the intersection with the horizontal projections of the corresponding straight lines at points 1, 2, 3, 4. Constructed the horizontal projections of the lines of intersection are extended until they intersect each other at point k, which is the horizontal projection of the point K belonging to the line of intersection of the two planes.The frontal projection of this point is on the trace Rv.

To construct the second point belonging to the intersection line, draw a second auxiliary plane Q. For the convenience of construction, the plane Q is drawn through the point C parallel to the plane R. Then, to construct horizontal projections of the lines of intersection of the plane Q with the plane of the triangle ABC and with the plane defined by parallel straight lines, it is sufficient find two points: c and 5 and draw straight lines through them parallel to the previously constructed projections of the intersection lines 12 and 34, since the plane Q ║ R. Continuing these lines until they intersect with each other, we obtain a horizontal projection l of point L belonging to the line of intersection of the given planes. The frontal projection l" of point L lies on the trace Q v and is constructed using the projection connection line. By connecting the projections of the same name of points K and L, the projections of the desired intersection line are obtained.

If we take a straight line in one of the intersecting planes and construct the point of intersection of this line with another plane, then this point will belong to the line of intersection of these planes, since it belongs to both given planes. Let's construct the second point in the same way; we can find the line of intersection of two planes, since two points are enough to construct a straight line. In Fig. 228 shows such a construction of the line of intersection of two planes defined by triangles.

For this construction, take one of the sides of the triangle and construct the point of intersection of this side with the plane of the other triangle. If this fails, take the other side of the same triangle, then the third. If this does not lead to finding the desired point, construct the points of intersection of the sides of the second triangle with the first.

In Fig. 228 the point of intersection of straight line EF with the plane of triangle ABC is constructed. To do this, an auxiliary horizontally projecting plane S is drawn through the straight line EF and a frontal projection of 1" to 2" is constructed of the line of intersection of this plane with the plane of the triangle ABC. The frontal projection 1"2" of the intersection line, intersecting with the frontal projection e"f" of the straight line EF, gives the frontal projection m" of the intersection point M. The horizontal projection m of the point M is found using the projection connection line. The second point belonging to the line of intersection of the planes of the given triangles , - point N is the point of intersection of straight line BC with the plane of the triangle DEF. A frontal-projecting plane R is drawn through straight line BC, and on plane H the intersection of horizontal projections of straight line BC and intersection line 34 gives point n - the horizontal projection of the desired point. The frontal projection is constructed with using a projection connection line. Visible sections of given triangles are determined using competing points for each projection plane separately. To do this, select a point on one of the projection planes, which is a projection of two competing points. Visibility is determined from the second projections of these points by comparing their coordinates.

For example, points 5 and 6 are the intersection points of the horizontal projections bc and de. On the frontal plane of projections, the projections of these points do not coincide. By comparing their Z coordinates, they find out that point 5 covers point 6, since the Z 5 coordinate is greater than the Z 6 coordinate. Therefore, to the left of point 5 the side DE is invisible.

I determine visibility on the frontal plane of projections using competing points 4 and 7, belonging to segments DE and BC, comparing their coordinates Y 4 and Y 7 Since Y 4 >Y 7, the side DE on the plane V is visible.

It should be noted that when constructing the point of intersection of a straight line with the plane of a triangle, the intersection point may be outside the plane of the triangle. In this case, by connecting the resulting points belonging to the intersection line, only that section of it that belongs to both triangles is outlined.

REVIEW QUESTIONS

1. What coordinates of a point determine its position in the V plane?

2. What do the Y coordinate and Z coordinate of a point determine?

3. How are the projections of a segment perpendicular to the projection plane H located on the diagram? Perpendicular to the projection plane V?

4. How are the horizontal and frontal projections located on the diagram?

5. Formulate the basic thesis about whether a point belongs to a straight line.

6. How to distinguish intersecting lines from crossing lines on a diagram?

7. What points are called competing?

8. How to determine which of two points is visible if their projections on the frontal plane of projections coincide?

9. Formulate the basic proposition about the parallelism of a straight line and a plane.

10. What is the procedure for constructing the point of intersection of a line with a general plane?

11. What is the procedure for constructing the line of intersection of two general planes?

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 a straight line with a plane when intersecting form common point, which is called the point of intersection of a line and a 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 was introduced three-dimensional space. Each straight line has its own equation, and each plane corresponds to its own given equation, each point has a certain number 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 intersection point of the 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 the system is set linear equations, 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 a 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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It is known that a straight line intersects a plane if it does not belong to this plane and is not parallel to it. Following the algorithm below, we find the point of intersection of the line a with a generic plane α defined by the traces h 0α , f 0α .

Algorithm

Via direct a we draw an auxiliary frontally projecting plane γ. The figure shows its traces h 0γ, f 0γ.
We construct projections of straight line AB along which planes α and γ intersect. In this problem, point B" = h 0α ∩ h 0γ, A"" = f 0α ∩ f 0γ. Points A" and B"" lie on the x-axis, their position is determined by the communication lines.
Direct a and AB intersect at the desired point K. Its horizontal projection K" = a" ∩ A"B". The frontal projection K"" lies on the straight line a"".

The solution algorithm will remain the same if pl. α will be given by parallel, crossing lines, a section of a figure, or other possible means.

Visibility of line a relative to plane α. Competing Points Method

Let us mark the frontal-competing points A and C in the drawing (Fig. below). We will assume that point A belongs to the area. α, and C lies on the line a. The frontal projections A"" and C"" coincide, but at the same time points A and C are removed from the plane of projections P 2 at different distances.
Let's find the horizontal projections A" and C". As can be seen in the figure, point C" is removed from plane P 2 at a greater distance than point A", which belongs to the square. α. Consequently, a section of straight line a"", located to the left of point K"", will be visible. Section a"" to the right of K"" is invisible. We mark it with a dashed line.
Let us mark horizontally competing points D and E in the drawing. We will assume that point D belongs to the square. α, and E lies on the line a. Horizontal projections D" and E" coincide, but at the same time points D and E are removed from the plane P 1 at different distances.
Let us determine the position of the frontal projections D"" and E"". As can be seen in the figure, point D"", located in the square. α, is removed from the plane P 1 at a greater distance than point E "", belonging to straight line a. Consequently, section a" located to the right of point K" will be invisible. We mark it with a dashed line. Section a" to the left of K" is visible.

Given a straight line: (1) and a plane: Ax + By + Cz + D = 0 (2).

Let's find the coordinates of the point of intersection of the line and the plane. If straight line (1) and plane (2) intersect, then the coordinates of the intersection point satisfy equations (1) and (2):

, .

Substituting the found value of t into (1), we obtain the coordinates of the intersection point.

1) If Am + Bn + Cp = 0, and Ax 0 + By 0 + Cz 0 + D ≠ 0, then t does not exist, i.e. a straight line and a plane do not have a single common point. They are parallel.

2) Am + Bn + Cp = 0 and Ax 0 + By 0 + Cz 0 + D = 0. In this case, t can take any values and , i.e. the straight line is parallel to the plane and has a common point with it, i.e. it lies in a plane.

Example 1. Find the point of intersection of a line with plane 3x – 3y + 2z – 5 = 0.

3(2t – 1) – 3(4t + 3) + 2 3t – 5 = 0 => -17=0, which is impossible for any t, i.e. a straight line and a plane do not intersect.

Example 2. Find the point of intersection of a line and planes: x + 2y – 4z + 1 = 0.

8t + 13 + 2(2t + 1) – 4(3t + 4) + 1 = 0, 0 + 0 = 0. This is true for any value of t, i.e. the straight line lies in the plane.

Example 3. Find the point of intersection of a line and plane 3x – y + 2z – 5 = 0.

3(5t + 7) – t – 4 + 2(4t + 5) – 5 = 0, 22t + 22 = 0, t = -1, x = 5(-1) + 7 = 2, y = -1 + 4 = 3, z = 4(-1) + 5 = 1, M(2, 3, 1) – the point of intersection of the line and the plane.

The angle between a straight line and a plane. Conditions for parallelism and perpendicularity of a straight line and a plane.

The angle between a straight line and a plane is called sharp corner q between a straight line and its projection onto a plane.

Let a straight line and a plane be given:

And .

Let the straight line intersect the plane and form an angle μ () with it. Then b = 90 0 – c or b = 90 0 + c is the angle between normal vector plane and the directing vector of the straight line. But . Means

(3).

a) If L P, then - the condition of perpendicularity of a straight line and a plane.

b) If L||P, then is the condition for parallelism of the line and the plane.

c) If the line is L||P and at the same time the point M0(x0, y0, z0) P, then the line lies in this plane. Analytically:

- conditions for belonging to a straight line and a plane.

Example. Given a straight line and point M 0 (1, 0, –2). Through point M 0 draw a plane perpendicular to this line. We look for the equation of the desired plane in the form: A(x – 1) + B(y – 0) + C(z + 2) = 0. B in this case , ,

5(x – 1) – 5y + 5(z + 2) = 0, - x – y + z + 3 = 0.

A bunch of planes.

A beam of planes is the set of all planes passing through a given straight line—the axis of the beam.

To define a bundle of planes, it is enough to specify its axis. Let the equation of this line be given in general form:

To compose a beam equation means to compose an equation from which, under an additional condition, one can obtain the equation of any plane of the beam, except for b.m. one. Let's multiply equation II by l and add it to equation I:

A 1 x + B 1 y + C 1 z + D 1 + l(A 2 x + B 2 y + C 2 z + D 2) = 0 (1) or

(A 1 + lA 2)x + (B 1 + lB 2)y + (C 1 + lC 2)z + (D 1 + lD 2) = 0 (2).

l – parameter – a number that can take real values. For any chosen value of l, equations (1) and (2) are linear, i.e. these are the equations of a certain plane.

1. Let us show that this plane passes through the beam axis L. Take an arbitrary point M 0 (x 0, y 0, z 0) L. Consequently, M 0 P 1 and M 0 P 2. Means:

3x – y + 2z + 9 + 17x + 17z – 51 = 0; 20x – y + 19z – 42 = 0.

Example 3 (E). Write an equation for a plane passing through a line perpendicular to the plane x – 2y + z + 5 = 0. ; 3x – 2y + z – 3 + l(x – 2z) = 0; (3 + l)x – 2y + (1 – 2 l)z – 3 = 0; ; ; l = 8; 11x – 2y – 15z – 3 = 0.