Hello, friends! Today we analyze the topic of descriptive geometry - intersection of a straight line with a plane and line visibility determination.

We take the task from the collection of Bogolyubov, 1989, p. 63, var. 1. We need to build a complex drawing of a triangle ABC and a straight line MN according to the given coordinates. Find the point of meeting (intersection) of the line with the opaque plane ABC. Determine the visible sections of the line.

Intersection of a straight line with a plane

1. Based on the coordinates of points A, B and C, we build a complex drawing of a triangle and a straight line NM. We start drawing with a horizontal projection. We find the coordinates of the projection points using auxiliary lines.

2. We get such a complex drawing.

3. To determine coordinates of the point of intersection of the line and the plane let's do the following.

a) Draw an auxiliary plane P through the straight line NM, i.e. on the frontal projection we draw a trace of the plane Pv, on the horizontal plane we lower the perpendicular Pn - the horizontal trace of the plane P.

b) We find the frontal projection of the line of intersection of the trace of the plane P with the triangle ABC. This is the segment d'e'. We find the horizontal projection along the communication lines to the intersection with the sides ab (t. d) and ac (t. e) of the triangle. We connect points d and e.

c) Together the intersection of de and nm will be the horizontal projection of the desired point intersection of a straight line with a plane k.

d) We draw a communication line from k to the intersection with d'e', we get a frontal projection of the point k'.

e) along the communication lines we find the profile projection of the point k''.

Coordinates of the point of intersection of a line and a plane K found. This point is also called the meeting point of the line and the plane.

Determining line visibility

For line visibility determination use the method competing points.

With regard to our drawing, the competing points will be:

- points: d' belonging to a'b' and e' belonging to n'm' (frontally competing),

- points: g belonging to bc and h owned nm (horizontally competing),

- points: l'' belonging to b''c'' and p'' belonging to n''m'' (profile competing).

Of the two competing points, the one with the highest height will be visible. The line of sight is limited by point K.

For a pair of points d’ and e’, the visibility is defined as follows: we lower the perpendicular to the intersection with ab and nm on the horizontal projection, we find points d and f. We see that the y-coordinate for the point f is greater than that of d → the point f is visible → the direct line nm is visible on the section f'k', and on the section k'm' it is invisible.

We argue similarly for a pair of points g and h: on the frontal projection, the z-coordinate of the point h’ is greater than that of g’ → the point h’ is visible, g’ is not → the straight line nm is visible on the segment hk, but invisible on the segment kn.

And for a pair of points l''p'': on the frontal projection, the x-coordinate is greater than that of the point p', which means it covers the point l'' on the profile projection → p'' is visible, l'' is not → the line segment n' 'k'' is visible, k''m'' is invisible.

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.

Example1. The plane is set by traces (Fig. 70)

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

2. We build the line of intersection MN given and auxiliary plane М "=h оα ∩ h оβ, N""= f оβ ∩ f оα (Fig. 72).

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

4. Visibility direct l in the case of specifying the plane by traces, we do not determine.

Example 2. 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. Search point To will simultaneously belong to the plane α and the line a.

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

Through a straight line l we draw an auxiliary plane of particular position, for example, horizontally projecting β π 1 . l" h оβ , h оβ is a collecting trace, f оβ х (Fig. 75).

2. We build the line of intersection MN given and auxiliary planes. M"=A"C"∩hoβ M""A""C"" and N"=B"C"∩hoβ N""B""C""(Fig. 76).

3. Building an intersection point To given line l with 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 with respect to Δ ABC with competing points.

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

Determine the line visibility l relative to the plane pi 1 . Notice 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"" , hence the point M visible relatively pi 1 . Dot 3 lies on l, therefore, from M"≡3" before TO", horizontal projection l" invisible. in plan view TO" visibility is reversed. Beyond Δ ABC straight l visible everywhere.

The line of intersection of two planes is a straight line. Consider first a special case (Fig. 3.9), when one of the intersecting planes is parallel to the horizontal projection plane (α π 1, f 0 α X). In this case, the line of intersection a, belonging to the plane α, will also be parallel to the plane π 1, (Fig. 3.9. a), i.e., it will coincide with the horizontal of the intersecting planes (a ≡ h).

If one of the planes is parallel to the frontal plane of projections (Fig. 3.9. b), then the line of intersection a belonging to this plane will be parallel to the plane π 2 and will coincide with the front of the intersecting planes (a ≡ f).

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Rice. 3.9. A special case of plane intersection general position with planes: a - horizontal level; b - frontal level

An example of constructing an intersection point (K) of a straight line a (AB) with a plane α (DEF) is shown in fig. 3.10. To do this, the line a is enclosed in an arbitrary plane β and the line of intersection of the planes α and β is determined.

In the example under consideration, the lines AB and MN belong to the same plane β and intersect at the point K, and since the line MN belongs to the given plane α (DEF), the point K is also the point of intersection of the line a (AB) with the plane α. (Fig. 3.11).

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Rice. 3.10. Construction of the point of intersection of a line with a plane

To solve such a problem on a complex drawing, it is necessary to be able to find the point of intersection of a line in general position with a plane in general position.

Consider an example of finding the point of intersection of the line AB with the plane of the triangle DEF shown in Fig. 3.11.

To find the point of intersection through the frontal projection of the straight line A 2 B 2, a front-projecting plane β was drawn that intersected the triangle at points M and N. On the frontal projection plane (π 2), these points are represented by the projections M 2 , N 2 . From the condition of belonging to a straight plane on the horizontal plane of projections (π 1), the horizontal projections of the obtained points M 1 N 1 are found. At the intersection of the horizontal projections of the lines A 1 B 1 and M 1 N 1, a horizontal projection of the point of their intersection (K 1) is formed. According to the line of communication and the conditions of belonging on the frontal plane of projections, the frontal projection of the intersection point (K 2) is found.

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Rice. 3.11. An example of determining the point of intersection of a line and a plane

The visibility of the segment AB relative to the triangle DEF is determined by the method of competing points.

Two points NEF and 1AB are considered on the π 2 plane. According to the horizontal projections of these points, it can be established that point N is located closer to the observer (Y N >Y 1) than point 1 (the direction of the line of sight is parallel to S). Consequently, the straight line AB, i.e., the part of the straight line AB (K 1) is closed by the DEF plane on the π 2 plane (its projection K 2 1 2 is shown by the dashed line). The visibility on the plane π 1 is established similarly.

Questions for self-control

1) What is the essence of the method of competing points?

2) What properties of a straight line do you know?

3) What is the algorithm for determining the point of intersection of a line and a plane?

4) What tasks are called positional?

5) Formulate the conditions for belonging to a straight plane.

We bring to your attention the journals published by the publishing house "Academy of Natural History"

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 straight line with a plane is as follows:
1) drawing an auxiliary plane ( The auxiliary plane is recommended to choose the one that will give the simplest graphic solution tasks) through this line;
2) finding the line of intersection of the auxiliary plane with the given plane;
3) determination of the point of intersection of a given straight line with the line of intersection of the planes, and, consequently, with the given plane.

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

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

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

Example 2. On (Fig.252) an example of finding the projections of the point of intersection of the line AB with the plane of general position using the horizontal h.
Example 3. Given: triangle ABC and line NM ; it is required to determine the point of their intersection (Fig.253, a).
Let's take as an auxiliary plane a horizontally projecting plane δ, then the horizontal projection og merges with the horizontal projection N 1 M 1 straight line NM and intersects the projections of the sides of the triangle at points E 1 and F 1 (Fig.253,b). Segment E 1 F 1 will be a horizontal projection of the line of intersection. Then we find the frontal projection of the intersection line: using vertical communication lines, we get points E 2 and F 2, we draw a straight line E 2 F 2 through them, which will be the frontal projection of the intersection line.
The line E 2 F 2 intersects the line N 2 M 2 at the point K 2 . Point K 2 will be a frontal projection of the point of intersection of the line MN with the line EF; the horizontal projection K 1 of this point is determined using a vertical communication line.
The point K (K 1 , K 2 ) will be the point of intersection of the given line MN with the given triangle ABC, as belonging to them at the same time, because the line MN intersects in it with the line EF lying in the plane of the triangle ABC.

Exercise 1
Construct a complex drawing of the triangle ABC given the coordinates of the vertices. Find the life size of the sides of the triangle and build it to life size. Using the same coordinates, construct a visual image
Exercise 2
According to the frontal projection of the polygon and the horizontal projections of two adjacent sides of it, 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 with it (

Construction of the point of intersection of a straight line with a projecting plane reduces to constructing a second point projection on the diagram, since one point projection 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. On fig. 224,a shows the construction of the point of intersection of the straight line EF with the front-projecting plane of the triangle ABC (perpendicular to the plane V) On the plane V, the triangle ABC is projected into the segment a "c" of the straight line, and the point k "will also lie on this line and be at the point the intersection of e "f" with a "c". A horizontal projection is built using a projection connection line. The visibility of a straight line relative to the plane of a triangle ABC is determined by relative position projections of the triangle ABC and the straight line EF on the plane V. The direction of view in fig. 224, as indicated by the 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 the point k "the projection of the straight line is above the projection of the triangle, therefore, this section is visible on the plane H.

On fig. 224, b, the straight line EF intersects the horizontal plane P. The frontal projection k "of the point K - the point of intersection of the straight line EF with the plane P - will be 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 the point K is found using the projection connection line.

Construction of a line of intersection of two planes is reduced to finding two points common to these two planes. This is enough to construct an intersection line, since the intersection line is a straight line, and a straight line is defined by two points. When a projecting plane intersects with a plane in general position, one of the projections of the intersection line coincides with the trace of the plane located in the plane of projections to which the projecting plane is perpendicular. On fig. 225, and the frontal projection m "n" of the intersection line MN coincides with the trace Pv of the front-projecting plane P, and in fig. 225b, 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.

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

Finding the projections of the point K is shown in fig. 226b. Constructions are performed in the following sequence.

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

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

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 with respect 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, the frontal projections of which coincide. The Y-coordinate of point 3, which lies on the line BC, is less than the Y-coordinate of point 4, which lies on the line EF. Consequently, point 4 is closer to the observer (the direction of view is indicated by an arrow) and the projection of the straight line is depicted on the visible plane V. The line passes in front of the triangle. To the left of the point K" the line is closed by the plane of the 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. Therefore, to the right of point 1 (up to point K), the line EF is invisible.

To construct a line of intersection of two planes in general position, auxiliary secant planes are used. This is shown in fig. 227 a. One plane is given by triangle ABC, the other is given by parallel lines EF and MN. The given planes (Fig. 227, a) are crossed 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 a horizontal plane. It intersects the given planes along straight lines 12 and 34, which at the intersection give the point K, which belongs to all three planes, and, consequently, 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.

On fig. 227b, the auxiliary plane R is given by the frontal wake. The frontal projections of the lines of intersection 1 "2" and 3"4 of the plane R with the given planes coincide with the frontal trace Rv of the plane R, since the plane R is perpendicular to the plane V, and everything that is in it (including the lines of intersection) is projected onto its frontal trace Rv. The 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 lines at points 1, 2, 3, 4. Constructed the horizontal projections of the intersection lines are extended until they intersect with 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 line of intersection, a second auxiliary plane Q is drawn. 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 given by parallel 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 is Q ║ R. Continuing these lines until they intersect with each other, one obtains a horizontal projection l of the point L belonging to the line of intersection of the given planes. The frontal projection l" of the point L lies on the trace Q v and is constructed using the line of the projection connection. By connecting the projections of the same name of the points K and L, the projections of the desired intersection line are obtained.

If we take a line in one of the intersecting planes and construct a 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 build the second point in the same way, we can find the line of intersection of two planes, since two points are enough to build a straight line. On fig. 228 shows such a construction of the line of intersection of two planes given by triangles.

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

On fig. 228 the point of intersection of the line EF with the plane of the triangle ABC is constructed. To do this, an auxiliary horizontally projecting plane S is drawn through the straight line EF and a frontal projection 1 "2" of the line of intersection of this plane with the plane of the triangle ABC is built. 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 - the point of intersection of the line BC with the plane of the triangle DEF. Through the line BC, a front-projecting plane R is drawn, and on the plane H, the intersection of the horizontal projections of the line BC and the line of intersection 34 gives point n - the horizontal projection of the desired point. Visible sections of given triangles are determined using competing points for each projection plane separately.To do this, a point is selected 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 projection plane, the projections of these points do not coincide. Comparing their Z coordinates, they find out that point 5 closes 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.

Visibility on the frontal plane of projections is determined using competing points 4 and 7 belonging to the 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 point of intersection may be outside the plane of the triangle. In this case, by connecting the obtained points belonging to the intersection line, only that part 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 is the Y coordinate and Z coordinate of a point?

3. How are the projections of the segment perpendicular to the plane of projections 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 main position about the belonging of a point to a straight line.

6. How to distinguish intersecting lines from intersecting ones in a diagram?

7. What points are called competing?

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

9. Formulate the main position 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 plane in general position?

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