Among the huge number of stereometric tasks in geometry textbooks, in various collections of tasks, textbooks for preparing for universities, tasks for finding the distance between skew lines are extremely rare. Perhaps this is due both to the narrowness of their practical application (relative to the school curriculum, in contrast to the "winning" tasks for calculating areas and volumes), and the complexity of this topic.

The practice of conducting the Unified State Examination shows that many students do not begin to complete tasks in geometry that are included in the examination paper at all. To ensure the successful completion of geometric tasks of an increased level of complexity, it is necessary to develop the flexibility of thinking, the ability to analyze the proposed configuration and isolate parts in it, the consideration of which allows you to find a way to solve the problem.

The school course involves the study of four ways to solve problems for finding the distance between intersecting lines. The choice of method is determined, first of all, by the characteristics of a particular task, the opportunities for choice provided by it, and, secondly, by the abilities and characteristics of the "spatial thinking" of a particular student. Each of these methods allows you to solve the most important part of the problem - the construction of a segment perpendicular to both intersecting lines (for the computational part of the problems, division into methods is not required).

The main methods for solving problems of finding the distance between skew lines

Finding the length of the common perpendicular of two intersecting lines, i.e. a segment with ends on these lines and perpendicular to each of these lines.

Finding the distance from one of the intersecting lines to a plane parallel to it passing through the other line.

Finding the distance between two parallel planes passing through given skew lines.

Finding the distance from a point that is the projection of one of the skew lines onto a plane perpendicular to it (the so-called "screen") to the projection of another line onto the same plane.

We will demonstrate all four methods on the following simplest task: "In a cube with an edge a find the distance between any edge and the diagonal of a face that does not intersect it." Answer: .

Picture 1

h skr is perpendicular to the plane of the side face containing the diagonal d and is perpendicular to the edge, so h scr and is the distance between the edge a and diagonal d.

Figure 2

The plane A is parallel to the edge and passes through the given diagonal, hence the given h scr is not only the distance from the edge to the plane A, but also the distance from the edge to the given diagonal.

Figure 3

Planes A and B are parallel and pass through two given skew lines, so the distance between these planes is equal to the distance between the two skew lines.

Figure 4

Plane A is perpendicular to the edge of the cube. When projected onto the diagonal A d this diagonal turns to one of the sides of the base of the cube. This h scr is the distance between the line containing the edge and the projection of the diagonal onto the plane C, and hence between the line containing the edge and the diagonal.

Let us dwell in more detail on the application of each method for polyhedra studied at school.

The application of the first method is quite limited: it is well used only in some problems, since it is rather difficult to determine and justify the exact location in the simplest problems, and the approximate location of the common perpendicular of two intersecting lines in complex problems. In addition, when finding the length of this perpendicular in complex problems, one may encounter insurmountable difficulties.

Problem 1. In a rectangular parallelepiped with dimensions a, b, h find the distance between the side edge and the diagonal of the base that does not intersect with it.

Figure 5

Let AHBD. Since A 1 A is perpendicular to the plane ABCD, then A 1 A AH.

AH is perpendicular to both of the two intersecting lines, therefore AH? is the distance between lines A 1 A and BD. In a right triangle ABD, knowing the lengths of the legs AB and AD, we find the height AH, using the formulas for calculating the area of ​​a right triangle. Answer:

Problem 2. In a regular 4-sided pyramid with a side edge L and base side a find the distance between the apothem and the side of the base that intersects the side face containing this apothem.

Figure 6

SHCD as an apothem, ADCD as ABCD is a square. Therefore, DH is the distance between lines SH and AD. DH is equal to half the side of CD. Answer:

The use of this method is also limited due to the fact that if you can quickly build (or find a ready-made) plane passing through one of the intersecting lines parallel to another line, then building a perpendicular from any point of the second line to this plane (inside the polyhedron) causes difficulties. However, in simple tasks, where the construction (or finding) of the indicated perpendicular does not cause difficulties, this method is the fastest and easiest, and therefore accessible.

Task 2. The solution of the problem already indicated above in this way does not cause any special difficulties.

Figure 7

Plane EFM is parallel to line AD, since AD ​​|| EF. Line MF lies in this plane, so the distance between line AD and plane EFM is equal to the distance between line AD and line MF. Let's do an OHAD. OHEF, OHMO, hence OH(EFM), hence OH is the distance between line AD and plane EFM, and hence the distance between line AD and line MF. Finding OH from triangle AOD.

Problem 3. In a rectangular parallelepiped with dimensions a,b and h find the distance between the side edge and the diagonal of the parallelepiped that does not intersect with it.

Figure 8

Line AA 1 is parallel to plane BB 1 D 1 D, B 1 D belongs to this plane, hence the distance from AA 1 to plane BB 1 D 1 D is equal to the distance between lines AA 1 and B 1 D. Draw AHBD. Also, AH B 1 B, hence AH(BB 1 D 1 D), hence AHB 1 D, i.e. AH is the desired distance. Find AH from right triangle ABD.

Answer:

Problem 4. In a regular hexagonal prism A:F 1 with height h and base side a find distance between lines:

Figure 9 Figure 10

a) AA 1 and ED 1.

Consider the plane E 1 EDD 1 . A 1 E 1 EE 1 , A 1 E 1 E 1 D 1 , therefore

A 1 E 1 (E 1 EDD 1). Also A 1 E 1 AA 1 . Therefore, A 1 E 1 is the distance from the line AA 1 to the plane E 1 EDD 1 . ED 1 (E 1 EDD 1)., therefore AE 1 is the distance from the straight line AA 1 to the straight line ED 1. We find A 1 E 1 from the triangle F 1 A 1 E 1 using the cosine theorem. Answer:

b) AF and diagonal BE 1.

Let us draw a line FH perpendicular to BE from the point F. EE 1 FH, FHBE, hence FH(BEE 1 B 1), hence FH is the distance between line AF and (BEE 1 B 1), and hence the distance between line AF and diagonal BE 1 . Answer:

METHOD III

The use of this method is extremely limited, since it is easier to build a plane parallel to one of the lines (method II) than two parallel planes, however, method III can be used in prisms if the intersecting lines belong to parallel faces, and also in cases where in a polyhedron it is easy to construct parallel sections containing given lines.

Task 4.

Figure 11

a) Planes BAA 1 B 1 and DEE 1 D 1 are parallel because AB || ED and AA 1 || EE1. ED 1 DEE 1 D 1 , AA 1 (BAA 1 B 1), therefore, the distance between the straight lines AA 1 and ED 1 is equal to the distance between the planes BAA 1 B 1 and DEE 1 D 1 . A 1 E 1 AA 1 , A 1 E 1 A 1 B 1 , therefore, A 1 E 1 BAA 1 B 1 . We prove similarly that A 1 E 1 (DEE 1 D 1). Thus, A 1 E 1 is the distance between the planes BAA 1 B 1 and DEE 1 D 1 , and hence between the lines AA 1 and ED 1 . Find A 1 E 1 from triangle A 1 F 1 E 1 , which is isosceles with angle A 1 F 1 E 1 equal to . Answer:

Figure 12

b) The distance between AF and diagonal BE 1 is similar.

Problem 5. In a cube with an edge a find the distance between two non-intersecting diagonals of two adjacent faces.

This problem is considered as a classic one in some manuals, but, as a rule, its solution is given by method IV, however, it is quite accessible for solution using method III.

Figure 13

Some difficulty in this problem is the proof that the diagonal A 1 C is perpendicular to both parallel planes (AB 1 D 1 || BC 1 D). B 1 CBC 1 and BC 1 A 1 B 1 , therefore, the line BC 1 is perpendicular to the plane A 1 B 1 C, and therefore BC 1 A 1 C. Also, A 1 CBD. Therefore, the line A 1 C is perpendicular to the plane BC 1 D. The computational part of the problem does not cause any particular difficulties, since h scr= EF is found as the difference between the cube diagonal and the heights of two identical regular pyramids A 1 AB 1 D 1 and CC 1 BD.

METHOD IV.

This method has a fairly wide application. For tasks of medium and increased difficulty, it can be considered the main one. There is no need to apply it only when one of the three previous methods works easier and faster, since in such cases method IV can only complicate the solution of the problem, or make it difficult to access. This method is very advantageous to use in the case of perpendicularity of intersecting lines, since there is no need to build a projection of one of the lines on the "screen"

L and base side a.

Figure 16

In this and similar problems, method IV leads to a solution faster than other methods, since by constructing a section that plays the role of a "screen" perpendicular to AC (triangle BDM), it is clear that further there is no need to build a projection of another line (BM) onto this screen. DH - desired distance. DH is found from triangle MDB using area formulas. Answer: .

"Distance between skew lines" - Theorem. Preparatory oral tasks. Find the distance between the line MN and the plane AA1D1D. Find the distance between line B1K and plane DD1C1C. OK=OO1?OM/O1M =a/3 (according to the Pythagorean theorem O1M=3/2?2, OM=1/2?2). Diagonal plane AA1C1C is perpendicular to line BD. The new positions of points B and N will be the points of lines AD and BM closest to each other.

"Lesson Speed ​​time distance" - Mathematical warm-up. The purpose of the lesson: to teach students to solve problems on the movement. Distance. How long does it take to walk 30 km at a constant speed of 5 km/h? Relationship between speed, time and distance. How many people went to the city? An airplane flies the distance from city A to city B in 1 hour and 20 minutes.

“Speed ​​time distance mathematics” - Reduce the sum of the numbers 5 and 65 by 2 times. Dunno went to the moon. Journey through the pages of a fairy tale book. Fizkultminutka. One left at 8 o'clock and the other at 10 o'clock. Summarizing. Is Laura right? -Laura solved the following problem: “500 km. A car will pass in 10 hours. Time. The key with the answer "38" opens the book:

"Dialogue direct speech" - What is the difference between direct speech and dialogue? For example: L. N. Tolstoy said: “We all need each other in the world.” Graphics of direct speech. A: "p." Task 3. Replace direct speech with dialogue. For example: "P?" - a. "P!" - a. Point out the correct diagrams for the following sentences. Dialog graphics. How to write direct speech and dialogue in writing?

"Sentences with direct speech" - Petronius, ancient Roman writer. Game "Find the mistake" (check). Author's words introducing direct speech: I reappeared and went to the house of Father Gerasim. A friend from the village came to visit me. Proposals with direct speech. Creative task. In writing, direct speech is enclosed in quotation marks. Read!" exclaimed Konstantin Georgievich Paustovsky.

"Distance and Scale" - Model of the atom in high magnification scale. On a map with a scale, the distance is 5 cm. If the scale is given by a fraction with a numerator of 1, then. Scale model of a fire engine. Algorithm for finding the distance on the ground: On the highway, the length of the route is 700 km. Finish the sentence: The distance between two cities is 400 km.

In this article, using the example of solving problem C2 from the Unified State Examination, the method of finding coordinates using the method is analyzed. Recall that lines are skew if they do not lie in the same plane. In particular, if one line lies in a plane, and the second line intersects this plane at a point that does not lie on the first line, then such lines are skew (see figure).

For finding distances between intersecting lines necessary:

1. Draw a plane through one of the skew lines that is parallel to the other skew line.
2. Drop a perpendicular from any point of the second straight line to the resulting plane. The length of this perpendicular will be the desired distance between the lines.

Let us analyze this algorithm in more detail using the example of solving problem C2 from the Unified State Examination in mathematics.

Distance between lines in space

Task. in a single cube ABCDA 1 B 1 C 1 D 1 find the distance between the lines BA 1 and D.B. 1 .

Rice. 1. Drawing for the task

Decision. Through the midpoint of the diagonal of the cube D.B. 1 (dot O) draw a line parallel to the line A 1 B. Points of intersection of a given line with edges BC and A 1 D 1 denote respectively N and M. Straight MN lies in the plane MNB 1 and parallel to the line A 1 B, which does not lie in this plane. This means that the direct A 1 B parallel to the plane MNB 1 on the basis of parallelism of a straight line and a plane (Fig. 2).

Rice. 2. The desired distance between the crossing lines is equal to the distance from any point of the selected line to the depicted plane

We are now looking for the distance from some point on the straight line A 1 B up to the plane MNB one . This distance, by definition, will be the desired distance between the skew lines.

To find this distance, we use the coordinate method. We introduce a rectangular Cartesian coordinate system so that its origin coincides with point B, the axis X was directed along the edge BA, axis Y- along the rib BC, axis Z- along the rib BB 1 (Fig. 3).

Rice. 3. We choose a rectangular Cartesian coordinate system as shown in the figure

We find the equation of the plane MNB 1 in this coordinate system. To do this, we first determine the coordinates of the points M, N and B 1: We substitute the obtained coordinates into the general equation of a straight line and obtain the following system of equations:

From the second equation of the system, we obtain from the third one, and then from the first we obtain. We substitute the obtained values ​​into the general equation of the straight line:

Note that otherwise the plane MNB 1 would pass through the origin. We divide both sides of this equation by and we get:

The distance from a point to a plane is determined by the formula.

DISTANCE BETWEEN RIGHTS IN SPACE The distance between two intersecting lines in space is the length of the common perpendicular drawn to these lines. If one of the two intersecting lines lies in a plane, and the other is parallel to this plane, then the distance between these lines is equal to the distance between the line and the plane. If two intersecting lines lie in parallel planes, then the distance between these lines is equal to the distance between the parallel planes.

Cube 1 In the unit cube A…D 1 find the distance between lines AA 1 and BC. Answer: 1.

Cube 2 In the unit cube A…D 1 find the distance between lines AA 1 and CD. Answer: 1.

Cube 3 In the unit cube A…D 1 find the distance between lines AA 1 and B 1 C 1. Answer: 1.

Cube 4 In the unit cube A…D 1 find the distance between lines AA 1 and C 1 D 1. Answer: 1.

Cube 5 In the unit cube A…D 1 find the distance between lines AA 1 and BC 1. Answer: 1.

Cube 6 In the unit cube A…D 1 find the distance between lines AA 1 and B 1 C. Answer: 1.

Cube 7 In the unit cube A…D 1 find the distance between lines AA 1 and CD 1. Answer: 1.

Cube 8 In the unit cube A…D 1 find the distance between lines AA 1 and DC 1. Answer: 1.

Cube 9 In the unit cube A…D 1 find the distance between lines AA 1 and CC 1. Answer:

Cube 10 In the unit cube A…D 1 find the distance between lines AA 1 and BD. Decision. Let O be the midpoint of BD. The desired distance is the length of the segment AO. It is equal to Answer:

Cube 11 In the unit cube A…D 1 find the distance between lines AA 1 and B 1 D 1. Answer:

Cube 12 In the unit cube A…D 1 find the distance between lines AA 1 and BD 1. Solution. Let P, Q be the midpoints of AA 1, BD 1. The desired distance is the length of the segment PQ. It is equal to Answer:

Cube 13 In the unit cube A…D 1 find the distance between lines AA 1 and BD 1. Answer:

Cube 14 In the unit cube A…D 1 find the distance by lines AB 1 and CD 1. Answer: 1.

Cube 15 In the unit cube A…D 1 find the distance between lines AB 1 and BC 1. Solution. The desired distance is equal to the distance between the parallel planes AB 1 D 1 and BDC 1. The diagonal A 1 C is perpendicular to these planes and is divided into three equal parts at the intersection points. Therefore, the desired distance is equal to the length of the segment EF and is equal to Answer:

Cube 16 In the unit cube A…D 1 find the distance between the lines AB 1 and A 1 C 1. The solution is similar to the previous one. Answer:

Cube 17 In the unit cube A…D 1 find the distance between lines AB 1 and BD. The solution is similar to the previous one. Answer:

Cube 18 In the unit cube A…D 1 find the distance by lines AB 1 and BD 1. Solution. The diagonal BD 1 is perpendicular to the plane of the equilateral triangle ACB 1 and intersects it at the center P of its inscribed circle. The desired distance is equal to the radius OP of this circle. OP = Answer:

Pyramid 1 In unit tetrahedron ABCD find the distance between lines AD and BC. Decision. The desired distance is equal to the length of the segment EF, where E, F are the midpoints of the edges AD, GF. In triangle DAG DA = 1, AG = DG = Answer: Therefore, EF =

Pyramid 2 In a regular pyramid SABCD, all edges of which are equal to 1, find the distance between lines AB and CD. Answer: 1.

Pyramid 3 In a regular pyramid SABCD, all edges of which are equal to 1, find the distance between lines SA and BD. Decision. The desired distance is equal to the height OH of the triangle SAO, where O is the midpoint of BD. In a right triangle SAO we have: SA = 1, AO = SO = Answer: Therefore, OH =

Pyramid 4 In a regular pyramid SABCD, all edges of which are equal to 1, find the distance between lines SA and BC. Decision. Plane SAD is parallel to line BC. Therefore, the desired distance is equal to the distance between the line BC and the plane SAD. It is equal to the height EH of the triangle SEF, where E, F are the midpoints of the edges BC, AD. In triangle SEF we have: EF = 1, SE = SF = Height SO is Therefore, EH = Answer:

Pyramid 5 In a regular 6th pyramid SABCDEF with base edges equal to 1, find the distance between lines AB and DE. Answer:

Pyramid 6 In the regular 6th pyramid SABCDEF, whose side edges are 2 and the base edges are 1, find the distance between lines SA and BC. Solution: Extend edges BC and AF until they intersect at point G. The common perpendicular to SA and BC is the altitude AH of triangle ABG. It is equal to Answer:

Pyramid 7 In the regular 6th pyramid SABCDEF, whose side edges are 2 and the base edges are 1, find the distance between lines SA and BF. Solution: The desired distance is the height GH of the triangle SAG, where G is the intersection point of BF and AD. In the triangle SAG we have: SA = 2, AG = 0.5, height SO is equal to From here we find GH = Answer:

Pyramid 8 In the regular 6th pyramid SABCDEF, whose side edges are 2 and the base edges are 1, find the distance between lines SA and CE. Solution: The desired distance is the height GH of the triangle SAG, where G is the intersection point of CE and AD. In the triangle SAG we have: SA = 2, AG = , the height SO is equal to From here we find GH = Answer:

Pyramid 9 In the regular 6th pyramid SABCDEF, whose side edges are 2 and the base edges are 1, find the distance between lines SA and BD. Solution: Line BD is parallel to plane SAE. The desired distance is equal to the distance between the line BD and this plane and is equal to the height PH of the triangle SPQ. In this triangle, the height SO is, PQ = 1, SP = SQ = From here we find PH = Answer:

Pyramid 10 In the regular 6th pyramid SABCDEF, whose lateral edges are 2 and the base edges are 1, find the distance between lines SA and BG, where G is the midpoint of edge SC. Solution: Draw a line through point G parallel to SA. Let Q denote the point of its intersection with the line AC. The desired distance is equal to the height QH of the right triangle ASQ, in which AS = 2, AQ = , SQ = From here we find QH = Answer: .

Prism 1 In a regular triangular prism ABCA 1 B 1 C 1, all edges of which are equal to 1, find the distance between the lines: BC and B 1 C 1. Answer: 1.

Prism 2 In a regular triangular prism ABCA 1 B 1 C 1, all edges of which are equal to 1, find the distance between the lines: AA 1 and BC. Answer:

Prism 3 In a regular triangular prism ABCA 1 B 1 C 1, all edges of which are equal to 1, find the distance between the lines: AA 1 and BC 1. Answer:

Prism 4 In a regular triangular prism ABCA 1 B 1 C 1, all edges of which are equal to 1, find the distance between the lines: AB and A 1 C 1. Answer: 1.

Prism 5 In a regular triangular prism ABCA 1 B 1 C 1, all edges of which are equal to 1, find the distance between the lines: AB and A 1 C. Solution: The desired distance is equal to the distance between the line AB and the plane A 1 B 1 C. Let us denote D and D 1 the midpoints of the edges AB and A 1 B 1. In a right triangle CDD 1, draw a height DE from the vertex D. It will be the desired distance. We have, DD 1 = 1, CD = Answer: Therefore, DE = , CD 1 = .

Prism 6 In a regular triangular prism ABCA 1 B 1 C 1, all edges of which are equal to 1, find the distance between the lines: AB 1 and BC 1. Solution: Let's build the prism to a 4-angled prism. The desired distance will be equal to the distance between the parallel planes AB 1 D 1 and BDC 1. It is equal to the height OH of the right triangle AOO 1, in which the Answer. This height is

Prism 7 In the correct 6th prism A…F 1, whose edges are equal to 1, find the distance between the lines: AB and A 1 B 1. Answer: 1.

Prism 8 In the regular 6th prism A…F 1, whose edges are equal to 1, find the distance between the lines: AB and B 1 C 1. Answer: 1.

Prism 9 In the regular 6th prism A…F 1, whose edges are equal to 1, find the distance between the lines: AB and C 1 D 1. Answer: 1.

Prism 10 In the correct 6th prism A…F 1, whose edges are equal to 1, find the distance between the lines: AB and DE. Answer: .

Prism 11 In the correct 6th prism A ... F 1, whose edges are equal to 1, find the distance between the lines: AB and D 1 E 1. Answer: 2.

Prism 12 In the regular 6th prism A…F 1, whose edges are equal to 1, find the distance between the lines: AA 1 and CC 1. Answer: .

Prism 13 In the correct 6th prism A ... F 1, whose edges are equal to 1, find the distance between the lines: AA 1 and DD 1. Answer: 2.

Prism 14 In the regular 6th prism A…F 1, whose edges are equal to 1, find the distance between the lines: AA 1 and B 1 C 1. Solution: Let's continue the sides B 1 C 1 and A 1 F 1 until they intersect at point G. Triangle A 1 B 1 G is equilateral. Its height A 1 H is the desired common perpendicular. Its length is equal. Answer: .

Prism 15 In the regular 6th prism A…F 1, whose edges are equal to 1, find the distance between the lines: AA 1 and C 1 D 1. Solution: The desired common perpendicular is the segment A 1 C 1. Its length is equal. Answer: .

Prism 16 In the correct 6th prism A…F 1, whose edges are equal to 1, find the distance between the lines: AA 1 and BC 1. Solution: The desired distance is the distance between the parallel planes ADD 1 and BCC 1. It is equal. Answer: .

Prism 17 In the regular 6th prism A…F 1, whose edges are equal to 1, find the distance between the lines: AA 1 and CD 1. Solution: The desired common perpendicular is the segment AC. Its length is equal. Answer: .

Prism 18 In the regular 6th prism A…F 1, whose edges are equal to 1, find the distance between the lines: AA 1 and DE 1. Solution: The desired common perpendicular is the segment A 1 E 1. Its length is equal. Answer: .

Prism 19 In the regular 6th prism A…F 1, whose edges are equal to 1, find the distance between the lines: AA 1 and BD 1. Solution: The desired common perpendicular is the segment AB. Its length is 1. Answer: 1.

Prism 20 In a regular 6th prism A…F 1, whose edges are equal to 1, find the distance between the lines: AA 1 and CE 1. Solution: The desired distance is the distance between the line AA 1 and the plane CEE 1. It is equal. Answer: .

Prism 21 In the correct 6th prism A…F 1, whose edges are equal to 1, find the distance between the lines: AA 1 and BE 1. Solution: The required distance is the distance between the line AA 1 and the plane BEE 1. It is equal. Answer: .

Prism 22 In the correct 6th prism A…F 1, whose edges are equal to 1, find the distance between the lines: AA 1 and CF 1. Solution: The desired distance is the distance between the line AA 1 and the plane CFF 1. It is equal. Answer: .

Prism 23 In the regular 6th prism A…F 1, whose edges are equal to 1, find the angle between the lines: AB 1 and DE 1. Solution: The desired distance is the distance between the parallel planes ABB 1 and DEE 1. The distance between them is equal. Answer: .

Prism 24 In the correct 6th prism A…F 1, whose edges are equal to 1, find the angle between the lines: AB 1 and CF 1. Solution: The desired distance is the distance between the line AB 1 and the plane CFF 1. It is equal. Answer:

Prism 25 In a regular 6th prism A…F 1, whose edges are equal to 1, find the distance between the lines: AB 1 and BC 1. Solution: Let O, O 1 be the centers of the faces of the prism. Planes AB 1 O 1 and BC 1 O are parallel. The plane ACC 1 A 1 is perpendicular to these planes. The desired distance d is equal to the distance between the lines AG 1 and GC 1. In the parallelogram AGC 1 G 1 we have AG = Answer: ; AG 1 = The height drawn to the side AA 1 is equal to 1. Therefore, d= . .

Prism 26 In the regular 6th prism A…F 1, whose edges are equal to 1, find the distance between the lines: AB 1 and BD 1. Solution: Consider the plane A 1 B 1 HG perpendicular to BD 1. The orthogonal projection onto this plane translates the line BD 1 to point H, and line AB 1 to line GB 1. Therefore, the desired distance d is equal to the distance from point H to line GB 1. In a right triangle GHB 1 we have GH = 1; Answer: B 1 H = . Therefore, d = .

Prism 27 In the regular 6th prism A…F 1, whose edges are equal to 1, find the distance between the lines: AB 1 and BE 1. Solution: Consider the plane A 1 BDE 1, perpendicular to AB 1. The orthogonal projection onto this plane translates the line AB 1 to point G, and the line BE 1 leaves in place. Therefore, the desired distance d is equal to the distance GH from the point G to the line BE 1. In a right triangle A 1 BE 1 we have A 1 B = ; A 1 E 1 =. Answer: Therefore, d = .