Construction of a straight line at a certain distance from a point. Determination of distances

Determination of distances

Distances from point to point and from point to line

Distance from point to point is determined by the length of the line segment connecting these points. As shown above, this problem can be solved either by the right-angled triangle method, or by replacing the projection planes by moving the segment to the position of the level line.

Distance from point to line measured by a segment of a perpendicular drawn from a point to a line. A segment of this perpendicular is depicted in full size on the projection plane if it is drawn to the projecting line. Thus, first the straight line must be transferred to the projecting position, and then a perpendicular should be lowered onto it from a given point. On fig. 1 shows the solution to this problem. For direct translation general position AB in the position of the direct level spend x14 IIA1 B1. Then AB is transferred to the projecting position by introducing an additional projection plane P5, for which a new projection axis x45 \ A4 B4 is carried out.

Picture 1

Similarly to points A and B, point M is projected onto the projection plane P5.

The projection K5 of the base K of the perpendicular dropped from the point M to the line AB, on the projection plane P5, will coincide with the corresponding projections of the points

A and B. The projection M5 K5 of the perpendicular MK is the natural value of the distance from the point M to the line AB.

In the system of projection planes P4 / P5, the perpendicular MK will be a level line, since it lies in a plane parallel to the plane of projections P5. Therefore, its projection M4 K4 onto the plane P4 is parallel to x45 , i.e. perpendicular to the projection A4 B4. These conditions determine the position of the projection K4 of the base of the perpendicular K, which is found by drawing a straight line from M4 parallel to x45 until it intersects with the projection A4 B4. The remaining projections of the perpendicular are found by projecting the point K on the plane of the projections P1 and P2.

Distance from point to plane

The solution to this problem is shown in Fig. 2. The distance from the point M to the plane (ABC) is measured by a segment of the perpendicular dropped from the point to the plane.

Figure 2

Since the perpendicular to the projecting plane is a level line, we translate into this position given plane, as a result of which on the newly introduced projection plane P4 we obtain a degenerate projection C4 B4 of the plane ABC. Next, we project the point M onto P4. The natural value of the distance from the point M to the plane is determined by the segment of the perpendicular

[MK]=[M4 K4]. The remaining projections of the perpendicular are constructed in the same way as in the previous problem, i.e. taking into account the fact that the segment MK in the system of projection planes P1 / P4 is a level line and its projection M1 K1 is parallel to the axis

x14.

Distance between two straight lines

The shortest distance between skew lines is measured by the segment of the common perpendicular to them, cut off by these lines. The problem is solved by choosing (as a result of two successive changes) the projection plane perpendicular to one of the intersecting lines. In this case, the desired segment of the perpendicular will be parallel to the selected projection plane and will be displayed on it without distortion. On fig. 3 shows two intersecting straight lines defined by segments AB and CD.

Figure 3

The straight lines are projected at the beginning onto the projection plane P4, parallel to one (any) of them, for example, AB, and perpendicular to P1.

On the plane of projections P4, the segment AB will be displayed without distortion. Then the segments are projected onto a new plane P5 perpendicular to the same straight line AB and the plane P4. On the plane of projections P5, the projection of the segment AB perpendicular to it degenerates into the point A5 =B5, and the desired value N5 M5 of the segment NM is perpendicular to C5 D5 and is depicted in full size. Using the appropriate communication lines, the projections of the segment MN are built on the initial

drawing. As shown earlier, the projection N4 M4 of the desired segment onto the plane P4 is parallel to the x45 projection axis, since it is a level line in the system of projection planes P4 / P5.

The task of determining the distance D between two parallel lines AB to CD - special case the previous one (Fig. 4).

Figure 4

By a double replacement of the projection planes, the parallel lines are transferred to the projecting position, as a result of which on the projection plane P5 we will have two degenerate projections A5 = B5 and C5 = D5 of the lines AB and CD. The distance between them D will be equal to its natural value.

The distance from a straight line to a plane parallel to it is measured by a segment of the perpendicular dropped from any point on the straight line to the plane. Therefore, it is enough to transform the plane of general position into the position of the projecting plane, take a direct point, and the solution of the problem will be reduced to determining the distance from the point to the plane.

To determine the distance between parallel planes, it is necessary to translate them into a projecting position and construct a perpendicular to the degenerate projections of the planes, the segment of which between them will be the required distance.

155*. Determine the actual size of the segment AB of a straight line in general position (Fig. 153, a).

Decision. As you know, the projection of a straight line segment on any plane is equal to the segment itself (taking into account the scale of the drawing), if it is parallel to this plane

(Fig. 153, b). It follows from this that by converting the drawing it is necessary to achieve the parallelism of this segment pl. V or pl. H or supplement the system V, H with another plane perpendicular to the square. V or to pl. H and at the same time parallel to the given segment.

On fig. 153, c shows the introduction of an additional plane S, perpendicular to the square. H and parallel to the given segment AB.

The projection a s b s is equal to the natural value of the segment AB.

On fig. 153, d shows another method: the segment AB is rotated around a straight line passing through point B and perpendicular to the square. H, to a position parallel

sq. V. In this case, point B remains in place, and point A occupies a new position A 1 . Horizon in new position. projection a 1 b || x axis. The projection a "1 b" is equal to the natural value of the segment AB.

156. Pyramid SABCD is given (Fig. 154). Determine the natural size of the pyramid edges AS and CS using the method of changing the projection planes, and the edges BS and DS using the rotation method, and take the axis of rotation perpendicular to the square. H.

157*. Determine the distance from point A to the straight line BC (Fig. 155, a).

Decision. The distance from a point to a line is measured by a segment of a perpendicular drawn from a point to a line.

If the line is perpendicular to any plane (Fig. 155.6), then the distance from the point to the line is measured by the distance between the projection of the point and projection point straight line on this plane. If a straight line occupies a general position in the V, H system, then in order to determine the distance from a point to a straight line by changing the projection planes, two more additional planes must be introduced into the V, H system.

First (Fig. 155, c) we enter the square. S, parallel to the segment BC (the new axis S/H is parallel to the projection bс), and we construct the projections b s c s and a s . Then (Fig. 155, d) we introduce another square. T perpendicular to line BC (new T/S axis perpendicular to b s c s). We build projections of a straight line and a point - with t (b t) and a t. The distance between points a t and c t (b t) is equal to the distance l from point A to the line BC.

On fig. 155e, the same task is accomplished by the rotation method in its form, which is called the parallel movement method. First, the line BC and point A, keeping their mutual position unchanged, turn around some (not indicated in the drawing) line perpendicular to the square. H, so that the straight line BC is parallel to the square. V. This is equivalent to moving points A, B, C in planes parallel to the square. H. At the same time, the horizon. projection given system(BC + A) does not change either in magnitude or in configuration, only its position relative to the x-axis changes. Set up a horizon. the projection of the straight line BC parallel to the x axis (position b 1 c 1) and determine the projection a 1, setting aside c 1 1 1 \u003d c-1 and a 1 1 1 \u003d a-1, and a 1 1 1 ⊥ c 1 1 1. Drawing straight lines b "b" 1, a "a" 1, c "c" 1 parallel to the x axis, we find the front on them. projections b "1, a" 1, c "1. Next, we move the points B 1, C 1 and A 1 in planes parallel to square V (also without changing their relative position), so as to get B 2 C 2 ⊥ area H. In this case, the projection of the straight line to the front will be perpendicular to axes x,b 2 c "2 \u003d b" 1 c "1, and to build the projection a" 2, you need to take b "2 2" 2 \u003d b "1 2" 1, draw 2 "a" 2 ⊥ b "2 c" 2 and put aside a" 2 2" 2 \u003d a" 1 2" 1. Now, by swiping from 1 to 2 and a 1 a 2 || x 1 we get the projections b 2 c 2 and a 2 and the desired distance l from point A to the line BC. You can determine the distance from A to BC by turning the plane defined by point A and the straight line BC around the horizontal of this plane to the position T || sq. H (Fig. 155, e).

In the plane given by point A and the straight line BC, we draw a horizontal line A-1 (Fig. 155, g) and rotate point B around it. Point B moves to square. R (given in the drawing following R h), perpendicular to A-1; at point O is the center of rotation of point B. Now we determine the natural value of the radius of rotation of VO, (Fig. 155, c). In the required position, i.e. when pl. T defined by point A and line BC will become || sq. H, point B will turn out on R h at a distance Ob 1 from point O (there may be another position on the same track R h, but on the other side of O). Point b 1 is the horizon. the projection of point B after moving it to position B 1 in space, when the plane defined by point A and the straight line BC has taken position T.

Having drawn (Fig. 155, and) the straight line b 1 1, we get the horizon. projection of the straight line BC, already located || sq. H is in the same plane as A. In this position, the distance from a to b 1 1 is equal to the desired distance l. The plane P, in which the given elements lie, can be combined with the square. H (Fig. 155, j), turning the square. P around her horizon. trace. Having passed from setting the plane by the point A and the line BC to setting the lines BC and A-1 (Fig. 155, l), we find the traces of these lines and draw traces P ϑ and P h through them. We are building (Fig. 155, m) combined with the square. H position front. trace - P ϑ0 .

Draw the horizon through point a. frontal projection; the combined frontal passes through point 2 on the trace Р h parallel to Р ϑ0 . Point A 0 - combined with pl. H is the position of point A. Similarly, we find the point B 0 . Direct sun in combined with pl. H position passes through point B 0 and point m (horizontal trace of a straight line).

The distance from the point A 0 to the straight line B 0 C 0 is equal to the desired distance l.

It is possible to carry out the indicated construction by finding only one trace P h (Fig. 155, n and o). The whole construction is similar to turning around the horizontal (see Fig. 155, f, c, i): the trace P h is one of the horizontal lines of the square. R.

Of the methods for converting a drawing given to solve this problem, the method of rotation around a horizontal or frontal is preferable.

158. Pyramid SABC is given (Fig. 156). Determine distances:

a) from top B of the base to its side AC by the method of parallel movement;

b) from the top S of the pyramid to the sides BC and AB of the base by means of rotation around the horizontal;

c) from the top S to the side AC of the base by changing the projection planes.

159. Given a prism (Fig. 157). Determine distances:

a) between the edges AD and CF by changing the projection planes;

b) between ribs BE and CF by rotation around the front;

c) between the edges AD and BE by the method of parallel movement.

160. Determine the actual size of the quadrilateral ABCD (Fig. 158) by combining with the square. N. Use only the horizontal trace of the plane.

161*. Determine the distance between the intersecting lines AB and CD (Fig. 159, a) and construct projections of the common perpendicular to them.

Decision. The distance between the crossing lines is measured by the segment (MN) of the perpendicular to both lines (Fig. 159, b). Obviously, if one of the lines is placed perpendicular to any square. T then

the segment MN of the perpendicular to both lines will be parallel to the square. Its projection on this plane will display the required distance. Projection right angle maenad MN n AB on the square. T also turns out to be a right angle between m t n t and a t b t , since one of the sides of the right angle AMN, namely MN. parallel to square. T.

On fig. 159, c and d, the desired distance l is determined by the method of changing the projection planes. First, we introduce an additional square. projections S, perpendicular to the square. H and parallel to the straight line CD (Fig. 159, c). Then we introduce another additional square. T, perpendicular to the square. S and perpendicular to the same line CD (Fig. 159, d). Now you can build a projection of the common perpendicular by drawing m t n t from the point c t (d t) perpendicular to the projection a t b t . Points m t and n t are projections of the points of intersection of this perpendicular with lines AB and CD. From the point m t (Fig. 159, e) we find m s on a s b s: the projection m s n s should be parallel to the T / S axis. Further, from m s and n s we find m and n on ab and cd, and from them m "and n" on a "b" and c "d".

On fig. 159, in shows the solution to this problem by the method of parallel movements. First, we put the straight line CD parallel to the square. V: projection c 1 d 1 || X. Next, we move the lines CD and AB from positions C 1 D 1 and A 1 B 1 to positions C 2 B 2 and A 2 B 2 so that C 2 D 2 is perpendicular to H: projection c "2 d" 2 ⊥ x. The segment of the desired perpendicular is located || sq. H, and, therefore, m 2 n 2 expresses the desired distance l between AB and CD. We find the position of the projections m "2, and n" 2 on a "2 b" 2 and c "2 d" 2, then the projections and m 1 and m "1, n 1 and n" 1, finally, the projections m "and n ", m and n.

162. Pyramid SABC is given (Fig. 160). Determine the distance between the edge SB and the side AC of the base of the pyramid and construct projections of the common perpendicular to SB and AC, using the method of changing projection planes.

163. Pyramid SABC is given (Fig. 161). Determine the distance between the edge SH and the side BC of the base of the pyramid and construct projections of the common perpendicular to SX and BC using the parallel displacement method.

164*. Determine the distance from point A to the plane in cases where the plane is given: a) by the triangle BCD (Fig. 162, a); b) traces (Fig. 162, b).

Decision. As you know, the distance from a point to a plane is measured by the magnitude of the perpendicular drawn from the point to the plane. This distance is projected onto any square. life-size projections, if the given plane is perpendicular to the square. projections (Fig. 162, c). This situation can be achieved by converting the drawing, for example, by changing the square. projections. Let's introduce the square. S (Fig. 16ts, d), perpendicular to the square. triangle BCD. To do this, we spend in the square. triangle horizontal B-1 and position the axis of projections S perpendicular to the projection b-1 horizontal. We build projections of a point and a plane - a s and a segment c s d s . The distance from a s to c s d s is equal to the desired distance l of the point to the plane.

On rio. 162, d the method of parallel movement is applied. We move the entire system until the B-1 horizontal of the plane becomes perpendicular to the V plane: the projection b 1 1 1 must be perpendicular to the x-axis. In this position, the plane of the triangle will become front-projecting, and the distance l from point A to it will turn out to be square. V without distortion.

On fig. 162b the plane is given by traces. We introduce (Fig. 162, e) an additional square. S, perpendicular to the square. P: the S/H axis is perpendicular to P h . The rest is clear from the drawing. On fig. 162, well the problem is solved with the help of one displacement: pl. P goes into position P 1, that is, it becomes front-projecting. Track. P 1h is perpendicular to the x-axis. We build a front in this position of the plane. the trace of the horizontal is the point n "1, n 1. The trace P 1ϑ will pass through P 1x and n 1. The distance from a" 1 to P 1ϑ is equal to the desired distance l.

165. Pyramid SABC is given (see fig. 160). Determine the distance from point A to the face SBC of the pyramid using the parallel displacement method.

166. Pyramid SABC is given (see fig. 161). Determine the height of the pyramid using the parallel displacement method.

167*. Determine the distance between the intersecting lines AB and CD (see Fig. 159, a) as the distance between parallel planes drawn through these lines.

Decision. On fig. 163, and planes P and Q are shown parallel to each other, of which pl. Q is drawn through CD parallel to AB, and pl. P - through AB parallel to the square. Q. The distance between such planes is considered to be the distance between the skew lines AB and CD. However, you can restrict yourself to building only one plane, for example Q, parallel to AB, and then determine the distance at least from point A to this plane.

On fig. 163c shows plane Q through CD parallel to AB; in projections held with "e" || a"b" and se || ab. Using the method of changing square. projections (Fig. 163, c), we introduce an additional square. S, perpendicular to the square. V and at the same time

perpendicular to square. Q. To draw the S / V axis, we take the frontal D-1 in this plane. Now we draw S / V perpendicular to d "1" (Fig. 163, c). Pl. Q will be displayed on the square. S as a straight line with s d s . The rest is clear from the drawing.

168. Pyramid SABC is given (see Fig. 160). Determine the distance between the edges SC and AB. Apply: 1) method of changing the area. projections, 2) a method of parallel movement.

169*. Determine the distance between parallel planes, one of which is given by straight lines AB and AC, and the other by straight lines DE and DF (Fig. 164, a). Also perform construction for the case when the planes are given by traces (Fig. 164, b).

Decision. The distance (Fig. 164, c) between parallel planes can be determined by drawing a perpendicular from any point of one plane to another plane. On fig. 164, g introduced an additional square. S perpendicular to the square. H and to both given planes. The S.H axis is perpendicular to the horizon. projection of a horizontal line drawn in one of the planes. We build a projection of this plane and points In another plane on Sq. 5. The distance of the point d s to the line l s a s is equal to the desired distance between parallel planes.

On fig. 164, d another construction is given (according to the method of parallel movement). In order for the plane expressed by the intersecting lines AB and AC to be perpendicular to the square. V, horizon. we set the horizontal projection of this plane perpendicular to the x-axis: 1 1 2 1 ⊥ x. Distance between front. the projection d "1 of the point D and the straight line a" 1 2 "1 (frontal projection of the plane) is equal to the desired distance between the planes.

On fig. 164, e shows the introduction of an additional square. S, perpendicular to pl.H and to the given planes P and Q (the S/H axis is perpendicular to the traces P h and Q h). We construct traces Р s , and Q s . The distance between them (see Fig. 164, c) is equal to the desired distance l between the planes P and Q.

On fig. 164, g shows the movement of the planes P 1 n Q 1, to the position P 1 and Q 1 when the horizon. the traces turn out to be perpendicular to the x-axis. Distance between new front. traces P 1ϑ and Q 1ϑ is equal to the desired distance l.

170. Given a parallelepiped ABCDEFGH (Fig. 165). Determine the distances: a) between the bases of the parallelepiped - l 1; b) between faces ABFE and DCGH - l 2 ; c) between the ADHE and BCGF-l 3 faces.

These tasks include: tasks for determining distances from a point to a straight line, to a plane, to a surface; between parallel and intersecting lines; between parallel planes, etc.

All these tasks are united by three circumstances:

First of all, insofar as the shortest distance between such figures is a perpendicular, then all of them are reduced to the construction of mutually perpendicular lines and planes.

Secondly, in each of these problems it is necessary to determine the natural length of the segment, that is, to solve the second main metric problem.

third, these are complex tasks, they are solved in several stages, and at each stage a separate, small specific task is solved.

Let's consider the solution of one of these problems.

Task: Determine distance from point M up to line general position a(Figure 4-26).

Algorithm:

Stage 1: The distance from a point to a line is a perpendicular. Since the direct a- general position, then to construct a perpendicular to it, it is necessary to solve a problem similar to that given on pages M4-4 of this module, that is, first through the point M hold a plane S, perpendicular a. We set this plane, as usual, hÇ f, wherein h1^ a 1, a f2^ a 2

Stage 2: To construct a perpendicular, you need to find a second point for it. This will be the point To belonging to the line a. To find it, you need to solve the positional problem, that is, find the point of intersection of the line a with plane S. We solve 1GPZ according to the third algorithm (Fig. 4-28):

We introduce a plane - an intermediary G, G^^ P 1 , GÉ aÞ Г 1 = a 1;

- GÇ S = b, G^^ P 1Þ b 1 (1 1 2 1) = Г 1 , bÌ SÞ b 2 (1 2 2 2)Ì S2.

- b 2З a 2 = K 2Þ K 1.

Stage 3: Finding the actual size MK right triangle method

The complete solution of the problem is shown in fig. 4-30.

Algorithmic notation of the solution:

1. S^a,S = hЗ f = M, h 1^ a 1 , f 2^ a 2 .

2. We introduce a plane - an intermediary G,

- G^^ P 1 , GÉ aÞ Г 1 = a 1 ;

- GÇ S = b, G^^ P 1Þ b 1 (1 1 2 1) = Г 1 , bÌ SÞ b 2 (1 2 2 2)Ì S2.

- b 2З a 2 = K 2Þ K 1 .

3. Finding the actual size MK.

Findings:

1. The solution of all metric problems is reduced to solving the first basic metric problem - on the mutual perpendicularity of a line and a plane.

2. When determining distances between geometric shapes the second main metric problem is always used - to determine the natural size of the segment.

3. A plane tangent to a surface at one point can be defined by two intersecting lines, each of which is tangent to the given surface.

test questions

1. What tasks are called metric?

2. What two main metric tasks do you know?

3. What is more advantageous to set a plane perpendicular to a straight line in general position?

4. What is the name of the plane perpendicular to one of the level lines?

5. What is the name of a plane perpendicular to one of the projecting lines?

6. What is called a plane tangent to the surface?

It is required to determine the distance from a point to a line. Overall plan problem solving:

- through given point draw a plane perpendicular to a given line;

- find the meeting point of the line

with a plane;

- determine the natural value of the distance.

Through a given point we draw a plane perpendicular to the line AB. The plane is set by the intersecting horizontal and frontal, the projections of which are built according to the perpendicularity algorithm (inverse problem).

Find the meeting point of the line AB with the plane. This is typical task about the intersection of a straight line with a plane (see section "Intersection of a straight line with a plane").

Plane perpendicularity

Planes are mutually perpendicular if one of them contains a line perpendicular to the other plane. Therefore, to draw a plane perpendicular to another plane, you must first draw a perpendicular to the plane, and then draw the desired plane through it. On the diagram, the plane is given by two intersecting straight lines, one of which is perpendicular to the plane ABC.

If the planes are given by traces, then the following cases are possible:

- if two perpendicular planes are projecting, then their collective traces are mutually perpendicular;

- a plane in general position and a projecting plane are perpendicular if the collective trace of the projecting plane is perpendicular to the same-name trace of the plane in general position;

- if like traces of two planes in general position are perpendicular, then the planes are not perpendicular to each other.

Method for replacing projection planes

	projection plane replacements
lies in the fact that the planes
sections are replaced by other flat
	so that	geometric
object in new system planes
projections began to take a private -by
position, which makes it possible to simplify the re-
problem solving. On a spatial scale
ket shows the replacement of the plane V by
new V 1 . It is also shown
point A on the original planes
projections and a new projection plane
V1. When replacing projection planes
the orthogonality of the system is preserved.

Let's transform the spatial layout into a planar layout by rotating the planes along the arrows. We get three projection planes combined into one plane.

Then we remove the projection planes and
		projections
From the plot of the point follows the rule: when
replacing V with V 1 in order to
		frontal
point, it is necessary from the new axis
set aside the applicate point taken from
the previous system of planes
shares. Similarly, one can prove
	replacing H with H 1 is necessary
set the ordinate of the point.

The first typical problem of the method of replacing projection planes

The first typical task of the method of replacing projection planes is the transformation of a line in general position, first into a level line, and then into a projecting line. This problem is one of the main ones, as it is used in solving other problems, for example, in determining the distance between parallel and skew lines, in determining dihedral angle etc.

We make the change V → V 1 .
the axis is drawn parallel to the horizontal
		projections.
frontal projection direct, for
				postpone
point apps. New frontal
the projection of a straight line is an HB straight line.
The straight line itself becomes a frontal.
The angle α ° is determined.

We make the replacement H → H 1. Draw a new axis perpendicular front projection straight. We are building a new horizontal projection straight line, for which we set aside the ordinates of the straight line taken from the previous system of projection planes from the new axis. The line becomes a horizontally projecting line and “degenerates” into a point.

The distance from a point to a line is the length of the perpendicular from the point to the line. AT descriptive geometry it is defined graphically according to the algorithm below.

Algorithm

1. The straight line is transferred to a position in which it will be parallel to any projection plane. To do this, apply the methods of transformation of orthogonal projections.
2. Draw a perpendicular from a point to a line. At the core this construction is the right angle projection theorem.
3. The length of a perpendicular is determined by converting its projections or using the right triangle method.

The following figure is a complex drawing of point M and line b, given by the segment CD. You need to find the distance between them.

According to our algorithm, the first thing to do is to move the line to the position parallel to the plane projections. It is important to understand that after the transformations, the actual distance between the point and the line should not change. That is why it is convenient to use the plane replacement method here, which does not involve moving figures in space.

The results of the first stage of constructions are shown below. The figure shows how an additional frontal plane P 4 is introduced parallel to b. In the new system (P 1 , P 4) points C"" 1 , D"" 1 , M"" 1 are at the same distance from the X 1 axis as C"", D"", M"" from the axis x.

Performing the second part of the algorithm, from M"" 1 we lower the perpendicular M"" 1 N"" 1 to the line b"" 1, since the right angle MND between b and MN is projected onto the plane P 4 in full size. We determine the position of the point N" along the communication line and draw the projection M"N" of the segment MN.

On the final stage it is necessary to determine the value of the segment MN by its projections M"N" and M"" 1 N"" 1 . For this we build right triangle M"" 1 N"" 1 N 0 , whose leg N"" 1 N 0 is equal to the difference (Y M 1 – Y N 1) of the removal of points M" and N" from the X 1 axis. The length of the hypotenuse M"" 1 N 0 of the triangle M"" 1 N"" 1 N 0 corresponds to the desired distance from M to b.

The second way to solve

• Parallel to CD we introduce a new frontal plane П 4 . It intersects P 1 along the X 1 axis, and X 1 ∥C"D". In accordance with the method of replacing planes, we determine the projections of the points C "" 1, D"" 1 and M"" 1, as shown in the figure.
• Perpendicular to C "" 1 D "" 1 we build an additional horizontal plane P 5 on which the straight line b is projected to the point C" 2 \u003d b" 2.
• The distance between the point M and the straight line b is determined by the length of the segment M "2 C" 2 marked in red.

Related tasks: