PROBLEMS C2 OF THE UNIFORM STATE EXAMINATION IN MATHEMATICS TO FIND THE DISTANCE FROM A POINT TO A PLANE

Kulikova Anastasia Yurievna

5th year student, Department of Math. analysis, algebra and geometry EI KFU, Russian Federation, Republic of Tatarstan, Elabuga

Ganeeva Aigul Rifovna

scientific supervisor, Ph.D. ped. Sciences, Associate Professor EI KFU, Russian Federation, Republic of Tatarstan, Elabuga

In recent years, tasks on calculating the distance from a point to a plane have appeared in Unified State Examination tasks in mathematics. In this article, using the example of one problem, various methods for finding the distance from a point to a plane are considered. The most suitable method can be used to solve various problems. Having solved a problem using one method, you can check the correctness of the result using another method.

Definition. The distance from a point to a plane not containing this point is the length of the perpendicular segment drawn from this point to the given plane.

Task. Given a rectangular parallelepiped ABWITHD.A. 1 B 1 C 1 D 1 with sides AB=2, B.C.=4, A.A. 1 =6. Find the distance from the point D to plane ACD 1 .

1 way. Using definition. Find the distance r( D, ACD 1) from point D to plane ACD 1 (Fig. 1).

Figure 1. First method

Let's carry out D.H.⊥AC, therefore, by the theorem of three perpendiculars D 1 H⊥AC And (DD 1 H)⊥AC. Let's carry out direct D.T. perpendicular D 1 H. Straight D.T. lies in a plane DD 1 H, hence D.T.⊥A.C.. Hence, D.T.⊥ACD 1.

ADC let's find the hypotenuse AC and height D.H.

From a right triangle D 1 D.H. let's find the hypotenuse D 1 H and height D.T.

Answer: .

Method 2.Volume method (use of an auxiliary pyramid). A problem of this type can be reduced to the problem of calculating the height of a pyramid, where the height of the pyramid is the required distance from a point to a plane. Prove that this height is the required distance; find the volume of this pyramid in two ways and express this height.

Note that with this method there is no need to construct a perpendicular from a given point to a given plane.

A cuboid is a parallelepiped all of whose faces are rectangles.

AB=CD=2, B.C.=AD=4, A.A. 1 =6.

The required distance will be the height h pyramids ACD 1 D, lowered from the top D on the base ACD 1 (Fig. 2).

Let's calculate the volume of the pyramid ACD 1 D two ways.

When calculating, in the first way we take ∆ as the base ACD 1 then

When calculating in the second way, we take ∆ as the base ACD, Then

Let us equate the right-hand sides of the last two equalities and obtain

Figure 2. Second method

From right triangles ACD, ADD 1 , CDD 1 find the hypotenuse using the Pythagorean theorem

ACD

Calculate the area of ​​the triangle ACD 1 using Heron's formula

Answer: .

3 way. Coordinate method.

Let a point be given M(x 0 ,y 0 ,z 0) and plane α , given by the equation ax+by+cz+d=0 in a rectangular Cartesian coordinate system. Distance from point M to the plane α can be calculated using the formula:

Let's introduce a coordinate system (Fig. 3). Origin of coordinates at a point IN;

Straight AB- axis X, straight Sun- axis y, straight BB 1 - axis z.

Figure 3. Third method

B(0,0,0), A(2,0,0), WITH(0,4,0), D(2,4,0), D 1 (2,4,6).

Let ax+by+ cz+ d=0 – plane equation ACD 1 . Substituting the coordinates of points into it A, C, D 1 we get:

Plane equation ACD 1 will take the form

Answer: .

4 way. Vector method.

Let us introduce the basis (Fig. 4) , .

Figure 4. Fourth method

This article talks about determining the distance from a point to a plane. Let's analyze it using the coordinate method, which will allow us to find the distance from a given point in three-dimensional space. To reinforce this, let’s look at examples of several tasks.

The distance from a point to a plane is found using the known distance from a point to a point, where one of them is given, and the other is a projection onto a given plane.

When a point M 1 with a plane χ is specified in space, then a straight line perpendicular to the plane can be drawn through the point. H 1 is their common point of intersection. From this we obtain that the segment M 1 H 1 is a perpendicular drawn from point M 1 to the plane χ, where point H 1 is the base of the perpendicular.

Definition 1

The distance from a given point to the base of a perpendicular drawn from a given point to a given plane is called.

The definition can be written in different formulations.

Definition 2

Distance from point to plane is the length of the perpendicular drawn from a given point to a given plane.

The distance from point M 1 to the χ plane is determined as follows: the distance from point M 1 to the χ plane will be the smallest from a given point to any point on the plane. If point H 2 is located in the χ plane and is not equal to point H 2, then we obtain a right triangle of the form M 2 H 1 H 2 , which is rectangular, where there is a leg M 2 H 1, M 2 H 2 – hypotenuse. This means that it follows that M 1 H 1< M 1 H 2 . Тогда отрезок М 2 H 1 is considered inclined, which is drawn from point M 1 to the plane χ. We have that the perpendicular drawn from a given point to the plane is less than the inclined one drawn from the point to the given plane. Let's look at this case in the figure below.

Distance from a point to a plane - theory, examples, solutions

There are a number of geometric problems whose solutions must contain the distance from a point to a plane. There may be different ways to identify this. To resolve, use the Pythagorean theorem or similarity of triangles. When, according to the condition, it is necessary to calculate the distance from a point to a plane, given in a rectangular coordinate system of three-dimensional space, it is solved by the coordinate method. This paragraph discusses this method.

According to the conditions of the problem, we have that a point in three-dimensional space with coordinates M 1 (x 1, y 1, z 1) with a plane χ is given; it is necessary to determine the distance from M 1 to the plane χ. Several solution methods are used to solve this problem.

First way

This method is based on finding the distance from a point to a plane using the coordinates of point H 1, which are the base of the perpendicular from point M 1 to the plane χ. Next, you need to calculate the distance between M 1 and H 1.

To solve the problem in the second way, use the normal equation of a given plane.

Second way

By condition, we have that H 1 is the base of the perpendicular, which was lowered from point M 1 to the plane χ. Then we determine the coordinates (x 2, y 2, z 2) of point H 1. The required distance from M 1 to the χ plane is found by the formula M 1 H 1 = (x 2 - x 1) 2 + (y 2 - y 1) 2 + (z 2 - z 1) 2, where M 1 (x 1, y 1, z 1) and H 1 (x 2, y 2, z 2). To solve, you need to know the coordinates of point H 1.

We have that H 1 is the point of intersection of the χ plane with the line a, which passes through the point M 1 located perpendicular to the χ plane. It follows that it is necessary to compile an equation for a straight line passing through a given point perpendicular to a given plane. It is then that we will be able to determine the coordinates of point H 1. It is necessary to calculate the coordinates of the point of intersection of the line and the plane.

Algorithm for finding the distance from a point with coordinates M 1 (x 1, y 1, z 1) to the χ plane:

Definition 3

• draw up an equation of straight line a passing through point M 1 and at the same time
• perpendicular to the χ plane;
• find and calculate the coordinates (x 2 , y 2 , z 2) of point H 1, which are points
• intersection of line a with plane χ;
• calculate the distance from M 1 to χ using the formula M 1 H 1 = (x 2 - x 1) 2 + (y 2 - y 1) 2 + z 2 - z 1 2.

Third way

In a given rectangular coordinate system O x y z there is a plane χ, then we obtain a normal equation of the plane of the form cos α · x + cos β · y + cos γ · z - p = 0. From here we obtain that the distance M 1 H 1 with the point M 1 (x 1 , y 1 , z 1) drawn to the plane χ, calculated by the formula M 1 H 1 = cos α x + cos β y + cos γ z - p . This formula is valid, since it was established thanks to the theorem.

Theorem

If a point M 1 (x 1, y 1, z 1) is given in three-dimensional space, having a normal equation of the plane χ of the form cos α x + cos β y + cos γ z - p = 0, then calculating the distance from the point to plane M 1 H 1 is obtained from the formula M 1 H 1 = cos α · x + cos β · y + cos γ · z - p, since x = x 1, y = y 1, z = z 1.

Proof

The proof of the theorem comes down to finding the distance from a point to a line. From this we obtain that the distance from M 1 to the χ plane is the modulus of the difference between the numerical projection of the radius vector M 1 with the distance from the origin to the χ plane. Then we get the expression M 1 H 1 = n p n → O M → - p. The normal vector of the plane χ has the form n → = cos α, cos β, cos γ, and its length is equal to one, n p n → O M → is the numerical projection of the vector O M → = (x 1, y 1, z 1) in the direction determined by the vector n → .

Let's apply the formula for calculating scalar vectors. Then we obtain an expression for finding a vector of the form n → , O M → = n → · n p n → O M → = 1 · n p n → O M → = n p n → O M → , since n → = cos α , cos β , cos γ · z and O M → = (x 1 , y 1 , z 1) . The coordinate form of writing will take the form n → , O M → = cos α · x 1 + cos β · y 1 + cos γ · z 1 , then M 1 H 1 = n p n → O M → - p = cos α · x 1 + cos β · y 1 + cos γ · z 1 - p . The theorem has been proven.

From here we get that the distance from the point M 1 (x 1, y 1, z 1) to the plane χ is calculated by substituting cos α · x + cos β · y + cos γ · z - p = 0 into the left side of the normal equation of the plane instead of x, y, z coordinates x 1, y 1 and z 1, relating to point M 1, taking the absolute value of the obtained value.

Let's look at examples of finding the distance from a point with coordinates to a given plane.

Example 1

Calculate the distance from the point with coordinates M 1 (5, - 3, 10) to the plane 2 x - y + 5 z - 3 = 0.

Solution

Let's solve the problem in two ways.

The first method starts with calculating the direction vector of the line a. By condition, we have that the given equation 2 x - y + 5 z - 3 = 0 is a general plane equation, and n → = (2, - 1, 5) is the normal vector of the given plane. It is used as a direction vector of a straight line a, which is perpendicular to a given plane. It is necessary to write down the canonical equation of a line in space passing through M 1 (5, - 3, 10) with a direction vector with coordinates 2, - 1, 5.

The equation will become x - 5 2 = y - (- 3) - 1 = z - 10 5 ⇔ x - 5 2 = y + 3 - 1 = z - 10 5.

Intersection points must be determined. To do this, gently combine the equations into a system to move from the canonical to the equations of two intersecting lines. Let's take this point as H 1. We get that

x - 5 2 = y + 3 - 1 = z - 10 5 ⇔ - 1 · (x - 5) = 2 · (y + 3) 5 · (x - 5) = 2 · (z - 10) 5 · ( y + 3) = - 1 · (z - 10) ⇔ ⇔ x + 2 y + 1 = 0 5 x - 2 z - 5 = 0 5 y + z + 5 = 0 ⇔ x + 2 y + 1 = 0 5 x - 2 z - 5 = 0

After which you need to enable the system

x + 2 y + 1 = 0 5 x - 2 z - 5 = 0 2 x - y + 5 z - 3 = 0 ⇔ x + 2 y = 1 5 x - 2 z = 5 2 x - y + 5 z = 3

Let us turn to the Gaussian system solution rule:

1 2 0 - 1 5 0 - 2 5 2 - 1 5 3 ~ 1 2 0 - 1 0 - 10 - 2 10 0 - 5 5 5 ~ 1 2 0 - 1 0 - 10 - 2 10 0 0 6 0 ⇒ ⇒ z = 0 6 = 0 , y = - 1 10 10 + 2 z = - 1 , x = - 1 - 2 y = 1

We get that H 1 (1, - 1, 0).

We calculate the distance from a given point to the plane. We take points M 1 (5, - 3, 10) and H 1 (1, - 1, 0) and get

M 1 H 1 = (1 - 5) 2 + (- 1 - (- 3)) 2 + (0 - 10) 2 = 2 30

The second solution is to first bring the given equation 2 x - y + 5 z - 3 = 0 to normal form. We determine the normalizing factor and get 1 2 2 + (- 1) 2 + 5 2 = 1 30. From here we derive the equation of the plane 2 30 · x - 1 30 · y + 5 30 · z - 3 30 = 0. The left side of the equation is calculated by substituting x = 5, y = - 3, z = 10, and you need to take the distance from M 1 (5, - 3, 10) to 2 x - y + 5 z - 3 = 0 modulo. We get the expression:

M 1 H 1 = 2 30 5 - 1 30 - 3 + 5 30 10 - 3 30 = 60 30 = 2 30

Answer: 2 30.

When the χ plane is specified by one of the methods in the section on methods for specifying a plane, then you first need to obtain the equation of the χ plane and calculate the required distance using any method.

Example 2

In three-dimensional space, points with coordinates M 1 (5, - 3, 10), A (0, 2, 1), B (2, 6, 1), C (4, 0, - 1) are specified. Calculate the distance from M 1 to plane A B C.

Solution

First you need to write down the equation of the plane passing through the given three points with coordinates M 1 (5, - 3, 10), A (0, 2, 1), B (2, 6, 1), C (4, 0, - 1) .

x - 0 y - 2 z - 1 2 - 0 6 - 2 1 - 1 4 - 0 0 - 2 - 1 - 1 = 0 ⇔ x y - 2 z - 1 2 4 0 4 - 2 - 2 = 0 ⇔ ⇔ - 8 x + 4 y - 20 z + 12 = 0 ⇔ 2 x - y + 5 z - 3 = 0

It follows that the problem has a solution similar to the previous one. This means that the distance from point M 1 to plane A B C has a value of 2 30.

Answer: 2 30.

Finding the distance from a given point on a plane or to a plane to which they are parallel is more convenient by applying the formula M 1 H 1 = cos α · x 1 + cos β · y 1 + cos γ · z 1 - p. From this we obtain that the normal equations of planes are obtained in several steps.

Example 3

Find the distance from a given point with coordinates M 1 (- 3, 2, - 7) to the coordinate plane O x y z and the plane given by the equation 2 y - 5 = 0.

Solution

The coordinate plane O y z corresponds to an equation of the form x = 0. For the O y z plane it is normal. Therefore, it is necessary to substitute the values ​​x = - 3 into the left side of the expression and take the absolute value of the distance from the point with coordinates M 1 (- 3, 2, - 7) to the plane. We get a value equal to - 3 = 3.

After the transformation, the normal equation of the plane 2 y - 5 = 0 will take the form y - 5 2 = 0. Then you can find the required distance from the point with coordinates M 1 (- 3, 2, - 7) to the plane 2 y - 5 = 0. Substituting and calculating, we get 2 - 5 2 = 5 2 - 2.

Answer: The required distance from M 1 (- 3, 2, - 7) to O y z has a value of 3, and to 2 y - 5 = 0 has a value of 5 2 - 2.

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Determining the distance between: 1 - point and plane; 2 - straight and flat; 3 - planes; 4 - crossing straight lines are considered together, since the solution algorithm for all these problems is essentially the same and consists of geometric constructions that need to be performed to determine the distance between a given point A and plane α. If there is any difference, it consists only in the fact that in cases 2 and 3, before starting to solve the problem, you should mark an arbitrary point A on the straight line m (case 2) or plane β (case 3). distances between crossing lines, we first enclose them in parallel planes α and β and then determine the distance between these planes.

Let us consider each of the noted cases of problem solving.

1. Determining the distance between a point and a plane.

The distance from a point to a plane is determined by the length of a perpendicular segment drawn from a point to the plane.

Therefore, the solution to this problem consists of sequentially performing the following graphical operations:

1) from point A we lower the perpendicular to the plane α (Fig. 269);

2) find the point M of intersection of this perpendicular with the plane M = a ∩ α;

3) determine the length of the segment.

If the plane α is in general position, then in order to lower a perpendicular onto this plane, it is necessary to first determine the direction of the horizontal and frontal projections of this plane. Finding the meeting point of this perpendicular with the plane also requires additional geometric constructions.

The solution to the problem is simplified if the plane α occupies a particular position relative to the projection planes. In this case, both the projection of the perpendicular and the finding of the point of its meeting with the plane are carried out without any additional auxiliary constructions.

EXAMPLE 1. Determine the distance from point A to the frontally projecting plane α (Fig. 270).

SOLUTION. Through A" we draw the horizontal projection of the perpendicular l" ⊥ h 0α, and through A" - its frontal projection l" ⊥ f 0α. We mark the point M" = l" ∩ f 0α . Since AM || π 2, then [A" M"] == |AM| = d.

From the example considered, it is clear how simply the problem is solved when the plane occupies a projecting position. Therefore, if a general position plane is specified in the source data, then before proceeding with the solution, the plane should be moved to a position perpendicular to any projection plane.

EXAMPLE 2. Determine the distance from point K to the plane specified by ΔАВС (Fig. 271).

1. We transfer the plane ΔАВС to the projecting position *. To do this, we move from the system xπ 2 /π 1 to x 1 π 3 /π 1: the direction of the new x 1 axis is chosen perpendicular to the horizontal projection of the horizontal plane of the triangle.

2. Project ΔABC onto a new plane π 3 (the ΔABC plane is projected onto π 3, in [ C " 1 B " 1 ]).

3. Project point K onto the same plane (K" → K" 1).

4. Through the point K" 1 we draw (K" 1 M" 1)⊥ the segment [C" 1 B" 1]. The required distance d = |K" 1 M" 1 |

The solution to the problem is simplified if the plane is defined by traces, since there is no need to draw projections of level lines.

EXAMPLE 3. Determine the distance from point K to the plane α, specified by the tracks (Fig. 272).

* The most rational way to transfer the triangle plane to the projecting position is to replace the projection planes, since in this case it is enough to construct only one auxiliary projection.

SOLUTION. We replace the plane π 1 with the plane π 3, for this we draw a new axis x 1 ⊥ f 0α. On h 0α we mark an arbitrary point 1" and determine its new horizontal projection on the plane π 3 (1" 1). Through the points X α 1 (X α 1 = h 0α 1 ∩ x 1) and 1" 1 we draw h 0α 1. We determine the new horizontal projection of the point K → K" 1. From point K" 1 we lower the perpendicular to h 0α 1 and mark the point of its intersection with h 0α 1 - M" 1. The length of the segment K" 1 M" 1 will indicate the required distance.

2. Determining the distance between a straight line and a plane.

The distance between a line and a plane is determined by the length of a perpendicular segment dropped from an arbitrary point on the line to the plane (see Fig. 248).

Therefore, the solution to the problem of determining the distance between straight line m and plane α is no different from the examples discussed in paragraph 1 for determining the distance between a point and a plane (see Fig. 270 ... 272). As a point, you can take any point belonging to line m.

3. Determination of the distance between planes.

The distance between the planes is determined by the size of the perpendicular segment dropped from a point taken on one plane to another plane.

From this definition it follows that the algorithm for solving the problem of finding the distance between planes α and β differs from a similar algorithm for solving the problem of determining the distance between line m and plane α only in that line m must belong to plane α, i.e., in order to determine the distance between planes α and β follows:

1) take a straight line m in the α plane;

2) select an arbitrary point A on line m;

3) from point A, lower the perpendicular l to the plane β;

4) determine point M - the meeting point of the perpendicular l with the plane β;

5) determine the size of the segment.

In practice, it is advisable to use a different solution algorithm, which will differ from the one given only in that, before proceeding with the first step, the planes should be transferred to the projection position.

Including this additional operation in the algorithm simplifies the execution of all other points without exception, which ultimately leads to a simpler solution.

EXAMPLE 1. Determine the distance between planes α and β (Fig. 273).

SOLUTION. We move from the system xπ 2 /π 1 to x 1 π 1 /π 3. With respect to the new plane π 3, the planes α and β occupy a projecting position, therefore the distance between the new frontal traces f 0α 1 and f 0β 1 is the desired one.

In engineering practice, it is often necessary to solve the problem of constructing a plane parallel to a given plane and removed from it at a given distance. Example 2 below illustrates the solution to such a problem.

EXAMPLE 2. It is required to construct projections of a plane β parallel to a given plane α (m || n), if it is known that the distance between them is d (Fig. 274).

1. In the α plane, draw arbitrary horizontal lines h (1, 3) and front lines f (1,2).

2. From point 1 we restore the perpendicular l to the plane α(l" ⊥ h", l" ⊥ f").

3. On the perpendicular l we mark an arbitrary point A.

4. Determine the length of the segment - (the position indicates on the diagram the metrically undistorted direction of the straight line l).

5. Lay out the segment = d on the straight line (1"A 0) from point 1".

6. Mark on the projections l" and l" points B" and B", corresponding to point B 0.

7. Through point B we draw the plane β (h 1 ∩ f 1). To β || α, it is necessary to comply with the condition h 1 || h and f 1 || f.

4. Determining the distance between intersecting lines.

The distance between intersecting lines is determined by the length of the perpendicular contained between the parallel planes to which the intersecting lines belong.

In order to draw mutually parallel planes α and β through intersecting straight lines m and f, it is sufficient to draw through point A (A ∈ m) a straight line p parallel to straight line f, and through point B (B ∈ f) a straight line k parallel to straight m . The intersecting lines m and p, f and k define the mutually parallel planes α and β (see Fig. 248, e). The distance between the planes α and β is equal to the required distance between the crossing lines m and f.

Another way can be proposed for determining the distance between intersecting lines, which consists in the fact that, using some method of transforming orthogonal projections, one of the intersecting lines is transferred to the projecting position. In this case, one projection of the line degenerates into a point. The distance between the new projections of crossing lines (point A" 2 and segment C" 2 D" 2) is the required one.

In Fig. 275 shows a solution to the problem of determining the distance between crossing lines a and b, given segments [AB] and [CD]. The solution is performed in the following sequence:

1. Transfer one of the crossing lines (a) to a position parallel to the plane π 3; To do this, move from the system of projection planes xπ 2 /π 1 to the new x 1 π 1 /π 3, the x 1 axis is parallel to the horizontal projection of straight line a. Determine a" 1 [A" 1 B" 1 ] and b" 1.

2. By replacing the plane π 1 with the plane π 4, we translate the straight line

and to position a" 2, perpendicular to the plane π 4 (the new x 2 axis is drawn perpendicular to a" 1).

3. Construct a new horizontal projection of straight line b" 2 - [ C" 2 D" 2 ].

4. The distance from point A" 2 to straight line C" 2 D" 2 (segment (A" 2 M" 2 ] (is the required one.

It should be borne in mind that the transfer of one of the crossing lines to the projecting position is nothing more than the transfer of the planes of parallelism, in which the lines a and b can be enclosed, also to the projecting position.

In fact, by moving line a to a position perpendicular to the plane π 4, we ensure that any plane containing line a is perpendicular to the plane π 4, including the plane α defined by lines a and m (a ∩ m, m || b ). If we now draw a line n, parallel to a and intersecting line b, then we obtain the plane β, which is the second plane of parallelism, which contains the intersecting lines a and b. Since β || α, then β ⊥ π 4 .

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Goals:

• generalization and systematization of students’ knowledge and skills;
• development of skills to analyze, compare, draw conclusions.

Equipment:

• multimedia projector;
• computer;
• sheets with problem texts

PROGRESS OF THE CLASS

I. Organizational moment

II. Knowledge updating stage(slide 2)

We repeat how the distance from a point to a plane is determined

III. Lecture(slides 3-15)

In this lesson we will look at various ways to find the distance from a point to a plane.

First method: step-by-step computational

Distance from point M to plane α:
– equal to the distance to the plane α from an arbitrary point P lying on a straight line a, which passes through the point M and is parallel to the plane α;
– is equal to the distance to the plane α from an arbitrary point P lying on the plane β, which passes through the point M and is parallel to the plane α.

We will solve the following problems:

№1. In cube A...D 1, find the distance from point C 1 to plane AB 1 C.

It remains to calculate the value of the length of the segment O 1 N.

№2. In a regular hexagonal prism A...F 1, all edges of which are equal to 1, find the distance from point A to the plane DEA 1.

Next method: volume method.

If the volume of the pyramid ABCM is equal to V, then the distance from point M to the plane α containing ∆ABC is calculated by the formula ρ(M; α) = ρ(M; ABC) =
When solving problems, we use the equality of volumes of one figure, expressed in two different ways.

Let's solve the following problem:

№3. Edge AD of pyramid DABC is perpendicular to the base plane ABC. Find the distance from A to the plane passing through the midpoints of the edges AB, AC and AD, if.

When solving problems coordinate method the distance from point M to plane α can be calculated using the formula ρ(M; α) = , where M(x 0; y 0; z 0), and the plane is given by the equation ax + by + cz + d = 0

Let's solve the following problem:

№4. In a unit cube A...D 1, find the distance from point A 1 to plane BDC 1.

Let's introduce a coordinate system with the origin at point A, the y-axis will run along edge AB, the x-axis along edge AD, and the z-axis along edge AA 1. Then the coordinates of the points B (0; 1; 0) D (1; 0; 0;) C 1 (1; 1; 1)
Let's create an equation for a plane passing through points B, D, C 1.

Then – dx – dy + dz + d = 0 x + y – z – 1= 0. Therefore, ρ =

The following method that can be used to solve problems of this type is method of support problems.

The application of this method consists in the use of known reference problems, which are formulated as theorems.

Let's solve the following problem:

№5. In a unit cube A...D 1, find the distance from point D 1 to plane AB 1 C.

Let's consider the application vector method.

№6. In a unit cube A...D 1, find the distance from point A 1 to plane BDC 1.

So, we looked at various methods that can be used to solve this type of problem. The choice of one method or another depends on the specific task and your preferences.

IV. Group work

Try solving the problem in different ways.

№1. The edge of the cube A...D 1 is equal to . Find the distance from vertex C to plane BDC 1.

№2. In a regular tetrahedron ABCD with an edge, find the distance from point A to the plane BDC

№3. In a regular triangular prism ABCA 1 B 1 C 1 all edges of which are equal to 1, find the distance from A to the plane BCA 1.

№4. In a regular quadrilateral pyramid SABCD, all edges of which are equal to 1, find the distance from A to the plane SCD.

V. Lesson summary, homework, reflection