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TASKS C2 OF THE UNIFIED STATE EXAM IN MATHEMATICS FOR FINDING 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 for calculating the distance from a point to a plane have appeared in USE assignments 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. To solve various problems, you can use the most appropriate method. Having solved the problem with one method, another method can check the correctness of the result.

Definition. The distance from a point to a plane that does not contain this point is the length of the segment of the perpendicular dropped from this point to the given plane.

A task. Given a rectangular parallelepiped BUTBFROMDA 1 B 1 C 1 D 1 with sides AB=2, BC=4, AA 1=6. Find the distance from a point D up to the plane ACD 1 .

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

Figure 1. First way

Let's spend D.H.⊥AC, therefore, by the theorem on three perpendiculars D 1 H⊥AC And (DD 1 H)⊥AC. Let's spend direct DT perpendicular D 1 H. Straight DT lies in the plane DD 1 H, hence DT⊥AC. Consequently, DT⊥ACD 1.

BUTDC find the hypotenuse AC and height D.H.

From a right triangle D 1 D.H. find the hypotenuse D 1 H and height DT

Answer: .

2 way.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 desired distance from a point to a plane. Prove that this height is the desired 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 cuboid all of whose faces are rectangles.

AB=CD=2, BC=AD=4, AA 1 =6.

The desired distance will be the height h pyramids ACD 1 D, dropped from the top D on the ground ACD 1 (Fig. 2).

Calculate the volume of the pyramid ACD 1 D two ways.

Calculating, in the first way, we take ∆ as the basis ACD 1 , then

Calculating, in the second way, we take ∆ as the basis ACD, then

Equate the right-hand sides of the last two equalities, we get

Figure 2. The second way

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

ACD

Calculate the area of ​​a 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 rectangular Cartesian coordinates. Distance from point M to the plane α can be calculated by the formula:

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

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

Figure 3. The third way

B(0,0,0), BUT(2,0,0), FROM(0,4,0), D(2,4,0), D 1 (2,4,6).

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

Plane equation ACD 1 will take the form

Answer: .

4 way. vector method.

We introduce the basis (Fig. 4) , .

Figure 4. The fourth way

, Competition "Presentation for the lesson"

Class: 11

Presentation for the lesson

Back forward

Attention! The slide preview is for informational purposes only and may not represent the full extent of the presentation. If you are interested in this work, please download the full version.

Goals:

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

Equipment:

• multimedia projector;
• a computer;
• task sheets

STUDY PROCESS

I. Organizational moment

II. The stage of updating knowledge(slide 2)

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

III. Lecture(slides 3-15)

In the 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:
– is equal to the distance to the plane α from an arbitrary point P lying on the 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 tasks:

№1. In the cube A ... D 1 find the distance from the point C 1 to the 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 V, then the distance from the 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 the volumes of one figure, expressed in two different ways.

Let's solve the following problem:

№3. The edge AD of the pyramid DABC is perpendicular to the plane of the base 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 the point M to the plane α can be calculated by 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 the unit cube A…D 1 find the distance from point A 1 to plane BDC 1 .

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

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

The following method, which can be used in solving problems of this type - method of reference tasks.

The application of this method consists in the application of well-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 the point D 1 to the plane AB 1 C.

Consider Application vector method.

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

So, we have considered various methods that can be used in solving this type of problem. The choice of one or another method depends on the specific task and your preferences.

IV. Group work

Try to solve the problem in different ways.

№1. The edge of the cube А…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 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 quadrangular pyramid SABCD, all edges of which are equal to 1, find the distance from A to the plane SCD.

V. Lesson summary, homework, reflection

Consider some plane π and an arbitrary point M 0 in space. Let's choose for the plane unit normal vector n s start at some point M 1 ∈ π, and let p(M 0 ,π) be the distance from the point M 0 to the plane π. Then (Fig. 5.5)

p(M 0 ,π) = | pr n M 1 M 0 | = |nM 1 M 0 |, (5.8)

since |n| = 1.

If the plane π is given in rectangular coordinate system with its general equation Ax + By + Cz + D = 0, then its normal vector is the vector with coordinates (A; B; C) and as the unit normal vector we can choose

Let (x 0 ; y 0 ; z 0) and (x 1 ; y 1 ; z 1) be the coordinates of points M 0 and M 1 . Then the equality Ax 1 + By 1 + Cz 1 + D = 0 is satisfied, since the point M 1 belongs to the plane, and you can find the coordinates of the vector M 1 M 0 : M 1 M 0 = (x 0 -x 1; y 0 -y 1; z 0 -z 1). writing down scalar product nM 1 M 0 in coordinate form and transforming (5.8), we obtain

since Ax 1 + By 1 + Cz 1 = - D. So, to calculate the distance from a point to a plane, you need to substitute the coordinates of the point into the general equation of the plane, and then divide the absolute value of the result by a normalizing factor equal to the length of the corresponding normal vector.

This article talks about determining the distance from a point to a plane. let's analyze the coordinate method, which will allow us to find the distance from a given point in three-dimensional space. To consolidate, consider examples of several tasks.

The distance from a point to a plane is found by means of a 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 given in space, then a straight line perpendicular to the plane can be drawn through the point. H 1 is a common point of their intersection. From here we get that the segment M 1 H 1 is a perpendicular, which was drawn from the point M 1 to the plane χ, where the point H 1 is the base of the perpendicular.

Definition 1

They call the distance from a given point to the base of the perpendicular, which was drawn from a given point to a given plane.

The definition can be written in different formulations.

Definition 2

Distance from point to plane called the length of the perpendicular, which was drawn from a given point to a given plane.

The distance from the point M 1 to the plane χ is defined as follows: the distance from the point M 1 to the plane χ will be the smallest from a given point to any point in the plane. If the point H 2 is located in the χ plane and is not equal to the point H 2, then we get 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. Hence, this implies that M 1 H 1< M 1 H 2 . Тогда отрезок М 2 H 1 is considered inclined, which is drawn from the point M 1 to the plane χ. We have that the perpendicular drawn from a given point to a plane is less than the inclined one drawn from a point to a given plane. Consider 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. Ways to detect this may be different. To resolve, use the Pythagorean theorem or the 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, they solve using the coordinate method. This paragraph deals with this method.

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

First way

This method is based on finding the distance from a point to a plane using the coordinates of the point H 1, which are the base of the perpendicular from the 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, the normal equation of a given plane is used.

Second way

By condition, we have that H 1 is the base of the perpendicular, which was lowered from the point M 1 to the plane χ. Then we determine the coordinates (x 2, y 2, z 2) of the point H 1. The desired distance from M 1 to the χ plane is found by the formula M 1 H 1 \u003d (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 the point H 1.

We have that H 1 is the point of intersection of the plane χ with the line a, which passes through the point M 1 located perpendicular to the plane χ. It follows that it is necessary to formulate the equation of a straight line passing through a given point perpendicular to a given plane. It is then that we can determine the coordinates of the 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

• compose the equation of a straight line a passing through the 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 the point H 1, which are points
• intersection of the line a with the 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 get 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 is 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 χ plane 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 derived 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 is reduced to finding the distance from a point to a line. From here we get 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 the notation 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 into the left side of the normal equation of the plane cos α x + cos β y + cos γ z - p = 0 instead of x, y, z coordinates x 1 , y 1 and z1 relating to the point M 1 , taking the absolute value of the obtained value.

Consider 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 will start by 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 directing vector of the straight line a, which is perpendicular to the given plane. You should write the canonical equation of a straight line in space passing through M 1 (5, - 3, 10) with a direction vector with coordinates 2, - 1, 5.

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

Intersection points should be defined. To do this, gently combine the equations into a system for the transition 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

Then 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 rule for solving the system according to Gauss:

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 a plane. We take points M 1 (5, - 3, 10) and H 1 (1, - 1, 0) and get

M 1 H 1 \u003d (1 - 5) 2 + (- 1 - (- 3)) 2 + (0 - 10) 2 \u003d 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 \u003d 5, y \u003d - 3, z \u003d 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 \u003d 2 30 5 - 1 30 - 3 + 5 30 10 - 3 30 \u003d 60 30 \u003d 2 30

Answer: 2 30 .

When the χ plane is given by one of the methods of the plane definition methods section, then you first need to obtain the equation of the χ plane and calculate the desired distance using any method.

Example 2

Points with coordinates M 1 (5 , - 3 , 10) , A (0 , 2 , 1) , B (2 , 6 , 1) , C (4 , 0 , - 1) are set in three-dimensional space. Calculate the distance from M 1 to the 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 , - one) .

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

It follows that the problem has a solution similar to the previous one. Hence, the distance from the point M 1 to the plane A B C is 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 here we get that the normal equations of the 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 \u003d - 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 the 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 desired 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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