PROBLEMS C2 OF THE UNIFIED STATE EXAM IN MATH FOR FINDING THE DISTANCE FROM A POINT TO A PLANE

Kulikova Anastasia Yurievna

5th year student, department of mat. analysis, algebra and geometry EI KFU, RF, Republic of Tatarstan, Elabuga

Ganeeva Aigul Rifovna

scientific adviser, Ph.D. ped. Sci., Associate Professor, EI KFU, RF, Republic of Tatarstan, Elabuga

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

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

Task. Given a rectangular parallelepiped ABWITHDA 1 B 1 C 1 D 1 with sides AB=2, BC=4, AA 1 = 6. Find the distance from the point D to plane ASD 1 .

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

Figure 1. First method

We will carry out DH⊥AS, therefore, by the theorem about three perpendiculars D 1 H⊥AS and (DD 1 H)⊥AS... We will carry out straight DT perpendicular D 1 H... Straight DT lies in the plane DD 1 H, hence DT⊥AC... Hence, DT⊥ASD 1.

ADC find the hypotenuse AS and height DH

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

Answer: .

Method 2.Volume method (using an auxiliary pyramid). The 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.

Rectangular parallelepiped - a parallelepiped, all of whose faces are rectangles.

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

The desired distance is the height h pyramids ACD 1 D dropped from the top D on the basis ACD 1 (fig. 2).

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

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

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

Equating the right-hand sides of the last two equalities, we obtain

Figure 2. Second method

Of right-angled triangles ASD, ADD 1 , CDD 1 find the hypotenuses using the Pythagorean theorem

ACD

We calculate the area of ​​the triangle ASD 1 using Heron's formula

Answer: .

Method 3. Coordinate method.

Let a point be given M(x 0 ,y 0 ,z 0) and the 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 by the formula:

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

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

Figure 3. The 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 one . Substituting the coordinates of the points into it A, C, D 1 we get:

Plane equation ACD 1 will take the form

Answer: .

Method 4. Vector method.

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

Figure 4. Fourth method

, Competition "Presentation for the lesson"

Class: 11

Lesson presentation

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Goals:

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

Equipment:

• multimedia projector;
• computer;
• worksheets with texts of tasks

PROCESS OF THE LESSON

I. Organizational moment

II. Knowledge update 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 the 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.

Let's solve the following tasks:

№1. In the 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.

The 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 ∆ABS 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 ribs AB, AC and AD, if.

When solving problems coordinate method the distance from point M to 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 A1 to plane BDC 1.

We introduce a coordinate system with the origin at point A, the y-axis will run along the AB edge, the x-axis along the AD edge, and the z-axis along the AA 1 edge. Then the coordinates of points B (0; 1; 0) D (1; 0; 0;) C 1 (1; 1; 1)
Let's 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 next method that can be used when solving problems of this type is - method of support tasks.

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

Let's solve the following problem:

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

Consider the application vector method.

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

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

IV. Working in groups

Try to solve the problem in different ways.

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

№2. Find the distance from point A to plane BDC in a regular tetrahedron ABCD with an edge

№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 rectangular pyramid SABCD with all edges equal to 1, find the distance from A to the SCD plane.

V. Lesson summary, homework, reflection

This article talks about determining the distance from a point to a plane. we will parse the method of coordinates, 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 specified in space, then a straight line perpendicular to the plane can be drawn through the point. H 1 is the common point of their intersection. Hence, we find that the segment M 1 H 1 is a perpendicular drawn from the point M 1 to the χ plane, where the point H 1 is the base of the perpendicular.

Definition 1

The distance from a given point to the base of the perpendicular is called, 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 point M 1 to the plane χ is determined as follows: the distance from point M 1 to the plane χ will be the smallest from a given point to any point on the plane. If point Н 2 is located in the plane χ and is not equal to point Н 2, then we get a right-angled triangle of the form М 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 to be inclined, which is drawn from the point M 1 to the plane χ. We have that the perpendicular drawn from a given point to the plane is less inclined, which is drawn from a point to a given plane. Consider this case in the figure below.

Distance from point to plane - theory, examples, solutions

There are a number of geometric problems whose solutions must contain the distance from a point to a plane. The way to detect this can be different. For resolution, 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, it is decided by the method of coordinates. This clause discusses this method.

By the condition of the problem, we have that a point of 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. There are several ways to solve this problem.

The 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 χ plane. 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 χ plane. Then we determine the coordinates (x 2, y 2, z 2) of the point H 1. The required distance from М 1 to the plane χ 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 it, you need to find out the coordinates of the point H 1.

We have that H 1 is the point of intersection of the χ plane with the straight line a, which passes through the point M 1, which is perpendicular to the χ plane. Hence it follows that it is necessary to draw up an equation of 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 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

• draw up the equation of the 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 point H 1, which are points
• intersection of the straight line a with the plane χ;
• calculate the distance from М 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.

The 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. Hence 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 by virtue of the theorem.

Theorem

If a point M 1 (x 1, y 1, z 1) in three-dimensional space is given, having a normal equation of the plane χ of the form cos α x + cos β y + cos γ z - p = 0, then the calculation of the distance from the point to plane M 1 H 1 is produced 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. Hence, we find 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, npn → OM → is the numerical projection of the vector OM → = (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 →, OM → = n → npn → OM → = 1 npn → OM → = npn → OM →, since n → = cos α, cos β, cos γ z and OM → = (x 1, y 1, z 1). The coordinate notation will take the form n →, OM → = cos α x 1 + cos β y 1 + cos γ z 1, then M 1 H 1 = npn → OM → - p = cos α x 1 + cos β Y 1 + cos γ z 1 - p. The theorem is proved.

Hence, we obtain 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-hand 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 z 1, relating to point M 1, taking the absolute value of the obtained value.

Let's 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 starts with calculating the direction vector of the straight line a. By hypothesis, we have that the given equation 2 x - y + 5 z - 3 = 0 is an equation of the plane of general form, 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. 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 take the form x - 5 2 = y - (- 3) - 1 = z - 10 5 ⇔ x - 5 2 = y + 3 - 1 = z - 10 5.

The intersection points should be determined. To do this, it is gentle to combine the equations into a system for the transition from the canonical to the equations of two intersecting straight lines. We will 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 allow 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 Gaussian:

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 obtain

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 its normal form. Determine the normalizing factor and get 1 2 2 + (- 1) 2 + 5 2 = 1 30. From this we derive the equation of the plane 2 30 x - 1 30 y + 5 30 z - 3 30 = 0. The calculation of the left side of the equation is performed by substitution 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 for specifying the plane, 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 specified in three-dimensional space. 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, - one) .

x - 0 y - 2 z - 1 2 - 0 6 - 2 1 - 1 4 - 0 0 - 2 - 1 - 1 = 0 ⇔ xy - 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 the plane A B C has a value of 2 30.

Answer: 2 30.

Finding the distance from a given point on the plane or to the 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 the planes are obtained in several actions.

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 plane O y z, it is normal. Therefore, it is necessary to substitute the value x = - 3 in the left side of the expression and take the modulus of the distance value from the point with coordinates M 1 (- 3, 2, - 7) to the plane. We get the value equal to - 3 = 3.

After transformation, the normal equation of the plane 2 y - 5 = 0 will get 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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Consider in space some plane π and an arbitrary point M 0. Let us choose for the plane unit normal vector n with the beginning at some point М 1 ∈ π, and let p (М 0, π) be the distance from point М 0 to the plane π. Then (fig.5.5)

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

since | n | = 1.

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

Let (x 0; y 0; z 0) and (x 1; y 1; z 1) coordinates of points M 0 and M 1. Then the equality Ax 1 + By 1 + Cz 1 + D = 0 holds, since the point M 1 belongs to the plane, and we 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.