, Competition "Presentation for the lesson"
Class: 11
Presentation for the lesson
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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;
- 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 when 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. The result of the lesson, homework, reflection
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 USE assignments in mathematics at last years there are problems to calculate the distance from a point to a plane. In this article, using the example of one task, we consider various methods finding the distance from a point to a plane. 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.
Task. Dan cuboid ABWITHDA 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. Hence, DT⊥ACD 1.
ADC 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 when 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 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 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), 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 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
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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.
