, Competition "Presentation for the lesson"

Class: 11

Presentation for the lesson

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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

Let us consider a certain plane π and an arbitrary point M 0 in space. Let's choose for the plane unit normal vector n with the beginning 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)

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

since |n| = 1.

If the π plane 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 we can choose

Let (x 0 ; y 0 ; z 0) and (x 1 ; y 1 ; z 1) be the coordinates of the 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 the coordinates of the vector M 1 M 0 can be found: M 1 M 0 = (x 0 -x 1; y 0 -y 1 ; z 0 -z 1 ). Recording 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.

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