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Any plane in Cartesian system coordinates can be given by the equation `Ax + By + Cz + D = 0`, where at least one of the numbers `A`, `B`, `C` is non-zero. Let a point `M (x_0;y_0;z_0)` be given, let’s find the distance from it to the plane `Ax + By + Cz + D = 0`.
Let the line passing through the point `M` perpendicular to the plane `alpha`, intersects it at point `K` with coordinates `(x; y; z)`. Vector `vec(MK)` is perpendicular to the `alpha` plane, as is the vector `vecn` `(A;B;C)`, i.e., the vectors `vec(MK)` and `vecn` collinear, `vec(MK)= λvecn`.
Since `(x-x_0;y-y_0;z-z-0)` and `vecn(A,B,C)`, then `x-x_0=lambdaA`, `y-y_0=lambdaB`, `z-z_0=lambdaC`.
Point `K` lies in the `alpha` plane (Fig. 6), its coordinates satisfy the equation of the plane. We substitute `x=x_0+lambdaA`, `y=y_0+lambdaB`, `z=z_0+lambdaC` into the equation `Ax+By+Cz+D=0`, we get
`A(x_0+lambdaA)+(B(y_0+lambdaB)+C(z_0+lambdaC)+D=0`,
whence `lambda=-(Ax_0+By_0+Cz_0+D)/(A^2+B^2+C^2)`.
Find the length of the vector `vec(MK)`, which is equal to the distance from the point `M(x_0;y_0;z_0)` to the plane `Ax + By + Cz + D` `|vec(MK)|=|lambdavecn|=|lambda|*sqrt(A^2+B^2+C^2)`.
So, the distance `h` from the point `M(x_0;y_0;z_0)` to the plane `Ax + By + Cz + D = 0` is as follows
`h=(|Ax_0+By_0+Cz_0+D|)/(sqrt(A^2+B^2+C^2))`.
Using the geometric method of finding the distance from point `A` to the plane `alpha`, find the base of the perpendicular `A A^"`, lowered from point `A` to the plane `alpha`. If point `A^"` is located outside the section of the plane `alpha` specified in the problem, then through point `A` draw a straight line `c`, parallel to the plane`alpha`, and choose a more convenient point `C` on it, orthographic projection which `C^"` belongs to this section of the `alpha` plane. Length of segment `C C^"`will be equal to the required distance from point `A`to the `alpha` plane.
In the right hexagonal prism`A...F_1`, all edges of which are equal to `1`, find the distance from point `B` to the plane `AF F_1`.
Let `O` be the center of the lower base of the prism (Fig. 7). The straight line `BO` is parallel to the straight line `AF` and, therefore, the distance from the point `B` to the plane `AF F_1` is equal to the distance `OH` from the point `O` to the plane `AF F_1`. In the triangle `AOF` we have `AO=OF=AF=1`. The height `OH` of this triangle is `(sqrt3)/2`. Therefore, the required distance is `(sqrt3)/2`.
Let's show another way (auxiliary volume method) finding the distance from a point to a plane. It is known that the volume of the pyramid `V` , the area of its base `S`and height length `h`are related by the formula `h=(3V)/S`. But the length of the height of a pyramid is nothing more than the distance from its top to the plane of the base. Therefore, to calculate the distance from a point to a plane, it is enough to find the volume and area of the base of some pyramid with the apex at this point and with the base lying in this plane.
Dana correct prism`A...D_1`, in which `AB=a`, `A A_1=2a`. Find the distance from the intersection point of the diagonals of the base `A_1B_1C_1D_1` to the plane `BDC_1`.
Consider the tetrahedron `O_1DBC_1` (Fig. 8). The required distance `h` is the length of the height of this tetrahedron, lowered from the point `O_1` to the plane of the face `BDC_1` . To find it, it is enough to know the volume `V`tetrahedron `O_1DBC_1` and area triangle `DBC_1`. Let's calculate them. Note that straight line `O_1C_1` perpendicular to the plane `O_1DB`, because it is perpendicular to `BD` and `B B_1` . This means that the volume of the tetrahedron is `O_1DBC_1` equals
Back forward
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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
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 class we will look at various ways finding 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 can be used to solve problems of this type – method supporting tasks.
Application this method consists in the application 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 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
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 given point three-dimensional space. To reinforce this, let’s look at examples of several tasks.
Yandex.RTB R-A-339285-1
The distance from a point to a plane is found by known distance from point to 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 through the point you can draw perpendicular to the plane direct. H 1 is common point their intersections. 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
Call the distance from a given point to the base of a perpendicular drawn from a given point to given plane.
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 get right triangle type 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 geometric problems, the solutions of which must contain the distance from the point to the 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, specified in rectangular system coordinates of three-dimensional space are 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 normal equation 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 the normal equation of the plane type cosα · 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 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 the distance from the point to the plane M 1 H 1 is calculated 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 calculation formula 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 χ plane is calculated by substituting into left side 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, related to point M 1, taking absolute value 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 an equation of the plane general view, 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. Should be written down canonical equation a straight 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. This point let's take 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 coordinate plane O x y z and 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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