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Searching the distance from the point to the plane is a frequent task that occurs when solving various tasks of analytical geometry, for example, to this problem, you can reduce the distance between two cross-lived straight or between the straight and parallel plane.

Consider the plane $ β $ and the point $ M_0 $ with coordinates $ (x_0; y_0; z_0) $, not belonging to the $ β $ plane.

Definition 1.

The shortest distance between the point and the plane will be perpendicular, lowered from the $ m_0 $ point to the plane $ β $.

Figure 1. Distance from point, to the plane. Author24 - Student Internet Exchange

The following is considered to find the distance from the point to the plane by the coordinate method.

Displays the formula for the coordinate method for searching the distance from the point to the plane in space

Perpendicular from point $ M_0 $ intersecting with a $ β $ plane at $ M_1 point with coordinates $ (x_1; y_1; z_1) $, lies on a straight line, whose guide vector is the normal vector of the plane $ β $. In this case, the length of the unit vector $ n $ is equal to one. Accordingly, this distance from $ β $ to the $ M_0 point will be:

$ ρ \u003d | \\ VEC (N) \\ CDOT \\ VEC (M_1M_0) | \\ Left (1 \\ Right) $, where $ \\ vec (m_1m_0) $ is a normal plane vector $ β $, and $ \\ vec (n) $ - single normal vector of the plane under consideration.

In the case when the equation of the plane is specified in the total form $ AX + BY + CZ + D \u003d 0 $, the coordinates of the normal plane vector are the coefficients of the equation $ \\ (a; b; c \\) $, and the single normal vector in this case has coordinates calculated by the following equation:

$ \\ vec (n) \u003d \\ frac (\\ (a; b; c \\)) (\\ sqrt (a ^ 2 + b ^ 2 + c ^ 2)) \\ left (2 \\ Right) $.

Now you can find the coordinates of the normal vector $ \\ VEC (M_1M_0) $:

$ \\ VEC (m_0m_1) \u003d \\ (x_0 - x_1; y_0-y_1; z_0-z_1 \\) \\ left (3 \\ right) $.

Also express the coefficient of $ D $ using the coordinates of the point lying in the $ β plane:

$ D \u003d ax_1 + by_1 + cz_1 $

The coordinates of the single normal vector from the equality $ (2) $ can be substituted into the plane equation $ β $, then we have:

$ ρ \u003d \\ FRAC (| A (x_0 -x_1) + b (y_0-y_1) + c (z_0-z_1) |) (\\ sqrt (a ^ 2 + b ^ 2 + C ^ 2)) \u003d \\ FRAC ( | Ax_0 + by_0 + cz_0- (ax_1 + by_1 + cz_1) |) (\\ sqrt (a ^ 2 + b ^ 2 + C ^ 2)) \u003d \\ FRAC (AX_0 + BY_0 + CZ_0 + D) (\\ SQRT (A ^ 2 + B ^ 2 + C ^ 2)) \\ Left (4 \\ Right) $

Equality $ (4) $ is the formula for finding the distance from the point to the plane in space.

The overall algorithm for finding the distance from the point $ M_0 $ to the plane

1. If the equation of the plane is not specified in general form, it is necessary to bring it to the common one for a start.
2. After that, it is necessary to express from the general plane equation normal vector of this plane through a point $ M_0 $ and a point belonging to the specified plane, for this you need to use the equality $ (3) $.
3. The next step is to search for the coordinates of the unit normal vector of the plane by the formula $ (2) $.
4. Finally, it is possible to begin searching the distance from the point to the plane, this is carried out by calculating the scalar product of the vectors of $ \\ vec (n) $ and $ \\ vec (m_1m_0) $.

This article tells about determining the distance from the point to the plane. We will analyze the method of coordinates, which will allow the distance from a given point of the three-dimensional space. For consolidation, consider examples of several tasks.

The distance from the point to the plane is located by means of a known distance from the point to the point, where one of them is specified, and the other is a projection on a given plane.

When a point M 1 is set in the space with a plane χ, then perpendicular plane can be carried out through the point. H 1 is a common point of their intersection. From here we obtain that the segment M 1 H 1 is a perpendicular, which carried out 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 carried out from a given point to the specified plane.

The definition can be recorded in different formulations.

Definition 2.

Distance from point to plane They call the length of the perpendicular, which spent from a given point to a given plane.

The distance from the point M 1 to the plane χ is determined as follows: The distance from the point M 1 to the plane χ will be the smallest of the specified point to any point of the plane. If the point H 2 is located in the plane χ and is not equal to the point H 2, then we obtain a rectangular triangle of the m 2 H 1 H 2 which is rectangular, where there is a catat m 2 H 1, m 2 H 2 - hypotenuse. So, hence it follows that M 1 H 1< M 1 H 2 . Тогда отрезок М 2 H 1 it is considered oblique, which is carried out from the point M 1 to the plane χ. We have that perpendicular, carried out from a given point to the plane, is less inclined, which is carried out from the point to the specified plane. Consider this case in the figure below.

Distance from point to plane - Theory, examples, solutions

There are a number of geometric tasks whose solutions must contain the distance from the point to the plane. Methods for identifying this can be different. For permission, the theorem of Pythagora or the similarity of triangles is used. When, by condition, it is necessary to calculate the distance from the point to the plane, specified in the rectangular coordinate system of the three-dimensional space, solve the coordinate method. This item considers this method.

By the problem of the problem, we have that the point of three-dimensional space with coordinates M 1 (x 1, y 1, z 1) is given to the plane χ, it is necessary to determine the distance from M 1 to the plane χ. To solve several ways to solve.

First method

This method is based on the distance from the point to the 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, it is necessary to calculate the distance between M 1 and H 1.

To solve the problem, the normal equation of the specified plane is used in the second method.

Second way

By condition, we have that H 1 is the basis 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) points H 1. The desired distance from M 1 to the plane χ is located according to 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, it is necessary to know the coordinates of the point H 1.

We have that H 1 is the point of intersection of the plane χ with direct A, which passes through the point M 1 located perpendicular to the plane χ. It follows that it is necessary to compile the equation of direct passing through the specified point perpendicular to the specified plane. It is then that we can determine the coordinates of the point H 1. It is necessary to calculate the coordinates of the intersection of the direct and plane.

Algorithm for finding the distance from the point with coordinates M 1 (x 1, y 1, z 1) to the plane χ:

Definition 3.

• make the equation of direct A passing through the point M 1 and at the same time
• perpendicular to the plane χ;
• find and calculate the coordinates (x 2, y 2, z 2) points H 1, which are points
• intersection direct a with a plane χ;
• calculate the distance from M 1 to χ using the formula M 1 H 1 \u003d (x 2 - x 1) 2 + (y 2 - y 1) 2 + z 2 - z 1 2.

Third way

In a given rectangular coordinate system, the coordinates of x in Z has a plane χ, then we obtain the normal equation of the plane of the form cos α · x + cos β · y + cos γ · z - p \u003d 0. From here we obtain that the distance M 1 H 1 with a point M 1 (x 1, y 1, z 1), carried out on the plane χ calculated by the formula M 1 H 1 \u003d Cos α · x + cos β · y + cos γ · z - p. This formula is valid, since this is established thanks to the theorem.

Theorem

If point M 1 (x 1, y 1, z 1) is specified in a three-dimensional space having a normal equation of the plane of the form Cos α · x + cos β · y + cos γ · z - p \u003d 0, then calculating the distance from point to The M 1 H 1 plane is made from the formula M 1 H 1 \u003d cos α · x + cos β · y + cos γ · z - p, since x \u003d x 1, y \u003d y 1, z \u003d z 1.

Evidence

The proof of the theorem is reduced to finding the distance from the point to direct. From here we obtain that the distance from M 1 to the plane χ is the module of the difference in the numerical projection of the radius vector M 1 with distance from the origin to the plane χ. Then we obtain the expression M 1 H 1 \u003d n p n → o m → - p. The normal vector of the plane χ has the form n → \u003d cos α, cos β, cos γ, and its length is equal to one, NPN → OM → - the numerical projection of the vector OM → \u003d (X 1, Y 1, Z 1) in the direction determined by the vector N →.

Apply the formula for calculating scalar vectors. Then we obtain an expression to find the vector of the form N →, Om → \u003d N → · NPN → OM → \u003d 1 · NPN → OM → \u003d NPN → OM →, since n → \u003d cos α, cos β, cos γ · z and om → \u003d (x 1, y 1, z 1). The coordinate form of the recording will take the form n →, om → \u003d cos α · x 1 + cos β · y 1 + cos γ · z 1, then M 1 H 1 \u003d NPN → OM → - P \u003d COS α · x 1 + cos β · Y 1 + cos γ · z 1 - p. Theorem is proved.

From here we obtain that the distance from the point M 1 (x 1, y 1, z 1) to the plane χ is calculated using the substitution to the left part of the normal equation of the COS α · x + cos β · y + cos γ · z - p \u003d 0 instead of x, y, z coordinates x 1, y 1 and z 1. belonging to the point M 1, taking the absolute value of the obtained value.

Consider examples of finding the distance from the point with the coordinates to the specified plane.

Example 1.

Calculate the distance from the point with the coordinates M 1 (5, - 3, 10) to the plane 2 x - y + 5 z - 3 \u003d 0.

Decision

We will solve the problem in two ways.

The first method will begin with the calculation of the guide vector direct a. By condition, we have that the given equation 2 x - y + 5 z - 3 \u003d 0 is the equation of the plane of the general form, and n → \u003d (2, - 1, 5) is a normal vector of the specified plane. It is used as the guide vector direct A, which is perpendicular to the specified plane. The canonical equation is to be recorded in space, passing through M 1 (5, - 3, 10) with a guide vector with coordinates 2, - 1, 5.

The equation will receive the form x - 5 2 \u003d y - (- 3) - 1 \u003d z - 10 5 ⇔ x - 5 2 \u003d y + 3 - 1 \u003d z - 10 5.

You should define the intersection points. To do this, gently combine the equations into the system to transition from the canonical to the equations of two intersecting straight lines. This point will take it out of 1. We get that

x - 5 2 \u003d y + 3 - 1 \u003d z - 10 5 ⇔ - 1 · (x - 5) \u003d 2 · (y + 3) 5 · (x - 5) \u003d 2 · (z - 10) 5 · ( y + 3) \u003d - 1 · (z - 10) ⇔ ⇔ x + 2 y + 1 \u003d 0 5 x - 2 z - 5 \u003d 0 5 y + z + 5 \u003d 0 ⇔ x + 2 y + 1 \u003d 0 5 x - 2 z - 5 \u003d 0

After which it is necessary to resolve the system

x + 2 y + 1 \u003d 0 5 x - 2 z - 5 \u003d 0 2 x - y + 5 z - 3 \u003d 0 ⇔ x + 2 y \u003d 1 5 x - 2 z \u003d 5 2 x - y + 5 z \u003d 3.

Let us turn to the rule of the system in 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 \u003d 0 6 \u003d 0, y \u003d - 1 10 · 10 + 2 · z \u003d - 1, x \u003d - 1 - 2 · y \u003d 1

We obtain that H 1 (1, - 1, 0).

We produce distance calculations from the specified point to the plane. We take the point 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 of the solution is to begin to bring the specified equation 2 x - y + 5 z - 3 \u003d 0 to normal form. We determine the normalizing multiplier and we obtain 1 2 2 + (- 1) 2 + 5 2 \u003d 1 30. Hence the equation of the plane 2 30 · X - 1 30 · Y + 5 30 · Z - 3 30 \u003d 0. The calculation of the left part of the equation is made by the substitution of 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 \u003d 0 by module. We get 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 plane χ is set in one of the methods of section methods of setting the plane, then it is necessary to begin to obtain the equation of the χ plane and calculate the desired distance using any method.

Example 2.

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

Decision

To begin, it is necessary to record the equation of the plane passing through the three points with the 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 \u003d 0 ⇔ xy - 2 z - 1 2 4 0 4 - 2 - 2 \u003d 0 ⇔ ⇔ - 8 x + 4 y - 20 z + 12 \u003d 0 ⇔ 2 x - y + 5 z - 3 \u003d 0

From here it follows that the task has a similar solution to the previous one. It means that the distance from the point M 1 to the plane A in C has a value of 2 30.

Answer: 2 30.

Finding the distance from a given point on the plane or to the plane with which they are parallel, more convenient, applying the formula M 1 H 1 \u003d cos α · x 1 + cos β · y 1 + cos γ · z 1 - p. From here we obtain that the normal equations of the planes are obtained in several actions.

Example 3.

Find the distance from the specified point with coordinates M 1 (- 3, 2, - 7) to the coordinate plane of the x in Z and the plane given by the equation 2 y - 5 \u003d 0.

Decision

The coordinate plane o in Z corresponds to the equation of the species x \u003d 0. For the plane o z, it is normal. Therefore, it is necessary to substitute in the left of the expression value x \u003d - 3 and take the module of the distance value from the point with the coordinates M 1 (- 3, 2, - 7) to the plane. We obtain a value equal to 3 \u003d 3.

After conversion, the normal equation of the plane 2 y - 5 \u003d 0 will be viewed y - 5 2 \u003d 0. Then you can find the desired distance from the point with the coordinates M 1 (- 3, 2, - 7) to the plane 2 y - 5 \u003d 0. Substituting and calculating, we obtain 2 - 5 2 \u003d 5 2 - 2.

Answer: The desired distance from M 1 (- 3, 2, - 7) to o in Z has a value of 3, and to 2 y - 5 \u003d 0 is 5 2 - 2.

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Consider in the space some plane π and an arbitrary point m 0. Choose for a plane unit normal vector N S. beginning At some point M 1 ∈ π, and let P (m 0, π) be a distance from point m 0 to the plane π. Then (Fig. 5.5)

p (m 0, π) \u003d | Pr n M 1 m 0 | \u003d | nm 1 m 0 |, (5.8)

since | n | \u003d 1.

If the plane π is set in rectangular coordinate system with its common equation AX + BY + CZ + D \u003d 0, then its normal vector is the vector with coordinates (a; b; c) and as a single normal vector you can choose

Let (x 0; Y 0; z 0) and (x 1; y 1; z 1) the coordinates of the points M 0 and M 1. Then the equality AX 1 + BY 1 + CZ 1 + D \u003d 0 is made, since the point M 1 belongs to the plane, and the coordinates of the vector M 1 m 0: M 1 m 0 \u003d (x 0 -x 1; y 0 -y 1; z 0 -z 1). Writing scalar product nm 1 m 0 in coordinate form and converting (5.8), we get

since AX 1 + BY 1 + CZ 1 \u003d - D. So to calculate the distance from the point to the plane you need to substitute the coordinates of the point in the general equation of the plane, and then the absolute value of the result is divided into a normalizing multiplier, equal to the length of the corresponding normal vector.