The distance from the given point to the plane. Determining the distance between a point and a plane, a line and a plane, between planes and skew lines. Distance from a point to a plane

Conditions for parallelism and perpendicularity

1°. Complanarity condition for two planes

Let two planes be given:

A 1 x + B 1 y + C 1 z + D 1 = 0, n 1 = {A 1 ; B 1 ; C 1 } ≠ 0 ;(1)

A 2 x + B 2 y + C 2 z + D 2 = 0, n 2 = {A 2 ; B 2 ; C 2 } ≠ 0 .(2)

When are they coplanar (i.e., parallel or the same)? Obviously, this will be if and only if their normal vectors are collinear. Applying the complanarity criterion, we obtain

Suggestion 1. Two planes are coplanar if and only if the cross product of their normal vectors is equal to the zero vector:

[n 1 , n 2 ] = 0 .

2°. Condition of coincidence of two planes

Suggestion 2. Planes (1) and (2) coincide if and only if all four of their coefficients are proportional, i.e., there exists a number λ such that

A 2 = λ A 1 , B 2 = λ B 1 , C 2 = λ C 1 , D 2 = λ D 1 . (3)

Proof. Let conditions (3) be satisfied. Then the equation of the second plane can be written as follows:

λ A 1 x + λ B 1 y + λ C 1 z + λ D 1 = 0.

λ ≠ 0, otherwise it would be A 2 = B 2 = C 2 = D 2 = 0, which contradicts the condition n 2 ≠ 0 . Therefore, the last equation is equivalent to equation (1), which means that the two planes are the same.

Let now, on the contrary, it is known that the given planes coincide. Then their normal vectors are collinear, i.e., there exists a number λ such that

A 2 = λ A 1 , B 2 = λ B 1 , C 2 = λ C 1 .

Equation (2) can now be rewritten as:

λ A 1 x + λ B 1 y + λ C 1 z + D 2 = 0.

Multiplying equation (1) by λ, we get equivalent equation first plane (because λ ≠ 0):

λ A 1 x + λ B 1 y + λ C 1 z + λ D 1 = 0.

Take some point x 0 , y 0 , z 0) from the first (and hence the second) plane and substitute its coordinates into the last two equations; we get the correct equalities:

λ A 1 x 0 + λ B 1 y 0 + λ C 1 z 0 + D 2 = 0 ;

λ A 1 x 0 + λ B 1 y 0 + λ C 1 z 0 + λ D 1 = 0.

Subtracting from the top bottom, we get D 2 − λ D 1 = 0, i.e. D 2 = λ D 1, QED.

3°. The condition of perpendicularity of two planes

Obviously, for this it is necessary and sufficient that the normal vectors be perpendicular.

Suggestion 3. Two planes are perpendicular if and only if the dot product of the normal vectors is zero:

(n 1 , n 2) = 0 .

Let the plane equation be given

Ax + By + cz + D = 0, n = {A; B; C} ≠ 0 ,

and dot M 0 = (x 0 , y 0 , z 0). We derive the formula for the distance from a point to a plane:

Take an arbitrary point Q = (x 1 , y 1 , z 1) lying in the given plane. Its coordinates satisfy the plane equation:

Ax 1 + By 1 + cz 1 + D = 0.

Note now that the desired distance d is equal to the absolute value of the vector projection to the direction of the vector n (here we take the projection as numerical value, not as a vector). Next, apply the formula to calculate the projection:

A similar formula is valid for the distance d from the point M 0 = (x 0 , y 0) plane to the straight line given by the general equation Ax + By + C = 0.

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

AT 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 problem, various methods for finding the distance from a point to a plane are considered. 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.

A task. Dan cuboid BUTBFROMDA 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, Consequently DT⊥AC. Consequently, DT⊥ACD 1.

BUTDC 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 AT;

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

Figure 3. The third way

B(0,0,0), BUT(2,0,0), FROM(0,4,0), D(2,4,0), D 1 (2,4,6).

Let ax+by+ cz+ d=0 – plane equation ACD one . 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

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 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 given in space, then a straight line perpendicular to the plane can be drawn through the point. H 1 is a common point of their intersection. From here we get that the segment M 1 H 1 is a perpendicular, which was drawn 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 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 the point M 1 to the plane χ is defined as follows: the distance from the point M 1 to the plane χ will be the smallest from a given point to any point in the plane. If the point H 2 is located in the χ plane and is not equal to the point H 2, then we get a right triangle of the form M 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 inclined, which is drawn from the point M 1 to the plane χ. We have that the perpendicular drawn from a given point to a plane is less than the inclined one drawn from a point to a given plane. Consider this case in the figure below.

Distance from a point to a plane - theory, examples, solutions

There are a number of geometric problems whose solutions must contain the distance from a point to a plane. Ways to detect this may be different. To resolve, 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, they solve using the coordinate method. This paragraph deals with this method.

According to the condition of the problem, we have that a point in 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. Several solutions are used to solve.

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 plane χ. 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 plane χ. Then we determine the coordinates (x 2, y 2, z 2) of the point H 1. The desired distance from M 1 to the χ plane is found by 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, you need to know the coordinates of the point H 1.

We have that H 1 is the point of intersection of the plane χ with the line a, which passes through the point M 1 located perpendicular to the plane χ. It follows that it is necessary to formulate the equation of a straight line passing through a given point perpendicular to a given plane. It is then that we can 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

compose the equation of a 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 the point H 1, which are points
intersection of the line a with the 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 a normal equation of the plane of the form cos α · x + cos β · y + cos γ · z - p = 0 . From here we get 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 is established thanks to the theorem.

Theorem

If a point M 1 (x 1 , y 1 , z 1) is given in three-dimensional space, having a normal equation of the χ plane of the form cos α x + cos β y + cos γ z - p = 0, then calculating the distance from the point to plane M 1 H 1 is derived 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. From here we get 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 formula for calculating 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 the notation 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 plane χ is calculated by substituting into the left 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 z1 relating to the point M 1 , taking the absolute value of the obtained value.

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 will start by calculating the direction vector of the line a . By condition, we have that the given equation 2 x - y + 5 z - 3 = 0 is a general plane equation, and n → = (2 , - 1 , 5) is the normal vector of the given plane. It is used as a directing vector for the straight line a, which is perpendicular to the 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 look like x - 5 2 = y - (- 3) - 1 = z - 10 5 ⇔ x - 5 2 = y + 3 - 1 = z - 10 5 .

Intersection points should be defined. To do this, gently combine the equations into a system for the transition from the canonical to the equations of two intersecting lines. Let's 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 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 rule for solving the system according to 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 = 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 a plane. We take points 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 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 \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 = 0 modulo. We get the 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 χ plane is given by one of the methods of the plane definition methods section, 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 set in three-dimensional space. Calculate the distance from M 1 to the 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 ⇔ x y - 2 z - 1 2 4 0 4 - 2 - 2 = 0 ⇔ ⇔ - 8x + 4y - 20z + 12 = 0 ⇔ 2x - y + 5z - 3 = 0

It follows that the problem has a solution similar to the previous one. Hence, the distance from the point M 1 to the plane A B C is 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 here we get that the normal equations of the planes are obtained in several steps.

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 2y - 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 \u003d - 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 the 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 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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The plane in space is given by the equation 3x-4y+2z+5=0, find the distance from it to the point M(3;-2;6).
Given:
$$ x_0 = 3, \quad y_0 = -2, \quad z_0 = 6 $$

$$ A = 3, \quad B = -4, \quad C = 2, \quad D = 5 $$
Solution:
To solve the problem, we use the formula for finding the distance from a point to a plane, which is equal to the length of the perpendicular dropped from this point to the plane:

$$ p = (| A \cdot x_0 + B \cdot y_0 + C \cdot z_0 + D|) \over \sqrt((A^2 + B^2 + C^2)) $$

where A, B, C, D are the coefficients of the plane equation, and x0, y0, z0 are the point coordinates.

Let's make a substitution:

$$ \frac(|3 \cdot 3 + (-4) \cdot (-2)+2 \cdot 6 + 5 |)( \sqrt((3^2 + (-4)^2 + 2^2) ) ) = \frac(|9+8+12+5|)(\sqrt((9+16+4))) =6.314$$ (linear units)
Answer:
Given a cube ABCDA1B1C1D1 with an edge equal to 1 cm. Calculate the distance from point A1 to the plane defined by points B, D and C1.
Solution:
To solve the problem, we apply the coordinate method. The origin of the coordinate system is located at point A. The x-axis is compatible with the edge AD, the y-axis is compatible with the edge AB, the z-axis is compatible with the edge AA1.

Then the coordinates of point A1 (0;0;1), points B (0; 1; 0), D (1; 0; 0), C1(1; 1; 1). Putting the coordinates of each of the points into the general equation for the plane A·x+B·y+C·z+D=0, we obtain a system of three equations, solving which we find the coefficients and the equation of the plane x+y-z-1=0.

$$ p = \frac( |A \cdot x_0 + B \cdot y_0 + C \cdot z_0 + D| )( \sqrt((A^2 + B^2 + C^2)) ) $$, substitution :

$$ p = \frac( |1 \cdot 0 + 1 \cdot 0 - 1 \cdot 1 - 1| )( \sqrt((1+1+1)) ) = 1.155 cm$$
Answer:
$$ R = 1.155 cm $$
Find the distance then of the point M (2; 4; -7) to the XOY plane.
Solution:
The XOY plane equation is special case, its equation is z=0. Let's apply the formula:

$$ p = \frac( | A \cdot x_0 + B \cdot y_0 + C \cdot z_0 + D| )( (A^2 + B^2 + C^2) ) $$ , where A=0, B =0, С=1, D=0, x0=2, y0=4, z0=-7.

Let's make a substitution:

$$ p = \frac( |0 \cdot 2 + 0 \cdot 4 + 1 \cdot (-7)) + 0| )( \sqrt((0^2 + 0^2 + 1^2)) ) = 7$$
Answer:
The plane is determined by a frame of three points with coordinates in the rectangular system A1 (0;2;1), B1(2;6;1), C1(4;0;-1). Determine at what distance from it is a point with coordinates M (5; -3; 10).
Solution:
In order to determine the distance from a point to a plane, we use the formula

$$ p= \frac( |A \cdot x_0 + B \cdot y_0 + C \cdot z_0 + D| )( sqrt( (A^2 + B^2 + C^2) ) ) $$

To use it, it is necessary to derive the equation of the plane defined by points A1, B1 and C1. General form of this equation A·x+B·y+C·z+D=0. Using one of the methods for deriving the equation of the plane (a system of equations with coordinates of points or a determinant), we find the equation of the plane, we get $$2x-y+5z-3=0$$.

We substitute the obtained coefficients of the equation and the coordinates of the point into the formula:

$$ p = \frac( |A \cdot x_0 + B \cdot y_0 + C \cdot z_0 + D| )( \sqrt( (A^2 + B^2 + C^2) ) ) = \frac( | 2 \cdot 5 - (-3) + 5 \cdot 10 - 3|)( \sqrt( (2^2 + (-1)^2 + 5^2) ) ) = $10.95
Answer:
Find the distance from the 4x-6y-4z+7=0 plane to the origin of the O coordinate system.
Given:
$$ x_0 = 0, \quad y_0 = 0, \quad z_0 = 0 $$

$$ A = 4, \quad B = -6, \quad C = -4, \quad D = 7 $$
Solution:
Coordinates of the origin of the coordinate system O(0;0;0). Let's use the formula:

$$ p= \frac( |A \cdot x_0 + B \cdot y_0 + C \cdot z_0 + D| )( sqrt( (A^2 + B^2 + C^2) ) ) $$ For plane $$4 x-6y-4z+7=0$$,

$$ A=4, $$
$$ B=-6, $$
$$ C=-4, $$
$$ D=7. $$

Substitute the values:

$$ p = \frac( |A \cdot x_0 + B \cdot y_0 + C \cdot z_0 + D| )( \sqrt( (A^2 + B^2 + C^2) ) ) = \frac( | 4 \cdot 0 - 6 \cdot 0 - 4 \cdot 0 + 7|)( \sqrt( (4^2 + (-6)^2 + (-4)^2) ) ) = 0.85 $$
Answer:

Let there be a plane . Let's draw a normal
through the origin O. Let
are the angles formed by the normal with coordinate axes.
. Let is the length of the normal segment
before crossing the plane. Assuming that the direction cosines of the normal are known , we derive the equation of the plane .

Let
) is an arbitrary point of the plane. The unit normal vector has coordinates. Let's find the projection of the vector
to normal.

Since the point M belongs to the plane, then

This is the equation for a given plane, called normal .

Distance from point to plane

Let a plane be given ,M*
- a point in space d is its distance from the plane.

Definition. deviation points M* from the plane is called the number ( + d), if M* lies on the other side of the plane where the positive direction of the normal points , and number (- d) if the point is located on the other side of the plane:

Theorem. Let the plane with unit normal given by the normal equation:

Let M*
– point of space Deviation t. M* from the plane is given by the expression

Proof. projection t.
* denote the normal Q. Point Deviation M* from the plane is

Rule. To find deviation t. M* from the plane, you need to substitute the coordinates t in the normal equation of the plane. M* . The distance from a point to a plane is .

Reduction of the general equation of the plane to normal form

Let the same plane be given by two equations:

General Equation,

normal equation.

Since both equations define the same plane, their coefficients are proportional:

We square the first three equalities and add:

From here we find is the normalizing factor:

. (10)

Multiplying the general equation of the plane by the normalizing factor, we obtain the normal equation of the plane:

Examples of tasks on the topic "Plane".

Example 1 Compose the equation of the plane passing through a given point
(2,1,-1) and parallel to the plane.

Solution. Normal to plane :
. Since the planes are parallel, the normal is also the normal to the desired plane . Using the equation of a plane passing through a given point (3), we obtain for the plane the equation:

Answer:

Example 2 The base of the perpendicular dropped from the origin to the plane , is a point
. Find the equation of the plane .

Solution. Vector
is the normal to the plane . Dot M 0 belongs to the plane. You can use the equation of a plane passing through a given point (3):

Answer:

Example 3 Build Plane passing through the points

and perpendicular to the plane :.

Therefore, for some point M (x, y, z) belonged to the plane , it is necessary that three vectors
were coplanar:

=0.

It remains to open the determinant and bring the resulting expression to the form of the general equation (1).

Example 4 Plane given by the general equation:

Find point deviation
from a given plane.

Solution. We bring the equation of the plane to normal form.

Substitute into the resulting normal equation the coordinates of the point M*.

Answer:
.

Example 5 Whether the segment intersects the plane.

Solution. To cut AB crossed the plane, deviations and from the plane must have different signs:

Example 6 The intersection of three planes at one point.

The system has only decision, therefore, three planes have one common point.

Example 7 Finding the bisectors of a dihedral angle formed by two given planes.

Let and - deviation of some point
from the first and second planes.

On one of the bisectoral planes (corresponding to the angle in which the origin of coordinates lies), these deviations are equal in magnitude and sign, and on the other, they are equal in magnitude and opposite in sign.

This is the equation of the first bisectoral plane.

This is the equation of the second bisectoral plane.

Example 8 Finding the Location of Two Data Points and relative to the dihedral angles formed by these planes.