Equation of a plane passing through the point m0. Equation of a plane that passes through a given line and a given point. Examples of composing the equation of a plane passing through a given point and a straight line

Three points in space that do not lie on the same straight line define a single plane. Let's create an equation for a plane that passes through three given points M 1 (X 1 ; at 1 ; z 1), M 2 (X 2 ; at 2 ; z 2), M 3 (X 3 ; at 3 ; z 3). Let's take an arbitrary point on the plane M(X; at; z) and compose vectors = ( x – x 1 ; at– at 1 ; z–z 1), = (X 2 - X 1 ; at 2 – at 1 ; z 2 – z 1), = (X 3 - X 1 ; at 3 – at 1 ; z 3 – z 1). These vectors lie in the same plane, therefore they are coplanar. Using the condition of coplanarity of three vectors (their mixed product is equal to zero), we obtain ∙ ∙ = 0, that is

= 0. (3.5)

Equation (3.5) is called equation of a plane passing through three given points.

Mutual arrangement planes in space

Angle between planes

Let two planes be given

A 1 X + IN 1 at + WITH 1 z + D 1 = 0,

A 2 X + IN 2 at + WITH 2 z + D 2 = 0.

Behind angle between planes we take the angle φ between any two vectors perpendicular to them (which gives two angles, acute and obtuse, complementing each other to π). Since normal vectors of planes = ( A 1 , IN 1 , WITH 1) and = ( A 2 , IN 2 , WITH 2) are perpendicular to them, then we get

cosφ = .

Condition for perpendicularity of two planes

If two planes are perpendicular, then the normal vectors of these planes are also perpendicular and their scalar product is equal to zero: ∙ = 0. This means that the condition for the perpendicularity of two planes is

A 1 A 2 + IN 1 IN 2 + WITH 1 WITH 2 = 0.

Condition for parallelism of two planes

If the planes are parallel, then their normal vectors will also be parallel. Then the coordinates of the normal vectors of the same name are proportional. This means that the condition for parallel planes is

= = .

Distance from pointM 0 (x 0 , y 0 , z 0) to plane Oh + Wu + Сz + D = 0.

Distance from point M 0 (x 0 , y 0 , z 0) to plane Ax + Wu + Сz + D= 0 is the length of the perpendicular drawn from this point to the plane and is found by the formula

d = .

Example 1. R(– 1, 2, 7) perpendicular to the vector = (3, – 1, 2).

Solution

According to equation (3.1) we obtain

3(x + 1) – (y – 2) + 2(z – 7) = 0,

3X – at + 2z – 9 = 0.

Example 2. Write an equation for a plane passing through a point M(2; – 3; – 7) parallel to plane 2 X – 6at – 3z + 5 = 0.

Solution

Vector = (2; – 6; – 3) perpendicular to the plane is also perpendicular to parallel to the plane. This means that the desired plane passes through the point M(2; – 3; – 7) perpendicular to the vector = (2; – 6; – 3). Let us find the equation of the plane using formula (3.1):

2(X - 2) – 6(y + 3) – 3(z + 7) = 0,

2X – 6at – 3z – 43 = 0.

Example 3. Find the equation of the plane passing through the points M 1 (2; 3; – 1) and M 2 (1; 5; 3)perpendicular to plane 3 X – at + 3z + 15 = 0.

Solution

Vector = (3; – 1; 3) perpendicular to the given plane will be parallel to the desired plane. Thus, the plane passes through the points M 1 and M 2 is parallel to the vector .

Let M(x; y; z) arbitrary point of the plane, then vectors = ( X – 2; at – 3; z+ 1), = (– 1; 2; 4), = (3; – 1; 3) are coplanar, which means their mixed product is zero:

= 0.

Let's calculate the determinant by expanding over the elements of the first row:

(X – 2) – (at – 3) + (z + 1) = 0,

10(X - 2) – (– 15)(y – 3) + (– 5)(z + 1) = 0,

2(X - 2) + 3(y – 3) – (z + 1) = 0,

2x + 3at – z– 14 = 0 – plane equation.

Example 4. Write an equation for a plane passing through the origin perpendicular to planes 2 X – at + 5z+ 3 = 0 and X + 3at – z – 7 = 0.

Solution

Let be the normal vector of the desired plane. By condition, the plane is perpendicular to these planes, which means and , where = (2; – 1; 5), = (1; 3; – 1). This means that we can take as a vector vector product vectors and , that is = × .

= = – 14 + 7 + 7 .

Substituting the coordinates of the vector into the equation of the plane passing through the origin Oh + Wu + Сz= 0, we get

– 14X + 7at + 7z = 0,

2X – at – z = 0.

Self-test questions

1 Write general equation plane.

2 What is geometric meaning coefficients for X, y, z in the general equation of the plane?

3 Write down the equation of the plane passing through the point M 0 (x 0 ; y 0 ; z 0) perpendicular to the vector = ( A; IN; WITH).

4 Write down the equation of the plane in segments along the axes and indicate the geometric meaning of the parameters included in it.

5 Write down the equation of the plane passing through the points M 1 (X 1 ; at 1 ; z 1), M 2 (X 2 ; at 2 ; z 2), M 3 (X 3 ; at 3 ; z 3).

6 Write down the formula used to find the angle between two planes.

7 Write down the conditions for parallelism of two planes.

8 Write down the condition of perpendicularity of two planes.

9 Write down the formula that calculates the distance from a point to a plane.

Tasks for independent decision

1 Write an equation for a plane passing through a point M(2; – 1; 1) perpendicular to the vector = (1; – 2; 3). ( Answer: X – 2at + 3z – 7 = 0)

2 Dot R(1; – 2; – 2) is the base of the perpendicular drawn from the origin to the plane. Write an equation for this plane. ( Answer: X – 2at – 2z – 9 = 0)

3 Given two points M 1 (2; – 1; 3) and M 2 (– 1; 2; 4). Write an equation for a plane passing through a point M 1 is perpendicular to the vector . ( Answer: 3X – 3at – z – 6 = 0)

4 Write an equation for a plane passing through three points M 1 (3; – 1; 2), M 2 (4; – 1; – 1), M 3 (2; 0; 2). (Answer: 3X + 3at + z – 8 = 0)

5 M 1 (3; – 1; 2) and M 2 (2; 1; 3) parallel to the vector = (3; – 1; 4). ( Answer: 9X + 7at – 5z – 10 = 0)

6 Write an equation for a plane passing through a point M 1 (2; 3; – 4) parallel to the vectors = (3; 1; – 1) and = (1; – 2; 1). ( Answer: X + at + 7z + 14 = 0)

7 Write an equation for a plane passing through a point M(1; – 1; 1) perpendicular to planes 2 X – at + z– 1 = 0 and X + 2at – z + 1 = 0. (Answer: X – 3at – 5z + 1 = 0)

8 Write an equation for a plane passing through the points M 1 (1; 0; 1) and M 2 (1; 2; – 3) perpendicular to the plane X – at + z – 1 = 0. (Answer: X + 2at + z – 2 = 0)

9 Find the angle between planes 4 X – 5at + 3z– 1 = 0 and X – 4at – z + 9 = 0. (Answer: φ = arccos0.7)

10 Find the distance from a point M(2; – 1; – 1) to plane 16 X – 12at + 15z – 4 = 0. (Answer: d = 1)

11 Find the intersection point of three planes 5 X + 8at – z – 7 = 0, X + 2at + 3z – 1 = 0, 2X – 3at + 2z – 9 = 0. (Answer: (3; – 1; 0))

12 Write an equation for a plane that passes through the points M 1 (1; – 2; 6) and M 2 (5; – 4; 2) and cuts off equal segments on the axes Oh And OU. (Answer: 4X + 4at + z – 2 = 0)

13 Find the distance between planes X + 2at – 2z+ 2 = 0 and 3 X + 6at – 6z – 4 = 0. (Answer: d = )

Lecture 5. Solving problems on the topic "Analytical geometry in space"

1. Create an equation for a plane passing through a point M 0 (1, -2, 5) parallel to plane 7 x-y-2z-1=0.

Solution. Let us denote by R given plane, let R 0 – the desired parallel plane passing through the point M 0 (1, -2, 5).

Consider the normal (perpendicular) vector plane R. The coordinates of the normal vector are the coefficients of the variables in the plane equation 
.

Since the plane R And R 0 are parallel, then the vector perpendicular to the plane R 0 , i.e. - normal vector of the plane R 0 .

Equation of a plane passing through a point M 0 (x 0 , y 0 , z 0) with normal
:

Substitute the coordinates of the point M 0 and normal vectors into equation (1):

Opening the brackets, we get the general equation of the plane (final answer):

2. Compose canonical and parametric equations of a line passing through a point M 0 (-2, 3, 0) parallel to the straight line
.

Solution. Let us denote by L given straight line, let L 0 – the desired parallel line passing through the point M 0 (-2,3,0).

Guide vector straight L(non-zero vector parallel to this line) is also parallel to the line L 0 . Therefore, the vector is the direction vector of the line L 0 .

Direction vector coordinates are equal to the corresponding denominators in the canonical equations of a given line

.

Canonical equations of a line in space passing through a point M 0 (x 0 , y 0 , z {l, m, n}

. (2)

Substitute the coordinates of the point M 0 and direction vector into equation (2) and obtain the canonical equations of the line:

.

Parametric equations a straight line in space passing through a point M 0 (x 0 , y 0 , z 0) parallel to a non-zero vector {l, m, n), have the form:

(3)

Substitute the coordinates of the point M 0 and direction vector into equations (3) and obtain the parametric equations of the straight line:

3. Find a point
, symmetrical to the point
, relative to: a) straight
b) planes

Solution. a) Let’s create an equation for the perpendicular plane P, projecting point
to this line:

To find
we use the condition of perpendicularity of the given straight line and the projecting plane. Direct vector
perpendicular to the plane  vector
is the normal vector
to the plane  The equation of a plane perpendicular to a given line has the form or

Let's find the projection R points M to the straight line. Dot R is the point of intersection of a straight line and a plane, i.e. its coordinates must simultaneously satisfy both the equations of the line and the equation of the plane. Let's solve the system:

.

To solve it, we write the equation of the line in parametric form:

Substituting expressions for
into the equation of the plane, we get:

From here we find The found coordinates are the coordinates of the middle R line segment connecting a point
and a point symmetrical to it

IN school course geometry a theorem was formulated.

The coordinates of the middle of a segment are equal to half the sum of the corresponding coordinates of its ends.

Finding the coordinates of the point
from the formulas for the coordinates of the midpoint of the segment:

We get: So,
.

Solution. b) To find a point symmetrical to a point
relative to a given plane P, drop a perpendicular from the point
to this plane. Let's create an equation of a straight line with a direction vector
, passing through the point
:

Perpendicularity between a line and a plane means that the direction vector of the line is perpendicular to the plane 
. Then the equation of the straight line projecting the point
to a given plane, has the form:

Having solved the equations together
And
let's find the projection R points
to the plane. To do this, we rewrite the equations of the straight line in parametric form:

Let's substitute these values
into the equation of the plane: Similar to step a), using formulas for the coordinates of the middle of the segment, we find the coordinates of the symmetric point
:

Those.
.

4. Write an equation for a plane passing a) through a straight line
parallel to the vector
; b) through two intersecting lines
And
(having previously proven that they intersect); c) through two parallel lines
And
; d) through direct
and period
.

Solution. a) Since the given straight line lies in the desired plane, and the desired plane is parallel to the vector , then the normal vector of the plane will be perpendicular to the direction vector of the line
and vector .

Therefore, as the normal vector of the plane, we can choose the vector product of vectors And :

We get the coordinates of the normal vector of the plane
.

Let's find a point on a line. Equating the ratios in the canonical equations of the straight line to zero:

,

we find
,
,
. The given straight line passes through the point
, therefore, the plane also passes through the point
. Using the equation of a plane passing through this point perpendicular to the vector , we obtain the equation of the plane , or , or, finally,
.

Solution. b) Two lines in space can intersect, cross or be parallel. Given straight lines

And
(4)

are not parallel, since their direction vectors
And
not collinear:
.

How to check that lines intersect? You can solve system (4) of 4 equations with 3 unknowns. If the system has a unique solution, then we obtain the coordinates of the point of intersection of the lines. However, to solve our problem - constructing a plane in which both lines lie, the point of their intersection is not needed. Therefore, it is possible to formulate a condition for the intersection of two non-parallel lines in space without finding the intersection point.

If two non-parallel lines intersect, then the direction vectors
,
and connecting points lying on straight lines
And
vector lie in the same plane, i.e. coplanar  the mixed product of these vectors is equal to zero:

. (5)

We equate the ratios in the canonical equations of lines to zero (or to 1 or any number)

And
,

and find the coordinates of points on straight lines. The first line passes through the point
, and the second straight line – through the point
. The direction vectors of these lines are respectively equal
And
. We get

Equality (5) is satisfied, therefore, the given lines intersect. This means that there is a single plane passing through these two lines.

Let's move on to the second part of the problem - drawing up the equation of the plane.

As the normal vector of the plane, you can choose the vector product of their direction vectors And :

Coordinates of the normal vector of the plane
.

We found out that straight
goes through
, therefore, the desired plane also passes through this point. We get the equation of the plane, or
or, finally,
.

c) Since they are straight
And
are parallel, then the vector product of their direction vectors cannot be chosen as a normal vector; it will be equal to the zero vector.

Let's determine the coordinates of the points
And
, through which these lines pass. Let
And
, Then
,
. Let's calculate the coordinates of the vector. Vector
lies in the desired plane and is non-collinear to the vector , then as its normal vector you can choose the cross product of a vector
and the direction vector of the first straight line
:

So,
.

The plane passes through the line
, which means it passes through the point
. We get the equation of the plane: , or .

d) Equating the ratios in the canonical equations of the straight line to zero
, we find
,
,
. Therefore, the line passes through the point
.

Let's calculate the coordinates of the vector. Vector
belongs to the desired plane, as its normal vector choose the vector product of the direction vector of the straight line
and vector
:

Then the plane equation has the form: , or .

This article contains the information necessary to solve the problem of composing the equation of a plane passing through a given line and given point. After solving this problem in general view We will give detailed solutions to examples of composing the equation of a plane that passes through a given line and point.

Page navigation.

Finding the equation of a plane passing through a given line and a given point.

Let in three-dimensional space Oxyz is fixed, a line a and a point not lying on the line a are given. Let us set ourselves the task: to obtain the equation of the plane passing through the line a and the point M 3.

First, we will show that there is a single plane for which we need to construct an equation.

Let us recall two axioms:

• a single plane passes through three different points in space that do not lie on the same straight line;
• if two distinct points of a line lie in a certain plane, then all points of this line lie in this plane.

From these statements it follows that a unique plane can be drawn through a straight line and a point not lying on it. Thus, in the problem we have posed, a single plane passes through straight line a and point M 3, and we need to write the equation of this plane.

Now let's start finding the equation of a plane passing through a given straight line a and point .

If straight line a is given by indicating the coordinates of two different points M 1 and M 2 lying on it, then our task is reduced to finding the equation of the plane passing through three given points M 1, M 2 and M 3.

If straight line a is given differently, then we first have to find the coordinates of two points M 1 and M 2 lying on line a, and after that write down the equation of the plane passing through three points M 1, M 2 and M 3, which will be the desired equation of the plane passing through the line a and the point M 3.

Let's figure out how to find the coordinates of two different points M 1 and M 2 lying on a given line a.

In a rectangular coordinate system in space, any straight line corresponds to some equations of a straight line in space. We will assume that the method of specifying a straight line a in the problem statement allows us to obtain its parametric equations of a straight line in space of the form . Then, having accepted , we have the point , lying on the line a. By giving the parameter a real value other than zero, from the parametric equations of line a we can calculate the coordinates of point M 2, which also lies on line a and different from point M 1.

After this, we will only have to write the equation of a plane passing through three different and not lying on the same straight line points and , in the form .

So, we have obtained the equation of a plane passing through a given line a and a given point M 3 not lying on the line a.

Examples of composing the equation of a plane passing through a given point and a straight line.

We will show solutions to several examples in which we will analyze the considered method of finding the equation of a plane passing through a given straight line and a given point.

Let's start with the simplest case.

Example.

Solution.

Let us take two different points on the coordinate line Ox, for example, and .

Now we get the equation of a plane passing through three points M 1, M 2 and M 3:

This equation is the desired general equation of the plane passing through the given straight line Ox and the point .

Answer:

.

If it is known that the plane passes through a given point and a given line, and you need to write an equation of the plane in segments or a normal equation of the plane, then you should first obtain the general equation of the given plane, and from it proceed to the equation of the plane of the required type.

Example.

Compose normal equation plane that passes through the line and period .

Solution.

First, let's write the general equation of a given plane. To do this, find the coordinates of two different points lying on a straight line . The parametric equations of this line have the form . Let point M 1 correspond to the value, and point M 2 -. We calculate the coordinates of points M 1 and M 2:

Now we can write the general equation of a line passing through a point and direct :

It remains to obtain the required form of the plane equation by multiplying both sides of the resulting equation by a normalizing factor .

Answer:

.

So, finding the equation of a plane passing through a given point and a given line depends on finding the coordinates of two different points lying on a given line. This is often the main difficulty in solving such problems. In conclusion, we will analyze the solution to the example by composing the equation of a plane passing through a given point and a straight line, which is determined by the equations of two intersecting planes.

Example.

In the rectangular coordinate system Oxyz, a point and a straight line a are given, which is the line of intersection of two planes And . Write the equation of the plane passing through the line a and the point M 3.

With this online calculator you can find the equation of a plane passing through a given point and parallel to the given plane. A detailed solution with explanations is given. To find the equation of a plane, enter the coordinates of the point and the coefficients of the equation of the plane in the cells and click on the "Solve" button.

×

Warning

Clear all cells?

Close Clear

Data entry instructions. Numbers are entered as integers (examples: 487, 5, -7623, etc.), decimals (ex. 67., 102.54, etc.) or fractions. The fraction must be entered in the form a/b, where a and b (b>0) are integers or decimals. Examples 45/5, 6.6/76.4, -7/6.7, etc.

Equation of a plane passing through a given point and parallel to a given plane - theory, examples and solutions

Let a point be given M 0 (x 0 , y 0 , z 0) and plane equation

All parallel planes have collinear normal vectors. Therefore, to construct a plane parallel to (1) passing through the point M 0 (x 0 , y 0 , z 0) must be taken as the normal vector of the desired plane, the normal vector n=(A, B, C) plane (1). Next you need to find such a value D, at which point M 0 (x 0 , y 0 , z 0) satisfied the plane equation (1):

Substituting the value D from (3) to (1), we get:

Equation (5) is the equation of the plane passing through the point M 0 (x 0 , y 0 , z 0) and parallel to the plane (1).

Find the equation of the plane passing through the point M 0 (1, −6, 2) and parallel to the plane:

Substituting point coordinates M 0 and the coordinates of the normal vector in (3), we obtain.

Let us consider the plane Q in space. Its position is completely determined by specifying the vector N perpendicular to this plane and some fixed point lying in the Q plane. The vector N perpendicular to the Q plane is called the normal vector of this plane. If we denote by A, B and C the projections of the normal vector N, then

Let us derive the equation of the plane Q passing through a given point and having a given normal vector . To do this, consider a vector connecting a point with an arbitrary point on the Q plane (Fig. 81).

For any position of point M on the plane Q, the vector MHM is perpendicular to the normal vector N of the plane Q. Therefore, the scalar product We write the scalar product in terms of projections. Since , and is a vector, then

and therefore

We have shown that the coordinates of any point in the Q plane satisfy equation (4). It is easy to see that the coordinates of points not lying on the Q plane do not satisfy this equation (in the latter case). Consequently, we have obtained the required equation of the plane Q. Equation (4) is called the equation of the plane passing through a given point. It is of the first degree relative to the current coordinates

So, we have shown that every plane corresponds to an equation of the first degree with respect to the current coordinates.

Example 1. Write the equation of a plane passing through a point perpendicular to the vector.

Solution. Here . Based on formula (4) we obtain

or, after simplification,

By giving the coefficients A, B and C of equation (4) different values, we can obtain the equation of any plane passing through the point . The set of planes passing through a given point is called a bundle of planes. Equation (4), in which the coefficients A, B and C can take any values, is called the equation of a bunch of planes.

Example 2. Create an equation for a plane passing through three points (Fig. 82).

Solution. Let's write the equation for a bunch of planes passing through the point