Write an equation of the plane using 3 points. Equation of a plane. How to write an equation of a plane? Mutual arrangement of planes. Tasks

You can set different ways(one point and a vector, two points and a vector, three points, etc.). It is with this in mind that the equation of the plane can have different kinds. Also, subject to certain conditions, planes can be parallel, perpendicular, intersecting, etc. We'll talk about this in this article. We will learn how to create a general equation of a plane and more.

Normal form of equation

Let's say there is a space R 3 that has a rectangular XYZ coordinate system. Let us define the vector α, which will be released from the initial point O. Through the end of the vector α we draw a plane P, which will be perpendicular to it.

Let us denote an arbitrary point on P as Q = (x, y, z). Let's sign the radius vector of point Q with the letter p. In this case, the length of the vector α is equal to р=IαI and Ʋ=(cosα,cosβ,cosγ).

This is a unit vector that is directed to the side, like the vector α. α, β and γ are the angles that are formed between the vector Ʋ and the positive directions of the space axes x, y, z, respectively. The projection of any point QϵП onto the vector Ʋ is a constant value that is equal to p: (p,Ʋ) = p(p≥0).

The above equation makes sense when p=0. The only thing is that the plane P in this case will intersect the point O (α = 0), which is the origin of coordinates, and the unit vector Ʋ released from the point O will be perpendicular to P, despite its direction, which means that the vector Ʋ is determined with accurate to the sign. The previous equation is the equation of our plane P, expressed in vector form. But in coordinates it will look like this:

P here is greater than or equal to 0. We have found the equation of the plane in space in normal form.

General equation

If we multiply the equation in coordinates by any number that is not equal to zero, we obtain an equation equivalent to this one, defining that very plane. It will look like this:

Here A, B, C are numbers that are simultaneously different from zero. This equation is called the general plane equation.

Equations of planes. Special cases

Equation in general view may be modified subject to additional conditions. Let's look at some of them.

Let's assume that the coefficient A is 0. This means that this plane is parallel to the given Ox axis. In this case, the form of the equation will change: Ву+Cz+D=0.

Similarly, the form of the equation will change under the following conditions:

• Firstly, if B = 0, then the equation will change to Ax + Cz + D = 0, which will indicate parallelism to the Oy axis.
• Secondly, if C=0, then the equation will be transformed into Ax+By+D=0, which will indicate parallelism to the given Oz axis.
• Thirdly, if D=0, the equation will look like Ax+By+Cz=0, which will mean that the plane intersects O (the origin).
• Fourth, if A=B=0, then the equation will change to Cz+D=0, which will prove parallel to Oxy.
• Fifthly, if B=C=0, then the equation becomes Ax+D=0, which means that the plane to Oyz is parallel.
• Sixth, if A=C=0, then the equation will take the form Ву+D=0, that is, it will report parallelism to Oxz.

Type of equation in segments

In the case when the numbers A, B, C, D are different from zero, the form of equation (0) can be as follows:

x/a + y/b + z/c = 1,

in which a = -D/A, b = -D/B, c = -D/C.

We get as a result. It is worth noting that this plane will intersect the Ox axis at a point with coordinates (a,0,0), Oy - (0,b,0), and Oz - (0,0,c).

Taking into account the equation x/a + y/b + z/c = 1, it is not difficult to visually imagine the placement of the plane relative to a given coordinate system.

Normal vector coordinates

The normal vector n to the plane P has coordinates that are coefficients general equation of a given plane, that is, n (A, B, C).

In order to determine the coordinates of the normal n, it is enough to know the general equation of a given plane.

When using an equation in segments, which has the form x/a + y/b + z/c = 1, as when using a general equation, you can write the coordinates of any normal vector of a given plane: (1/a + 1/b + 1/ With).

It is worth noting that the normal vector helps solve a variety of problems. The most common ones include problems that involve proving the perpendicularity or parallelism of planes, problems of finding angles between planes or angles between planes and straight lines.

Type of plane equation according to the coordinates of the point and normal vector

A nonzero vector n perpendicular to a given plane is called normal for a given plane.

Let us assume that in the coordinate space (rectangular coordinate system) Oxyz are given:

• point Mₒ with coordinates (xₒ,yₒ,zₒ);
• zero vector n=A*i+B*j+C*k.

It is necessary to create an equation for a plane that will pass through the point Mₒ perpendicular to the normal n.

We choose any arbitrary point in space and denote it M (x y, z). Let the radius vector of any point M (x,y,z) be r=x*i+y*j+z*k, and the radius vector of the point Mₒ (xₒ,yₒ,zₒ) - rₒ=xₒ*i+yₒ *j+zₒ*k. Point M will belong to a given plane if the vector MₒM is perpendicular to vector n. Let us write the orthogonality condition using the scalar product:

[MₒM, n] = 0.

Since MₒM = r-rₒ, the vector equation of the plane will look like this:

This equation can have another form. To do this, the properties of the scalar product are used, and the transformation is left-hand side equations = - . If we denote it as c, we get the following equation: - c = 0 or = c, which expresses the constancy of the projections onto the normal vector of the radius vectors of given points that belong to the plane.

Now we can get the coordinate form of writing the vector equation of our plane = 0. Since r-rₒ = (x-xₒ)*i + (y-yₒ)*j + (z-zₒ)*k, and n = A*i+B *j+С*k, we have:

It turns out that we have an equation for a plane passing through a point perpendicular to the normal n:

A*(x- xₒ)+B*(y- yₒ)C*(z-zₒ)=0.

Type of plane equation according to the coordinates of two points and a vector collinear to the plane

Let us define two arbitrary points M′ (x′,y′,z′) and M″ (x″,y″,z″), as well as a vector a (a′,a″,a‴).

Now we can create an equation for a given plane, which will pass through the existing points M′ and M″, as well as any point M with coordinates (x, y, z) in parallel given vector A.

In this case, the vectors M′M=(x-x′;y-y′;z-z′) and M″M=(x″-x′;y″-y′;z″-z′) must be coplanar with the vector a=(a′,a″,a‴), which means that (M′M, M″M, a)=0.

So, our plane equation in space will look like this:

Type of equation of a plane intersecting three points

Let's say we have three points: (x′,y′,z′), (x″,y″,z″), (x‴,y‴,z‴), which do not belong to the same line. It is necessary to write the equation of a plane passing through given three points. The theory of geometry claims that this kind of plane really exists, but it is the only one and unique. Since this plane intersects the point (x′,y′,z′), the form of its equation will be as follows:

Here A, B, C are different from zero at the same time. Also, the given plane intersects two more points: (x″,y″,z″) and (x‴,y‴,z‴). In this regard, the following conditions must be met:

Now we can create a homogeneous system with unknowns u, v, w:

In our case x,y or z acts as an arbitrary point that satisfies equation (1). Given equation (1) and the system of equations (2) and (3), the system of equations indicated in the figure above is satisfied by the vector N (A,B,C), which is non-trivial. That is why the determinant of this system is equal to zero.

Equation (1) that we have obtained is the equation of the plane. It passes through 3 points exactly, and this is easy to check. To do this, we need to expand our determinant into the elements in the first row. From the existing properties of the determinant it follows that our plane simultaneously intersects three initially given points (x′,y′,z′), (x″,y″,z″), (x‴,y‴,z‴). That is, we have solved the task assigned to us.

Dihedral angle between planes

A dihedral angle represents a spatial geometric figure, formed by two half-planes that emanate from one straight line. In other words, this is the part of space that is limited by these half-planes.

Let's say we have two planes with the following equations:

We know that the vectors N=(A,B,C) and N¹=(A¹,B¹,C¹) are perpendicular according to the given planes. In this regard, the angle φ between the vectors N and N¹ is equal to the angle (dihedral) that is located between these planes. Scalar product has the form:

NN¹=|N||N¹|cos φ,

precisely because

cosφ= NN¹/|N||N¹|=(AA¹+BB¹+CC¹)/((√(A²+B²+C²))*(√(A¹)²+(B¹)²+(C¹)²)).

It is enough to take into account that 0≤φ≤π.

In fact, two planes that intersect form two angles (dihedral): φ 1 and φ 2. Their sum is equal to π (φ 1 + φ 2 = π). As for their cosines, their absolute values ​​are equal, but they differ in sign, that is, cos φ 1 = -cos φ 2. If in equation (0) we replace A, B and C with the numbers -A, -B and -C, respectively, then the equation that we get will determine the same plane, the only one, the angle φ in the equation cos φ= NN 1 /| N||N 1 | will be replaced by π-φ.

Equation of a perpendicular plane

Planes between which the angle is 90 degrees are called perpendicular. Using the material presented above, we can find the equation of a plane perpendicular to another. Let's say we have two planes: Ax+By+Cz+D=0 and A¹x+B¹y+C¹z+D=0. We can say that they will be perpendicular if cosφ=0. This means that NN¹=AA¹+BB¹+CC¹=0.

Parallel plane equation

Two planes that do not contain common points are called parallel.

The condition (their equations are the same as in the previous paragraph) is that the vectors N and N¹, which are perpendicular to them, are collinear. This means that the following proportionality conditions are met:

A/A¹=B/B¹=C/C¹.

If the proportionality conditions are extended - A/A¹=B/B¹=C/C¹=DD¹,

this indicates that these planes coincide. This means that the equations Ax+By+Cz+D=0 and A¹x+B¹y+C¹z+D¹=0 describe one plane.

Distance to plane from point

Let's say we have a plane P, which is given by equation (0). It is necessary to find the distance to it from a point with coordinates (xₒ,yₒ,zₒ)=Qₒ. To do this, you need to bring the equation of the plane P into normal form:

(ρ,v)=р (р≥0).

IN in this caseρ (x,y,z) is the radius vector of our point Q located on P, p is the length of the perpendicular P that was released from the zero point, v is the unit vector, which is located in the direction a.

The difference ρ-ρº radius vector of some point Q = (x, y, z), belonging to P, as well as the radius vector of a given point Q 0 = (xₒ, уₒ, zₒ) is such a vector, the absolute value of the projection of which onto v equals the distance d that needs to be found from Q 0 = (xₒ,уₒ,zₒ) to P:

D=|(ρ-ρ 0 ,v)|, but

(ρ-ρ 0 ,v)= (ρ,v)-(ρ 0 ,v) =р-(ρ 0 ,v).

So it turns out

d=|(ρ 0 ,v)-р|.

Thus, we will find the absolute value of the resulting expression, that is, the desired d.

Using the parameter language, we get the obvious:

d=|Ахₒ+Вуₒ+Czₒ|/√(А²+В²+С²).

If set point Q 0 is on the other side of the plane P, like the origin of coordinates, then between the vector ρ-ρ 0 and v is therefore located:

d=-(ρ-ρ 0 ,v)=(ρ 0 ,v)-р>0.

In the case when the point Q 0, together with the origin of coordinates, is located on the same side of P, then the angle created is acute, that is:

d=(ρ-ρ 0 ,v)=р - (ρ 0 , v)>0.

As a result, it turns out that in the first case (ρ 0 ,v)>р, in the second (ρ 0 ,v)<р.

Tangent plane and its equation

The tangent plane to the surface at the point of contact Mº is a plane containing all possible tangents to the curves drawn through this point on the surface.

With this type of surface equation F(x,y,z)=0, the equation of the tangent plane at the tangent point Mº(xº,yº,zº) will look like this:

F x (xº,yº,zº)(x- xº)+ F x (xº, yº, zº)(y- yº)+ F x (xº, yº,zº)(z-zº)=0.

If you specify the surface in explicit form z=f (x,y), then the tangent plane will be described by the equation:

z-zº =f(xº, yº)(x- xº)+f(xº, yº)(y- yº).

Intersection of two planes

In the coordinate system (rectangular) Oxyz is located, two planes П′ and П″ are given, which intersect and do not coincide. Since any plane located in a rectangular coordinate system is determined by a general equation, we will assume that P′ and P″ are given by the equations A′x+B′y+C′z+D′=0 and A″x+B″y+ С″z+D″=0. In this case, we have the normal n′ (A′,B′,C′) of the plane P′ and the normal n″ (A″,B″,C″) of the plane P″. Since our planes are not parallel and do not coincide, these vectors are not collinear. Using the language of mathematics, we can write this condition as follows: n′≠ n″ ↔ (A′,B′,C′) ≠ (λ*A″,λ*B″,λ*C″), λϵR. Let the straight line that lies at the intersection of P′ and P″ be denoted by the letter a, in this case a = P′ ∩ P″.

a is a straight line consisting of the set of all points of the (common) planes P′ and P″. This means that the coordinates of any point belonging to line a must simultaneously satisfy the equations A′x+B′y+C′z+D′=0 and A″x+B″y+C″z+D″=0. This means that the coordinates of the point will be a partial solution of the following system of equations:

As a result, it turns out that the (general) solution of this system of equations will determine the coordinates of each of the points of the line, which will act as the intersection point of P′ and P″, and determine the straight line a in the Oxyz (rectangular) coordinate system in space.

In this material, we will look at how to find the equation of a plane if we know the coordinates of three different points that do not lie on the same straight line. To do this, we need to remember what a rectangular coordinate system is in three-dimensional space. To begin, we will introduce the basic principle of this equation and show exactly how to use it to solve specific problems.

Yandex.RTB R-A-339285-1

First, we need to remember one axiom, which sounds like this:

Definition 1

If three points do not coincide with each other and do not lie on the same straight line, then in three-dimensional space only one plane passes through them.

In other words, if we have three different points whose coordinates do not coincide and which cannot be connected by a straight line, then we can determine the plane passing through it.

Let's say we have a rectangular coordinate system. Let's denote it O x y z. It contains three points M with coordinates M 1 (x 1, y 1, z 1), M 2 (x 2, y 2, z 2), M 3 (x 3, y 3, z 3), which cannot be connected straight line. Based on these conditions, we can write down the equation of the plane we need. There are two approaches to solving this problem.

1. The first approach uses the general plane equation. In letter form, it is written as A (x - x 1) + B (y - y 1) + C (z - z 1) = 0. With its help, you can define in a rectangular coordinate system a certain alpha plane that passes through the first given point M 1 (x 1, y 1, z 1). It turns out that the normal vector of the plane α will have coordinates A, B, C.

Definition of N

Knowing the coordinates of the normal vector and the coordinates of the point through which the plane passes, we can write down the general equation of this plane.

This is what we will proceed from in the future.

Thus, according to the conditions of the problem, we have the coordinates of the desired point (even three) through which the plane passes. To find the equation, you need to calculate the coordinates of its normal vector. Let's denote it n → .

Let us remember the rule: any non-zero vector of a given plane is perpendicular to the normal vector of the same plane. Then we have that n → will be perpendicular to the vectors composed of the original points M 1 M 2 → and M 1 M 3 → . Then we can denote n → as a vector product of the form M 1 M 2 → · M 1 M 3 → .

Since M 1 M 2 → = (x 2 - x 1, y 2 - y 1, z 2 - z 1) and M 1 M 3 → = x 3 - x 1, y 3 - y 1, z 3 - z 1 (proofs of these equalities are given in the article devoted to calculating the coordinates of a vector from the coordinates of points), then it turns out that:

n → = M 1 M 2 → × M 1 M 3 → = i → j → k → x 2 - x 1 y 2 - y 1 z 2 - z 1 x 3 - x 1 y 3 - y 1 z 3 - z 1

If we calculate the determinant, we will obtain the coordinates of the normal vector n → we need. Now we can write down the equation we need for a plane passing through three given points.

2. The second approach to finding the equation passing through M 1 (x 1, y 1, z 1), M 2 (x 2, y 2, z 2), M 3 (x 3, y 3, z 3), is based on such a concept as coplanarity of vectors.

If we have a set of points M (x, y, z), then in a rectangular coordinate system they define a plane for given points M 1 (x 1, y 1, z 1), M 2 (x 2, y 2, z 2 ) , M 3 (x 3 , y 3 , z 3) only in the case when the vectors M 1 M → = (x - x 1 , y - y 1 , z - z 1) , M 1 M 2 → = ( x 2 - x 1 , y 2 - y 1 , z 2 - z 1) and M 1 M 3  → = (x 3 - x 1 , y 3 - y 1 , z 3 - z 1) will be coplanar.

In the diagram it will look like this:

This will mean that the mixed product of the vectors M 1 M → , M 1 M 2 → , M 1 M 3 → will be equal to zero: M 1 M → · M 1 M 2 → · M 1 M 3 → = 0 , since this is the main condition of coplanarity: M 1 M → = (x - x 1, y - y 1, z - z 1), M 1 M 2 → = (x 2 - x 1, y 2 - y 1, z 2 - z 1 ) and M 1 M 3 → = (x 3 - x 1, y 3 - y 1, z 3 - z 1).

Let us write the resulting equation in coordinate form:

After we calculate the determinant, we can obtain the plane equation we need for three points that do not lie on the same line M 1 (x 1, y 1, z 1), M 2 (x 2, y 2, z 2), M 3 (x 3 , y 3 , z 3) .

From the resulting equation, you can go to the equation of the plane in segments or to the normal equation of the plane, if the conditions of the problem require it.

In the next paragraph we will give examples of how the approaches we have indicated are implemented in practice.

Examples of problems for composing an equation of a plane passing through 3 points

Previously, we identified two approaches that can be used to find the desired equation. Let's look at how they are used to solve problems and when you should choose each one.

Example 1

There are three points that do not lie on the same line, with coordinates M 1 (- 3, 2, - 1), M 2 (- 1, 2, 4), M 3 (3, 3, - 1). Write an equation for the plane passing through them.

Solution

We use both methods alternately.

1. Find the coordinates of the two vectors we need M 1 M 2 → , M 1 M 3 → :

M 1 M 2 → = - 1 - - 3 , 2 - 2 , 4 - - 1 ⇔ M 1 M 2 → = (2 , 0 , 5) M 1 M 3 → = 3 - - 3 , 3 - 2 , - 1 - - 1 ⇔ M 1 M 3 → = 6 , 1 , 0

Now let's calculate their vector product. We will not describe the calculations of the determinant:

n → = M 1 M 2 → × M 1 M 3 → = i → j → k → 2 0 5 6 1 0 = - 5 i → + 30 j → + 2 k →

We have a normal vector of the plane that passes through the three required points: n → = (- 5, 30, 2) . Next, we need to take one of the points, for example, M 1 (- 3, 2, - 1), and write down the equation for the plane with vector n → = (- 5, 30, 2). We get that: - 5 (x - (- 3)) + 30 (y - 2) + 2 (z - (- 1)) = 0 ⇔ - 5 x + 30 y + 2 z - 73 = 0

This is the equation we need for a plane that passes through three points.

2. Let's take a different approach. Let us write the equation for a plane with three points M 1 (x 1, y 1, z 1), M 2 (x 2, y 2, z 2), M 3 (x 3, y 3, z 3) in the following form:

x - x 1 y - y 1 z - z 1 x 2 - x 1 y 2 - y 1 z 2 - z 1 x 3 - x 1 y 3 - y 1 z 3 - z 1 = 0

Here you can substitute data from the problem statement. Since x 1 = - 3, y 1 = 2, z 1 = - 1, x 2 = - 1, y 2 = 2, z 2 = 4, x 3 = 3, y 3 = 3, z 3 = - 1, as a result we get:

x - x 1 y - y 1 z - z 1 x 2 - x 1 y 2 - y 1 z 2 - z 1 x 3 - x 1 y 3 - y 1 z 3 - z 1 = x - (- 3) y - 2 z - (- 1) - 1 - (- 3) 2 - 2 4 - (- 1) 3 - (- 3) 3 - 2 - 1 - (- 1) = = x + 3 y - 2 z + 1 2 0 5 6 1 0 = - 5 x + 30 y + 2 z - 73

We got the equation we needed.

Answer:- 5 x + 30 y + 2 z - 73 .

But what if the given points still lie on the same line and we need to create a plane equation for them? Here it must be said right away that this condition will not be entirely correct. An infinite number of planes can pass through such points, so it is impossible to calculate a single answer. Let us consider such a problem to prove the incorrectness of such a formulation of the question.

Example 2

We have a rectangular coordinate system in three-dimensional space, in which three points are placed with coordinates M 1 (5, - 8, - 2), M 2 (1, - 2, 0), M 3 (- 1, 1, 1) . It is necessary to create an equation of the plane passing through it.

Solution

Let's use the first method and start by calculating the coordinates of two vectors M 1 M 2 → and M 1 M 3 →. Let's calculate their coordinates: M 1 M 2 → = (- 4, 6, 2), M 1 M 3 → = - 6, 9, 3.

The cross product will be equal to:

M 1 M 2 → × M 1 M 3 → = i → j → k → - 4 6 2 - 6 9 3 = 0 i ⇀ + 0 j → + 0 k → = 0 →

Since M 1 M 2 → × M 1 M 3 → = 0 →, then our vectors will be collinear (re-read the article about them if you forgot the definition of this concept). Thus, the initial points M 1 (5, - 8, - 2), M 2 (1, - 2, 0), M 3 (- 1, 1, 1) are on the same line, and our problem has infinitely many options answer.

If we use the second method, we will get:

x - x 1 y - y 1 z - z 1 x 2 - x 1 y 2 - y 1 z 2 - z 1 x 3 - x 1 y 3 - y 1 z 3 - z 1 = 0 ⇔ x - 5 y - (- 8) z - (- 2) 1 - 5 - 2 - (- 8) 0 - (- 2) - 1 - 5 1 - (- 8) 1 - (- 2) = 0 ⇔ ⇔ x - 5 y + 8 z + 2 - 4 6 2 - 6 9 3 = 0 ⇔ 0 ≡ 0

From the resulting equality it also follows that the given points M 1 (5, - 8, - 2), M 2 (1, - 2, 0), M 3 (- 1, 1, 1) are on the same line.

If you want to find at least one answer to this problem from the infinite number of its options, then you need to follow these steps:

1. Write down the equation of the line M 1 M 2, M 1 M 3 or M 2 M 3 (if necessary, look at the material about this action).

2. Take a point M 4 (x 4, y 4, z 4), which does not lie on the straight line M 1 M 2.

3. Write down the equation of a plane that passes through three different points M 1, M 2 and M 4 that do not lie on the same line.

If you notice an error in the text, please highlight it and press Ctrl+Enter

In order for a single plane to be drawn through any three points in space, it is necessary that these points do not lie on the same straight line.

Consider the points M 1 (x 1, y 1, z 1), M 2 (x 2, y 2, z 2), M 3 (x 3, y 3, z 3) in the general Cartesian coordinate system.

In order for an arbitrary point M(x, y, z) to lie in the same plane with points M 1, M 2, M 3, it is necessary that the vectors be coplanar.

Definition 2.1.

Two lines in space are called parallel if they lie in the same plane and do not have common points.

If two lines a and b are parallel, then, as in planimetry, write a || b. In space, lines can be placed so that they do not intersect or are parallel. This case is special for stereometry.

Definition 2.2.

Lines that do not have common points and are not parallel are called intersecting.

Theorem 2.1.

Through a point outside a given line it is possible to draw a line parallel to the given one, and only one.

25.Sign of parallelism between a line and a plane

Theorem

If a line that does not belong to a plane is parallel to some line in this plane, then it is parallel to the plane itself.

Proof

Let α be a plane, a a line not lying in it, and a1 a line in the α plane parallel to line a. Let us draw the plane α1 through the lines a and a1. Planes α and α1 intersect along straight line a1. If line a intersected plane α, then the intersection point would belong to line a1. But this is impossible, since the lines a and a1 are parallel. Consequently, line a does not intersect the plane α, and therefore is parallel to the plane α. The theorem has been proven.

27.Existence of a plane parallel to a given plane

Theorem

Through a point outside a given plane it is possible to draw a plane parallel to the given one, and only one.

Proof

Let us draw in this plane α any two intersecting lines a and b. Through a given point A we draw lines a1 and b1 parallel to them. The plane β passing through the lines a1 and b1, according to the theorem on the parallelism of planes, is parallel to the plane α.

Suppose that another plane β1 passes through point A, also parallel to the plane α. Let us mark some point C on the plane β1 that does not lie in the plane β. Let us draw the plane γ through points A, C and some point B of the plane α. This plane will intersect planes α, β and β1 along straight lines b, a and c. Lines a and c do not intersect line b, since they do not intersect the plane α. Therefore, they are parallel to line b. But in the γ plane only one line parallel to line b can pass through point A. which contradicts the assumption. The theorem has been proven.

28.Properties of parallel planes th

29.

Perpendicular lines in space. Two lines in space are called perpendicular if the angle between them is 90 degrees. c. m. k. k. m. c. k. Intersecting. Crossbreeding.

Theorem 2 1st PROPERTY OF PERPENDICULAR LINES AND PLANES. If a plane is perpendicular to one of two parallel lines, then it is also perpendicular to the other.
Proof: Let a 1 and a 2 - 2 be parallel lines and a plane perpendicular to the line a 1. Let us prove that this plane is perpendicular to the line a 2. Let us draw an arbitrary line x 2 in the plane through the point A 2 of the intersection of the straight line a 2 with the plane. Let us draw in the plane through point A 1 the intersection of line a 1 with line x 1 parallel to line x 2. Since line a 1 is perpendicular to the plane, then lines a 1 and x 1 are perpendicular. And by Theorem 1, the intersecting straight lines a 2 and x 2 parallel to them are also perpendicular. Thus, line a 2 is perpendicular to any line x 2 in the plane. And this (by definition) means that straight line a 2 is perpendicular to the plane. The theorem has been proven. See also support task No. 2.
Theorem 3 2nd PROPERTY OF PERPENDICULAR LINES AND PLANES. Two lines perpendicular to the same plane are parallel.
Proof: Let a and b be 2 straight lines perpendicular to the plane. Let us assume that lines a and b are not parallel. Let us choose a point C on line b that does not lie in the plane. Let us draw a line b 1 through point C, parallel to line a. Line b 1 is perpendicular to the plane according to Theorem 2. Let B and B 1 be the points of intersection of lines b and b 1 with the plane. Then straight line BB 1 is perpendicular to the intersecting lines b and b 1. And this is impossible. We have arrived at a contradiction. The theorem has been proven.

33.Perpendicular, lowered from a given point on a given plane, is a segment connecting a given point with a point on the plane and lying on a straight line perpendicular to the plane. The end of this segment lying in the plane is called base of the perpendicular.
Inclined drawn from a given point to a given plane is any segment connecting a given point with a point on the plane that is not perpendicular to the plane. The end of a segment lying in a plane is called inclined base. A segment connecting the bases of a perpendicular to an inclined one drawn from the same point is called oblique projection.

AB is perpendicular to the α plane.
AC – oblique, CB – projection.

Statement of the theorem

If a straight line drawn on a plane through the base of an inclined line is perpendicular to its projection, then it is perpendicular to the inclined one.

Proof

Let AB- perpendicular to plane α, A.C.- inclined and c- a straight line in the α plane passing through the point C and perpendicular to the projection B.C.. Let's make a direct CK parallel to the line AB. Straight CK is perpendicular to the plane α (since it is parallel AB), and therefore any straight line of this plane, therefore, CK perpendicular to a straight line c. Let's draw through parallel lines AB And CK plane β (parallel lines define a plane, and only one). Straight c perpendicular to two intersecting lines lying in the β plane, this is B.C. according to the condition and CK by construction, it means that it is perpendicular to any line belonging to this plane, which means it is perpendicular to the line A.C..

In order for a single plane to be drawn through any three points in space, it is necessary that these points do not lie on the same straight line.

Consider the points M 1 (x 1, y 1, z 1), M 2 (x 2, y 2, z 2), M 3 (x 3, y 3, z 3) in the general Cartesian coordinate system.

In order for an arbitrary point M(x, y, z) to lie in the same plane with points M 1, M 2, M 3, it is necessary that the vectors be coplanar.

(
) = 0

Thus,

Equation of a plane passing through three points:

Equation of a plane given two points and a vector collinear to the plane.

Let the points M 1 (x 1,y 1,z 1),M 2 (x 2,y 2,z 2) and the vector be given
.

Let's create an equation for a plane passing through the given points M 1 and M 2 and an arbitrary point M (x, y, z) parallel to the vector .

Vectors
and vector
must be coplanar, i.e.

(
) = 0

Plane equation:

Equation of a plane using one point and two vectors,

collinear to the plane.

Let two vectors be given
And
, collinear planes. Then for an arbitrary point M(x, y, z) belonging to the plane, the vectors
must be coplanar.

Plane equation:

Equation of a plane by point and normal vector .

Theorem. If a point M is given in space 0 (X 0 , y 0 , z 0 ), then the equation of the plane passing through the point M 0 perpendicular to the normal vector (A, B, C) has the form:

A(x – x 0 ) + B(y – y 0 ) + C(z – z 0 ) = 0.

Proof. For an arbitrary point M(x, y, z) belonging to the plane, we compose a vector. Because vector is the normal vector, then it is perpendicular to the plane, and, therefore, perpendicular to the vector
. Then the scalar product

= 0

Thus, we obtain the equation of the plane

The theorem has been proven.

Equation of a plane in segments.

If in the general equation Ax + Bi + Cz + D = 0 we divide both sides by (-D)

,

replacing
, we obtain the equation of the plane in segments:

The numbers a, b, c are the intersection points of the plane with the x, y, z axes, respectively.

Equation of a plane in vector form.

Where

- radius vector of the current point M(x, y, z),

A unit vector having the direction of a perpendicular dropped onto a plane from the origin.

,  and  are the angles formed by this vector with the x, y, z axes.

p is the length of this perpendicular.

In coordinates, this equation looks like:

xcos + ycos + zcos - p = 0.

Distance from a point to a plane.

The distance from an arbitrary point M 0 (x 0, y 0, z 0) to the plane Ax+By+Cz+D=0 is:

Example. Find the equation of the plane, knowing that point P(4; -3; 12) is the base of the perpendicular dropped from the origin to this plane.

So A = 4/13; B = -3/13; C = 12/13, we use the formula:

A(x – x 0 ) + B(y – y 0 ) + C(z – z 0 ) = 0.

Example. Find the equation of a plane passing through two points P(2; 0; -1) and

Q(1; -1; 3) perpendicular to the plane 3x + 2y – z + 5 = 0.

Normal vector to the plane 3x + 2y – z + 5 = 0
parallel to the desired plane.

We get:

Example. Find the equation of the plane passing through points A(2, -1, 4) and

B(3, 2, -1) perpendicular to the plane X + at + 2z – 3 = 0.

The required equation of the plane has the form: A x+B y+C z+ D = 0, normal vector to this plane (A, B, C). Vector
(1, 3, -5) belongs to the plane. The plane given to us, perpendicular to the desired one, has a normal vector (1, 1, 2). Because points A and B belong to both planes, and the planes are mutually perpendicular, then

So the normal vector (11, -7, -2). Because point A belongs to the desired plane, then its coordinates must satisfy the equation of this plane, i.e. 112 + 71 - 24 +D= 0;D= -21.

In total, we get the equation of the plane: 11 x - 7y – 2z – 21 = 0.

Example. Find the equation of the plane, knowing that point P(4, -3, 12) is the base of the perpendicular dropped from the origin to this plane.

Finding the coordinates of the normal vector
= (4, -3, 12). The required equation of the plane has the form: 4 x – 3y + 12z+ D = 0. To find the coefficient D, we substitute the coordinates of point P into the equation:

16 + 9 + 144 + D = 0

In total, we get the required equation: 4 x – 3y + 12z – 169 = 0

Example. The coordinates of the vertices of the pyramid are given: A 1 (1; 0; 3), A 2 (2; -1; 3), A 3 (2; 1; 1),

Find the length of edge A 1 A 2.

Find the angle between edges A 1 A 2 and A 1 A 4.

Find the angle between edge A 1 A 4 and face A 1 A 2 A 3.

First we find the normal vector to the face A 1 A 2 A 3 as a cross product of vectors
And
.

= (2-1; 1-0; 1-3) = (1; 1; -2);

Let's find the angle between the normal vector and the vector
.

-4 – 4 = -8.

The desired angle  between the vector and the plane will be equal to  = 90 0 - .

Find the area of ​​face A 1 A 2 A 3.

Find the volume of the pyramid.

Find the equation of the plane A 1 A 2 A 3.

Let's use the formula for the equation of a plane passing through three points.

2x + 2y + 2z – 8 = 0

x + y + z – 4 = 0;

When using the computer version “ Higher mathematics course” you can run a program that will solve the above example for any coordinates of the vertices of the pyramid.

To start the program, double-click on the icon:

In the program window that opens, enter the coordinates of the vertices of the pyramid and press Enter. In this way, all decision points can be obtained one by one.

Note: To run the program, you must have the Maple program ( Waterloo Maple Inc.) installed on your computer, any version starting with MapleV Release 4.

Suppose we need to find the equation of a plane passing through three given points that do not lie on the same line. Denoting their radius vectors by and the current radius vector by , we can easily obtain the required equation in vector form. In fact, the vectors must be coplanar (they all lie in the desired plane). Therefore, the vector-scalar product of these vectors must be equal to zero:

This is the equation of a plane passing through three given points, in vector form.

Moving on to the coordinates, we get the equation in coordinates:

If three given points lay on the same straight line, then the vectors would be collinear. Therefore, the corresponding elements of the last two lines of the determinant in equation (18) would be proportional and the determinant would be identically equal to zero. Consequently, equation (18) would become identical for any values ​​of x, y and z. Geometrically, this means that through each point in space there passes a plane in which the three given points lie.

Remark 1. The same problem can be solved without using vectors.

Denoting the coordinates of the three given points, respectively, we will write the equation of any plane passing through the first point:

To obtain the equation of the desired plane, it is necessary to require that equation (17) be satisfied by the coordinates of two other points:

From equations (19), it is necessary to determine the ratio of two coefficients to the third and enter the found values ​​into equation (17).

Example 1. Write an equation for a plane passing through the points.

The equation of the plane passing through the first of these points will be:

The conditions for the plane (17) to pass through two other points and the first point are:

Adding the second equation to the first, we find:

Substituting into the second equation, we get:

Substituting into equation (17) instead of A, B, C, respectively, 1, 5, -4 (numbers proportional to them), we obtain:

Example 2. Write an equation for a plane passing through the points (0, 0, 0), (1, 1, 1), (2, 2, 2).

The equation of any plane passing through the point (0, 0, 0) will be]

The conditions for the passage of this plane through points (1, 1, 1) and (2, 2, 2) are:

Reducing the second equation by 2, we see that to determine two unknowns, there is one equation with

From here we get . Now substituting the value of the plane into the equation, we find:

This is the equation of the desired plane; it depends on arbitrary

quantities B, C (namely, from the relation i.e. there are an infinite number of planes passing through three given points (three given points lie on the same straight line).

Remark 2. The problem of drawing a plane through three given points that do not lie on the same line can be easily solved in general form if we use determinants. Indeed, since in equations (17) and (19) the coefficients A, B, C cannot be simultaneously equal to zero, then, considering these equations as a homogeneous system with three unknowns A, B, C, we write a necessary and sufficient condition for the existence of a solution of this system, different from zero (Part 1, Chapter VI, § 6):

Having expanded this determinant into the elements of the first row, we obtain an equation of the first degree with respect to the current coordinates, which will be satisfied, in particular, by the coordinates of the three given points.

You can also verify this latter directly by substituting the coordinates of any of these points instead of . On the left side we get a determinant in which either the elements of the first row are zeros or there are two identical rows. Thus, the equation constructed represents a plane passing through the three given points.