It can be specified in 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 forms. Also, under certain conditions, the planes can be parallel, perpendicular, intersecting, etc. We will talk about this in this article. We will learn how to write the general equation of the plane and not only.

Normal form of the equation

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

Denote by P an arbitrary point Q=(x, y, z). We will sign the radius vector of the point Q with the letter p. The length of the vector α is p=IαI and Ʋ=(cosα,cosβ,cosγ).

This is a unit vector that points sideways, just like the vector α. α, β and γ are the angles that form between the vector Ʋ and the positive directions of the space axes x, y, z, respectively. The projection of some point QϵП onto the vector Ʋ is a constant value equal to р: (р,Ʋ) = р(р≥0).

This 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, and the unit vector Ʋ released from the point O will be perpendicular to P, regardless of its direction, which means that the vector Ʋ is determined from sign-accurate. The previous equation is the equation of our P plane, 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 a plane in space in its normal form.

General Equation

If we multiply the equation in coordinates by any number that is not equal to zero, we get an equation equivalent to the given one, which determines that same plane. It will look like this:

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

Plane equations. Special cases

The equation in general form can be modified in the presence of additional conditions. Let's consider some of them.

Assume that the coefficient A is 0. This means that the given plane is parallel to the given axis Ox. 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 С=0, then the equation is transformed into Ах+Ву+D=0, which will indicate parallelism to the given axis Oz.
• 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.
• Fifth, 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 non-zero, the form of equation (0) can be as follows:

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

in which a \u003d -D / A, b \u003d -D / B, c \u003d -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 easy to visually represent 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 the coefficients of the general equation of the given plane, that is, n (A, B, C).

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

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

It should be noted that the normal vector helps to solve various problems. The most common are tasks that consist in proving the perpendicularity or parallelism of planes, problems in finding angles between planes or angles between planes and lines.

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

A non-zero vector n perpendicular to a given plane is called normal (normal) for a given plane.

Suppose 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 compose an equation for a plane that will pass through the point Mₒ perpendicular to the normal n.

In space, we choose any arbitrary point and denote it by 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. The point M will belong to the given plane if the vector MₒM is perpendicular to the vector n. We write the orthogonality condition using the scalar product:

[MₒM, n] = 0.

Since MₒM \u003d r-rₒ, the vector equation of the plane will look like this:

This equation can take another form. To do this, the properties of the scalar product are used, and the left side of the equation is transformed. = - . If denoted as c, then the following equation will be obtained: - c \u003d 0 or \u003d c, which expresses the constancy of the projections onto the normal vector of the radius vectors of the given points that belong to the plane.

Now you 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+C*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.

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

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

Now we can compose an equation for a given plane, which will pass through the available points M′ and M″, as well as any point M with coordinates (x, y, z) parallel to the 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 equation of a plane in space will look like this:

Type of the equation of a plane intersecting three points

Suppose we have three points: (x′, y′, z′), (x″,y″,z″), (x‴,y‴,z‴), which do not belong to the same straight line. It is necessary to write the equation of the plane passing through the given three points. The theory of geometry claims that this kind of plane really exists, only it is the only one and inimitable. 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 compose a homogeneous system with unknowns u, v, w:

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

Equation (1), which we have obtained, is the equation of the plane. It passes exactly through 3 points, and this is easy to check. To do this, we need to expand our determinant over the elements in the first row. It follows from the existing properties of the determinant 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 set before us.

Dihedral angle between planes

A dihedral angle is 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), which is between these planes. The 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 suffices to take into account that 0≤φ≤π.

In fact, two planes that intersect form two (dihedral) angles: 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 signs, 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 angle φ in the equation cos φ= NN 1 /| N||N 1 | will be replaced by π-φ.

Perpendicular plane equation

Planes are called perpendicular if the angle between them is 90 degrees. Using the material outlined 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 state that they will be perpendicular if cosφ=0. This means that NN¹=AA¹+BB¹+CC¹=0.

Parallel plane equation

Parallel are two planes that do not contain common points.

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 conditions of proportionality are satisfied:

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 the 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)=p (p≥0).

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

The difference ρ-ρº of the radius vector of some point Q \u003d (x, y, z) belonging to P, as well as the radius vector of a given point Q 0 \u003d (xₒ, yₒ, zₒ) is such a vector, the absolute value of the projection of which on v is equal to the distance d, which must be found from Q 0 \u003d (xₒ, yₒ, 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)-p|.

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

Using the language of parameters, we get the obvious:

d=|Axₒ+Vuₒ+Czₒ|/√(A²+B²+C²).

If the given point Q 0 is on the other side of the plane P, as well as the origin, then between the vector ρ-ρ 0 and v is therefore:

d=-(ρ-ρ 0 ,v)=(ρ 0 ,v)-p>0.

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

d \u003d (ρ-ρ 0, v) \u003d p - (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 the plane containing all possible tangents to the curves drawn through this point on the surface.

With this form of the surface equation F (x, y, z) \u003d 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 the 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 P′ plane and the normal n″ (A″, B″, C″) of the P″ plane. 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 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 (common) planes П′ and П″. This means that the coordinates of any point belonging to the 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 particular 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 straight line, which will act as the intersection point of П′ and П″, and determine the straight line a in the coordinate system Oxyz (rectangular) in space.

1. Find the equation of a plane passing through a given point parallel to two given (non-collinear) vectors

Note: 1way . Take an arbitrary point of the plane M (x, y, z). The vectors will be coplanar because they are in parallel planes. Therefore, their mixed product
Writing this condition in coordinates, we obtain the equation of the desired plane:

It is more convenient to calculate this determinant by expansion in the first row.

2 way . Vectors
parallel to the desired plane. Therefore, a vector equal to the vector product of vectors
perpendicular to this plane , i.e.
and
. Vector is the normal vector of the plane . If a
and
, then the vector is found according to the formula:

Plane equation find by point
and normal vector

2. Find the equation of a plane passing through two given points parallel to a given vector
.(
non-collinear).

Note: 1 way. Let M (x, y, z) be an arbitrary point of the plane. Then the vectors and
are located in parallel planes, therefore, they are coplanar, i.e. their mixed product
Writing this condition in coordinates, we obtain the equation of the desired plane .

2 way . The normal vector to the desired plane will be equal to the vector product of vectors
, i.e.
or in coordinates:

Desired plane equation found by the normal vector and point
(or point
) by formula (2.1.1)

(see example 1 point 2.2).

3. Find the equation of a plane passing through a point
parallel to the plane 2x – 6y – 3z +5 =0.

Note: normal vector we find from the general equation of the given plane 2x – 6y – 3z +5 =0 (2.2.1).
Vector is perpendicular to a given plane, therefore, it is perpendicular to any plane parallel to it. Vector can be taken as the normal vector of the desired plane. Compose the equation of the desired plane by the point
and normal vector
(see example 1 point 2.2).

Answer:

4. Write an equation for a plane passing through a point
perpendicular to the line of intersection of the planes 2x + y - 2z + 1 = 0 and

x + y + z - 5 = 0.

Note: 1 way. Vectors perpendicular to each of their planes (the coordinates of the vectors are found from the general equations of the planes, formula (2.2.1)) are perpendicular to the line of their intersection and, therefore, are parallel to the desired plane. The desired plane passes through the point
parallel to two vectors
(see task 1 point 5).

The equation of the desired plane has the form:

Expanding the third-order determinant in the first row, we obtain the desired equation.

2 way. Compose the equation of the plane by the point
and normal vector by formula (2.2.1). normal vector is equal to the cross product of vectors
,those.
Since the vectors
are perpendicular to the line of intersection of the planes, then the vector parallel to the line of intersection of the planes and perpendicular to the desired plane.

Vectors (see formula 2.2.1), then

Compose the equation of the plane by the point
and normal vector

(see example 1 point 2.2)

Answer:

5. Find the equation of the plane passing through the points
and
perpendicular to the plane 3x – y + 3z +15 = 0.

Note: 1 way. Let us write out the coordinates of the normal vector of the given n flatness

3x - y + 3z +15 = 0:
Since the planes are perpendicular, the vector parallel to the desired plane Compose the equation of the desired plane
which is parallel to the vector and goes through the points
(see the solution of problem 2 point 5; 1 method).

Calculating the determinant, we obtain the equation of the desired plane

10x + 15y - 5z - 70 =0
2x + 3y – z – 14 =0.

2 way. Compose the equation of the desired plane by point
and the normal vector
Vector

We compose the equation of the desired plane .

10(x - 2) +15(y - 3) - 5(z + 1) = 0;

10x + 15y - 5z - 70 = 0 (see problem 2 point 5; 2nd method). Divide both sides of the equation by 5.

2x + 3y - z - 14 = 0.

Answer: 2x + 3y - z - 14 = 0.

6. Write an equation for a plane passing through points

and

Note: Let us compose the equation of a plane passing through three points (see example 1, clause 2.3, formula 2.3.1).

Expanding the determinant, we get

Answer:

Comment. To check the correctness of the calculation of the determinant, it is recommended to substitute the coordinates of these points through which the plane passes into the resulting equation. There must be an identity; otherwise, an error was made in the calculations.

7. Write an equation for a plane passing through a point
parallel to the x-plane – 4y + 5z + 1 = 0.

Note: From the general equation of a given plane
x – 4y + 5z + 1 = 0 find the normal vector
(formula 2.2.1). Vector perpendicular to the desired plane
Compose the equation of the plane by the point
and normal vector
(see example 1; clause 2.2):

x - 4y + 5z + 15 = 0.

Answer: x - 4y + 5z + 15 = 0.

8. Write an equation for a plane passing through a point
parallel to the vectors

Note: See the solution of the problem 1 point 5. We solve the problem in one of the indicated ways.

Answer: x - y - z - 1 = 0.

9. Write an equation for a plane passing through a point
perpendicular to the line of intersection of the planes 3x - 2y - z + 1 = 0 and x - y - z = 0.

Note: See the solution of problem 4 point 5. We solve the problem in one of the indicated ways.

Answer: x + 2y - z - 8 = 0.

10. Find the equation of the plane passing through the points

perpendicular to the plane 3x – y – 4z = 0.

Note: See the solution of problem 5 point 5.

Answer: 9x - y + 7z - 40 = 0.

11. Find the equation of the plane passing through the points

parallel to the straight line defined by points A (5; –2; 3) and B (6; 1; 0).

Note: The desired plane is parallel to the line AB, therefore, it is parallel to the vector
Desired plane equation we find, as in task 2, paragraph 5 (one of the ways).

Answer: 3x - 4y - 3z +4 = 0.

12. Point P (2; -1; -2) serves as the base of the perpendicular dropped from the origin to the plane. Write an equation for this plane.

Note: Normal vector to the desired plane is the vector
Find its coordinates. P (2; -1; -2) and O(0; 0; 0)

those.
Compose the equation of the plane by point and normal vector
(see example 1, paragraph 2.2).

Answer: 2x - y - 2z - 9 = 0.

13. Write an equation for a plane passing through a point
parallel to the plane: a) xoy; b) yoz; c) xoz.

Note: Vector
- the unit vector of the axis oz is perpendicular to the xoy plane, therefore, it is perpendicular to the desired plane
We compose the equation of the plane at point A (0; -1; 2) and

= (0; 0; 1), because
(see solution of problem 3, item 5).
z - 2 = 0.

We solve problems b) and c) in a similar way.

b)
where
(1; 0; 0).

in)
where (0; 1; 0).

y + 1 = 0.

Answer: a) z - 2 = 0; b) x = 0; c) y + 1 = 0.

14. Write an equation for a plane passing through points
and

B (2; 1; –1) perpendicular to the plane: a) xoy; b) xoz.

Note: The normal vector of the xoy plane is the vector

= (0; 0; 1) is the unit vector of the oz axis. Compose the equation of a plane passing through two points
and B (2; 1; –1) and perpendicular to the plane having the normal vector
(0; 0; 1), using one of the methods for solving problem 5 of paragraph 5.
y - 1 = 0.

Similarly for problem b):
where = (0; 1; 0).

Answer: a) y - 1 = 0; b) x + z - 1 = 0.

15. Write an equation for a plane passing through points
and

B (2; 3; –1) parallel to the oz axis.

Note: On the oz axis, you can take the unit vector = (0; 0; 1). The solution of the problem is similar to the solution of problem 2 point 5 (by any means).

Answer: x - y + 1 = 0.

16. Write an equation for a plane passing through the ox axis and a point

Note: Plane
passes through the x-axis, and hence also through the point O(0; 0; 0). On the ox axis, you can take the unit vector = (1; 0; 0). We compose the equation of the desired plane using two points A(2; –1; 6) and O(0; 0; 0) and the vector parallel to the plane. (See the solution of problem 2 point 5).

Answer: 6y + z = 0.

17. At what value of A will the planes Ax + 2y - 7z - 1 \u003d 0 and 2x - y + 2z \u003d 0 be perpendicular?

Note: From the general equations of planes

Ax + 2y - 7z - 1 = 0 and
2x – y + 2z = 0 normal vectors

= (A; 2; -7) and
= (2; –1; 2) (2.2.1). The condition of perpendicularity of two planes (2.6.1).

Answer: A = 8.

18. At what value A of the plane 2x + 3y - 6z - 23 = 0 and

4x + Ay - 12z + 7 = 0 will be parallel?

Note:
2x + 3y - 6z - 23 = 0 and
4x + Ay - 12y + 7 = 0

= (2; 3; -6) and
= (4;A; –12) (2.2.1). Because
(2.5.1)

Answer: A = 6.

19. Find the angle between two planes 2x + y + z + 7 = 0 and x - 2y + 3z = 0.

Note:
2x + y + z + 7 = 0 and
x – 2y + 3z = 0

= (2; 1; 1) and
= (1; –2; 3)

(2.4.1)

Answer:

20. Compose the canonical equations of a straight line passing through a point

A (1; 2; -3) parallel to the vector =(1; –2; 1).

Note: See the solution of the example of point 3.1.

Answer:

21. Compose parametric equations of a straight line passing through a point

A (–2; 3; 1) parallel to the vector =(3; –1; 2).

Note: See the solution of the example of point 3.2.

Answer:
.

22. Compose canonical and parametric equations of a straight line passing through points A (1; 0; -2) and B (1; 2; -4).

Note: See the solution of example 1 of paragraph 3.3.

Answer: a)
b)

23. Compose canonical and parametric equations of a straight line defined as the intersection of two planes x - 2y + 3z - 4 = 0 and 3x + 2y - 5z - 4 = 0.

Note: See example 1 point 3.4. Let z = 0, then the x and y coordinates of the point
find from the solution of the system

Hence the point
, lying on the desired line, has coordinates

(2; -1; 0). To find the direction vector of the desired straight line from the general equations of the planes
x – 2y +3z – 4 = 0 and
3x + 2y - 5z - 4 = 0

find normal vectors =(1; -2; 3) and
=(3; 2; –5).

The canonical equations of the line are found from the point
(2; -1; 0) and direction vector

(See formula (3.1.1)).

The parametric equations of the straight line can be found by the formula (3.2.1) or from the canonical equations:
We have:

Answer:
;
.

24. Through the dot
(2; -3; -4) draw a line parallel to a line

.

Note: Canonical Equations of the Required Line find by point
and direction vector Because
then for the direction vector straight you can take the direction vector straight L. Further, see the solution of problem 23, paragraph 5 or example 1, paragraph 3.4.

Answer:

25. Triangle vertices A (–5; 7; 1), B (2; 4; –1), and C (–1; 3; 5) are given. Find the equation for the median of triangle ABC drawn from vertex B.

Note: We find the coordinates of the point M from the condition AM = MC (BM is the median of triangle ABC).

FROM we leave the canonical equations of the straight line BM at two points B (2; 4; –1) and
(See example 1 point 3.3).

Answer:

26. Compose canonical and parametric equations of a straight line passing through a point
(–1; –2; 2) parallel to the x axis.

Note: Vector
– the unit vector of the x-axis is parallel to the required straight line. Therefore, it can be taken as the directing vector of the straight line
= (1; 0; 0). Compose the equations of a straight line by a point

(–1; –2: 2) and the vector = (1; 0; 0) (see example point 3.1 and example 1 point 3.2).

Answer:
;

27. Compose canonical equations of a straight line passing through a point
(3; –2; 4) perpendicular to the plane 5x + 3y – 7z + 1 = 0.

Note: From the general equation of the plane
5x + 3y – 7z + 1 = 0 find the normal vector = (5; 3; -7). According to the condition, the desired line
hence the vector
those. vector is the direction vector of the straight line L: = (5; 3; -7). We compose the canonical equations of a straight line by a point
(3; –2; 4) and direction vector

= (5; 3; -7). (See example point 3.1).

Answer:

28. Compose the parametric equations of the perpendicular dropped from the origin to the plane 4x - y + 2z - 3 = 0.

Note: Let us compose the equation of the desired perpendicular, i.e. straight line perpendicular to the plane
4x – y + 2z – 3 = 0 and passing through the point O (0; 0; 0). (See the solution of problem 27 point 5 and example 1 point 3.2).

Answer:

29. Find the point of intersection of a line
and plane

x - 2y + z - 15 = 0.

Note: To find the point M of the intersection of a line

L:
and plane

x - 2y + z - 15 = 0, it is necessary to solve the system of equations:

;

To solve the system, we transform the canonical equations of the straight line to parametric equations. (See problem 23 point 5).

Answer:

30. Find the projection of the point M (4; -3; 1) onto the plane x + 2y - z - 3 = 0.

Note: The projection of point M onto the plane will be point P - point p intersection of the perpendicular dropped from the point M to the plane
and planes Let's compose the parametric equations of the MP perpendicular. (See the solution of problem 28, paragraph 5).

Let's find the point P - the point of intersection of the line MP and the plane (See the solution of problem 29 point 5).

Answer:

31. Find the projection of the point A (1; 2; 1) on a straight line

Note: Projection of point A onto line L:
is t points At the intersection of the line L and the plane
which passes through point A and is perpendicular to line L. From the canonical equations of the line L, we write out the direction vector =(3; -1; 2). Plane perpendicular to line L, so
So the vector can be taken as the normal vector of the plane
= (3; –1; 2). Compose the equation of the plane point A(1; 2; 1) and = (3; –1; 2) (see example 1 point 2.2):
3(x - 1) - 1(y - 2) + 2(z - 1) = 0

3x - y + 2z - 3 = 0. Find the point B at the intersection of the line and the plane (see problem 29, paragraph 5):

Answer:

32. Draw a line through the point M (3; -1; 0) parallel to two planes x - y + z - 3 = 0 and x + y + 2z - 3 = 0.

Note: planes
x – y + z – 3 = 0 and
x + y + 2z - 3 = 0 are not parallel, because condition (2.5.1) is not satisfied:
planes
intersect. The desired line L, parallel to the planes
parallel to the line of intersection of these planes. (See the solution of problems 24 and 23 point 5).

Answer:

33. Write an equation for a plane passing through two lines

Note:1 way. Compose the equation of the desired plane by point
lying on a straight line , and the normal vector . Vector will be equal to the vector product of the directing vectors of the lines
, which we find from the canonical equations of lines
(formula 3.1.1): = (7; 3; 5) and

= (5; 5; –3)

Point coordinates
find from the canonical equations of the straight line

We compose the equation of the plane by point
and the normal vector =(–34; 46; 20) (see example 1 point 2.2)
17x - 23y - 10z + 36 = 0.

2 way. Finding direction vectors = (7; 3; 5) and = (5; 5; –3) from the canonical equations of lines
Point
(0; 2; –1) we find from the equation

. Take an arbitrary point on the plane

M (x; y; z). Vectors
are coplanar, therefore
From this condition we obtain the equation of the plane:

Answer: 17x - 23y - 10z +36 = 0.

34. Write an equation for a plane passing through a point
(2; 0; 1) and a straight line

Note: Let us first make sure that the point
on this straight line Ezhit:
Point
and direction vector we find from the canonical equations of the straight line
:
(1; -1; -1) and

= (1; 2; -1). Normal vector of the desired plane
We find the coordinates of the normal vector, knowing the coordinates =(1; 2; -1) and

= (1; 1; 2):

We compose the equation of the plane by the point
(2; 0; 1) and the normal vector = (–5; 3; 1):

–5(x – 2) + 3(y – 0) + 1(z – 1) = 0.

Answer: 5x - 3y - z - 9 = 0.

To get the general equation of the plane, we analyze the plane passing through a given point.

Let there be three coordinate axes already known to us in space - Ox, Oy and Oz. Hold the sheet of paper so that it remains flat. The plane will be the sheet itself and its continuation in all directions.

Let P arbitrary plane in space. Any vector perpendicular to it is called normal vector to this plane. Naturally, we are talking about a non-zero vector.

If any point of the plane is known P and some vector of the normal to it, then by these two conditions the plane in space is completely determined(through a given point, there is only one plane perpendicular to a given vector). The general equation of the plane will look like:

So, there are conditions that set the equation of the plane. To get it self plane equation, which has the above form, we take on the plane P arbitrary point M with variable coordinates x, y, z. This point belongs to the plane only if vector perpendicular to the vector(Fig. 1). For this, according to the condition of perpendicularity of vectors, it is necessary and sufficient that the scalar product of these vectors be equal to zero, that is

The vector is given by condition. We find the coordinates of the vector by the formula :

.

Now, using the dot product formula of vectors , we express the scalar product in coordinate form:

Since the point M(x; y; z) is chosen arbitrarily on the plane, then the last equation is satisfied by the coordinates of any point lying on the plane P. For point N, not lying on a given plane, , i.e. equality (1) is violated.

Example 1 Write an equation for a plane passing through a point and perpendicular to a vector.

Solution. We use formula (1), look at it again:

In this formula, the numbers A , B and C vector coordinates and numbers x0 , y0 and z0 - point coordinates.

The calculations are very simple: we substitute these numbers into the formula and get

We multiply everything that needs to be multiplied and add up just numbers (which are without letters). Result:

.

The required equation of the plane in this example turned out to be expressed by the general equation of the first degree with respect to variable coordinates x, y, z arbitrary point of the plane.

So, an equation of the form

called the general equation of the plane .

Example 2 Construct in a rectangular Cartesian coordinate system the plane given by the equation .

Solution. To construct a plane, it is necessary and sufficient to know any three of its points that do not lie on one straight line, for example, the points of intersection of the plane with the coordinate axes.

How to find these points? To find the point of intersection with the axis Oz, you need to substitute zeros instead of x and y in the equation given in the problem statement: x = y= 0 . Therefore, we get z= 6 . Thus, the given plane intersects the axis Oz at the point A(0; 0; 6) .

In the same way, we find the point of intersection of the plane with the axis Oy. At x = z= 0 we get y= −3 , that is, a point B(0; −3; 0) .

And finally, we find the point of intersection of our plane with the axis Ox. At y = z= 0 we get x= 2 , that is, a point C(2; 0; 0) . According to the three points obtained in our solution A(0; 0; 6) , B(0; −3; 0) and C(2; 0; 0) we build the given plane.

Consider now special cases of the general equation of the plane. These are cases when certain coefficients of equation (2) vanish.

1. When D= 0 equation defines a plane passing through the origin, since the coordinates of a point 0 (0; 0; 0) satisfy this equation.

2. When A= 0 equation defines a plane parallel to the axis Ox, since the normal vector of this plane is perpendicular to the axis Ox(its projection on the axis Ox equals zero). Similarly, when B= 0 plane axis parallel Oy, and when C= 0 plane parallel to axis Oz.

3. When A=D= 0 equation defines a plane passing through the axis Ox because it is parallel to the axis Ox (A=D= 0). Similarly, the plane passes through the axis Oy, and the plane through the axis Oz.

4. When A=B= 0 equation defines a plane parallel to the coordinate plane xOy because it is parallel to the axes Ox (A= 0) and Oy (B= 0). Likewise, the plane is parallel to the plane yOz, and the plane - the plane xOz.

5. When A=B=D= 0 equation (or z= 0) defines the coordinate plane xOy, since it is parallel to the plane xOy (A=B= 0) and passes through the origin ( D= 0). Similarly, the equation y= 0 in space defines the coordinate plane xOz, and the equation x= 0 - coordinate plane yOz.

Example 3 Compose the equation of the plane P passing through the axis Oy and point .

Solution. So the plane passes through the axis Oy. So in her equation y= 0 and this equation has the form . To determine the coefficients A and C we use the fact that the point belongs to the plane P .

Therefore, among its coordinates there are those that can be substituted into the equation of the plane, which we have already derived (). Let's look at the coordinates of the point again:

M0 (2; −4; 3) .

Among them x = 2 , z= 3 . We substitute them into the general equation and get the equation for our particular case:

2A + 3C = 0 .

We leave 2 A on the left side of the equation, we transfer 3 C to the right side and get

A = −1,5C .

Substituting the found value A into the equation , we get

or .

This is the equation required in the example condition.

Solve the problem on the equations of the plane yourself, and then look at the solution

Example 4 Determine the plane (or planes if more than one) with respect to the coordinate axes or coordinate planes if the plane(s) is given by the equation .

Solutions to typical problems that occur in tests - in the manual "Problems on a plane: parallelism, perpendicularity, intersection of three planes at one point" .

Equation of a plane passing through three points

As already mentioned, a necessary and sufficient condition for constructing a plane, in addition to one point and a normal vector, are also three points that do not lie on one straight line.

Let there be given three different points , and , not lying on the same straight line. Since these three points do not lie on one straight line, the vectors and are not collinear, and therefore any point of the plane lies in the same plane with the points , and if and only if the vectors , and coplanar, i.e. if and only if the mixed product of these vectors equals zero.

Using the mixed product expression in coordinates, we obtain the plane equation

(3)

After expanding the determinant, this equation becomes an equation of the form (2), i.e. the general equation of the plane.

Example 5 Write an equation for a plane passing through three given points that do not lie on a straight line:

and to determine a particular case of the general equation of the line, if any.

Solution. According to formula (3) we have:

Normal equation of the plane. Distance from point to plane

The normal equation of a plane is its equation, written in the form

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 one 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 a common Cartesian coordinate system.

In order for an arbitrary point M(x, y, z) to lie in the same plane as the points M 1 , M 2 , M 3 , the vectors must be coplanar.

(
) = 0

In this way,

Equation of a plane passing through three points:

Equation of a plane with respect to 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
.

Let us compose the equation of the 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 with respect to one point and two vectors,

collinear 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:

Plane equation 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) looks like:

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 - 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 + Wu + Cz + D \u003d 0, divide both parts by (-D)

,

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

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

Plane equation in vector form.

where

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

A unit vector that has the direction of the perpendicular dropped to the 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 has the form:

xcos + ycos + zcos - p = 0.

The 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 + Vy + Cz + D \u003d 0 is:

Example. Find the equation of the plane, knowing that the 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, 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) is 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 the points A(2, -1, 4) and

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

The desired plane equation has the form: A x+ B y+C z+ D = 0, the 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 the 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 desired equation of the plane has the form: 4 x – 3y + 12z+ D = 0. To find the coefficient D, we substitute the coordinates of the point Р into the equation:

16 + 9 + 144 + D = 0

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

Example. Given the coordinates of the pyramid vertices A 1 (1; 0; 3), A 2 (2; -1; 3), A 3 (2; 1; 1),

Find the length of the edge A 1 A 2 .

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

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

First, 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);

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 А 1 А 2 А 3 .

We 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 PC version of “ Course of higher mathematics” you can run a program that will solve the above example for any coordinates of the pyramid vertices.

Double-click the icon to launch the program:

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

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

Plane equation. How to write an equation for a plane?
Mutual arrangement of planes. Tasks

Spatial geometry is not much more complicated than "flat" geometry, and our flights in space begin with this article. In order to understand the topic, one must have a good understanding of vectors, in addition, it is desirable to be familiar with the geometry of the plane - there will be many similarities, many analogies, so the information will be digested much better. In a series of my lessons, the 2D world opens with an article Equation of a straight line on a plane. But now Batman has stepped off the flat screen TV and is launching from the Baikonur Cosmodrome.

Let's start with drawings and symbols. Schematically, the plane can be drawn as a parallelogram, which gives the impression of space:

The plane is infinite, but we have the opportunity to depict only a piece of it. In practice, in addition to the parallelogram, an oval or even a cloud is also drawn. For technical reasons, it is more convenient for me to depict the plane in this way and in this position. The real planes, which we will consider in practical examples, can be arranged as you like - mentally take the drawing in your hands and twist it in space, giving the plane any slope, any angle.

Notation: it is customary to designate planes in small Greek letters, apparently so as not to confuse them with straight on the plane or with straight in space. I'm used to using the letter . In the drawing, it is the letter "sigma", and not a hole at all. Although, a holey plane, it is certainly very funny.

In some cases, it is convenient to use the same Greek letters with subscripts to designate planes, for example, .

It is obvious that the plane is uniquely determined by three different points that do not lie on the same straight line. Therefore, three-letter designations of planes are quite popular - according to the points belonging to them, for example, etc. Often letters are enclosed in parentheses: so as not to confuse the plane with another geometric figure.

For experienced readers, I will give shortcut menu:

• How to write an equation for a plane using a point and two vectors?
• How to write an equation for a plane using a point and a normal vector?

and we will not languish in long waits:

General equation of the plane

The general equation of the plane has the form , where the coefficients are simultaneously non-zero.

A number of theoretical calculations and practical problems are valid both for the usual orthonormal basis and for the affine basis of space (if oil is oil, return to the lesson Linear (non) dependence of vectors. Vector basis). For simplicity, we will assume that all events occur in an orthonormal basis and a Cartesian rectangular coordinate system.

And now let's train a little spatial imagination. It's okay if you have it bad, now we'll develop it a little. Even playing on nerves requires practice.

In the most general case, when the numbers are not equal to zero, the plane intersects all three coordinate axes. For example, like this:

I repeat once again that the plane continues indefinitely in all directions, and we have the opportunity to depict only part of it.

Consider the simplest equations of planes:

How to understand this equation? Think about it: “Z” ALWAYS, for any values ​​of “X” and “Y” is equal to zero. This is the equation of the "native" coordinate plane. Indeed, formally the equation can be rewritten as follows: , from where it is clearly visible that we don’t care, what values ​​“x” and “y” take, it is important that “z” is equal to zero.

Similarly:
is the equation of the coordinate plane ;
is the equation of the coordinate plane.

Let's complicate the problem a little, consider a plane (here and further in the paragraph we assume that the numerical coefficients are not equal to zero). Let's rewrite the equation in the form: . How to understand it? "X" is ALWAYS, for any value of "y" and "z" is equal to a certain number. This plane is parallel to the coordinate plane. For example, a plane is parallel to a plane and passes through a point.

Similarly:
- the equation of the plane, which is parallel to the coordinate plane;
- the equation of a plane that is parallel to the coordinate plane.

Add members: . The equation can be rewritten like this: , that is, "Z" can be anything. What does it mean? "X" and "Y" are connected by a ratio that draws a certain straight line in the plane (you will recognize equation of a straight line in a plane?). Since Z can be anything, this line is "replicated" at any height. Thus, the equation defines a plane parallel to the coordinate axis

Similarly:
- the equation of the plane, which is parallel to the coordinate axis;
- the equation of the plane, which is parallel to the coordinate axis.

If the free terms are zero, then the planes will directly pass through the corresponding axes. For example, the classic "direct proportionality":. Draw a straight line in the plane and mentally multiply it up and down (since “z” is any). Conclusion: the plane given by the equation passes through the coordinate axis.

We conclude the review: the equation of the plane passes through the origin. Well, here it is quite obvious that the point satisfies the given equation.

And, finally, the case that is shown in the drawing: - the plane is friends with all coordinate axes, while it always “cuts off” a triangle that can be located in any of the eight octants.

Linear inequalities in space

In order to understand the information, it is necessary to study well linear inequalities in the plane because many things will be similar. The paragraph will be of a brief overview with a few examples, since the material is quite rare in practice.

If the equation defines a plane, then the inequalities
ask half-spaces. If the inequality is not strict (the last two in the list), then the solution of the inequality, in addition to the half-space, includes the plane itself.

Example 5

Find the unit normal vector of the plane .

Solution: A unit vector is a vector whose length is one. Let's denote this vector by . It is quite clear that the vectors are collinear:

First, we remove the normal vector from the equation of the plane: .

How to find the unit vector? To find the unit vector, you need every vector coordinate divided by vector length.

Let's rewrite the normal vector in the form and find its length:

According to the above:

Answer:

Check: , which was required to check.

Readers who have carefully studied the last paragraph of the lesson, probably noticed that the coordinates of the unit vector are exactly the direction cosines of the vector:

Let's digress from the disassembled problem: when you are given an arbitrary non-zero vector, and by the condition it is required to find its direction cosines (see the last tasks of the lesson Dot product of vectors), then you, in fact, also find a unit vector collinear to the given one. In fact, two tasks in one bottle.

The need to find a unit normal vector arises in some problems of mathematical analysis.

We figured out the fishing of the normal vector, now we will answer the opposite question:

How to write an equation for a plane using a point and a normal vector?

This rigid construction of a normal vector and a point is well known by a darts target. Please stretch your hand forward and mentally select an arbitrary point in space, for example, a small cat in a sideboard. Obviously, through this point, you can draw a single plane perpendicular to your hand.

The equation of a plane passing through a point perpendicular to the vector is expressed by the formula: