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 plane equation can have different forms. Also, under certain conditions, the 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 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.

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

The equation in general form can be modified in the presence of additional conditions. Let's look at 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 = -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 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 an equation in segments, which has the form x/a + y/b + z/c = 1, as well 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 left side of the equation is transformed. = - . 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 that will pass through the existing 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 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 is 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 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) that is located 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 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 this case, ρ (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 a given 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 there is therefore:

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 created angle 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 (common) planes П′ and П″. 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 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 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.

Equation of a plane. How to write an equation of 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. To master the topic, you need to have a good understanding of vectors, in addition, it is advisable 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 left the flat TV screen and is launching from the Baikonur Cosmodrome.

Let's start with drawings and symbols. Schematically, the plane can be drawn in the form of a parallelogram, which creates 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 exactly this way and in exactly this position. Real planes, which we will consider in practical examples, can be located in any way - mentally take the drawing in your hands and rotate it in space, giving the plane any inclination, any angle.

Designations: planes are usually denoted in small Greek letters, apparently so as not to confuse them with straight line on a plane or with straight line 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 lower subscripts to designate planes, for example, .

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

For experienced readers, I will give quick access menu:

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

and we will not languish in long waits:

General plane equation

The general equation of the plane has the form , where the coefficients are not equal to zero at the same time.

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

Now let’s practice our spatial imagination a little. It’s okay if yours is bad, now we’ll develop it a little. Even playing on nerves requires training.

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.

Let's 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 you can clearly see that we don’t care what values ​​“x” and “y” take, it is important that “z” is equal to zero.

Likewise:
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 values ​​of “Y” and “Z”, 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.

Likewise:
– equation of a plane that is parallel to the coordinate plane;
– equation of a plane that is parallel to the coordinate plane.

Let's add members: . The equation can be rewritten as follows: , that is, “zet” can be anything. What does it mean? “X” and “Y” are connected by the relation, which draws a certain straight line in the plane (you will find out equation of a line in a plane?). Since “z” can be anything, this straight line is “replicated” at any height. Thus, the equation defines a plane parallel to the coordinate axis

Likewise:
– equation of a plane that is parallel to the coordinate axis;
– equation of a plane that 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 defined by the equation passes through the coordinate axis.

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

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

Linear inequalities in space

To understand the information you need to study well linear inequalities in the plane, because many things will be similar. The paragraph will be of a brief overview nature with several 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, also 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 us denote this vector by . It is absolutely clear that the vectors are collinear:

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

How to find a unit vector? In order to find the unit vector, you need every divide the vector coordinate by the vector length.

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

According to the above:

Answer:

Verification: what was required to be verified.

Readers who 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 take a break from the problem at hand: when you are given an arbitrary non-zero vector, and according to the condition it is required to find its direction cosines (see the last problems of the lesson Dot product of vectors), then you, in fact, find a unit vector collinear to this one. Actually two tasks in one bottle.

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

We’ve figured out how to fish out a normal vector, now let’s answer the opposite question:

How to create an equation of a plane using a point and a normal vector?

This rigid construction of a normal vector and a point is well known to the dartboard. Please stretch your hand forward and mentally select an arbitrary point in space, for example, a small cat in the 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:

Let the points M 1, M 2, M 3 not lie on the same line. As is known, three such points uniquely define a certain plane p (Fig. 199).

Let us derive the equation of the plane R. Let M be an arbitrary point in space. Obviously, point M belongs to the plane R if and only if the vectors

\(\overrightarrow(M_(1)M)\), \(\overrightarrow(M_(1)M_2)\), \(\overrightarrow(M_(1)M_3)\) are coplanar. A necessary and sufficient condition for the coplanarity of three vectors is that their mixed product is equal to zero (§ 23*, Theorem 2). Therefore, the equation of a plane passing through three points that do not lie on the same line can be written as follows:

(\(\overrightarrow(M_(1)M)\), \(\overrightarrow(M_(1)M_2)\), \(\overrightarrow(M_(1)M_3)\)) = 0. (1)

If the points M 1, M 2 and M 3 are given coordinates in some rectangular Cartesian coordinate system, then equation (1) can be written in coordinates.

Let M 1 ( x 1 ; y 1 ; z 1), M 2 ( X 2 ; at 2 ; z 2), M 3 ( X 3 ; at 3 ; z 3) - point data. Let us denote the coordinates of an arbitrary point M on the plane p by x, y And z. Let's find the coordinates of the vectors included in equation (1):

\(\overrightarrow(M_(1)M)\) = ( x - x 1 ; y - y 1 ; z - z 1),

\(\overrightarrow(M_(1)M_2)\) = ( x 2 -x 1 ; y 2 -y 1 ; z 2 - z 1),

\(\overrightarrow(M_(1)M_3)\) = ( x 3 -x 1 ; at 3 -y 1 ; z 3 - z 1).

The mixed product of three vectors is equal to the third-order determinant, the lines of which contain the coordinates of the vectors. Therefore, equation (1) in coordinates has the form

$$ \begin(vmatrix) x-x_1 & y-y_1 & z-z_1 \\ x_2-x_1 & y_2-y_1 & z_2-z_1 \\ x_3-x_1 & y_3-y_1 & z_3-z_1 \end(vmatrix)= 0 \;\; (2)$$

Let us find the equation of a plane passing through three points A ( A; 0; 0), B(0; b; 0), C(0; 0; With), which A =/= 0, b =/= 0, c=/= 0. These points lie on the coordinate axes (Fig. 200).

Assuming in equation (2) x 1 = A, at 1 = 0, z 1 = 0, x 2 = 0, at 2 = b, z 2 = 0, x 3 = 0, at 3 = 0, z 3 = With, we get

$$ \begin(vmatrix) x-a & y & z \\ -a & b & 0 \\ -a & 0 & c \end(vmatrix)=0 $$

Expanding the determinant into the elements of the first row, we obtain the equation

bc(x - a) + acy + abz = 0

bcx + asu + abz = abc,

x / a + y / b + z / c = 1. (3)

Equation (3) is called equation of the plane in segments, since the numbers a, b And With indicate which segments the plane cuts off on the coordinate axes.

Task. Write the equation of the plane passing through the points M 1 (-1; 4; -1), M 2 (-13; 2; -10), M 3 (6; 0; 12). Simplify the resulting equation. Obtain the equation of a given plane in segments.

Equation (2) in this case is written as follows:

$$ \begin(vmatrix) x+1 & y-4 & z+1 \\ -12 & -2 & -9 \\ 7 & -4 & 13 \end(vmatrix)=0 $$

This is the equation of this plane. Expanding the determinant along the first row, we get

62(X+ 1) +93(y- 4)+ 62 (z + 1) = 0,

2x + 3y + 2z - 12 = 0.

Dividing term by term by 12 and moving the free term of the equation to the right side, we obtain the equation of this plane in segments

$$ \frac(x)(-6)+\frac(y)(4)+\frac(z)(6)=1 $$

From the equation it is clear that this plane cuts off segments on the coordinate axes whose lengths are equal to 6, 4 and 6, respectively. Axis Oh intersects the plane at a point with a negative abscissa, the axis OU- at a point with a positive ordinate, axis Oz- at a point with a positive applicate.

To obtain the general equation of a plane, let us 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. Every vector perpendicular to it is called normal vector to this plane. Naturally, we are talking about a non-zero vector.

If any point on the plane is known P and some normal vector to it, then by these two conditions the plane in space is completely defined(through a given point you can draw a single plane perpendicular to the given vector). The general equation of the plane will be:

So, the conditions that define the equation of the plane are. To get yourself plane equation, having the above form, 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 specified by condition. We find the coordinates of the vector using the formula :

.

Now, using the scalar product of vectors formula , 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 a 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 the vector.

Solution. Let’s use formula (1) and look at it again:

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

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 a 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 general plane equation .

Example 2. Construct in a rectangular Cartesian coordinate system a 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 the same 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 for 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, the 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.

Let's now consider special cases of the general equation of the plane. These are cases when certain coefficients of equation (2) become zero.

1. When D= 0 equation defines a plane passing through the origin, since the coordinates of the 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 onto the axis Ox equal to zero). Similarly, when B= 0 plane parallel to the axis Oy, and when C= 0 plane parallel to axis Oz.

3. When A=D= 0 equation defines a plane passing through the axis Ox, since 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, since it is parallel to the axes Ox (A= 0) and Oy (B= 0). Similarly, the plane is parallel to the plane yOz, and the plane is 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). Likewise, Eq. y = 0 in space defines the coordinate plane xOz, and the equation x = 0 - coordinate plane yOz.

Example 3. Create an equation of the plane P, passing through the axis Oy and period.

Solution. So the plane passes through the axis Oy. Therefore, in her equation y= 0 and this equation has the form . To determine the coefficients A And C let's take advantage of the fact that the point belongs to the plane P .

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

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 .

Leave 2 A on the left side of the equation, move 3 C to the right side and we 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 plane equation problem yourself, and then look at the solution

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

Solutions to typical problems that occur during tests are in the textbook “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 the normal vector, are also three points that do not lie on the same line.

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

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

(3)

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

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

and determine a special case of the general equation of a line, if one occurs.

Solution. According to formula (3) we have:

Normal plane equation. Distance from point to plane

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

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

Note: 1 way . Let's take an arbitrary point of the plane M (x, y, z). The vectors will be coplanar since they are located 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 expanding along the first line.

Method 2 . Vectors
parallel to the desired plane. Therefore, a vector equal to the cross product of vectors
perpendicular to this plane , i.e.
And
. Vector is a normal vector of the plane . If
And
, then the vector is found by 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 on the plane. Then the vectors and
are located in parallel planes, therefore, coplanar, i.e. their mixed product
Having written this condition in coordinates, we obtain the equation of the desired plane .

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

Equation of the desired plane found by normal vector and point
(or point
) by formula (2.1.1)

(see example 1 point 2.2).

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

Note: normal vector we find from the general equation of this plane 2x – 6y – 3z +5 =0 (2.2.1).
Vector 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. Let's create an equation for the desired plane based on the point
and normal vector
(see example 1, paragraph 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 its planes (the coordinates of the vectors are found from the general equations of planes, formula (2.2.1)) are perpendicular to the line of their intersection and, therefore, 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 along the first line, we obtain the required equation.

Method 2. Let's create an equation of the plane based on a point
and normal vector according to formula (2.2.1). Normal vector equal to the vector product of vectors
,those.
Since 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

Let's create an equation of the plane based on a point
and normal vector

(see example 1 clause 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 down the coordinates of the normal vector of a given n glossiness

3x – y + 3z +15 = 0:
Since the planes are perpendicular, then the vector parallel to the desired plane Let's compose the equation of the desired plane
which is parallel to the vector and passes through the points
(see solution to problem 2, point 5; method 1).

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

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

Method 2. Let's 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; method 2). 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 the points

And

Note: Let's create an equation for a plane passing through three points (see example 1, paragraph 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. The result should be an identity; otherwise, there is an error 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
Let's create an equation of the plane based on a point
and normal vector
(see example 1; paragraph 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 solution to problem 1, point 5. We solve the problem using one of the indicated methods.

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 to problem 4, point 5. We solve the problem using one of the indicated methods.

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 solution to problem 5, point 5.

Answer: 9x – y +7z – 40 = 0.

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

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

Note: The desired plane is parallel to line AB, therefore it is parallel to the vector
Equation of the desired plane we find, as in problem 2 of paragraph 5 (by one of the methods).

Answer: 3x – 4y – 3z +4 = 0.

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

Note: Normal vector to the desired plane is the vector
Let's find its coordinates. P (2; –1; –2) and O(0; 0; 0)

those.
Let's create an 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 axis vector 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 to problem 3, point 5).
z – 2 = 0.

We solve problems b) and c) similarly.

b)
Where
(1; 0; 0).

V)
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 the 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) – unit vector of the oz axis. Let's create an equation for a plane passing through two points
and B (2; 1; –1) and perpendicular to the plane having a 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 the points
And

B (2; 3; –1) parallel to the oz axis.

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

Answer: x – y + 1 = 0.

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

Note: Plane
passes through the axis ox, therefore, through the point O(0; 0; 0). On the ox axis we can take a 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 solution to problem 2, point 5).

Answer: 6y + z = 0.

17. At what value of A will the planes Ax + 2y – 7z – 1 = 0 and 2x – y + 2z = 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). Condition for 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 canonical equations of a line passing through a point

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

Note: See the solution to the example of paragraph 3.1.

Answer:

21. Write parametric equations for a line passing through a point

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

Note: See the solution to the example in paragraph 3.2.

Answer:
.

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

Note: See the solution to example 1 of clause 3.3.

Answer: A)
b)

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

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

Therefore, 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 planes
x – 2y +3z – 4 = 0 and
3x + 2y – 5z – 4 = 0

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

We find the canonical equations of a straight line from a point
(2; –1; 0) and the direction vector

(See formula (3.1.1)).

Parametric equations of a straight line can be found using formula (3.2.1) or from canonical equations:
We have:

Answer:
;
.

24. Through the point
(2; –3; –4) draw a line parallel to the line

.

Note: Canonical equations of the desired line let's find by point
and direction vector Because
then for the direction vector straight you can take a direction vector straight L. Next, see the solution to problem 23, paragraph 5 or example 1, paragraph 3.4.

Answer:

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

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

WITH Let us leave the canonical equations of the straight line BM for two points B (2; 4; –1) and
(See example 1 point 3.3).

Answer:

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

Note: Vector
– the unit vector axisox is parallel to the desired line. Therefore, it can be taken as the directing vector of the straight line
= (1; 0; 0). Let's compose the equations of a straight line from a point

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

Answer:
;

27. Compose canonical equations of a 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 line L: = (5; 3; -7). We compose the canonical equations of a straight line from a point
(3; –2; 4) and direction vector

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

Answer:

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

Note: Let's create an equation for 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 to problem 27, paragraph 5 and example 1, paragraph 3.2).

Answer:

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

x - 2y + z - 15 = 0.

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

L:
and planes

x – 2y + z – 15 = 0, we need to solve the system of equations:

;

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

Answer:

30. Find the projection of 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 a perpendicular dropped from point M to a plane
and planes Let's compose the parametric equations of the perpendicular MR. (See the solution to problem 28, point 5).

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

Answer:

31. Find the projection of point A(1; 2; 1) onto the line

Note: Projection of point A onto line L:
is t points B intersection of straight line L and plane
which passes through point A and is perpendicular to line L. From the canonical equations of straight line L we write out the direction vector =(3; –1; 2). Plane is perpendicular to line L, therefore,
So the vector can be taken as the normal vector of the plane
= (3; –1; 2). Let's create an equation of the plane at point A(1; 2; 1) and = (3; –1; 2) (see example 1, paragraph 2.2):
3(x – 1) – 1(y – 2) + 2(z – 1) = 0

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

Answer:

32. Through point M (3; –1; 0) draw a straight line 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 met:
Planes
intersect. The required straight line L, parallel to the planes
parallel to the line of intersection of these planes. (See solution to problems 24 and 23, paragraph 5).

Answer:

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

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

= (5; 5; –3)

Point coordinates
we 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, paragraph 2.2)
17x – 23y – 10z + 36 = 0.

Method 2. Finding direction vectors = (7; 3; 5) and = (5; 5; –3) from the canonical equations of lines
Full stop
(0; 2; –1) is found from the equation

. Let's 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 straight line

Note: Let's make sure first of all that the point
on this straight line hedges:
Full stop
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 a plane from a point
(2; 0; 1) and normal vector = (–5; 3; 1):

–5(x – 2) + 3(y – 0) + 1(z – 1) = 0.

Answer: 5x – 3y – z – 9 = 0.