Consider the following figure:

It depicts a certain function y = f(x), which is differentiable at point a. Point M with coordinates (a; f(a)) is marked. A secant MR is drawn through an arbitrary point P(a + ∆x; f(a + ∆x)) of the graph.

If now point P is shifted along the graph to point M, then straight line MR will rotate around point M. In this case, ∆x will tend to zero. From here we can formulate the definition of a tangent to the graph of a function.

Tangent to the graph of a function

The tangent to the graph of a function is the limiting position of the secant as the increment of the argument tends to zero. It should be understood that the existence of the derivative of the function f at the point x0 means that at this point on the graph there is tangent to him.

At the same time slope the tangent will be equal to the derivative of this function at this point f’(x0). This is geometric meaning derivative. The tangent to the graph of a function f differentiable at point x0 is a certain straight line passing through the point (x0;f(x0)) and having an angular coefficient f’(x0).

Tangent equation

Let's try to obtain the equation of the tangent to the graph of some function f at point A(x0; f(x0)). The equation of a straight line with slope k has the following form:

Since our slope coefficient is equal to the derivative f’(x0), then the equation will take the following form: y = f’(x0)*x + b.

Now let's calculate the value of b. To do this, we use the fact that the function passes through point A.

f(x0) = f’(x0)*x0 + b, from here we express b and get b = f(x0) - f’(x0)*x0.

We substitute the resulting value into the tangent equation:

y = f’(x0)*x + b = f’(x0)*x + f(x0) - f’(x0)*x0 = f(x0) + f’(x0)*(x - x0).

y = f(x0) + f’(x0)*(x - x0).

Let's consider next example: find the equation of the tangent to the graph of the function f(x) = x 3 - 2*x 2 + 1 at point x = 2.

2. f(x0) = f(2) = 2 2 - 2*2 2 + 1 = 1.

3. f’(x) = 3*x 2 - 4*x.

4. f’(x0) = f’(2) = 3*2 2 - 4*2 = 4.

5. Substitute the obtained values ​​into the tangent formula, we get: y = 1 + 4*(x - 2). Opening the brackets and bringing similar terms we get: y = 4*x - 7.

Answer: y = 4*x - 7.

General scheme for composing the tangent equation to the graph of the function y = f(x):

1. Determine x0.

2. Calculate f(x0).

3. Calculate f’(x)

Equation of a plane. How to write an equation of a plane?
Mutual position 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. I don't care technical reasons it is more convenient to depict the plane in exactly this way and in exactly this position. Real planes that we will consider in practical examples, can be positioned 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, the holey plane is certainly quite funny.

In some cases, it is convenient to use the same symbols to designate planes. greek letters with subscripts, 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 create an equation of 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 won't languish 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 valid both for the usual orthonormal basis and for affine basis 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 system coordinates

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 very general case, when the numbers are not 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 given equation? Think about it: “Z” is ALWAYS equal to zero, for any values ​​of “X” and “Y”. This equation is "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:
– equation of the coordinate plane;
– equation of the coordinate plane.

Let's complicate the problem a little, consider a plane (here and further in the paragraph we assume that numerical odds 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 free members 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: plane, given 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 friends with everyone coordinate axes, while it always “cuts off” the 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's denote given vector through . It is absolutely clear that the vectors are collinear:

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

How to find 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. latest tasks 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. It is obvious that 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:

You can set 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 various types. Also, subject to certain conditions, planes can be parallel, perpendicular, intersecting, etc. We'll talk about this in this article. We will learn how to create a general equation of a plane and more.

Normal form of equation

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

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

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

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

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

General equation

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

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

Equations of planes. Special cases

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

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

Type of equation in segments

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

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

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

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

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

Normal vector coordinates

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

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

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

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

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

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

Let us assume that in coordinate space (rectangular coordinate system) Oxyz 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 the vector n. Let us write the orthogonality condition using the scalar product:

[MₒM, n] = 0.

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

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

Now you can get the coordinate view of the record vector equation our plane = 0. Since r-rₒ = (x-xₒ)*i + (y-yₒ)*j + (z-zₒ)*k, and n = A*i+B*j+C*k, we 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 specify two arbitrary points M′ (x′,y′,z′) and M″ (x″,y″,z″), as well as a vector a (a′,a″,a‴).

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

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

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

Type of equation of a plane intersecting three points

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

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

Now we can compose homogeneous system with unknown u, v, w:

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

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

Dihedral angle between planes

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

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

We know that the vectors N=(A,B,C) and N¹=(A¹,B¹,C¹) are perpendicular according to 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. Dot 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, angle φ in cos equationφ=NN 1 /|N||N 1 | will be replaced by π-φ.

Equation of a perpendicular plane

Planes with an angle of 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. And this means that they are fulfilled following conditions proportionality:

A/A¹=B/B¹=C/C¹.

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

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

Distance to plane from point

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

(ρ,v)=р (р≥0).

IN in this caseρ (x,y,z) is the radius vector of our point Q located on P, p is the length of the perpendicular P that was released from 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, yₒ, zₒ) is such a vector, absolute value whose projection onto v is equal to the distance d, which 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|.

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

Using the parameter language, we get the obvious:

d=|Ахₒ+Вуₒ+Czₒ|/√(А²+В²+С²).

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

d=-(ρ-ρ 0 ,v)=(ρ 0 ,v)-р>0.

In the case when the point Q 0, together with the origin of coordinates, is located on the same side of P, then the 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 the (common) planes P′ and P″. This means that the coordinates of any point belonging to line a must simultaneously satisfy the equations A′x+B′y+C′z+D′=0 and A″x+B″y+C″z+D″=0. This means that the coordinates of the point will be a partial solution of the following system of equations:

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

In this lesson we will look at how to use the determinant to create plane equation. If you don’t know what a determinant is, go to the first part of the lesson - “Matrices and determinants”. Otherwise, you risk not understanding anything in today’s material.

Equation of a plane using three points

Why do we need a plane equation at all? It's simple: knowing it, we can easily calculate angles, distances and other crap in problem C2. In general, you can’t do without this equation. Therefore, we formulate the problem:

Task. Three points are given in space that do not lie on the same line. Their coordinates:

M = (x 1, y 1, z 1);
N = (x 2, y 2, z 2);
K = (x 3, y 3, z 3);

You need to create an equation for the plane passing through these three points. Moreover, the equation should look like:

Ax + By + Cz + D = 0

where the numbers A, B, C and D are the coefficients that, in fact, need to be found.

Well, how to get the equation of a plane if only the coordinates of the points are known? The easiest way is to substitute the coordinates into the equation Ax + By + Cz + D = 0. You get a system of three equations that can be easily solved.

Many students find this solution extremely tedious and unreliable. Last year's Unified State Examination in mathematics showed that the likelihood of making a computational error is really high.

Therefore, the most advanced teachers began to look for simpler and more elegant solutions. And they found it! True, the technique obtained rather relates to higher mathematics. Personally, I had to rummage through the entire Federal List of Textbooks to make sure that we have the right to use this technique without any justification or evidence.

Equation of a plane through a determinant

Enough of the lyrics, let's get down to business. To begin with, a theorem about how the determinant of a matrix and the equation of the plane are related.

Theorem. Let the coordinates of three points through which the plane must be drawn be given: M = (x 1, y 1, z 1); N = (x 2, y 2, z 2); K = (x 3, y 3, z 3). Then the equation of this plane can be written through the determinant:

As an example, let's try to find a pair of planes that actually occur in problems C2. Look how quickly everything is calculated:

A 1 = (0, 0, 1);
B = (1, 0, 0);
C 1 = (1, 1, 1);

We compose a determinant and equate it to zero:

We expand the determinant:

a = 1 1 (z − 1) + 0 0 x + (−1) 1 y = z − 1 − y;
b = (−1) 1 x + 0 1 (z − 1) + 1 0 y = −x;
d = a − b = z − 1 − y − (−x ) = z − 1 − y + x = x − y + z − 1;
d = 0 ⇒ x − y + z − 1 = 0;

As you can see, when calculating the number d, I “combed” the equation a little so that the variables x, y and z were in the correct sequence. That's it! The plane equation is ready!

Task. Write an equation for a plane passing through the points:

A = (0, 0, 0);
B 1 = (1, 0, 1);
D 1 = (0, 1, 1);

We immediately substitute the coordinates of the points into the determinant:

We expand the determinant again:

a = 1 1 z + 0 1 x + 1 0 y = z;
b = 1 1 x + 0 0 z + 1 1 y = x + y;
d = a − b = z − (x + y ) = z − x − y;
d = 0 ⇒ z − x − y = 0 ⇒ x + y − z = 0;

So, the equation of the plane is obtained again! Again, at the last step we had to change the signs in it to get a more “beautiful” formula. It is not at all necessary to do this in this solution, but it is still recommended - to simplify the further solution of the problem.

As you can see, composing the equation of a plane is now much easier. We substitute the points into the matrix, calculate the determinant - and that’s it, the equation is ready.

This could end the lesson. However, many students constantly forget what is inside the determinant. For example, which line contains x 2 or x 3, and which line contains just x. To really get this out of the way, let's look at where each number comes from.

Where does the formula with the determinant come from?

So, let’s figure out where such a harsh equation with a determinant comes from. This will help you remember it and apply it successfully.

All planes that appear in Problem C2 are defined by three points. These points are always marked on the drawing, or even indicated directly in the text of the problem. In any case, to create an equation we will need to write down their coordinates:

M = (x 1, y 1, z 1);
N = (x 2, y 2, z 2);
K = (x 3, y 3, z 3).

Let's consider another point on our plane with arbitrary coordinates:

T = (x, y, z)

Take any point from the first three (for example, point M) and draw vectors from it to each of the three remaining points. We get three vectors:

MN = (x 2 − x 1 , y 2 − y 1 , z 2 − z 1 );
MK = (x 3 − x 1 , y 3 − y 1 , z 3 − z 1 );
MT = (x − x 1 , y − y 1 , z − z 1 ).

Now let's make a square matrix from these vectors and equate its determinant to zero. The coordinates of the vectors will become rows of the matrix - and we will get the very determinant that is indicated in the theorem:

This formula means that the volume of a parallelepiped built on the vectors MN, MK and MT is equal to zero. Therefore, all three vectors lie in the same plane. In particular, an arbitrary point T = (x, y, z) is exactly what we were looking for.

Replacing points and lines of a determinant

Determinants have several great properties that make it even easier solution to problem C2. For example, it doesn’t matter to us from which point we draw the vectors. Therefore, the following determinants give the same plane equation as the one above:

You can also swap the lines of the determinant. The equation will remain unchanged. For example, many people like to write a line with the coordinates of the point T = (x; y; z) at the very top. Please, if it is convenient for you:

Some people are confused by the fact that one of the lines contains variables x, y and z, which do not disappear when substituting points. But they shouldn’t disappear! Substituting the numbers into the determinant, you should get this construction:

Then the determinant is expanded according to the diagram given at the beginning of the lesson, and the standard equation of the plane is obtained:

Ax + By + Cz + D = 0

Take a look at an example. It's the last one in today's lesson. I will deliberately swap the lines to make sure that the answer will give the same equation of the plane.

Task. Write an equation for a plane passing through the points:

B 1 = (1, 0, 1);
C = (1, 1, 0);
D 1 = (0, 1, 1).

So, we consider 4 points:

B 1 = (1, 0, 1);
C = (1, 1, 0);
D 1 = (0, 1, 1);
T = (x, y, z).

First, let's create a standard determinant and equate it to zero:

We expand the determinant:

a = 0 1 (z − 1) + 1 0 (x − 1) + (−1) (−1) y = 0 + 0 + y;
b = (−1) 1 (x − 1) + 1 (−1) (z − 1) + 0 0 y = 1 − x + 1 − z = 2 − x − z;
d = a − b = y − (2 − x − z ) = y − 2 + x + z = x + y + z − 2;
d = 0 ⇒ x + y + z − 2 = 0;

That's it, we got the answer: x + y + z − 2 = 0.

Now let's rearrange a couple of lines in the determinant and see what happens. For example, let’s write a line with the variables x, y, z not at the bottom, but at the top:

We again expand the resulting determinant:

a = (x − 1) 1 (−1) + (z − 1) (−1) 1 + y 0 0 = 1 − x + 1 − z = 2 − x − z;
b = (z − 1) 1 0 + y (−1) (−1) + (x − 1) 1 0 = y;
d = a − b = 2 − x − z − y;
d = 0 ⇒ 2 − x − y − z = 0 ⇒ x + y + z − 2 = 0;

We got exactly the same plane equation: x + y + z − 2 = 0. This means that it really does not depend on the order of the rows. All that remains is to write down the answer.

So, we are convinced that the equation of the plane does not depend on the sequence of lines. We can carry out similar calculations and prove that the equation of the plane does not depend on the point whose coordinates we subtract from other points.

In the problem considered above, we used the point B 1 = (1, 0, 1), but it was quite possible to take C = (1, 1, 0) or D 1 = (0, 1, 1). In general, any point with known coordinates lying on the desired plane.

13.Angle between planes, distance from a point to a plane.

Let the planes α and β intersect along a straight line c.
The angle between planes is the angle between perpendiculars to the line of their intersection drawn in these planes.

In other words, in the α plane we drew a straight line a perpendicular to c. In the β plane - straight line b, also perpendicular to c. The angle between planes α and β is equal to the angle between straight lines a and b.

Note that when two planes intersect, four angles are actually formed. Do you see them in the picture? As the angle between the planes we take spicy corner.

If the angle between the planes is 90 degrees, then the planes perpendicular,

This is the definition of perpendicularity of planes. When solving problems in stereometry, we also use sign of perpendicularity of planes:

If plane α passes through the perpendicular to plane β, then planes α and β are perpendicular.

distance from point to plane

Consider point T, defined by its coordinates:

T = (x 0 , y 0 , z 0)

Also consider the plane α, given by the equation:

Ax + By + Cz + D = 0

Then the distance L from point T to plane α can be calculated using the formula:

In other words, we substitute the coordinates of the point into the equation of the plane, and then divide this equation by the length of the normal vector n to the plane:

The resulting number is the distance. Let's see how this theorem works in practice.

We have already derived the parametic equations of a straight line on a plane, let's get the parametric equations of a straight line, which is defined in a rectangular coordinate system in three-dimensional space.

Let a rectangular coordinate system be fixed in three-dimensional space Oxyz. Let us define a straight line in it a(see the section on methods for defining a line in space), indicating the direction vector of the line and the coordinates of some point on the line . We will start from these data when drawing up parametric equations of a straight line in space.

Let be an arbitrary point in three-dimensional space. If we subtract from the coordinates of the point M corresponding point coordinates M 1, then we will get the coordinates of the vector (see the article finding the coordinates of a vector from the coordinates of the points of its end and beginning), that is, .

Obviously, the set of points defines a line A if and only if the vectors and are collinear.

Let us write down the necessary and sufficient condition for the collinearity of vectors And : , where is some real number. The resulting equation is called vector-parametric equation of the line in a rectangular coordinate system Oxyz in three-dimensional space. The vector-parametric equation of a straight line in coordinate form has the form and represents parametric equations of the line a. The name “parametric” is not accidental, since the coordinates of all points on the line are specified using the parameter.

Let us give an example of parametric equations of a straight line in a rectangular coordinate system Oxyz in space: . Here

15.Angle between a straight line and a plane. The point of intersection of a line with a plane.

Every first degree equation with respect to coordinates x, y, z

Ax + By + Cz +D = 0 (3.1)

defines a plane, and vice versa: any plane can be represented by equation (3.1), which is called plane equation.

Vector n(A, B, C) orthogonal to the plane is called normal vector plane. In equation (3.1), the coefficients A, B, C are not equal to 0 at the same time.

Special cases of equation (3.1):

1. D = 0, Ax+By+Cz = 0 - the plane passes through the origin.

2. C = 0, Ax+By+D = 0 - the plane is parallel to the Oz axis.

3. C = D = 0, Ax + By = 0 - the plane passes through the Oz axis.

4. B = C = 0, Ax + D = 0 - the plane is parallel to the Oyz plane.

Equations of coordinate planes: x = 0, y = 0, z = 0.

A straight line in space can be specified:

1) as a line of intersection of two planes, i.e. system of equations:

A 1 x + B 1 y + C 1 z + D 1 = 0, A 2 x + B 2 y + C 2 z + D 2 = 0; (3.2)

2) by its two points M 1 (x 1, y 1, z 1) and M 2 (x 2, y 2, z 2), then the straight line passing through them is given by the equations:

3) the point M 1 (x 1, y 1, z 1) belonging to it, and the vector a(m, n, p), collinear to it. Then the straight line is determined by the equations:

. (3.4)

Equations (3.4) are called canonical equations of the line.

Vector a called direction vector straight.

We obtain parametric equations of the line by equating each of the relations (3.4) to the parameter t:

x = x 1 +mt, y = y 1 + nt, z = z 1 + rt. (3.5)

Solving system (3.2) as a system of linear equations for unknowns x And y, we arrive at the equations of the line in projections or to given equations of the straight line:

x = mz + a, y = nz + b. (3.6)

From equations (3.6) we can go to the canonical equations, finding z from each equation and equating the resulting values:

.

From general equations (3.2) you can go to canonical ones in another way, if you find any point on this line and its direction vector n= [n 1 , n 2 ], where n 1 (A 1, B 1, C 1) and n 2 (A 2 , B 2 , C 2 ) - normal vectors of given planes. If one of the denominators m, n or r in equations (3.4) turns out to be equal to zero, then the numerator of the corresponding fraction must be set equal to zero, i.e. system

is equivalent to the system ; such a straight line is perpendicular to the Ox axis.

System is equivalent to the system x = x 1, y = y 1; the straight line is parallel to the Oz axis.

Example 1.15. Write an equation for the plane, knowing that point A(1,-1,3) serves as the base of a perpendicular drawn from the origin to this plane.

Solution. According to the problem conditions, the vector OA(1,-1,3) is a normal vector of the plane, then its equation can be written as
x-y+3z+D=0. Substituting the coordinates of point A(1,-1,3) belonging to the plane, we find D: 1-(-1)+3×3+D = 0 Þ D = -11. So x-y+3z-11=0.

Example 1.16. Create an equation for a plane passing through the Oz axis and forming an angle of 60 degrees with the plane 2x+y-z-7=0.

Solution. The plane passing through the Oz axis is given by the equation Ax+By=0, where A and B do not simultaneously vanish. Let B not
equals 0, A/Bx+y=0. Using the cosine formula for the angle between two planes

.

Solving the quadratic equation 3m 2 + 8m - 3 = 0, we find its roots
m 1 = 1/3, m 2 = -3, from where we get two planes 1/3x+y = 0 and -3x+y = 0.

Example 1.17. Compose the canonical equations of the line:
5x + y + z = 0, 2x + 3y - 2z + 5 = 0.

Solution. The canonical equations of the line have the form:

Where m, n, p- coordinates of the directing vector of the straight line, x 1 , y 1 , z 1- coordinates of any point belonging to a line. A straight line is defined as the line of intersection of two planes. To find a point belonging to a line, one of the coordinates is fixed (the easiest way is to set, for example, x=0) and the resulting system is solved as a system of linear equations with two unknowns. So, let x=0, then y + z = 0, 3y - 2z+ 5 = 0, hence y=-1, z=1. We found the coordinates of the point M(x 1, y 1, z 1) belonging to this line: M (0,-1,1). The direction vector of a straight line is easy to find, knowing the normal vectors of the original planes n 1 (5,1,1) and n 2 (2,3,-2). Then

The canonical equations of the line have the form: x/(-5) = (y + 1)/12 =
= (z - 1)/13.

Example 1.18. In the beam defined by the planes 2x-y+5z-3=0 and x+y+2z+1=0, find two perpendicular planes, one of which passes through the point M(1,0,1).

Solution. The equation of the beam defined by these planes has the form u(2x-y+5z-3) + v(x+y+2z+1)=0, where u and v do not vanish simultaneously. Let us rewrite the beam equation as follows:

(2u +v)x + (- u + v)y + (5u +2v)z - 3u + v = 0.

In order to select a plane from the beam that passes through point M, we substitute the coordinates of point M into the equation of the beam. We get:

(2u+v)×1 + (-u + v)×0 + (5u + 2v)×1 -3u + v =0, or v = - u.

Then we find the equation of the plane containing M by substituting v = - u into the beam equation:

u(2x-y +5z - 3) - u (x + y +2z +1) = 0.

Because u¹0 (otherwise v=0, and this contradicts the definition of a beam), then we have the equation of the plane x-2y+3z-4=0. The second plane belonging to the beam must be perpendicular to it. Let us write down the condition for the orthogonality of planes:

(2u+ v)×1 + (v - u)×(-2) + (5u +2v)×3 = 0, or v = - 19/5u.

This means that the equation of the second plane has the form:

u(2x -y+5z - 3) - 19/5 u(x + y +2z +1) = 0 or 9x +24y + 13z + 34 = 0