Reducing the general plane equation to normal form. Normal plane equation. Parametric equations of a line in space
Normal plane equation.
The general plane equation of the form is called normal plane equation, if the vector length 
equal to one, that is, 
, And .
You can often see that the normal equation of a plane is written as . Here are the direction cosines of the normal vector of a given plane of unit length, that is, and p– a non-negative number equal to the distance from the origin to the plane.
Normal equation of a plane in a rectangular coordinate system Oxyz defines a plane that is removed from the origin by a distance p in the positive direction of the normal vector of this plane 
. If p=0, then the plane passes through the origin.
Let us give an example of a normal plane equation.
Let the plane be specified in a rectangular coordinate system Oxyz general plane equation of the form 
. This general equation of the plane is the normal equation of the plane. Indeed, the normal vector of this plane is 
has a length equal to one, since 
.
The equation of a plane in normal form allows you to find the distance from a point to a plane.
Distance from a point to a plane.
The distance from a point to a plane is the smallest of the distances between this point and the points of the plane. It is known that distance from a point to a plane is equal to the length of the perpendicular drawn from this point to the plane.
If and the origin of coordinates lie on different sides of the plane, in the opposite case. The distance from a point to a plane is
Mutual arrangement of planes. Conditions for parallelism and perpendicularity of planes.
Distance between parallel planes
Related concepts
Planes are parallel , If
or 
(Vector product)
Planes are perpendicular, If
Or 
. (Scalar product)
Straight in space. Different types of straight line equations.
Equations of a straight line in space - initial information.
Equation of a straight line on a plane Oxy is a linear equation in two variables x And y, which is satisfied by the coordinates of any point on a line and is not satisfied by the coordinates of any other points. With a straight line in three-dimensional space the situation is a little different - there is no linear equation with three variables x, y And z, which would be satisfied only by the coordinates of the points of a line specified in a rectangular coordinate system Oxyz. Indeed, an equation of the form , where x, y And z are variables, and A, B, C And D– some real numbers, and A, IN And WITH are not equal to zero at the same time, represents general plane equation. Then the question arises: “How can a straight line be described in a rectangular coordinate system? Oxyz»?
The answer to this is contained in the following paragraphs of the article.
The equations of a straight line in space are the equations of two intersecting planes.
Let us recall one axiom: if two planes in space have a common point, then they have a common straight line on which all the common points of these planes are located. Thus, a straight line in space can be defined by specifying two planes intersecting along this straight line.
Let us translate the last statement into the language of algebra.
Let a rectangular coordinate system be fixed in three-dimensional space Oxyz and it is known that the straight line a is the line of intersection of two planes and, which correspond to the general equations of the plane of the form and, respectively. Since it's straight a is the set of all common points of the planes and, then the coordinates of any point on the line a will simultaneously satisfy both the equation and the equation, the coordinates of no other points will simultaneously satisfy both equations of the planes. Therefore, the coordinates of any point on the line a in a rectangular coordinate system Oxyz represent particular solution to a system of linear equations kind 
, and the general solution to the system of equations 
determines the coordinates of each point on a line a, that is, defines a straight line a.
So, a straight line in space in a rectangular coordinate system Oxyz can be given by a system of equations of two intersecting planes 
.
Here is an example of defining a straight line in space using a system of two equations - 
.
Describing a straight line with the equations of two intersecting planes is excellent for finding the coordinates of the intersection point of a line and a plane, and also when finding the coordinates of the point of intersection of two lines in space.
We recommend further study of this topic by referring to the article equations of a line in space - equations of two intersecting planes. It provides more detailed information, analyzes in detail the solutions to typical examples and problems, and also shows a way to move to equations of a straight line in a space of a different type.
It should be noted that there are different ways to define a line in space, and in practice, a straight line is often defined not by two intersecting planes, but by the directing vector of the straight line and a point lying on this straight line. In these cases, it is easier to obtain canonical and parametric equations of a line in space. We'll talk about them in the following paragraphs.
Parametric equations of a line in space.
Parametric equations of a line in space look like 
,
Where x 1 ,y 1 And z 1 – coordinates of some point on the line, a x , a y And a z (a x , a y And a z are not equal to zero at the same time) - corresponding coordinates of the directing vector of the straight line, a is some parameter that can take any real value.
For any value of the parameter, using the parametric equations of a line in space, we can calculate a triple of numbers,
it will correspond to some point on the line (hence the name of this type of line equation). For example, when
from the parametric equations of a straight line in space we obtain the coordinates x 1
, y 1
And z 1
: 
.
As an example, consider a straight line defined by parametric equations of the form 
. This line passes through a point, and the direction vector of this line has coordinates.
We recommend continuing to study the topic by referring to the article parametric equations of a line in space. It shows the derivation of parametric equations of a line in space, examines special cases of parametric equations of a line in space, provides graphic illustrations, provides detailed solutions to characteristic problems, and indicates the connection between parametric equations of a line and other types of equations of a line.
Canonical equations of a straight line in space.
Having resolved each of the parametric straight line equations of the form 
regarding the parameter, it is easy to go to canonical equations of a straight line in space kind 
.
The canonical equations of a line in space determine a line passing through a point 
, and the direction vector of the straight line is the vector 
. For example, the equations of a straight line in canonical form 
correspond to a line passing through a point in space with coordinates, the direction vector of this line has coordinates.
It should be noted that one or two of the numbers in the canonical equations of a line can be equal to zero (all three numbers cannot be equal to zero at the same time, since the direction vector of a line cannot be zero). Then a notation of the form 
is considered formal (since the denominators of one or two fractions will have zeros) and should be understood as 
, Where.
If one of the numbers in the canonical equations of a line is equal to zero, then the line lies in one of the coordinate planes, or in a plane parallel to it. If two of the numbers are zero, then the line either coincides with one of the coordinate axes or is parallel to it. For example, a line corresponding to the canonical equations of a line in space of the form 
, lies in the plane z=-2, which is parallel to the coordinate plane Oxy, and the coordinate axis Oy is determined by canonical equations.
For graphic illustrations of these cases, the derivation of the canonical equations of a line in space, detailed solutions of typical examples and problems, as well as the transition from the canonical equations of a line to other equations of a line in space, see the article canonical equations of a line in space.
General equation of a straight line. Transition from the general to the canonical equation.
| " |
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 cannot 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 all! 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.
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, subject to certain conditions, planes can be parallel, perpendicular, intersecting, etc. We'll talk about this in this article. We will learn how to create a general equation of a plane and more.
Normal form of equation
Let's say there is a space R 3 that has a rectangular XYZ coordinate system. Let us define the vector α, which will be released from the initial point O. Through the end of the vector α we draw a plane P, which will be perpendicular to it.
Let us denote an arbitrary point on P as Q = (x, y, z). Let's sign the radius vector of point Q with the letter p. In this case, the length of the vector α is equal to р=IαI and Ʋ=(cosα,cosβ,cosγ).
This is a unit vector that is directed to the side, like the vector α. α, β and γ are the angles that are formed between the vector Ʋ and the positive directions of the space axes x, y, z, respectively. The projection of any point QϵП onto the vector Ʋ is a constant value that is equal to p: (p,Ʋ) = p(p≥0).
The above equation makes sense when p=0. The only thing is that the plane P in this case will intersect the point O (α=0), which is the origin of coordinates, and the unit vector Ʋ released from the point O will be perpendicular to P, despite its direction, which means that the vector Ʋ is determined with accurate to the sign. The previous equation is the equation of our plane P, expressed in vector form. But in coordinates it will look like this:
P here is greater than or equal to 0. We have found the equation of the plane in space in normal form.
General equation
If we multiply the equation in coordinates by any number that is not equal to zero, we obtain an equation equivalent to this one, defining that very plane. It will look like this:
Here A, B, C are numbers that are simultaneously different from zero. This equation is called the general plane equation.
Equations of planes. Special cases
The equation in general form can be modified in the presence of additional conditions. Let's look at some of them.
Let's assume that coefficient A is 0. This means that this plane is parallel to the given Ox axis. In this case, the form of the equation will change: Ву+Cz+D=0.
Similarly, the form of the equation will change under the following conditions:
- Firstly, if B = 0, then the equation will change to Ax + Cz + D = 0, which will indicate parallelism to the Oy axis.
- Secondly, if C=0, then the equation will be transformed into Ax+By+D=0, which will indicate parallelism to the given Oz axis.
- Thirdly, if D=0, the equation will look like Ax+By+Cz=0, which will mean that the plane intersects O (the origin).
- Fourth, if A=B=0, then the equation will change to Cz+D=0, which will prove parallel to Oxy.
- Fifthly, if B=C=0, then the equation becomes Ax+D=0, which means that the plane to Oyz is parallel.
- Sixth, if A=C=0, then the equation will take the form Ву+D=0, that is, it will report parallelism to Oxz.
Type of equation in segments
In the case when the numbers A, B, C, D are different from zero, the form of equation (0) can be as follows:
x/a + y/b + z/c = 1,
in which a = -D/A, b = -D/B, c = -D/C.
We get as a result. It is worth noting that this plane will intersect the Ox axis at a point with coordinates (a,0,0), Oy - (0,b,0), and Oz - (0,0,c).
Taking into account the equation x/a + y/b + z/c = 1, it is not difficult to visually imagine the placement of the plane relative to a given coordinate system.
Normal vector coordinates
The normal vector n to the plane P has coordinates that are coefficients of the general equation of this 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 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 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, 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 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.
Let us consider the plane Q in space. Its position is completely determined by specifying the vector N perpendicular to this plane and some fixed point lying in the Q plane. The vector N perpendicular to the Q plane is called the normal vector of this plane. If we denote by A, B and C the projections of the normal vector N, then
Let us derive the equation of the plane Q passing through a given point and having a given normal vector . To do this, consider a vector connecting a point with an arbitrary point on the Q plane (Fig. 81).
For any position of point M on the plane Q, the vector MHM is perpendicular to the normal vector N of the plane Q. Therefore, the scalar product Let us write the scalar product in terms of projections. Since , and is a vector, then
and therefore
We have shown that the coordinates of any point in the Q plane satisfy equation (4). It is easy to see that the coordinates of points not lying on the Q plane do not satisfy this equation (in the latter case). Consequently, we have obtained the required equation of the plane Q. Equation (4) is called the equation of the plane passing through a given point. It is of the first degree relative to the current coordinates
So, we have shown that every plane corresponds to an equation of the first degree with respect to the current coordinates.
Example 1. Write the equation of a plane passing through a point perpendicular to the vector.
Solution. Here . Based on formula (4) we obtain
or, after simplification,
By giving the coefficients A, B and C of equation (4) different values, we can obtain the equation of any plane passing through the point . The set of planes passing through a given point is called a bundle of planes. Equation (4), in which the coefficients A, B and C can take any values, is called the equation of a bunch of planes.
Example 2. Create an equation for a plane passing through three points (Fig. 82).
Solution. Let's write the equation for a bunch of planes passing through the point
1. General equation of the plane
Definition. A plane is a surface all points of which satisfy the general equation: Ax + By + Cz + D = 0, where A, B, C are the coordinates of the vector
N = Ai + Bj + Ck is the normal vector to the plane. The following special cases are possible:
A = 0 – plane parallel to the Ox axis
B = 0 – plane is parallel to the Oy axis C = 0 – plane is parallel to the Oz axis
D = 0 – the plane passes through the origin
A = B = 0 – plane is parallel to the xOy plane A = C = 0 – plane is parallel to the xOz plane B = C = 0 – plane is parallel to the yOz plane A = D = 0 – plane passes through the Ox axis
B = D = 0 – the plane passes through the Oy axis C = D = 0 – the plane passes through the Oz axis
A = B = D = 0 – the plane coincides with the xОу plane A = C = D = 0 – the plane coincides with the xOz plane B = C = D = 0 – the plane coincides with the yOz plane
2. Surface equation in space
Definition. Any equation that relates the coordinates x, y, z of any point on a surface is an equation of that surface.
3. Equation of a plane passing through three points
In order for a single plane to be drawn through any three points in space, it is necessary that these points do not lie on the same straight line.
Consider points M1 (x1, y1, z1), M2 (x2, y2, z2), M3 (x3, y3, z3) in the general Cartesian system
coordinates |
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In order for an arbitrary point M (x, y, z) |
lay in the same plane with the points |
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M 1 , M 2 , M 3 it is necessary that the vectors M 1 M 2 , M 1 M 3 , M 1 M be coplanar, i.e. |
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M1 M = ( x − x1 ; y − y1 ; z − z1 ) |
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(M 1 M 2 , M 1 M 3 , M 1 M ) = 0. Thus, M 1 M 2 |
= ( x 2 − x 1 ; y 2 |
− y 1 ; z 2 − z 1) |
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M1 M 3 |
= ( x 3 − x 1 ; y 3 − y 1 ; z 3 − z 1) |
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x−x1 |
y−y1 |
z − z1 |
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Equation of a plane passing through three points: |
x 2 − x 1 |
y 2 − y 1 |
z 2 − z 1 |
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x 3 − x 1 |
y 3 − y 1 |
z 3 − z 1 |
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4. Equation of a plane based on two points and a vector collinear to the plane
Let points M1(x1, y1, z1), M2(x2, y2, z2) and vectors = (a 1, a 2, a 3) be given.
Let’s create an equation for a plane passing through these points M1 and M2 and an arbitrary
point M(x, y, z) parallel to vector a. |
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Vectors M1 M = ( x − x1 ; y − y1 ; z − z1 ) |
and vector a = (a , a |
must be |
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M 1M 2 = ( x 2 − x 1 ; y 2 − y 1 ; z 2 − z 1) |
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x−x1 |
y−y1 |
z − z1 |
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coplanar, i.e. (M 1 M, M 1 M 2, a) = 0. Plane equation: |
x 2 − x 1 |
y 2 − y 1 |
z 2 − z 1 |
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5. Equation of a plane using one point and two vectors collinear to the plane
Let two vectors a = (a 1, a 2, a 3) and b = (b 1,b 2,b 3), collinear planes, be given. Then for an arbitrary point M(x, y, z) belonging to the plane, the vectors a, b, MM 1 must be coplanar.
6. Equation of a plane by point and normal vector
Theorem. If a point M 0 (x 0 , y 0 , z 0 ) is given in space, then the equation of the plane passing through the point M 0 perpendicular to the normal vector N (A , B , C ) has the form: A (x − x 0 ) + B (y − y 0 ) + C (z − z 0 ) = 0 .
7. Equation of a plane in segments
If in the general equation Ax + By + Cz + D = 0 we divide both sides by (-D)
x− |
y − |
z − 1 = 0 , replacing − |
C , we obtain the equation of the plane |
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in segments: |
1 . The numbers a, b, c are the intersection points of the plane, respectively |
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with axes x, y, z.
8. Equation of a plane in vector form
r n = p, where r = xi + yj + zk is the radius vector of the current point M (x, y, z),
n = i cosα + j cos β + k cosγ - unit vector having the direction perpendicular,
lowered onto the plane from the origin. α, β and γ are the angles formed by this vector with the x, y, z axes. p is the length of this perpendicular. In coordinates, this equation looks like:
x cosα + y cos β + z cosγ − p = 0
9. Distance from point to plane
The distance from an arbitrary point M 0 (x 0 , y 0 , z 0 ) to the plane Ax + By + Cz + D = 0 is:
d = Ax0 + By0 + Cz0 + D
A2 + B2 + C 2
Example. Find the equation of the plane passing through points A(2,-1,4) and B(3,2,-1) perpendicular to the plane x + y + 2z − 3 = 0.
The required plane equation has the form: Ax + By + Cz + D = 0, normal vector to this plane n 1 (A,B,C). Vector AB (1,3,-5) belongs to the plane. The plane given to us,
perpendicular to the desired one has a normal vector n 2 (1,1,2). Because points A and B belong to both planes, and the planes are mutually perpendicular, then
n = AB × n |
− 5 |
− j |
− 5 |
11 i − 7 j − 2 k . |
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− 5 |
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Thus, the normal vector n is 1 (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 obtain the equation of the plane: 11x − 7 y − 2z − 21 = 0
10. Equation of a line in space
Both on the plane and in space, any line can be defined as a set of points whose coordinates in some coordinate system chosen in space satisfy the equation:
F(x, y, z) = 0. This equation is called the equation of a line in space.
In addition, a line in space can be defined differently. It can be considered as the line of intersection of two surfaces, each of which is specified by some equation.
Let F (x, y, z) = 0 and Ф (x, y, z) = 0 – equations of surfaces intersecting along the line L.
F(x, y, z) = 0
Then the pair of equations Ф (x, y, z) = 0 will be called the equation of a line in space.
11. Equation of a straight line in space given a point and a direction vector 0 = M 0 M .
Because vectors М 0 М and S are collinear, then the relation М 0 М = St is true, where t is a certain parameter. In total, we can write: r = r 0 + St.
Because If this equation is satisfied by the coordinates of any point on the line, then the resulting equation is a parametric equation of the line.
x = x0 + mt
This vector equation can be represented in coordinate form: y = y 0 + nt
z = z0 + pt
Transforming this system and equating the values of the parameter t, we obtain the canonical
equations of a straight line in space: |
x−x0 |
y−y0 |
z − z0 |
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Definition. The direction cosines of a straight line are the direction cosines of the vector S, which can be calculated using the formulas:
cosα = |
; cos β = |
; cosγ = |
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N2+p2 |
m 2 + n 2 + p 2 |
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From here we get: m: n: p = cosα: cos β: cosγ.
The numbers m, n, p are called the slopes of the line. Because S is a non-zero vector, then m, n and p cannot be zero at the same time, but one or two of these numbers can be zero. In this case, in the equation of the line, the corresponding numerators should be set equal to zero.
12. Equation of a line in space passing through two points
If on a straight line in space we mark two arbitrary points M 1 (x 1, y 1, z 1) and
M 2 (x 2 , y 2 , z 2 ), then the coordinates of these points must satisfy the straight line equation obtained above:
x 2 − x 1 |
y 2 − y 1 |
z 2 − z 1 |
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