Parametric equation of a plane passing through a point. Plane and line in space: general and parametric equation of the plane
– general equation planes in space
Normal vector of the plane
The normal vector of the plane is a nonzero vector orthogonal to each vector lying in the plane.
Equation of a plane passing through a point with a given normal vector
Is the equation of the plane passing through the point M0 with a given normal vector
Direction vectors of the plane
Two non-collinear vectors parallel to the plane are called direction vectors of the plane
Parametric plane equations
- parametric equation of the plane in vector form
- parametric equation of the plane in coordinates
Equation of a plane through a given point and two direction vectors
–Fixed point
- just a point lol
- coplanar, so their mixed product is equal to 0.
Equation of a plane passing through three given points
- equation of the plane through three points
Equation of the plane in line segments
- equation of the plane in segments
Proof
For the proof, we use the fact that our plane passes through A, B, C, and the normal vector
Substitute the coordinates of the point and the vector n into the equation of the plane with the normal vector
Divide everything into and get
So it goes.
Normal equation of the plane
Is the angle between ox and the normal vector to the plane, outgoing from O.
- the angle between oy and the normal vector to the plane, outgoing from O.
Is the angle between oz and the normal vector to the plane outgoing from O.
Is the distance from the origin to the plane.
Proof or some such bullshit
Opposite of D.
Similarly for the rest of the cosines. End.
Distance from point to plane
S point, plane
- oriented distance from point S to plane
If, then S and O lie on opposite sides of the plane
If, then S and O lie on the same side
Multiply by n 
The relative position of two straight lines in space
Angle between planes
When crossing, two pairs of vertical dihedral angles are formed, the smallest is called the angle between the planes
Straight in space
A straight line in space can be specified as
Intersection of two planes:
Parametric equations of a straight line
- parametric equation of a straight line in vector form
- parametric equation of a straight line in coordinates
Canonical equation
- the canonical equation of the straight line.
Equation of a straight line passing through two given points
- canonical equation of a straight line in vector form;
The relative position of two straight lines in space
The relative position of a straight line and a plane in space
Angle between line and plane
Distance from point to line in space
a is the direction vector of our line.
- an arbitrary point belonging to a given straight line
- the point to which we are looking for the distance.
Distance between two crossed lines
Distance between two parallel straight lines
М1 - point belonging to the first straight line
М2 - point belonging to the second straight line
Curves and surfaces of the second order
An ellipse is the set of points on the plane, the sum of the distances from which to two given points (foci) is a constant value.
Canonical Ellipse Equation
Replace with
Divide into
Ellipse properties
Intersection with coordinate axes
Origin of coordinates
Symmetry relative
An ellipse is a curve that lies in a limited part of the plane
An ellipse can be obtained from a circle by stretching or compressing it.
Parametric ellipse equation:
- directors
Hyperbola
A hyperbola is the set of points of the plane for which the modulus of the difference in distances to 2 given points (foci) is a constant value (2a)
We do everything the same as with the ellipse, we get
Replace with
Divide by
Hyperbola properties
;
- directors
Asymptote
An asymptote is a straight line, to which the curve approaches indefinitely, moving away to infinity.
Parabola
Parabot properties
Relationship between ellipse, hyperbola and parabola.
The relationship between these curves has an algebraic explanation: they are all given by equations of the second degree. In any coordinate system, the equations of these curves have the form: ax 2 + bxy + cy 2 + dx + ey + f = 0, where a, b, c, d, e, f are numbers
Converting Rectangular Cartesian Coordinate Systems
Parallel translation of the coordinate system
-O 'in the old coordinate system
–Coordinates of the point in the old coordinate system
–Coordinates of the point in the new coordinate system
The coordinates of the point in the new coordinate system.
Rotation in a rectangular Cartesian coordinate system
–New coordinate system
Matrix of the transition from the old basis to the new
- (under the first column I’
, under the second - j’
) the transition matrix from the basis I,j to the base I’
,j’
General case
Rotating the coordinate system
Rotating the coordinate system
Parallel translation of the origin
Option 1
Option 2
General equation of lines of the second order and its reduction to the canonical form
– general form second-order curve equations
Classification of curves of the second order
Ellipsoid
Ellipsoid sections
- ellipse
- ellipse
Ellipsoids of revolution
Ellipsoids of revolution are either flattened or elongated spheroids, depending on what we rotate around.
Single-strip hyperboloid
Sections of a single-strip hyperboloid
- hyperbola with real y-axis
- hyperbola with real axis oh
It turns out an ellipse for any h. So it goes.
Single-strip hyperboloids of revolution
A one-sheet hyperboloid of revolution can be obtained by rotating the hyperbola around its imaginary axis.
Two-sheeted hyperboloid
Sections of a two-sheet hyperboloid 
- hyperbola with action. Axisoz
- hyperbola with real axis oz
Cone 
- a pair of intersecting straight lines
- a pair of intersecting straight lines
Elliptical paraboloid 
- parabola
- parabola
Rotations
If, then the elliptical paraboloid is a surface of revolution formed by the rotation of the parabola around its axis of symmetry.
Hyperbolic paraboloid
Parabola
- parabola
h> 0 hyperbola with real axis parallel to oh
h<0 гипербола с действительной осью паралльной оу и мнимой ох
By a cylinder we mean a surface that will be obtained when a straight line moves in space, which does not change its direction, if the straight line moves relative to oz, then the equation of the cylinder is the equation of the section by the xoy plane.
Elliptical cylinder
Hyperbolic cylinder
Parabolic cylinder
Rectilinear generators of surfaces of the second order
Straight lines completely lying on the surface are called rectilinear generatrices of the surface.
Surfaces of revolution
Fuck you goof
Display
By displaying let's call the rule according to which each element of the set A is associated with one or more elements of the set B. If each is assigned a unique element of the set B, then the mapping is called unambiguous, otherwise ambiguous.
Transformation set is a one-to-one mapping of a set onto itself
Injection
Injection or one-to-one mapping of a set A onto a set B
(different elements of a correspond to different elements of B) for example y = x ^ 2
Surjection
Subjection or mapping of a set A onto a set B
For each B, there is at least one A (for example, a sine)
Each element of the set B corresponds to only one element of the set A. (for example, y = x)
One of the sub-items of the topic "Equation of a straight line on a plane" is the question of drawing up parametric equations of a straight line on a plane in a rectangular coordinate system. The article below discusses the principle of drawing up such equations with certain known data. Let us show how to pass from parametric equations to equations of a different kind; Let's analyze the solution of typical tasks.
A specific line can be determined by specifying a point that belongs to this line and the direction vector of the line.
Let's say we are given a rectangular coordinate system O x y. And also a straight line a with an indication of the point M 1 (x 1, y 1) lying on it and a direction vector of a given straight line are given a → = (a x, a y) . Let us give a description of the given line a using equations.
We use an arbitrary point M (x, y) and get the vector M 1 M →; calculate its coordinates by the coordinates of the start and end points: M 1 M → = (x - x 1, y - y 1). Let us describe the obtained: a straight line is given by a set of points M (x, y), passes through a point M 1 (x 1, y 1) and has a direction vector a → = (a x, a y) . The specified set defines a straight line only if the vectors M 1 M → = (x - x 1, y - y 1) and a → = (a x, a y) are collinear.
There is a necessary and sufficient condition for vectors collinearity, which in this case for vectors M 1 M → = (x - x 1, y - y 1) and a → = (a x, a y) can be written in the form of the equation:
M 1 M → = λ · a →, where λ is some real number.
Definition 1
The equation M 1 M → = λ · a → is called the vector-parametric equation of the line.
In coordinate form, it has the form:
M 1 M → = λ a → ⇔ x - x 1 = λ a x y - y 1 = λ a y ⇔ x = x 1 + a x λ y = y 1 + a y λ
The equations of the resulting system x = x 1 + a x · λ y = y 1 + a y · λ are called parametric equations of a straight line on a plane in a rectangular coordinate system. The essence of the name is as follows: the coordinates of all points of the straight line can be determined by parametric equations on the plane of the form x = x 1 + a x λ y = y 1 + a y λ when iterating over all real values of the parameter λ
According to the above, the parametric equations of a straight line on the plane x = x 1 + ax vector a → = (a x, a y) . Therefore, if the coordinates of some point of the straight line and the coordinates of its direction vector are given, then it is possible to immediately write down the parametric equations of a given straight line.
Example 1
It is necessary to compose parametric equations of a straight line on a plane in a rectangular coordinate system if the point M 1 (2, 3) belonging to it and its direction vector are given a → = (3, 1).
Solution
Based on the initial data, we get: x 1 = 2, y 1 = 3, a x = 3, a y = 1. Parametric equations will be as follows:
x = x 1 + a x λ y = y 1 + a y λ ⇔ x = 2 + 3 λ y = 3 + 1 λ ⇔ x = 2 + 3 λ y = 3 + λ
Let's clearly illustrate:
Answer: x = 2 + 3 λ y = 3 + λ
It should be noted: if the vector a → = (a x, a y) serves as the direction vector of the straight line a, and the points M 1 (x 1, y 1) and M 2 (x 2, y 2) belong to this straight line, then it can be determined by specifying parametric equations of the form: x = x 1 + ax λ y = y 1 + ay λ, as well as this option: x = x 2 + ax λ y = y 2 + ay λ.
For example, we are given a directing vector of a straight line a → = (2, - 1), as well as points М 1 (1, - 2) and М 2 (3, - 3), belonging to this line. Then the straight line is determined by the parametric equations: x = 1 + 2 · λ y = - 2 - λ or x = 3 + 2 · λ y = - 3 - λ.
Attention should be paid to the following fact: if a → = (a x, a y) is the direction vector of the straight line a, then its direction vector will be any of the vectors μ a → = (μ a x, μ a y), where μ ϵ R, μ ≠ 0.
Thus, the straight line a on the plane in a rectangular coordinate system can be defined by the parametric equations: x = x 1 + μax λ y = y 1 + μ a y λ for any value of μ other than zero.
Suppose the line a is given by the parametric equations x = 3 + 2 · λ y = - 2 - 5 · λ. Then a → = (2, - 5) - direction vector of this line. And also any of the vectors μ a → = (μ 2, μ - 5) = 2 μ, - 5 μ, μ ∈ R, μ ≠ 0 will become the direction vector for the given line. For clarity, consider a specific vector - 2 a → = (- 4, 10), it corresponds to the value μ = - 2. In this case, the given line can also be determined by the parametric equations x = 3 - 4 · λ y = - 2 + 10 · λ.
Transition from parametric equations of a straight line on a plane to other equations of a given straight line and vice versa
In solving some problems, the use of parametric equations is not the best option, then it becomes necessary to translate the parametric equations of a straight line into equations of a straight line of a different kind. Let's see how to do this.
Parametric equations of a straight line of the form x = x 1 + a x · λ y = y 1 + a y · λ will correspond to the canonical equation of a straight line on the plane x - x 1 a x = y - y 1 a y.
Let us solve each of the parametric equations with respect to the parameter λ, equate the right-hand sides of the obtained equalities, and obtain the canonical equation of the given straight line:
x = x 1 + a x λ y = y 1 + a y λ ⇔ λ = x - x 1 a x λ = y - y 1 a y ⇔ x - x 1 a x = y - y 1 a y
At the same time, it should not be embarrassing if a x or a y are equal to zero.
Example 2
It is necessary to make the transition from the parametric equations of the straight line x = 3 y = - 2 - 4 · λ to the canonical equation.
Solution
We write the given parametric equations in the following form: x = 3 + 0 λ y = - 2 - 4 λ
Let us express the parameter λ in each of the equations: x = 3 + 0 λ y = - 2 - 4 λ ⇔ λ = x - 3 0 λ = y + 2 - 4
Let us equate the right-hand sides of the system of equations and obtain the required canonical equation of a straight line on the plane:
x - 3 0 = y + 2 - 4
Answer: x - 3 0 = y + 2 - 4
In the case when it is necessary to write down the equation of a straight line of the form A x + B y + C = 0, while the parametric equations of a straight line on a plane are given, it is necessary to first make the transition to the canonical equation, and then to the general equation of the straight line. Let's write down the whole sequence of actions:
x = x 1 + ax λ y = y 1 + ay λ ⇔ λ = x - x 1 ax λ = y - y 1 ay ⇔ x - x 1 ax = y - y 1 ay ⇔ ⇔ ay (x - x 1) = ax (y - y 1) ⇔ A x + B y + C = 0
Example 3
It is necessary to write down the general equation of the straight line if the parametric equations defining it are given: x = - 1 + 2 λ y = - 3 λ
Solution
To begin with, let's make the transition to the canonical equation:
x = - 1 + 2 λ y = - 3 λ ⇔ λ = x + 1 2 λ = y - 3 ⇔ x + 1 2 = y - 3
The resulting proportion is identical to the equality - 3 · (x + 1) = 2 · y. Let's open the brackets and get the general equation of the line: - 3 x + 1 = 2 y ⇔ 3 x + 2 y + 3 = 0.
Answer: 3 x + 2 y + 3 = 0
Following the above logic of actions, in order to obtain an equation of a straight line with a slope, an equation of a straight line in segments or a normal equation of a straight line, it is necessary to obtain the general equation of a straight line, and from it to carry out a further transition.
Now we will consider the opposite action: writing the parametric equations of the straight line for another given form of the equations of this straight line.
The easiest transition: from a canonical equation to a parametric one. Let a canonical equation of the form: x - x 1 a x = y - y 1 a y be given. We take each of the relations of this equality to be equal to the parameter λ:
x - x 1 a x = y - y 1 a y = λ ⇔ λ = x - x 1 a x λ = y - y 1 a y
Let us solve the obtained equations for the variables x and y:
x = x 1 + a x λ y = y 1 + a y λ
Example 4
It is necessary to write down the parametric equations of a straight line if the canonical equation of a straight line on a plane is known: x - 2 5 = y - 2 2
Solution
Equate the parts of the known equation to the parameter λ: x - 2 5 = y - 2 2 = λ. From the obtained equality, we obtain the parametric equations of the straight line: x - 2 5 = y - 2 2 = λ ⇔ λ = x - 2 5 λ = y - 2 5 ⇔ x = 2 + 5 λ y = 2 + 2 λ
Answer: x = 2 + 5 λ y = 2 + 2 λ
When it is necessary to make the transition to parametric equations from a given general equation of a straight line, an equation of a straight line with a slope, or an equation of a straight line in segments, it is necessary to reduce the original equation to the canonical one, and then proceed to the parametric equations.
Example 5
It is necessary to write down the parametric equations of the straight line with the known general equation of this straight line: 4 x - 3 y - 3 = 0.
Solution
We transform the given general equation into an equation of the canonical form:
4 x - 3 y - 3 = 0 ⇔ 4 x = 3 y + 3 ⇔ ⇔ 4 x = 3 y + 1 3 ⇔ x 3 = y + 1 3 4
Let us equate both sides of the equality to the parameter λ and obtain the required parametric equations of the straight line:
x 3 = y + 1 3 4 = λ ⇔ x 3 = λ y + 1 3 4 = λ ⇔ x = 3 λ y = - 1 3 + 4 λ
Answer: x = 3 λ y = - 1 3 + 4 λ
Examples and problems with parametric equations of a straight line on a plane
Consider the most common types of problems using parametric equations of a straight line on a plane in a rectangular coordinate system.
- In problems of the first type, the coordinates of points are given, whether or not they belong to a straight line described by parametric equations.
The solution of such problems is based on the following fact: the numbers (x, y) determined from the parametric equations x = x 1 + ax λ y = y 1 + ay λ for some real value λ are the coordinates of a point belonging to the straight line described these parametric equations.
Example 6
It is necessary to determine the coordinates of a point that lies on the straight line defined by the parametric equations x = 2 - 1 6 · λ y = - 1 + 2 · λ at λ = 3.
Solution
Substitute the known value λ = 3 into the given parametric equations and calculate the required coordinates: x = 2 - 1 6 3 y = - 1 + 2 3 ⇔ x = 1 1 2 y = 5
Answer: 1 1 2 , 5
The following problem is also possible: let some point M 0 (x 0, y 0) be given on a plane in a rectangular coordinate system and it is necessary to determine whether this point belongs to a straight line described by parametric equations x = x 1 + ax λ y = y 1 + ay λ.
To solve a similar problem, it is necessary to substitute the coordinates of a given point into the known parametric equations of the straight line. If it is determined that such a value of the parameter λ = λ 0 is possible, at which both parametric equations are true, then the given point belongs to the given straight line.
Example 7
Points M 0 (4, - 2) and N 0 (- 2, 1) are set. It is necessary to determine whether they belong to the straight line defined by the parametric equations x = 2 · λ y = - 1 - 1 2 · λ.
Solution
Substitute the coordinates of the point М 0 (4, - 2) into the given parametric equations:
4 = 2 λ - 2 = - 1 - 1 2 λ ⇔ λ = 2 λ = 2 ⇔ λ = 2
We conclude that point М 0 belongs to a given straight line, since corresponds to the value λ = 2.
2 = 2 λ 1 = - 1 - 1 2 λ ⇔ λ = - 1 λ = - 4
Obviously, there is no such parameter λ to which the point N 0 will correspond. In other words, the given line does not pass through the point N 0 (- 2, 1).
Answer: point М 0 belongs to a given straight line; point N 0 does not belong to a given straight line.
- In problems of the second type, it is required to compose parametric equations of a straight line on a plane in a rectangular coordinate system. The simplest example of such a problem (with known coordinates of a point of a straight line and a direction vector) was considered above. Now let's look at examples in which you first need to find the coordinates of the direction vector, and then write down the parametric equations.
Point M 1 1 2, 2 3 is specified. It is necessary to compose parametric equations of the straight line passing through this point and parallel to the straight line x 2 = y - 3 - 1.
Solution
According to the condition of the problem, the straight line, the equation of which we have to get ahead of, is parallel to the straight line x 2 = y - 3 - 1. Then, as a directing vector of a straight line passing through a given point, it is possible to use a directing vector of a straight line x 2 = y - 3 - 1, which we write in the form: a → = (2, - 1). Now we know all the necessary data in order to compose the required parametric equations:
x = x 1 + ax λ y = y 1 + ay λ ⇔ x = 1 2 + 2 λ y = 2 3 + (- 1) λ ⇔ x = 1 2 + x λ y = 2 3 - λ
Answer: x = 1 2 + x · λ y = 2 3 - λ.
Example 9
Point М 1 (0, - 7) is set. It is necessary to write down the parametric equations of a straight line passing through this point perpendicular to the straight line 3 x - 2 y - 5 = 0.
Solution
As the directing vector of the straight line, the equation of which must be drawn up, it is possible to take the normal vector of the straight line 3 x - 2 y - 5 = 0. Its coordinates are (3, - 2). Let us write down the required parametric equations of the straight line:
x = x 1 + a x λ y = y 1 + a y λ ⇔ x = 0 + 3 λ y = - 7 + (- 2) λ ⇔ x = 3 λ y = - 7 - 2 λ
Answer: x = 3 λ y = - 7 - 2 λ
- In problems of the third type, it is required to make the transition from parametric equations of a given straight line to other types of equations that determine it. We considered the solution of such examples above, we will give one more.
A straight line on a plane in a rectangular coordinate system is given, determined by the parametric equations x = 1 - 3 4 · λ y = - 1 + λ. It is necessary to find the coordinates of any normal vector of this straight line.
Solution
To determine the desired coordinates of the normal vector, we will carry out the transition from parametric equations to the general equation:
x = 1 - 3 4 λ y = - 1 + λ ⇔ λ = x - 1 - 3 4 λ = y + 1 1 ⇔ x - 1 - 3 4 = y + 1 1 ⇔ ⇔ 1 x - 1 = - 3 4 y + 1 ⇔ x + 3 4 y - 1 4 = 0
The coefficients of the variables x and y give us the required coordinates of the normal vector. Thus, the normal vector of the line x = 1 - 3 4 · λ y = - 1 + λ has coordinates 1, 3 4.
Answer: 1 , 3 4 .
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So far, we have considered the equation of a surface in space with coordinate axes X, Y, Z in explicit or implicit form
It is possible to write the equations of the surface in parametric form, expressing the coordinates of its points in the form of functions of two independent variable parameters and
We will assume that these functions are single-valued, continuous and have continuous derivatives up to the second order in some range of parameters
If we substitute these coordinate expressions in terms of u and v into the left-hand side of equation (37), then we must obtain an identity with respect to u and V. Differentiating this identity with respect to the independent variables u and v, we will have
Considering these equations as two homogeneous equations with respect to and applying the algebraic lemma mentioned in, we obtain
where k is some coefficient of proportionality.
We believe that the factor k and at least one of the differences on the right-hand sides of the last formulas are nonzero.
For brevity, we denote the three differences written as follows:
As is known, the equation of the tangent plane to our surface at some point (x, y, z) can be written in the form
or, replacing with proportional quantities, we can rewrite the equation of the tangent plane as follows:
The coefficients in this equation are known to be proportional to the direction cosines of the normal to the surface.
The position of the variable point M on the surface is characterized by the values of the parameters u and v, and these parameters are usually called the coordinates of the surface points or coordinate parameters.
By assigning constant values to the parameters and and v, we obtain two families of lines on the surface, which we will call the coordinate lines of the surface: the coordinate lines along which only v changes, and the coordinate lines along which only and change. These two coordinate line families provide a coordinate grid on the surface.
As an example, consider a sphere with a center at the origin and radius R. The parametric equations of such a sphere can be written in the form
Coordinate lines represent in this case, obviously, the parallels and meridians of our sphere.
Digressing from the coordinate axes, we can characterize the surface with a variable radius vector going from a constant point O to a variable point M of our surface. The partial derivatives of this radius vector with respect to parameters will obviously give vectors directed along the tangents to the coordinate lines. The components of these vectors along the axes
will, according to and from here it is seen, that the coefficients in the equation of the tangent plane (39) are the components of the vector product This vector product is a vector perpendicular to the tangents, i.e., a vector directed along the normal to the surface. The square of the length of this vector is obviously expressed by the scalar product of the vector by itself, that is, more simply, by the square of this vector 1). In what follows, the unit normal vector to the surface will play an essential role, which we can obviously write in the form
Changing the order of the factors in the written vector product, we get the opposite direction for vector (40). In what follows, we will fix in a certain way the order of the factors, that is, we will in a certain way fix the direction of the normal to the surface.
Take some point M on the surface and draw through this point some curve (L) lying on the surface. This curve, generally speaking, is not a coordinate line, and both Well and v will change along it. The direction of the tangent to this curve will be determined by the vector if we assume that along (L) in the vicinity of the point the parameter v is a function of which has a derivative. From this it can be seen that the direction of the tangent to the curve drawn on the surface, at any point M of this curve, is fully characterized by the value at this point. In determining the Tangent Plane and deriving its equation (39), we assumed that functions (38) at the point under consideration and its vicinity have continuous partial derivatives and that at least one of the coefficients of equation (39) is nonzero at the point under consideration.
Any equation of the first degree 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 equation of the plane.
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 simultaneously equal to 0.
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 the 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), which belongs to it, and the vector a(m, n, p), collinear to it. Then the straight line is determined by the equations:
Equations (3.4) are called canonical equations of the line.
Vector a called directing vector of the straight line.
Parametric equations of a straight line we obtain by equating each of the relations (3.4) to the parameter t:
x = x 1 + mt, y = y 1 + nt, z = z 1 + р t. (3.5)
Solving system (3.2) as a system of linear equations with respect to unknowns x and y, we arrive at the equations of the straight line in projections or to the reduced equations of the straight line :
x = mz + a, y = nz + b. (3.6)
From equations (3.6), we can go over to canonical equations by finding z from each equation and equating the obtained values:
From the general equations (3.2), one can pass to the canonical and in another way, if we find some point of this straight 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.
The system is equivalent to the system x = x 1, y = y 1; the straight line is parallel to the Oz axis.
Example 1.15... Equate the plane, knowing that point A (1, -1.3) is the base of the perpendicular drawn from the origin to this plane.
Solution. By the condition of the problem, the vector OA(1, -1,3) is the 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... Make an equation for the plane passing through the Oz axis and forming an angle of 60 ° 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. According to the formula for the cosine of 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, whence we obtain two planes 1 / 3x + y = 0 and -3x + y = 0.
Example 1.17. Make the canonical equations of the straight line:
5x + y + z = 0, 2x + 3y - 2z + 5 = 0.
Solution. The canonical equations of the straight line are:
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 the straight line. A straight line is specified as the intersection line of two planes. To find a point belonging to a straight line, one of the coordinates is fixed (the easiest way is to put, 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, whence y = -1, z = 1. The coordinates of the point M (x 1, y 1, z 1), belonging to this line, we found: M (0, -1,1). The direction vector of the 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 straight line are: x / (- 5) = (y + 1) / 12 =
= (z - 1) / 13.
Vector and parametric equations of the plane. Let r 0 and r be the radius vectors of points М 0 and M, respectively. Then M 0 M = r - r 0, and condition (5.1) that point M belongs to a plane passing through point M 0 perpendicularly nonzero vector n (Fig.5.2, a), can be written using dot product as a ratio
n (r - r 0) = 0, (5.4)
which is called vector equation of the plane.
A fixed plane in space corresponds to a set of vectors parallel to it, i.e. space V 2. Let's choose in this space basis e 1, e 2, i.e. a pair of non-collinear vectors parallel to the plane in question, and a point M 0 on the plane. If the point M belongs to the plane, then this is equivalent to the fact that the vector M 0 M is parallel to it (Fig. 5.2, b), i.e. it belongs to the indicated space V 2. This means that there is expansion of the vector M 0 M in the basis e 1, e 2, i.e. there are numbers t 1 and t 2 for which M 0 M = t 1 e 1 + t 2 e 2. Writing the left side of this equation through the radius vectors r 0 and r points M 0 and M, respectively, we obtain vector parametric plane equation
r = r 0 + t 1 e 1 + t 2 e 2, t 1, t 1 ∈ R. (5.5)
To pass from the equality of the vectors in (5.5) to the equality of their coordinates, we denote by (x 0; y 0; z 0), (x; y; z) point coordinates M 0, M and through (e 1x; e 1y; e 1z), (e 2x; e 2y; e 2z) coordinates of vectors e 1, e 2. Equating the same coordinates of the vectors r and r 0 + t 1 e 1 + t 2 e 2, we obtain parametric plane equations
A plane passing through three points. Suppose that three points M 1, M 2 and M 3 do not lie on one straight line. Then there is a unique plane π to which these points belong. Let us find the equation of this plane, formulating the criterion for the membership of an arbitrary point M in a given plane π. Then we write this criterion through the coordinates of the points. The specified criterion is the description of the plane π as the set of those points M for which the vectors M 1 M 2, M 1 M 3 and M 1 M coplanar... The criterion for coplanarity of three vectors is their equality to zero. mixed work(see 3.2). The mixed product is calculated using determinant of the third order, the lines of which are the coordinates of the vectors in orthonormal basis... Therefore, if (xi; yx i; Zx i) are the coordinates of the points Mx i, i = 1, 2, 3, and (x; y; z) are the coordinates of the point M, then M 1 M = (х-x 1; y-y 1; zz 1), M 1 M 2 = (x 2 -x 1; y 2 -y 1; z 2 -z 1), M 1 M 3 = (x 3 -x 1; y 3 -y 1; z 3 -z 1) and the condition for the vanishing of the mixed product of these vectors has the form
Calculating the determinant, we get linear with respect to x, y, z the equation being the general equation of the desired plane... For example, if expand the determinant on the 1st line, then we get
This equality, after calculating the determinants and expanding the parentheses, is converted to the general equation of the plane.
Note that the coefficients of the variables in the last equation coincide with the coordinates vector product M 1 M 2 × M 1 M 3. This cross product, being the product of two noncollinear vectors parallel to the plane π, gives a nonzero vector perpendicular to π, i.e. her normal vector... So the appearance of the coordinates of the vector product as the coefficients of the general equation of the plane is quite natural.
Consider the following special case of a plane passing through three points. Points M 1 (a; 0; 0), M 2 (0; b; 0), M 3 (0; 0; c), abc ≠ 0, do not lie on one straight line and define a plane that cuts off segments on the coordinate axes of nonzero length (Figure 5.3). Here, "the lengths of the segments" mean the value of the nonzero coordinates of the radius vectors of the points M i, i = 1,2,3.
Since M 1 M 2 = (-a; b; 0), M 1 M 3 = (-a; 0; c), M 1 M = (x-a; y; z), then equation (5.7) takes the form
Having calculated the determinant, we find bc (x - a) + acy + abz = 0, divide the resulting equation by abc and transfer the free term to the right-hand side,
x / a + y / b + z / c = 1.
This equation is called the equation of the plane in segments.
Example 5.2. Let us find the general equation of a plane that passes through a point with coordinates (1; 1; 2) and cuts off segments of the same length from the coordinate axes.
The equation of a plane in segments, provided that it cuts off segments of equal length from the coordinate axes, say a ≠ 0, has the form x / a + y / b + z / c = 1. This equation must be satisfied by the coordinates (1; 1; 2) known point on the plane, i.e. the equality 4 / a = 1 holds. Therefore, a = 4 and the required equation is x + y + z - 4 = 0.
Normal equation of the plane. Consider some plane π in space. We fix for her unit normal vector n directed from origin"towards the plane", and denote by p the distance from the origin O of the coordinate system to the plane π (Fig. 5.4). If the plane passes through the origin of the coordinate system, then p = 0, and any of the two possible ones can be chosen as the direction for the normal vector n.
If the point M belongs to the plane π, then this is equivalent to the fact that vector orthographic projection OM per direction vector n is equal to p, i.e. the condition nOM = pr n OM = p is satisfied, since vector length n is equal to one.
We denote the coordinates of a point M by (x; y; z) and let n = (cosα; cosβ; cosγ) (recall that for a unit vector n its direction cosines cosα, cosβ, cosγ are simultaneously its coordinates). Writing the scalar product in the equality nOM = p in coordinate form, we obtain normal plane equation
xcosα + ycosbeta; + zcosγ - p = 0.
Similar to the case of a straight line on a plane, the general equation of a plane in space can be transformed into its normal equation by dividing by a normalizing factor.
For the plane equation Ax + By + Cz + D = 0, the normalizing factor is the number ± √ (A 2 + B 2 + C 2), the sign of which is chosen opposite to the sign of D. In absolute value, the normalizing factor is the length of the normal vector (A; B ; C) plane, and the sign corresponds to the desired direction of the unit normal vector of the plane. If the plane passes through the origin of the coordinate system, i.e. D = 0, then any sign of the normalizing factor can be chosen.
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