The alignment of the plane through the point is parallel to the straight line. The alignment of the area to pass through the given point is parallel to the given area online. Knowing the plane of the plane to pass through a given straight line and a given point

help someone else online calculator you can know the plane's alignment by passing through a given point that is parallel to the given plane. We hope to report a solution with explanations. In order to know the level of the area, enter the coordinates of the point and the coefficient of the level of the area in the box and click on the "Verishity" button.

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Instructions for entering data. Numbers are entered as whole numbers (apply: 487, 5, -7623 thin.), tenth numbers (eg. 67., 102.54 thin.) or fractions. The fraction is required to be typed at the sight of a / b, de a і b (b> 0) tsіlі or tens of numbers. Apply 45/5, 6.6/76.4, -7/6.7 thin.

Flatness of the plane to pass through a given point that is parallel to the given plane - theory, apply that solution

Let's set a point M 0 (x 0 , y 0 , z 0) that equal area

All parallel planes have norms and collenear vectors. Therefore, in order to encourage a plane parallel to (1) to pass through a point M 0 (x 0 , y 0 , z 0) it is necessary to take as the normal vector of the shukano plane, the normal vector n=(A, B, C) area (1). It is necessary to know such a meaning D, for which point M 0 (x 0 , y 0 , z 0) pleased with the flatness of the area (1):

Submitting a value D from (3) to (1), we take:

Level (5) is the level of the area that passes through the point M 0 (x 0 , y 0 , z 0) that parallel plane (1).

Know the level of the plane to pass through the point M 0 (1, −6, 2) that parallel plane:

Submitting point coordinates M 0 i coordinates of the normal vector (3), optional.

The plane Q is viewed in space. If we know through A, B and C the projections of the normal vector N, then

Let's take a look at the plane Q that passes through the given point and can be a normal vector . For which we can look at a vector that connects a point with a sufficient point of the plane Q (Fig. 81).

For any position of the point M on the plane Q, the MXM vector of perpendiculars to the normal vector N of the plane Q. Therefore, scalar tvir We write the scalar tvir through projections. Oskіlki , and the vector , then

i, otzhe,

We have shown that the coordinates, whether or not the points of the plane Q, are satisfied with the alignment (4). It is not important to note that the coordinate points, which do not lie on the plane Q, which are not satisfied with (in the rest of the fall). Also, we have taken off the level of the area Q. The level (4) is called the level of the area that passes through this point. Vono first step shodo streamline coordinates

Later, we showed that whether the planes are equal to the first stage of the current coordinates.

Example 1. Write a flat plane to pass through a point perpendicular to the vector .

Solution. Here. On the basis of formula (4) we take

otherwise, after asking forgiveness,

Giving coefficients A, B and C equal (4) different values, we can take into account the evenness of the plane that passes through the point. The cluster of planes that pass through this point is called the conjunctive plane. Rivnyannia (4), in which coefficients A, B and C can take on any value, are called equal links of planes.

Butt 2. Flatten the plane to pass through three points (Fig. 82).

Solution. Let's write the alignment of the links of the planes that pass through the point

Lecture 5

1. Flatten the plane to pass through the point M 0 (1, -2, 5) parallel to plane 7 x-y-2z-1=0.

Solution. Significantly through R given the area, let R 0 - Shukana is parallel to a plane that passes through a point M 0 (1, -2, 5).

Clearly normal (perpendicular) vector flats R. Coordinates of the normal vector є coefficients for changing plane planes 
.

Oskіlki plaschinі Rі R 0 parallel, then the vector perpendicular to the plane R 0 , then. - normal vector of the area R 0 .

Flatness of the plane to pass through the point M 0 (x 0 , y 0 , z 0) from the normal
:

Submit the coordinates of the point M 0 and normal vectors equalization (1):

Curving the arches, it is necessary to take a flat flat surface (residual wear):

2. Fold the canonical and parametric straight lines to pass through the point M 0 (-2, 3, 0) parallel straight
.

Solution. Significantly through L set straight, let it go L 0 - Shukana is parallel to a straight line to pass through a point M 0 (-2,3,0).

Direct vector straight L(non-zero vector, parallel to line) parallel also to line L 0 . Father, vector є direct vector of a straight line L 0 .

Direct vector coordinates equal to the canonical equals of the canonical equals of the given straight line

.

The canonical line is straight in the open space that passes through the point M 0 (x 0 , y 0 , z {l, m, n}

. (2)

Submit the coordinates of the point M 0 and direct vector equivalence (2) and prima facie canonical equivalence straight:

.

Parametric alignment of a straight line in a space that passes through a point M 0 (x 0 , y 0 , z 0) parallel to a non-zero vector {l, m, n), to look at:

(3)

Submit the coordinates of the point M 0 and direct vector at alignment (3) and obsessively parametric alignment straight lines:

3. Know the point
, symmetric point
shodo: a) straight
b) flat

Solution. a) Storing the alignment of the perpendicular plane P, which projects a point
straight ahead:

to know
vikoristovuєmo to the mind perpendicularity of the given straight line and the plane that the project is designing. Direct vector direct
perpendicular to the plane  vector
є normal vector
up to the plane  Alignment of the plane perpendicular to the given straight line

We know the projection R specks M straight. Krapka RІsnuє point of the cross line is straight and flat, tobto. її the coordinates of the fault are simultaneously satisfied with the alignment of the straight line, and the alignment of the plane. Check the system:

.

In order to untie її, we write down the alignment of the straight line with the parametric look:

Submitting Virazi for
in the plane of the plane, we take into account:

Found coordinates - all coordinates of the middle R v_drіzka
that symmetrical point

A high school course in geometry formulated a theorem.

The coordinates of the midpoint of the vіdrіzk are equal to the sum of the other coordinates of the yоgo kіntsіv.

We know the coordinates of the point
from formulas for the coordinates of the middle of the vіdrіzka:

Otrimuemo: Otzhe,
.

Solution. b) To know a point that is symmetrical to a point
schodo given area P, we drop the perpendicular from the point
on the square. Folding alignment of straight lines with a direct vector
, to pass through the point
:

The perpendicularity of the line and the plane means that the direct vector of the line perpendicular to the plane 
. To di straight lines that project a point
on a given plane, may look:

Virishivshi spіlno rivnyannia
і
we know the projection R specks
on the flat. For which we rewrite the straight lines in the parametric view:

Let's imagine the value
in the alignment of the plane: Similar to item a), vicorist formulas for the coordinates of the middle of the vіrіzka, we know the coordinates of the symmetrical point
:

Tobto.
.

4. Fold flat planes to pass a) through a straight line
parallel to the vector
; b) through two straight lines that intertwine
і
(before dovіvshi, scho stench peretinayutsya); c) through two parallel lines
і
; d) straight ahead
i point
.

Solution. a) The shards are given to lie straight in the plane, which is whispering, and the plane, which is whispering, is parallel to the vector , then the normal vector of the area will be perpendicular to the direct vector of the line
ta vector .

Also, as a normal vector of the area, you can choose the vector additional vector_v і :

We take away the coordinates of the normal vector of the area
.

Let's know the speck on a straight line. Equating sts in canonical equals straight to zero:

,

known
,
,
. Given a straight line to pass through a point
, also, the area can also pass through the point
. Vikoristovuyuchi planes, scho to pass through a given point perpendicular to the vector. , otrimuemo flatness of the area, or, or, obviously,
.

Solution. b) Two straight lines in the open space can intertwine, cross each other and become parallel. Direct tasks

і
(4)

not parallel, shards of their direct vectors
і
not collinear:
.

How to distort, what do they straighten up? It is possible to reverse the system (4) out of 4 equals out of 3 nevіdomimi. If the system has only one solution, we need to coordinate the points of the cross line. However, for the accomplishment of our task - to encourage the flats, in which to lie straight, the point of their cross is not needed. To this, it is possible to formulate the mental span of two non-parallel lines in the space without knowing the span point.

If two non-parallel lines intersect, then the lines of the vector
,
and back lying on straight points
і
vector to lie near one plane, tobto. complanarnі  zmіshani dobutok tsikh vektor_v dorivnyuє zero:

. (5)

We compare the ratio in the canonical straight lines to zero (and you can up to 1 or up to whatever number)

і
,

and is known to be a coordinate point on the lines. Persha straight line to pass through the point
, and the other is straight through the point
. Direct vectors
і
. Acceptable

Rivnist (5) is wicked, and the tasks are straightened out. Otzhe, іsnuіє єdina plaschina, scho to pass through і two lines.

Let's move on to the other part of the task - the folding of the flat area.

As a normal vector of the area, one can choose vector witwearїх direct vectors і :

Coordinates of the normal vector of the area
.

Mi z'yasuvali, sho straight
pass through
Otzhe, shukana area tezh to pass through the qiu point. We take away the flatness of the area, otherwise
abo, obviously,
.

c) So straight
і
parallel, then, as a normal vector, it is impossible to choose a vector supplement of їх direct vectors, which will be equal to the zero vector.

Significant coordinate point
і
through the yaks pass q straight. Come on
і
also
,
. Let's calculate the coordinates of the vector. Vector
lie near the required area and is not collinear to the vector , then the normal vector you can choose the vector tv_r of the vector
that of the direct vector of the first line
:

Otzhe,
.

The area to pass through a straight line
mean, won pass through the point
. We take the flatness of the area: , or .

d) Equating the blue of the canonical equals straight to zero
, we know
,
,
. Oh, go straight through the dot
.

Let's calculate the coordinates of the vector. Vector
lie on the shukaniy flat, like a normal vector vibero vector twir of direct vector direct
ta vector
:

Todi evenness of the area may look:, or.

This statute has selected information that is necessary for determining the alignment of the plane, which passes through a given straight line and a given point. After the rozv'yazannya tsgogo zavdannya at the infamous looking, we will bring up the solution of butts on the folding flatness of the area, so as to pass through a given straight line that point.

Navigation on the side.

Znakhodzhennya r_vnyannya flat, scho to pass through a given straight line and a given point.

Let Oxyz be fixed near the trivi-worldly space, a line a and a point are given that do not lie on line a. Let's put our own task: take a flat plane that passes through the straight line a i point M 3.

It is shown on the back that there is only one flat, equal to what it is necessary to lay down.

We guess two axioms:

• through three different points of space, not to lie on one straight line, to pass through a single plane;
• If two different points of the straight line lie near the singing plane, then all points of the straight line lie near this plane.

Z tsikh hardness is vibrating, that through a straight line and a point that does not lie on it, you can draw a single plane. In this way, in the tasks we have set, a single plane passes through the straight line a and point M 3, and we need to write the equal plane.

Now let's proceed to the level of the plane, which is to pass through the given straight line and point.

If the straight line a is given by entering the coordinates of two different points M 1 and M 2 that lie on it, then our task is reduced to the level of the plane that passes through three points M 1 , M 2 and M 3 .

If the straight line a is given otherwise, then we will occasionally know the coordinates of two points M 1 and M 2, which lie on the straight line a, and even then, write down the flatness of the plane, which will pass through three points M 1 M 2 and M 3 let's look at the flatness of the plane, which will pass through the straight line and the point M 3 .

Let's figure out how to know the coordinates of two different points M1 and M2, which lie on the given straight line a.

In a rectangular coordinate system in the space, be-like a straight line in the space. It is important to note that the method of creating a straight line in a mental problem allows you to take parametric alignment of a straight line out of sight . Todi, having accepted, maybe a speck , which lies on the line a . Having given the parameter vіdmіnne vіd zero dіisne value, from the parametric lines it is possible to calculate the coordinates of the point M 2 , which also lie on the line a and vіdminnoї vіd point M 1 .

Why should we be left with less to write a flat plane, so that we can pass through three different ones and not lie on one straight point and look .

Otzhe, we took away the flatness of the plane, which to pass through a given straight line and a given point M 3, not lying on a straight line a.

Apply a flattened plane to pass through a given point and a straight line.

It is shown that a number of applications have been developed, for some of them, the method of determining the level of the plane that passes through a given straight line and a given point.

Let's start from the simplest point.

butt.

Solution.

Take two different points on the coordinate line Ox, for example, i .

Now we take into account the level of the plane to pass through the three points M1, M2 and M3:

The purpose is to level the upper levels of the area, which pass through the given straight line Ox i point. .

Suggestion:

.

Assuming that the plane passes through a given point and a given straight line, it is necessary to write the plane at the vdrіzkah, or normally the plane is equal, next to the top of the head, take the edge of the plane of the given plane, and then go to the plane of the required plane.

butt.

Store normal planes that pass through a straight line i point .

Solution.

We will write on the back of the head the equal equality of the given area. For which we know the coordinates of two different points that lie on a straight line . Parametric alignment of straight lines may look . Let the point M1 show the value, and the point M2 -. We calculate the coordinate points M1 and M2:

Now we can lay down a straight line, like passing through a point that straight :

The necessary appearance of the leveling of the flat was lost, multiplying the offending parts of the removed leveling by the multiplier that normalizes. .

Suggestion:

.

Also, the significance of the flatness of the plane, which passes through a given point and a given straight line, rests on the significance of the coordinates of two different points, which lie on a given straight line. For whom, the main skladnіst in case of vyrіshennі podіbnih zavdan. At the visnovka, I’ll take the solution to the butt on the folding of the flat plane, which will pass through a given point and straight, so that the equal plane of the two flats, which overlap.

butt.

The rectangular coordinate system Oxyz is given a point and a straight line a , as a line of intersection of two planes і . Write the plane of the plane to pass through the straight line a and point M 3 .

Three points of space, which do not lie on one straight line, signify a single plane. Let's make the plane equal to pass through three given points M 1 (X 1 ; at 1 ; z 1), M 2 (X 2 ; at 2 ; z 2), M 3 (X 3 ; at 3 ; z 3). Take a good point on the flat M(X; at; z) i foldable vectori = ( x - x 1 ; at– at 1 ; z-z 1), = (X 2 - X 1 ; at 2 – at 1 ; z 2 -z 1), = (X 3 - X 1 ; at 3 – at 1 ; z 3 -z one). Qi vectors lie in the same plane, then, the stench is coplanar. Wikoristovuyuchi mentally complanarity of three vectors (six confusions dobutok to zero), it is taken ∙ ∙ = 0, then

= 0. (3.5)

Rivnyannia (3.5) is called equal to the plane that passes through three given points.

Mutually roztashuvannya areas near the space

Kut between flats

Give me two areas

BUT 1 X + At 1 at + W 1 z+D 1 = 0,

BUT 2 X + At 2 at + W 2 z+D 2 = 0.

Per cut between flats we take kut φ between two vectors perpendicular to them (which gives two kuti, gostry and stupid, which adds one to one to π). So, as normal vectors of planes = ( BUT 1 , At 1 , W 1) i = ( BUT 2 , At 2 , W 2) perpendicular to them, then we take

cosφ = .

Wash the perpendicularity of two planes

If two planes are perpendicular, then the normal vectors of these planes are also perpendicular and both scalar additions are equal to zero: ∙ = 0. Also, the intellectual perpendicularity of two planes is

BUT 1 BUT 2 + At 1 At 2 + W 1 W 2 = 0.

Umov parallelism of two planes

If the planes are parallel, then those normal vectors will be parallel. Same-dimensional coordinates of normal vectors in proportions. Otzhe, mental parallelism of the planes

= = .

Vіdstan vіd pointM 0 (x 0 , y 0 , z 0) up to the flat Oh + Wu + Сz+D = 0.

View of point M 0 (x 0 , y 0 , z 0) to Ah Square + Wu + Сz+D= 0 is called the length of the perpendicular drawn from the center of the point to the plane, i is behind the formula

d= .

example 1. R(-1, 2, 7) perpendicular to the vector = (3, - 1, 2).

Solution

Vіdpovіdno to іvnyannya (3.1) otrimuєemo

3(x + 1) – (y - 2) + 2(z- 7) = 0,

3X – at + 2z – 9 = 0.

butt 2. Fold flat planes to pass through a point M(2; – 3; – 7) parallel to plane 2 X – 6at – 3z + 5 = 0.

Solution

Vector = (2; – 6; – 3) perpendiculars to the plane of perpendiculars і to the parallel plane. So, it is necessary for the area to pass through the point M(2; – 3; – 7) perpendicular to the vector = (2; – 6; – 3). We know the equalization of the area for the formula (3.1):

2(X - 2) – 6(y + 3) – 3(z+ 7) = 0,

2X – 6at – 3z – 43 = 0.

example 3. Know the alignment of the plane to pass through the points M 1 (2; 3; - 1) and M 2 (1; 5; 3) perpendicular to plane 3 X – at + 3z + 15 = 0.

Solution

The vector = (3; – 1; 3) of perpendiculars to the given plane will be parallel to the jogging plane. In this order, the plane pass through the points M 1 ta M 2 parallel to the vector.

Come on M(x; y; z) is enough point of the plane, then the vectors = ( X – 2; at – 3; z+ 1), \u003d (-1; 2; 4), \u003d (3; - 1; 3) are coplanar, meaning that their deviations are equal to zero:

= 0.

Let's calculate the leader of the layout for the elements of the first row:

(X – 2) – (at – 3) + (z + 1) = 0,

10(X - 2) – (– 15)(y - 3) + (– 5)(z+ 1) = 0,

2(X - 2) + 3(y - 3) – (z+ 1) = 0,

2x + 3at – z- 14 = 0 - equalization of the area.

butt 4. Fold flat planes to pass through the cob of coordinates perpendicular to the planes. X – at + 5z+ 3 = 0 that X + 3at – z – 7 = 0.

Solution

Nehai is the normal vector of the shukano area. Behind the mind, the plane is perpendicular to these planes, hence, de = (2; - 1; 5), = (1; 3; - 1). Also, as a vector, you can take the vector extension of the vector in i , so = × .

= = – 14 + 7 + 7 .

Substituting the coordinates of the flatness vector to pass through the cob of coordinates Oh + Wu + Сz= 0, take

– 14X + 7at + 7z = 0,

2X – at – z = 0.

Food for self-verification

1 Write down the hot flatness of the area.

2 Yaky geometric zmist coefficients at X, y, z at the scorching equal area?

3 Write down the alignment of the area to pass through the point M 0 (x 0 ; y 0 ; z 0) perpendicular to the vector = ( BUT; At; W).

4 Write down the flatness of the plane at the rails along the axes and indicate the geometrical change of the parameters, which enters up to the new one.

5 Record the alignment of the area to pass through the points M 1 (X 1 ; at 1 ; z 1), M 2 (X 2 ; at 2 ; z 2), M 3 (X 3 ; at 3 ; z 3).

6 Write down the formula for knowing the cut between two planes.

7 Write down the parallelism of two planes.

8 Write down the perpendicularity of the two planes.

9 Write down the formula for which the number of points to the plane is calculated.

Task for independent vision

1 Fold flat planes to pass through a point M(2; – 1; 1) perpendicular to the vector = (1; – 2; 3). ( Vidpovid: X – 2at + 3z – 7 = 0)

2 Krapka R(1; - 2; - 2) - the basis of the perpendicular drawn from the cob of coordinates to the plane. Lay flat squares. ( Vidpovid: X – 2at – 2z – 9 = 0)

3 Given two points M 1 (2; – 1; 3) and M 2 (-1; 2; 4). Fold flat planes to pass through a point M 1 is perpendicular to the vector. ( Vidpovid: 3X – 3at – z – 6 = 0)

4 Fold the plane through three points. M 1 (3; – 1; 2), M 2 (4; – 1; – 1), M 3 (2; 0; 2). (Vidpovid: 3X + 3at + z – 8 = 0)

5 M 1 (3; – 1; 2) and M 2 (2; 1; 3) parallel to the vector = (3; - 1; 4). ( Vidpovid: 9X + 7at – 5z – 10 = 0)

6 Fold flat planes to pass through a point M 1 (2; 3; – 4) parallel to the vectors = (3; 1; – 1) and = (1; – 2; 1). ( Vidpovid: X + at + 7z + 14 = 0)

7 Fold flat planes to pass through a point M(1; – 1; 1) perpendicular to planes 2 X – at + z- 1 = 0 i X + 2at – z + 1 = 0. (Vidpovid: X – 3at – 5z + 1 = 0)

8 Fold flat planes to pass through points M 1 (1; 0; 1) that M 2 (1; 2; – 3) perpendicular to the plane X – at + z – 1 = 0. (Vidpovid: X + 2at + z – 2 = 0)

9 Know the cut between the flats 4 X – 5at + 3z- 1 = 0 i X – 4at – z + 9 = 0. (Vidpovid: φ = arccos0.7)

10 Know the location of the point M(2; – 1; – 1) up to area 16 X – 12at + 15z – 4 = 0. (Vidpovid: d = 1)

11 Find the cross point of three planes 5 X + 8at – z – 7 = 0, X + 2at + 3z – 1 = 0, 2X – 3at + 2z – 9 = 0. (Vidpovid: (3; – 1; 0))

12 Flatten the plane so as to pass through the points M 1 (1; - 2; 6) and M 2 (5; - 4; 2) Ohі OU. (Vidpovid: 4X + 4at + z – 2 = 0)

13 Know the difference between flats X + 2at – 2z+ 2 = 0 and 3 X + 6at – 6z – 4 = 0. (Vidpovid: d = )