Equation of a plane that passes through a given line and a given point. Equation of a plane passing through three given points Equation of a plane through a point and a parallel plane
Three points in space that do not lie on the same straight line define a single plane. Write an equation for a plane that passes 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 an arbitrary point on the plane M(X; at; z) and compose vectors = ( 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). These vectors lie in the same plane, hence they are coplanar. Using the condition of complanarity of three vectors (their mixed product is equal to zero), we obtain ∙ ∙ = 0, i.e.
= 0. (3.5)
Equation (3.5) is called the equation of a plane passing through three given points.
Mutual arrangement planes in space
Angle between planes
Let two planes be given
BUT 1 X + AT 1 at + FROM 1 z + D 1 = 0,
BUT 2 X + AT 2 at + FROM 2 z + D 2 = 0.
Per angle between planes we take the angle φ between any two vectors perpendicular to them (which gives two angles, acute and obtuse, complementing each other up to π). Since the normal vectors of the planes = ( BUT 1 , AT 1 , FROM 1) and = ( BUT 2 , AT 2 , FROM 2) are perpendicular to them, then we get
cosφ = 
.
The condition of perpendicularity of two planes
If two planes are perpendicular, then the normal vectors of these planes are also perpendicular and their scalar product is equal to zero: ∙ = 0. Hence, the condition for the perpendicularity of two planes is
BUT 1 BUT 2 + AT 1 AT 2 + FROM 1 FROM 2 = 0.
Condition of parallelism of two planes
If the planes are parallel, then their normal vectors will also be parallel. Then the like-named coordinates of the normal vectors are proportional. Hence, the condition for parallel planes is
= = .
Distance from pointM 0 (x 0 , y 0 , z 0) up to the plane Oh + Wu + Сz + D = 0.
Distance from point M 0 (x 0 , y 0 , z 0) to the plane Ah + Wu + Сz + D= 0 is the length of the perpendicular drawn from this point to the plane, and is found by the formula
d= 
.
Example 1 R(– 1, 2, 7) is perpendicular to the vector = (3, – 1, 2).
Solution
According to equation (3.1) we get
3(x + 1) – (y - 2) + 2(z- 7) = 0,
3X – at + 2z – 9 = 0.
Example 2 Write an equation for a plane passing through a point M(2; – 3; – 7) parallel to plane 2 X – 6at – 3z + 5 = 0.
Solution
Vector = (2; - 6; - 3) perpendicular to the plane is also perpendicular to the parallel plane. So the desired plane passes through the point M(2; – 3; – 7) perpendicular to the vector = (2; – 6; – 3). Let us find the equation of the plane by the formula (3.1):
2(X - 2) – 6(y + 3) – 3(z + 7) = 0,
2X – 6at – 3z – 43 = 0.
Example 3 Find the equation of a plane passing through points M 1 (2; 3; – 1) and M 2 (1; 5; 3) perpendicular to plane 3 X – at + 3z + 15 = 0.
Solution
Vector = (3; - 1; 3) perpendicular to the given plane will be parallel to the desired plane. So the plane passes through the points M 1 and M 2 parallel to the vector .
Let M(x; y; z) an arbitrary point of the plane, then the vectors = ( X – 2; at – 3; z+ 1), = (– 1; 2; 4), = (3; – 1; 3) are coplanar, so their mixed product is equal to zero:
= 0.
Calculate the determinant by expanding over 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 – plane equation.
Example 4 Write an equation for a plane passing through the origin perpendicular to the planes 2 X – at + 5z+ 3 = 0 and X + 3at – z – 7 = 0.
Solution
Let be the normal vector of the required plane. By condition, the plane is perpendicular to these planes, hence and , where = (2; – 1; 5), = (1; 3; – 1). So, as a vector, you can take the cross product of the vectors and , that is, = × .
= = – 14 + 7 + 7 .
Substituting the coordinates of the vector into the equation of the plane passing through the origin Oh + Wu + Сz= 0, we get
– 14X + 7at + 7z = 0,
2X – at – z = 0.
Questions for self-examination
1 Write down the general equation of the plane.
2 What is geometric meaning coefficients at X, y, z in the general equation of the plane?
3 Write the equation of the plane passing through the point M 0 (x 0 ; y 0 ; z 0) perpendicular to the vector = ( BUT; AT; FROM).
4 Write the equation of the plane in segments along the axes and indicate the geometric meaning of the parameters included in it.
5 Write the equation of the plane passing 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 finding the angle between two planes.
7 Write down the conditions for parallelism of two planes.
8 Write down the condition of perpendicularity of two planes.
9 Write down the formula by which the distance from a point to a plane is calculated.
Tasks for independent decision
1 Write an equation for a plane passing through a point M(2; – 1; 1) perpendicular to the vector = (1; – 2; 3). ( Answer: X – 2at + 3z – 7 = 0)
2 Dot R(1; - 2; - 2) is the base of the perpendicular drawn from the origin to the plane. Write an equation for this plane. ( Answer: X – 2at – 2z – 9 = 0)
3 Given two points M 1 (2; – 1; 3) and M 2 (– 1; 2; 4). Write an equation for a plane passing through a point M 1 is perpendicular to the vector . ( Answer: 3X – 3at – z – 6 = 0)
4 Write an equation for a plane passing through three points M 1 (3; – 1; 2), M 2 (4; – 1; – 1), M 3 (2; 0; 2). (Answer: 3X + 3at + z – 8 = 0)
5 M 1 (3; – 1; 2) and M 2 (2; 1; 3) parallel to the vector = (3; - 1; 4). ( Answer: 9X + 7at – 5z – 10 = 0)
6 Write an equation for a plane passing through a point M 1 (2; 3; – 4) parallel to the vectors = (3; 1; – 1) and = (1; – 2; 1). ( Answer: X + at + 7z + 14 = 0)
7 Write an equation for a plane passing through a point M(1; – 1; 1) perpendicular to planes 2 X – at + z– 1 = 0 and X + 2at – z + 1 = 0. (Answer: X – 3at – 5z + 1 = 0)
8 Write an equation for a plane passing through the points M 1 (1; 0; 1) and M 2 (1; 2; – 3) perpendicular to the plane X – at + z – 1 = 0. (Answer: X + 2at + z – 2 = 0)
9 Find angle between planes 4 X – 5at + 3z– 1 = 0 and X – 4at – z + 9 = 0. (Answer: φ = arccos0.7)
10 Find distance from point M(2; – 1; – 1) up to plane 16 X – 12at + 15z – 4 = 0. (Answer: d = 1)
11 Find the point of intersection of three planes 5 X + 8at – z – 7 = 0, X + 2at + 3z – 1 = 0, 2X – 3at + 2z – 9 = 0. (Answer: (3; – 1; 0))
12 Write an equation for a plane that passes through the points M 1 (1; – 2; 6) and M 2 (5; - 4; 2) and cuts off equal segments on the axes Oh and OU. (Answer: 4X + 4at + z – 2 = 0)
13 Find the distance between planes X + 2at – 2z+ 2 = 0 and 3 X + 6at – 6z – 4 = 0. (Answer: d = )
Lecture 5
1. Write an equation for a plane passing through a point M 0 (1, -2, 5) parallel to plane 7 x-y-2z-1=0.
Solution. Denote by R given plane, let R 0 is the desired parallel plane passing through the point M 0 (1, -2, 5).
Consider the normal (perpendicular) vector 
plane R. The coordinates of the normal vector are the coefficients of the variables in the equation of the plane 
.
Because planes R and R 0
are parallel, then the vector 
perpendicular to the plane R 0
, i.e. 
is the normal vector of the plane R 0
.
Equation of a plane passing through a point M 0 (x 0 ,
y 0 , z 0) with normal 
:
Substitute point coordinates M 0 and normal vectors 
into equation (1):
Expanding the brackets, we obtain the general equation of the plane (final answer):
2. Compose canonical and parametric equations of a straight line passing through a point M 0 (-2, 3, 0) parallel to line 
.
Solution. Denote by L given line, let L 0 is the desired parallel line passing through the point M 0 (-2,3,0).
Guide vector 
straight L(a non-zero vector parallel to this line) is also parallel to the line L 0
. Therefore, the vector 
is the direction vector of the line L 0
.
Guide vector coordinates 
are equal to the corresponding denominators in the canonical equations of the given line 
.
Canonical equations of a straight line in space passing through a point M 0 (x 0 , y 0 , z 
{l, m, n}
.
(2)
Substitute point coordinates M 0 and direction vector 
into equation (2) and obtain the canonical equations of the straight line:
.
Parametric equations of a straight line in space passing through a point M 0 (x 0 , y 0 , z 0) parallel to a non-zero vector 
{l, m, n), have the form:
(3)
Substitute point coordinates M 0 and direction vector 
into equations (3) and obtain the parametric equations of the straight line:
3. Find a point 
, symmetrical to the point 
, relative to: a) direct 
b) planes
Solution. a) Compose the equation of the perpendicular plane P projecting a point 
to this line:
To find 
we use the condition of perpendicularity of the given straight line and the projecting plane. Direction vector straight 
perpendicular to the plane vector 
is the normal vector 
to the plane The equation of a plane perpendicular to a given line has the form or 
Let's find the projection R points M straight. Dot R is the point of intersection of the line and the plane, i.e. its coordinates must simultaneously satisfy both the equations of a straight line and the equation of a plane. Let's solve the system:
.
To solve it, we write the equation of a straight line in a parametric form:
Substituting expressions for 
into the plane equation, we get:
From here we find Found coordinates - these are the coordinates of the middle R line segment connecting a point
and a point symmetrical to it 
AT school course geometry, a theorem was formulated.
The midpoints of a segment are half the sums of the corresponding coordinates of its ends.
Finding the coordinates of a point 
from the formulas for the coordinates of the middle of the segment:
We get: So 
.
Solution. b) To find a point symmetrical to a point 
relative to this plane P, drop the perpendicular from the point 
to this plane. Compose the equation of a straight line with a direction vector 
passing through the point
:
Perpendicularity of a line and a plane means that the direction vector of the line is perpendicular to the plane 
. Then the equation of the straight line projecting the point 
on a given plane, has the form:
By solving the equations together 
and 
find the projection R points 
to the plane. To do this, we rewrite the equations of the straight line in a parametric form:
Substitute these values 
into the equation of the plane: Similarly to item a), using the formulas for the coordinates of the middle of the segment, we find the coordinates of the symmetrical point 
:
Those. 
.
4. Write an equation for a plane passing through a) a straight line 
parallel to the vector 
; b) through two intersecting lines
and 
(previously proving that they intersect); c) through two parallel lines 
and 
; d) through a straight line 
and point 
.
Solution. a) Since the given line lies in the desired plane, and the desired plane is parallel to the vector 
, then the normal vector of the plane will be perpendicular to the directing vector of the line 
and vector 
.
Therefore, as a normal vector of the plane, one can choose the cross product of vectors 
and 
:
We get the coordinates of the normal vector of the plane 
.
Let's find a point on the line. Equating the ratios in the canonical equations of the straight line to zero:
,
find 
,
,
. A given line passes through a point 
, therefore, the plane also passes through the point 
. Using the equation of a plane passing through a given point perpendicular to the vector 
, we obtain the equation of the plane , or , or, finally, 
.
Solution. b) Two lines in space can intersect, cross or be parallel. Given straight lines
and 
(4)
are not parallel because their direction vectors 
and 
not collinear: 
.
How to check if lines intersect? It is possible to solve system (4) of 4 equations with 3 unknowns. If the system has a unique solution, then we get the coordinates of the point of intersection of the lines. However, to solve our problem - constructing a plane in which both lines lie, their intersection point is not needed. Therefore, it is possible to formulate the condition for the intersection of two non-parallel lines in space without finding the point of intersection.
If two non-parallel lines intersect, then the direction vectors 
,
and joining the points lying on the lines 
and 
vector lie in the same plane, i.e. coplanar the mixed product of these vectors is equal to zero:
. (5)
We equate the ratios in the canonical equations of lines to zero (or to 1 or any number)
and 
,
and find the coordinates of the points on the lines. The first line passes through the point 
, and the second straight line through the point 
. The direction vectors of these lines are respectively equal to 
and 
. We get
Equality (5) is satisfied, therefore, the given lines intersect. This means that there is only one plane passing through these two lines.
Let's move on to the second part of the problem - drawing up the equation of the plane.
As a normal vector of the plane, you can choose the cross product of their direction vectors 
and 
:
Plane normal vector coordinates 
.
We have found that the direct 
goes through 
, therefore, the desired plane also passes through this point. We get the equation of the plane, or 
or, finally, 
.
c) Since the lines 
and 
are parallel, then the vector product of their direction vectors cannot be chosen as a normal vector, it will be equal to the zero vector.
Determine the coordinates of the points 
and 
through which these lines pass. Let 
and 
, then 
,
. Let's calculate the coordinates of the vector . Vector 
lies in the desired plane and is non-collinear to the vector 
, then as its normal vector 
you can choose the cross product of a vector 
and direction vector of the first line 
:
So, 
.
The plane passes through a straight line 
, so it passes through the point 
. We get the equation of the plane: , or .
d) Equating the ratios in the canonical equations of the straight line to zero 
, we find 
,
,
. Therefore, the line passes through the point 
.
Let's calculate the coordinates of the vector . Vector 
belongs to the desired plane, as its normal vector 
choose the vector product of the directing vector of the straight line 
and vector 
:
Then the plane equation has the form: , or .
This article contains the information necessary to solve the problem of compiling the equation of a plane passing through a given line and a given point. After solving this problem in a general form, we will give detailed solutions of examples for compiling an equation for a plane that passes through a given line and point.
Page navigation.
Finding the equation of a plane passing through a given line and a given point.
Let in three-dimensional space fixed Oxyz , given a line a and a point not lying on the line a . Let's set ourselves the task: to obtain the equation of the plane passing through the line a and the point M 3.
Let us first show that there is a single plane whose equation we want to write.
Recall two axioms:
- through three different points of space that do not lie on one straight line, a single plane passes;
- if two distinct points of a line lie in a certain plane, then all points of this line lie in that plane.
From these statements it follows that through a line and a point not lying on it, one plane can be drawn. Thus, in the problem posed by us, a single plane passes through the line a and the point M 3 , and we need to write the equation of this plane.
Now let's start finding the equation of the plane passing through the given line a and the point .
If the line a is given by specifying the coordinates of two different points M 1 and M 2 lying on it, then our task is to find the equation of the plane passing through three given points M 1 , M 2 and M 3 .
If the line a is given differently, then we first have to find the coordinates of two points M 1 and M 2 lying on the line a, and after that write the equation of the plane passing through the three points M 1, M 2 and M 3, which will be the desired equation of the plane passing through the line a and the point M 3 .
Let's figure out how to find the coordinates of two different points M 1 and M 2 lying on a given line a.
In a rectangular coordinate system in space, any straight line corresponds to some equations of a straight line in space. We assume that the method of specifying the line a in the condition of the problem allows us to obtain its parametric equations of the line in the space of the form 
. Then, assuming , we have a point 
, lying on the line a . By giving the parameter a non-zero real value, from the parametric equations of the line a we can calculate the coordinates of the point M 2 , which also lies on the line a and is different from the point M 1 .
After that, we will only have to write the equation of the plane passing through three different and not lying on one straight point and , in the form 
.
So, we have obtained the equation of a plane passing through a given line a and a given point M 3 that does not lie on the line a.
Examples of compiling the equation of a plane passing through a given point and a straight line.
Let us show the solutions of several examples, in which we will analyze the considered method for finding the equation of a plane passing through a given line and a given point.
Let's start with the simplest case.
Example.
Solution.
Take two different points on the coordinate line Ox, for example, and .
Now we get the equation of a plane passing through three points M 1, M 2 and M 3: 
This equation is the desired general equation of the plane passing through the given line Ox and the point 
.
Answer:
.
If it is known that the plane passes through a given point and a given straight line, and it is required to write the equation of the plane in segments or the normal equation of the plane, then you should first obtain the general equation of the given plane, and from it proceed to the equation of the plane of the required form.
Example.
Write the normal equation for a plane passing through a straight line. 
and point 
.
Solution.
First, we write the general equation for a given plane. To do this, we find the coordinates of two different points lying on a straight line . The parametric equations of this line have the form 
. Let the point M 1 correspond to the value, and the point M 2 -. We calculate the coordinates of the points M 1 and M 2: 
Now we can write the general equation of a straight line passing through a point and direct :
It remains to obtain the required form of the plane equation by multiplying both parts of the resulting equation by the normalizing factor 
.
Answer:
.
So, finding the equation of a plane passing through a given point and a given straight line rests on finding the coordinates of two different points lying on a given straight line. This is often the main difficulty in solving such problems. In conclusion, we will analyze the solution of the example for compiling the equation of a plane passing through a given point and a straight line, which is determined by the equations of two intersecting planes.
Example.
In a rectangular coordinate system Oxyz given a point and a line a , which is the line of intersection of two planes 
and 
. Write the equation of the plane passing through the line a and the point M 3 .
With the help of this online calculator one can find the equation of a plane passing through a given point and parallel to the given plane. A detailed solution with explanations is given. To find the equation of the plane, enter the coordinates of the point and the coefficients of the equation of the plane into the cells and click on the "Solve" button.
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Equation of a plane passing through a given point and parallel to a given plane - theory, examples and solutions
Let a point be given M 0 (x 0 , y 0 , z 0) and the plane equation
All parallel planes have collinear normal vectors. Therefore, to construct a plane parallel to (1) passing through the point M 0 (x 0 , y 0 , z 0) you need to take as the normal vector of the desired plane, the normal vector n=(A, B, C) plane (1). Next, you need to find such a value D, at which the point M 0 (x 0 , y 0 , z 0) satisfied the plane equation (1):
Substituting value D from (3) to (1), we get:
Equation (5) is the equation of a plane passing through the point M 0 (x 0 , y 0 , z 0) and parallel to the plane (1).
Find the equation of a plane passing through a point M 0 (1, −6, 2) and parallel to the plane:
Substituting point coordinates M 0 and the coordinates of the normal vector in (3), we get.
Consider a plane Q in space. Its position is completely determined by specifying a vector N perpendicular to this plane and some fixed point lying in the plane Q. The vector N perpendicular to the plane Q 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 the given point and having the given normal vector . To do this, consider a vector connecting a point with an arbitrary point of the plane Q (Fig. 81).
For any position of the point M on the plane Q, the MXM vector is perpendicular to the normal vector N of the plane Q. Therefore, the scalar product Let's write the scalar product in terms of projections. Since , and vector , then
and hence
We have shown that the coordinates of any point of the Q plane satisfy equation (4). It is easy to see that the coordinates of points that do not lie on the plane Q do not satisfy this equation (in the latter case, ). Therefore, we have obtained the required equation of the plane Q. Equation (4) is called the equation of the plane passing through the given point. It is of the first degree relative to the current coordinates
So, we have shown that any 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 bunch of planes. Equation (4), in which the coefficients A, B and C can take on any values, is called the equation of a bundle of planes.
Example 2. Write an equation for a plane passing through three points, (Fig. 82).
Solution. Let us write the equation for a bunch of planes passing through a point
