Equation of a plane perpendicular to a given vector. Straight line
This article gives an idea of how to write the equation of a plane passing through a given point in three-dimensional space perpendicular to a given line. Let us analyze the above algorithm using the example of solving typical problems.
Finding the equation of a plane passing through a given point in space perpendicular to a given line
Let a three-dimensional space and a rectangular coordinate system O x y z be given in it. The point M 1 (x 1, y 1, z 1), the straight line a and the plane α passing through the point M 1 perpendicular to the straight line a are also given. It is necessary to write down the equation of the plane α.
Before proceeding to solve this problem, let's recall the geometry theorem from the program for grades 10 - 11, which reads:
Definition 1
A single plane passes through a given point in three-dimensional space and is perpendicular to a given line.
Now consider how to find the equation of this single plane passing through the starting point and perpendicular to the given line.
It is possible to write the general equation of a plane if the coordinates of a point belonging to this plane are known, as well as the coordinates of the normal vector of the plane.
By the condition of the problem, we are given the coordinates x 1, y 1, z 1 of the point M 1 through which the plane α passes. If we determine the coordinates of the normal vector of the plane α, then we will be able to write the desired equation.
The normal vector of the plane α, since it is non-zero and lies on the line a, perpendicular to the plane α, will be any directing vector of the line a. So, the problem of finding the coordinates of the normal vector of the plane α is transformed into the problem of determining the coordinates of the directing vector of the straight line a .
The determination of the coordinates of the directing vector of the straight line a can be carried out by different methods: it depends on the variant of setting the straight line a in the initial conditions. For example, if the line a in the condition of the problem is given by canonical equations of the form
x - x 1 a x = y - y 1 a y = z - z 1 a z
or parametric equations of the form:
x = x 1 + a x λ y = y 1 + a y λ z = z 1 + a z λ
then the directing vector of the straight line will have coordinates a x, a y and a z. In the case when the straight line a is represented by two points M 2 (x 2, y 2, z 2) and M 3 (x 3, y 3, z 3), then the coordinates of the direction vector will be determined as (x3 - x2, y3 - y2 , z3 – z2).
Definition 2
Algorithm for finding the equation of a plane passing through a given point perpendicular to a given line:
Determine the coordinates of the directing vector of the straight line a: a → = (a x, a y, a z) ;
We define the coordinates of the normal vector of the plane α as the coordinates of the directing vector of the straight line a:
n → = (A , B , C) , where A = a x , B = a y , C = a z;
We write the equation of the plane passing through the point M 1 (x 1, y 1, z 1) and having a normal vector n→=(A, B, C) in the form A (x - x 1) + B (y - y 1) + C (z - z 1) = 0. This will be the required equation of a plane that passes through a given point in space and is perpendicular to a given line.
The resulting general equation of the plane: A (x - x 1) + B (y - y 1) + C (z - z 1) \u003d 0 makes it possible to obtain the equation of the plane in segments or the normal equation of the plane.
Let's solve some examples using the algorithm obtained above.
Example 1
A point M 1 (3, - 4, 5) is given, through which the plane passes, and this plane is perpendicular to the coordinate line O z.
Solution
the direction vector of the coordinate line O z will be the coordinate vector k ⇀ = (0 , 0 , 1) . Therefore, the normal vector of the plane has coordinates (0 , 0 , 1) . Let's write the equation of a plane passing through a given point M 1 (3, - 4, 5) whose normal vector has coordinates (0, 0, 1) :
A (x - x 1) + B (y - y 1) + C (z - z 1) = 0 ⇔ ⇔ 0 (x - 3) + 0 (y - (- 4)) + 1 (z - 5) = 0 ⇔ z - 5 = 0
Answer: z - 5 = 0 .
Consider another way to solve this problem:
Example 2
A plane that is perpendicular to the line O z will be given by an incomplete general equation of the plane of the form С z + D = 0 , C ≠ 0 . Let's define the values of C and D: those for which the plane passes through a given point. Substitute the coordinates of this point in the equation C z + D = 0 , we get: C · 5 + D = 0 . Those. numbers, C and D are related by - D C = 5 . Taking C \u003d 1, we get D \u003d - 5.
Substitute these values into the equation C z + D = 0 and obtain the required equation for a plane perpendicular to the line O z and passing through the point M 1 (3, - 4, 5) .
It will look like: z - 5 = 0.
Answer: z - 5 = 0 .
Example 3
Write an equation for a plane passing through the origin and perpendicular to the line x - 3 = y + 1 - 7 = z + 5 2
Solution
Based on the conditions of the problem, it can be argued that the guiding vector of a given straight line can be taken as a normal vector n → of a given plane. Thus: n → = (- 3 , - 7 , 2) . Let's write the equation of a plane passing through the point O (0, 0, 0) and having a normal vector n → \u003d (- 3, - 7, 2) :
3 (x - 0) - 7 (y - 0) + 2 (z - 0) = 0 ⇔ - 3 x - 7 y + 2 z = 0
We have obtained the required equation for the plane passing through the origin perpendicular to the given line.
Answer:- 3x - 7y + 2z = 0
Example 4
Given a rectangular coordinate system O x y z in three-dimensional space, it contains two points A (2 , - 1 , - 2) and B (3 , - 2 , 4) . The plane α passes through the point A perpendicular to the line AB. It is necessary to compose the equation of the plane α in segments.
Solution
The plane α is perpendicular to the line A B, then the vector A B → will be the normal vector of the plane α. The coordinates of this vector are determined as the difference between the corresponding coordinates of points B (3, - 2, 4) and A (2, - 1, - 2):
A B → = (3 - 2 , - 2 - (- 1) , 4 - (- 2)) ⇔ A B → = (1 , - 1 , 6)
The general equation of the plane will be written in the following form:
1 x - 2 - 1 y - (- 1 + 6 (z - (- 2)) = 0 ⇔ x - y + 6 z + 9 = 0
Now we compose the desired equation of the plane in the segments:
x - y + 6 z + 9 = 0 ⇔ x - y + 6 z = - 9 ⇔ x - 9 + y 9 + z - 3 2 = 1
Answer:x - 9 + y 9 + z - 3 2 = 1
It should also be noted that there are problems whose requirement is to write an equation for a plane passing through a given point and perpendicular to two given planes. In general, the solution to this problem is to write an equation for a plane passing through a given point perpendicular to a given line, since two intersecting planes define a straight line.
Example 5
A rectangular coordinate system O x y z is given, in it is a point M 1 (2, 0, - 5) . The equations of two planes 3 x + 2 y + 1 = 0 and x + 2 z - 1 = 0 are also given, which intersect along the straight line a . It is necessary to compose an equation for a plane passing through the point M 1 perpendicular to the line a.
Solution
Let's determine the coordinates of the directing vector of the straight line a . It is perpendicular to both the normal vector n 1 → (3 , 2 , 0) of the n → (1 , 0 , 2) plane and the normal vector 3 x + 2 y + 1 = 0 of the x + 2 z - 1 = 0 plane.
Then the directing vector α → straight line a we take the vector product of vectors n 1 → and n 2 → :
a → = n 1 → × n 2 → = i → j → k → 3 2 0 1 0 2 = 4 i → - 6 j → - 2 k → ⇒ a → = (4 , - 6 , - 2 )
Thus, the vector n → = (4, - 6, - 2) will be the normal vector of the plane perpendicular to the line a. We write the desired equation of the plane:
4 (x - 2) - 6 (y - 0) - 2 (z - (- 5)) = 0 ⇔ 4 x - 6 y - 2 z - 18 = 0 ⇔ ⇔ 2 x - 3 y - z - 9 = 0
Answer: 2 x - 3 y - z - 9 = 0
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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 one straight line.
Consider the points M 1 (x 1, y 1, z 1), M 2 (x 2, y 2, z 2), M 3 (x 3, y 3, z 3) in a common Cartesian coordinate system.
In order for an arbitrary point M(x, y, z) to lie in the same plane as the points M 1 , M 2 , M 3 , the vectors must be coplanar.
(
)
= 0
Thus, 
Equation of a plane passing through three points:
Equation of a plane with respect to two points and a vector collinear to the plane.
Let the points M 1 (x 1, y 1, z 1), M 2 (x 2, y 2, z 2) and the vector 
.
Let us compose the equation of the plane passing through the given points M 1 and M 2 and an arbitrary point M (x, y, z) parallel to the vector 
.
Vectors 
and vector 
must be coplanar, i.e.
(
)
= 0
Plane equation:
Equation of a plane with respect to one point and two vectors,
collinear plane.
Let two vectors be given 
And 
, collinear planes. Then for an arbitrary point M(x, y, z) belonging to the plane, the vectors 
must be coplanar.
Plane equation:
Plane equation by point and normal vector .
Theorem.
If a point M is given in space 0
(X 0
, y 0
,
z 0
), then the equation of the plane passing through the point M 0
perpendicular to the normal vector
(A,
B,
C) looks like:
A(x – x 0 ) + B(y – y 0 ) + C(z – z 0 ) = 0.
Proof.
For an arbitrary point M(x, y, z) belonging to the plane, we compose a vector . Because vector
- the normal vector, then it is perpendicular to the plane, and, therefore, perpendicular to the vector 
. Then the scalar product
=
0
Thus, we obtain the equation of the plane
The theorem has been proven.
Equation of a plane in segments.
If in the general equation Ax + Wu + Cz + D \u003d 0, divide both parts by (-D)
,
replacing 
, we obtain the equation of the plane in segments:
The numbers a, b, c are the intersection points of the plane, respectively, with the x, y, z axes.
Plane equation in vector form.
Where
- radius-vector of the current point M(x, y, z),
A unit vector that has the direction of the perpendicular dropped to the plane from the origin.
, and are the angles formed by this vector with the x, y, z axes.
p is the length of this perpendicular.
In coordinates, this equation has the form:
xcos + ycos + zcos - p = 0.
The distance from a point to a plane.
The distance from an arbitrary point M 0 (x 0, y 0, z 0) to the plane Ax + Vu + Cz + D \u003d 0 is:
Example. Find the equation of the plane, knowing that the point P (4; -3; 12) is the base of the perpendicular dropped from the origin to this plane.
So A = 4/13; B = -3/13; C = 12/13, use the formula:
A(x – x 0 ) + B(y – y 0 ) + C(z – z 0 ) = 0.
Example. Find the equation of a plane passing through two points P(2; 0; -1) and
Q(1; -1; 3) is perpendicular to the plane 3x + 2y - z + 5 = 0.
Normal vector to the plane 3x + 2y - z + 5 = 0 
parallel to the desired plane.
We get:
Example. Find the equation of the plane passing through the points A(2, -1, 4) and
В(3, 2, -1) perpendicular to the plane X + at + 2z – 3 = 0.
The desired plane equation has the form: A x+B y+C z+ D = 0, the normal vector to this plane 
(A, B, C). Vector 
(1, 3, -5) belongs to the plane. The plane given to us, perpendicular to the desired one, has a normal vector 
(1, 1, 2). Because points A and B belong to both planes, and the planes are mutually perpendicular, then
So the normal vector 
(11, -7, -2). Because point A belongs to the desired plane, then its coordinates must satisfy the equation of this plane, i.e. 112 + 71 - 24 + D= 0; D= -21.
In total, we get the equation of the plane: 11 x - 7y – 2z – 21 = 0.
Example. Find the equation of the plane, knowing that the point P(4, -3, 12) is the base of the perpendicular dropped from the origin to this plane.
Finding the coordinates of the normal vector 
= (4, -3, 12). The desired equation of the plane has the form: 4 x
– 3y
+ 12z+ D = 0. To find the coefficient D, we substitute the coordinates of the point Р into the equation:
16 + 9 + 144 + D = 0
In total, we get the desired equation: 4 x – 3y + 12z – 169 = 0
Example. Given the coordinates of the pyramid vertices A 1 (1; 0; 3), A 2 (2; -1; 3), A 3 (2; 1; 1),
Find the length of the edge A 1 A 2 .
Find the angle between the edges A 1 A 2 and A 1 A 4.
Find the angle between the edge A 1 A 4 and the face A 1 A 2 A 3 .
First, find the normal vector to the face A 1 A 2 A 3 
as a cross product of vectors 
And 
.
=
(2-1;
1-0; 1-3) = (1; 1; -2);
Find the angle between the normal vector and the vector 
.
-4
– 4 = -8.
The desired angle between the vector and the plane will be equal to = 90 0 - .
Find the area of face A 1 A 2 A 3 .
Find the volume of the pyramid.
Find the equation of the plane А 1 А 2 А 3 .
We use the formula for the equation of a plane passing through three points.
2x + 2y + 2z - 8 = 0
x + y + z - 4 = 0;
When using the PC version of “ Course of higher mathematics” you can run a program that will solve the above example for any coordinates of the pyramid vertices.
Double-click the icon to launch the program:
In the program window that opens, enter the coordinates of the pyramid vertices and press Enter. Thus, all decision points can be obtained one by one.
Note: To run the program, you must have Maple ( Waterloo Maple Inc.) installed on your computer, any version starting with MapleV Release 4.
To get the general equation of the plane, we analyze the plane passing through a given point.
Let there be three coordinate axes already known to us in space - Ox, Oy And Oz. Hold the sheet of paper so that it remains flat. The plane will be the sheet itself and its continuation in all directions.
Let P arbitrary plane in space. Any vector perpendicular to it is called normal vector to this plane. Naturally, we are talking about a non-zero vector.
If any point of the plane is known P and some vector of the normal to it, then by these two conditions the plane in space is completely determined(through a given point, there is only one plane perpendicular to a given vector). The general equation of the plane will look like:
So, there are conditions that set the equation of the plane. To get it self plane equation, which has the above form, we take on the plane P arbitrary point M with variable coordinates x, y, z. This point belongs to the plane only if vector perpendicular to the vector(Fig. 1). For this, according to the condition of perpendicularity of vectors, it is necessary and sufficient that the scalar product of these vectors be equal to zero, that is
The vector is given by condition. We find the coordinates of the vector by the formula 
:
.
Now, using the dot product formula of vectors 
, we express the scalar product in coordinate form:
Since the point M(x; y; z) is chosen arbitrarily on the plane, then the last equation is satisfied by the coordinates of any point lying on the plane P. For point N, not lying on a given plane, , i.e. equality (1) is violated.
Example 1 Write an equation for a plane passing through a point and perpendicular to a vector.
Solution. We use formula (1), look at it again:
In this formula, the numbers A , B And C vector coordinates and numbers x0 , y0 And z0 - point coordinates.
The calculations are very simple: we substitute these numbers into the formula and get
We multiply everything that needs to be multiplied and add up just numbers (which are without letters). Result:
.
The required equation of the plane in this example turned out to be expressed by the general equation of the first degree with respect to variable coordinates x, y, z arbitrary point of the plane.
So, an equation of the form
called the general equation of the plane .
Example 2 Construct in a rectangular Cartesian coordinate system the plane given by the equation 
.
Solution. To construct a plane, it is necessary and sufficient to know any three of its points that do not lie on one straight line, for example, the points of intersection of the plane with the coordinate axes.
How to find these points? To find the point of intersection with the axis Oz, you need to substitute zeros instead of x and y in the equation given in the problem statement: x = y= 0 . Therefore, we get z= 6 . Thus, the given plane intersects the axis Oz at the point A(0; 0; 6) .
In the same way, we find the point of intersection of the plane with the axis Oy. At x = z= 0 we get y= −3 , that is, a point B(0; −3; 0) .
And finally, we find the point of intersection of our plane with the axis Ox. At y = z= 0 we get x= 2 , that is, a point C(2; 0; 0) . According to the three points obtained in our solution A(0; 0; 6) , B(0; −3; 0) and C(2; 0; 0) we build the given plane.
Consider now special cases of the general equation of the plane. These are cases when certain coefficients of equation (2) vanish.
1. When D= 0 equation 
defines a plane passing through the origin, since the coordinates of a point 0
(0; 0; 0) satisfy this equation.
2. When A= 0 equation 
defines a plane parallel to the axis Ox, since the normal vector of this plane is perpendicular to the axis Ox(its projection on the axis Ox equals zero). Similarly, when B= 0 plane 
axis parallel Oy, and when C= 0 plane 
parallel to axis Oz.
3. When A=D= 0 equation defines a plane passing through the axis Ox because it is parallel to the axis Ox (A=D= 0). Similarly, the plane passes through the axis Oy, and the plane through the axis Oz.
4. When A=B= 0 equation defines a plane parallel to the coordinate plane xOy because it is parallel to the axes Ox (A= 0) and Oy (B= 0). Likewise, the plane is parallel to the plane yOz, and the plane - the plane xOz.
5. When A=B=D= 0 equation (or z= 0) defines the coordinate plane xOy, since it is parallel to the plane xOy (A=B= 0) and passes through the origin ( D= 0). Similarly, the equation y= 0 in space defines the coordinate plane xOz, and the equation x= 0 - coordinate plane yOz.
Example 3 Compose the equation of the plane P passing through the axis Oy and point .
Solution. So the plane passes through the axis Oy. So in her equation y= 0 and this equation has the form . To determine the coefficients A And C we use the fact that the point belongs to the plane P .
Therefore, among its coordinates there are those that can be substituted into the equation of the plane, which we have already derived (). Let's look at the coordinates of the point again:
M0 (2; −4; 3) .
Among them x = 2 , z= 3 . We substitute them into the general equation and get the equation for our particular case:
2A + 3C = 0 .
We leave 2 A on the left side of the equation, we transfer 3 C to the right side and get
A = −1,5C .
Substituting the found value A into the equation , we get
or .
This is the equation required in the example condition.
Solve the problem on the equations of the plane yourself, and then look at the solution
Example 4 Determine the plane (or planes if more than one) with respect to the coordinate axes or coordinate planes if the plane(s) is given by the equation .
Solutions to typical problems that occur in tests - in the manual "Problems on a plane: parallelism, perpendicularity, intersection of three planes at one point" .
Equation of a plane passing through three points
As already mentioned, a necessary and sufficient condition for constructing a plane, in addition to one point and a normal vector, are also three points that do not lie on one straight line.
Let there be given three different points , and , not lying on the same straight line. Since these three points do not lie on one straight line, the vectors and are not collinear, and therefore any point of the plane lies in the same plane with the points , and if and only if the vectors , and 
coplanar, i.e. if and only if the mixed product of these vectors equals zero.
Using the mixed product expression in coordinates, we obtain the plane equation
(3)
After expanding the determinant, this equation becomes an equation of the form (2), i.e. the general equation of the plane.
Example 5 Write an equation for a plane passing through three given points that do not lie on a straight line:
and to determine a particular case of the general equation of the line, if any.
Solution. According to formula (3) we have:
Normal equation of the plane. Distance from point to plane
The normal equation of a plane is its equation, written in the form