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
In this way, 
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 + Vy + 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.
It can be specified in different ways (one point and a vector, two points and a vector, three points, etc.). It is with this in mind that the equation of the plane can have different forms. Also, under certain conditions, the planes can be parallel, perpendicular, intersecting, etc. We will talk about this in this article. We will learn how to write the general equation of the plane and not only.
Normal form of the equation
Let's say there is a space R 3 that has a rectangular coordinate system XYZ. We set the vector α, which will be released from the initial point O. Through the end of the vector α we draw the plane P, which will be perpendicular to it.
Denote by P an arbitrary point Q=(x, y, z). We will sign the radius vector of the point Q with the letter p. The length of the vector α is p=IαI and Ʋ=(cosα,cosβ,cosγ).
This is a unit vector that points sideways, just like the vector α. α, β and γ are the angles that form between the vector Ʋ and the positive directions of the space axes x, y, z, respectively. The projection of some point QϵП onto the vector Ʋ is a constant value equal to р: (р,Ʋ) = р(р≥0).
This equation makes sense when p=0. The only thing is that the plane P in this case will intersect the point O (α=0), which is the origin, and the unit vector Ʋ released from the point O will be perpendicular to P, regardless of its direction, which means that the vector Ʋ is determined from sign-accurate. The previous equation is the equation of our P plane, expressed in vector form. But in coordinates it will look like this:
P here is greater than or equal to 0. We have found the equation of a plane in space in its normal form.
General Equation
If we multiply the equation in coordinates by any number that is not equal to zero, we get an equation equivalent to the given one, which determines that same plane. It will look like this:
Here A, B, C are numbers that are simultaneously different from zero. This equation is referred to as the general plane equation.
Plane equations. Special cases
The equation in general form can be modified in the presence of additional conditions. Let's consider some of them.
Assume that the coefficient A is 0. This means that the given plane is parallel to the given axis Ox. In this case, the form of the equation will change: Ву+Cz+D=0.
Similarly, the form of the equation will change under the following conditions:
- Firstly, if B = 0, then the equation will change to Ax + Cz + D = 0, which will indicate parallelism to the Oy axis.
- Secondly, if С=0, then the equation is transformed into Ах+Ву+D=0, which will indicate parallelism to the given axis Oz.
- Thirdly, if D=0, the equation will look like Ax+By+Cz=0, which will mean that the plane intersects O (the origin).
- Fourth, if A=B=0, then the equation will change to Cz+D=0, which will prove parallel to Oxy.
- Fifth, if B=C=0, then the equation becomes Ax+D=0, which means that the plane to Oyz is parallel.
- Sixth, if A=C=0, then the equation will take the form Ву+D=0, that is, it will report parallelism to Oxz.
Type of equation in segments
In the case when the numbers A, B, C, D are non-zero, the form of equation (0) can be as follows:
x/a + y/b + z/c = 1,
in which a \u003d -D / A, b \u003d -D / B, c \u003d -D / C.
We get as a result It is worth noting that this plane will intersect the Ox axis at a point with coordinates (a,0,0), Oy - (0,b,0), and Oz - (0,0,c).
Taking into account the equation x/a + y/b + z/c = 1, it is easy to visually represent the placement of the plane relative to a given coordinate system.
Normal vector coordinates
The normal vector n to the plane P has coordinates that are the coefficients of the general equation of the given plane, that is, n (A, B, C).
In order to determine the coordinates of the normal n, it is sufficient to know the general equation of a given plane.
When using the equation in segments, which has the form x/a + y/b + z/c = 1, as well as when using the general equation, one can write the coordinates of any normal vector of a given plane: (1/a + 1/b + 1/ With).
It should be noted that the normal vector helps to solve various problems. The most common are tasks that consist in proving the perpendicularity or parallelism of planes, problems in finding angles between planes or angles between planes and lines.
View of the equation of the plane according to the coordinates of the point and the normal vector
A non-zero vector n perpendicular to a given plane is called normal (normal) for a given plane.
Suppose that in the coordinate space (rectangular coordinate system) Oxyz are given:
- point Mₒ with coordinates (xₒ,yₒ,zₒ);
- zero vector n=A*i+B*j+C*k.
It is necessary to compose an equation for a plane that will pass through the point Mₒ perpendicular to the normal n.
In space, we choose any arbitrary point and denote it by M (x y, z). Let the radius vector of any point M (x, y, z) be r=x*i+y*j+z*k, and the radius vector of the point Mₒ (xₒ,yₒ,zₒ) - rₒ=xₒ*i+yₒ *j+zₒ*k. The point M will belong to the given plane if the vector MₒM is perpendicular to the vector n. We write the orthogonality condition using the scalar product:
[MₒM, n] = 0.
Since MₒM \u003d r-rₒ, the vector equation of the plane will look like this:
This equation can take another form. To do this, the properties of the scalar product are used, and the left side of the equation is transformed. = - . If denoted as c, then the following equation will be obtained: - c \u003d 0 or \u003d c, which expresses the constancy of the projections onto the normal vector of the radius vectors of the given points that belong to the plane.
Now you can get the coordinate form of writing the vector equation of our plane = 0. Since r-rₒ = (x-xₒ)*i + (y-yₒ)*j + (z-zₒ)*k, and n = A*i+B *j+C*k, we have:
It turns out that we have an equation for a plane passing through a point perpendicular to the normal n:
A*(x-xₒ)+B*(y-yₒ)C*(z-zₒ)=0.
View of the plane equation according to the coordinates of two points and a vector collinear to the plane
We define two arbitrary points M′ (x′,y′,z′) and M″ (x″,y″,z″), as well as the vector a (a′,a″,a‴).
Now we can compose an equation for a given plane, which will pass through the available points M′ and M″, as well as any point M with coordinates (x, y, z) parallel to the given vector a.
In this case, the vectors M′M=(x-x′;y-y′;z-z′) and M″M=(x″-x′;y″-y′;z″-z′) must be coplanar with the vector a=(a′,a″,a‴), which means that (M′M, M″M, a)=0.
So, our equation of a plane in space will look like this:
Type of the equation of a plane intersecting three points
Suppose we have three points: (x′, y′, z′), (x″,y″,z″), (x‴,y‴,z‴), which do not belong to the same straight line. It is necessary to write the equation of the plane passing through the given three points. The theory of geometry claims that this kind of plane really exists, only it is the only one and inimitable. Since this plane intersects the point (x′, y′, z′), the form of its equation will be as follows:
Here A, B, C are different from zero at the same time. Also, the given plane intersects two more points: (x″,y″,z″) and (x‴,y‴,z‴). In this regard, the following conditions must be met:
Now we can compose a homogeneous system with unknowns u, v, w:
In our case, x, y or z is an arbitrary point that satisfies equation (1). Taking into account the equation (1) and the system of equations (2) and (3), the system of equations indicated in the figure above satisfies the vector N (A, B, C), which is non-trivial. That is why the determinant of this system is equal to zero.
Equation (1), which we have obtained, is the equation of the plane. It passes exactly through 3 points, and this is easy to check. To do this, we need to expand our determinant over the elements in the first row. It follows from the existing properties of the determinant that our plane simultaneously intersects three initially given points (x′, y′, z′), (x″,y″,z″), (x‴,y‴,z‴). That is, we have solved the task set before us.
Dihedral angle between planes
A dihedral angle is a spatial geometric figure formed by two half-planes that emanate from one straight line. In other words, this is the part of space that is limited by these half-planes.
Let's say we have two planes with the following equations:
We know that the vectors N=(A,B,C) and N¹=(A¹,B¹,C¹) are perpendicular according to the given planes. In this regard, the angle φ between the vectors N and N¹ is equal to the angle (dihedral), which is between these planes. The scalar product has the form:
NN¹=|N||N¹|cos φ,
precisely because
cosφ= NN¹/|N||N¹|=(AA¹+BB¹+CC¹)/((√(A²+B²+C²))*(√(A¹)²+(B¹)²+(C¹)²)).
It suffices to take into account that 0≤φ≤π.
In fact, two planes that intersect form two (dihedral) angles: φ 1 and φ 2 . Their sum is equal to π (φ 1 + φ 2 = π). As for their cosines, their absolute values are equal, but they differ in signs, that is, cos φ 1 =-cos φ 2. If in equation (0) we replace A, B and C with the numbers -A, -B and -C, respectively, then the equation that we get will determine the same plane, the only angle φ in the equation cos φ= NN 1 /| N||N 1 | will be replaced by π-φ.
Perpendicular plane equation
Planes are called perpendicular if the angle between them is 90 degrees. Using the material outlined above, we can find the equation of a plane perpendicular to another. Let's say we have two planes: Ax+By+Cz+D=0 and A¹x+B¹y+C¹z+D=0. We can state that they will be perpendicular if cosφ=0. This means that NN¹=AA¹+BB¹+CC¹=0.
Parallel plane equation
Parallel are two planes that do not contain common points.
The condition (their equations are the same as in the previous paragraph) is that the vectors N and N¹, which are perpendicular to them, are collinear. This means that the following conditions of proportionality are satisfied:
A/A¹=B/B¹=C/C¹.
If the proportionality conditions are extended - A/A¹=B/B¹=C/C¹=DD¹,
this indicates that these planes coincide. This means that the equations Ax+By+Cz+D=0 and A¹x+B¹y+C¹z+D¹=0 describe one plane.
Distance to plane from point
Let's say we have a plane P, which is given by equation (0). It is necessary to find the distance to it from the point with coordinates (xₒ,yₒ,zₒ)=Qₒ. To do this, you need to bring the equation of the plane P into normal form:
(ρ,v)=p (p≥0).
In this case, ρ(x,y,z) is the radius vector of our point Q located on P, p is the length of the perpendicular to P that was released from the zero point, v is the unit vector that is located in the a direction.
The difference ρ-ρº of the radius vector of some point Q \u003d (x, y, z) belonging to P, as well as the radius vector of a given point Q 0 \u003d (xₒ, yₒ, zₒ) is such a vector, the absolute value of the projection of which on v is equal to the distance d, which must be found from Q 0 \u003d (xₒ, yₒ, zₒ) to P:
D=|(ρ-ρ 0 ,v)|, but
(ρ-ρ 0 ,v)= (ρ,v)-(ρ 0 ,v) =р-(ρ 0 ,v).
So it turns out
d=|(ρ 0 ,v)-p|.
Thus, we will find the absolute value of the resulting expression, that is, the desired d.
Using the language of parameters, we get the obvious:
d=|Axₒ+Vuₒ+Czₒ|/√(A²+B²+C²).
If the given point Q 0 is on the other side of the plane P, as well as the origin, then between the vector ρ-ρ 0 and v is therefore:
d=-(ρ-ρ 0 ,v)=(ρ 0 ,v)-p>0.
In the case when the point Q 0, together with the origin, is located on the same side of P, then the angle created is acute, that is:
d \u003d (ρ-ρ 0, v) \u003d p - (ρ 0, v)>0.
As a result, it turns out that in the first case (ρ 0 ,v)> р, in the second (ρ 0 ,v)<р.
Tangent plane and its equation
The tangent plane to the surface at the point of contact Mº is the plane containing all possible tangents to the curves drawn through this point on the surface.
With this form of the surface equation F (x, y, z) \u003d 0, the equation of the tangent plane at the tangent point Mº (xº, yº, zº) will look like this:
F x (xº, yº, zº)(x- xº)+ F x (xº, yº, zº)(y-yº)+ F x (xº, yº, zº)(z-zº)=0.
If you specify the surface in explicit form z=f (x, y), then the tangent plane will be described by the equation:
z-zº = f(xº, yº)(x- xº)+f(xº, yº)(y-yº).
Intersection of two planes
In the coordinate system (rectangular) Oxyz is located, two planes П′ and П″ are given, which intersect and do not coincide. Since any plane located in a rectangular coordinate system is determined by the general equation, we will assume that P′ and P″ are given by the equations A′x+B′y+C′z+D′=0 and A″x+B″y+ С″z+D″=0. In this case, we have the normal n′ (A′, B′, C′) of the P′ plane and the normal n″ (A″, B″, C″) of the P″ plane. Since our planes are not parallel and do not coincide, these vectors are not collinear. Using the language of mathematics, we can write this condition as follows: n′≠ n″ ↔ (A′, B′, C′) ≠ (λ*A″,λ*B″,λ*C″), λϵR. Let the line that lies at the intersection of P′ and P″ be denoted by the letter a, in this case a = P′ ∩ P″.
a is a straight line consisting of the set of all points of (common) planes П′ and П″. This means that the coordinates of any point belonging to the line a must simultaneously satisfy the equations A′x+B′y+C′z+D′=0 and A″x+B″y+C″z+D″=0. This means that the coordinates of the point will be a particular solution of the following system of equations:
As a result, it turns out that the (general) solution of this system of equations will determine the coordinates of each of the points of the straight line, which will act as the intersection point of П′ and П″, and determine the straight line a in the coordinate system Oxyz (rectangular) in space.
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
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 plane n → (1 , 0 , 2) and the normal vector 3 x + 2 y + 1 = 0 of the plane x + 2 z - 1 = 0 .
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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