In space, a plane can be defined 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 various forms. Also, subject to certain conditions, planes can be parallel, perpendicular, intersecting, etc. We will talk about this in this article. We will learn how to make a general equation of the plane and not only.
Normal view of the equation
Suppose there is a space R 3 that has a rectangular coordinate system XYZ. We define the vector α, which will be released from the starting point O. Through the end of the vector α, we draw a plane P that is perpendicular to it.
We denote on an arbitrary point Q = (x, y, z). We write the radius vector of the point Q by the letter p. Moreover, the length of the vector α is equal to p = IαI and Ʋ = (cosα, cosβ, cosγ).
This is a unit vector that is directed to the side, as is the vector α. α, β, and γ are the angles that are formed between the vector Ʋ and the positive directions of the axes of the space x, y, z, respectively. The projection of any point QϵP onto the vector Ʋ is a constant value, which is equal to p: (p, Ʋ) = p (p≥0).
The indicated equation makes sense when p = 0. The only plane 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, despite its direction, which means that the vector Ʋ is defined with accuracy to the mark. The previous equation is the equation of our plane P, expressed in vector form. But in the coordinates, its appearance will be like this:
P here is greater than or equal to 0. We found the equation of the plane in space in normal form.
General equation
If we multiply the equation in coordinates by any number that is not equal to zero, we obtain an equation equivalent to this one defining the same plane. It will look like this:
Here A, B, C are numbers that are nonzero at the same time. This equation is referred to as a general plane equation.
Equations of the planes. Special cases
The equation in general form can be modified in the presence of additional conditions. Let's consider some of them.
Suppose that coefficient A is 0. This means that this plane is parallel to a given axis Ox. In this case, the form of the equation will change: Wu + 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 axis Oy.
- Secondly, if C = 0, then the equation is transformed into Ax + Vy + D = 0, which will indicate parallelism to the given axis Oz.
- Thirdly, if D = 0, the equation will look like Ax + Vy + Cz = 0, which will mean that the plane intersects O (origin).
- Fourth, if A = B = 0, then the equation will change to Cz + D = 0, which will prove parallelism to Oxy.
- Fifth, if B = C = 0, then the equation will become Ax + D = 0, which means that the plane to Oyz is parallel.
- Sixth, if A = C = 0, then the equation will take the form Vu + 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 nonzero, the form of equation (0) can be as follows:
x / a + y / b + z / s = 1,
in which a = -D / A, b = -D / B, c = -D / C.
We get the equation of the plane in segments as a result . It should be noted that this plane will intersect the Ox axis at the point with coordinates (a, 0,0), Oy - (0, b, 0), and Oz - (0,0, s).
Taking into account the equation x / a + y / b + z / c = 1, it is easy to visually visualize the location of the plane relative to a given coordinate system.
Normal vector coordinates
The normal vector n to the plane has coordinates that are the coefficients of the general equation of this plane, that is, n (A, B, C).
In order to determine the coordinates of the normal n, it is enough 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 using the general equation, we can write the coordinates of any normal vector of a given plane: (1 / a + 1 / b + 1 / with).
It is worth noting that a normal vector helps to solve a variety of problems. The most common are the tasks of proving the perpendicularity or parallelism of the planes, the problems of finding angles between planes or angles between planes and lines.
View of the plane equation according to the coordinates of the point and the normal vector
A nonzero vector n, perpendicular to a given plane, is called normal (normal) for a given plane.
Assume that in the coordinate space (rectangular coordinate system) Oxyz given:
- point Mₒ with coordinates (xₒ, yₒ, zₒ);
- zero vector n = A * i + B * j + C * k.
It is necessary to draw up an equation for the 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 (xy, 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 a point Mₒ (xₒ, yₒ, zₒ) - rₒ = xₒ * i + yₒ * j + zₒ * k. A point M will belong to a given plane if the vector MₒM is perpendicular to the vector n. We write the condition of orthogonality using the scalar product:
[MₒM, n] = 0.
Since ₒ = r – rₒ, the vector equation of the plane will look like this:
[r - rₒ, n] = 0.
This equation may have another form. For this, the properties of the scalar product are used, and the left side of the equation is transformed. [r - rₒ, n] = [r, n] - [rₒ, n]. If [rₒ, n] is denoted as c, then the following equation is obtained: [r, n] - c = 0 or [r, n] = c, which expresses the constancy of the projections onto the normal vector of radius vectors of given points that belong to the plane.
Now we can get the coordinate form of the vector equation of our plane [r - rₒ, n] = 0. Since r – r r = ( – ₒ) * i + ( – ) * j + (z – zₒ) * k, and n = A * i + B * j + C * k, we have:
It turns out that we have the equation of a plane passing through a point perpendicular to the normal n:
A * (x-xₒ) + B * (y – yₒ) * (z – zₒ) = 0.
View of the plane equation according to the coordinates of two points and a vector, collinear 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 draw up 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 a given vector a.
Moreover, the vectors M′M = {x-x ′; y-y ′; zz ′} and M ″ M = {x ″ -x ′; y ″ -y ′; z ″ -z ′} must be coplanar with the vector = ( ′, ″, ‴), which means that (′, ″ , ) = 0.
So, our equation of a plane in space will look like this:
View of the equation of a plane intersecting three points
Suppose we have three points: (x ′, y ′, z ′), (x ″, y ″, z ″), (x ‴, y ‴, z ‴) that do not belong to one line. It is necessary to write the equation of the plane passing through the given three points. The theory of geometry claims that such a plane really exists, only it is the one and only. Since this plane intersects the point (x, y, z), the form of its equation will be as follows:
Here A, B, C are non-zero at the same time. Also, the given plane intersects two more points: (x ″, y ″, z ″) and (x ‴, y ‴, z ‴). In this regard, such conditions must be met:
Now we can compose a homogeneous system of equations (linear) with unknowns u, v, w:
In our case, x, y, or z is an arbitrary point that satisfies equation (1). Given equation (1) and the system of equations (2) and (3), the system of equations shown in the figure above satisfies the vector N (A, B, C), which is nontrivial. That is why the determinant of this system is zero.
Equation (1), which we obtained, is the equation of the plane. Through 3 points, it definitely passes, and this is easy to verify. To do this, we need to expand our determinant according to the elements in the first line. It follows from the existing properties of the determinant that our plane simultaneously intersects the three initially given points (x, y, z), (x, y, z,), (x, y, z). That is, we have solved the task set before us.
The dihedral angle between the planes
The dihedral angle is a spatial geometric figure formed by two half-planes that come from one straight line. In other words, this is the part of space that is limited by these half-planes.
Suppose 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 located between these planes. The scalar product has the form:
NN¹ = | N || N¹ | cos φ,
precisely because
cosφ = NN¹ / | N || N¹ | = (AA¹ + BB¹ + CC¹) / ((√ (² + ² + ²)) * (√ (¹) ² + (¹) ² + (¹) ²)).
It is enough to take into account that 0≤φ≤π.
In fact, two planes that intersect form two angles (dihedral): φ 1 and φ 2 . Their sum is 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 we replace A, B and C in equation (0) 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
Perpendicular are called planes between which the angle is 90 degrees. Using the material described above, we can find the equation of a plane perpendicular to the other. Suppose we have two planes: Ax + Vu + Cz + D = 0 and A¹x + B¹u + 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 called two planes that do not contain common points.
The condition for parallel planes (their equations are the same as in the previous paragraph) is that the vectors N and N¹, which are perpendicular to them, are collinear. And this means that the following proportionality conditions are satisfied:
/ ¹ = / ¹ = / ¹.
If the proportionality conditions are extended - A / ¹ = / ¹ = / ¹ = DD¹,
this indicates that these planes are the same. And this means that the equations Ax + Vy + Cz + D = 0 and ¹ + ¹ + ¹z + D¹ = 0 describe one plane.
Distance to plane from point
Suppose we have a plane,, which is given by equation (0). It is necessary to find the distance from the point with coordinates (xₒ, yₒ, zₒ) = Qₒ to it. 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 P that was released from the zero point, v is the unit vector that is located in the direction a.
The difference ρ-ρº of the radius vector of some point Q = (x, y, z) belonging to , as well as the radius vector of a given point Q 0 = (xₒ, yₒ, zₒ) is a vector whose absolute value of the projection onto v is equal to the distance d, which must be found from Q 0 = (xₒ, yₒ, zₒ) to :
D = | (ρ-ρ 0 , v) |, but
(ρ-ρ 0 , v) = (ρ, v) - (ρ 0 , v) = p– (ρ 0 , v).
So it turns out
d = | (ρ 0 , v) -p |.
Now it is clear that in order to calculate the distance d from Q 0 to the plane , you need to use the normal form of the plane equation, at the same time transfer to the left side p, and substitute (xₒ, ₒ, zₒ) instead of x, y, z.
Thus, we will find the absolute value of the resulting expression, that is, the desired d.
Using the parameter language, we get the obvious:
d = | Axₒ + Wuₒ + Czₒ | / √ (² + ² + ²).
If the given point Q 0 is located on the other side of the plane , as is the origin, then there is an obtuse angle between the vector ρ-ρ 0 and v , therefore:
d = - (ρ-ρ 0 , v) = (ρ 0 , v) -> 0.
In the case when the point Q 0 together with the origin is located on the same side from P, the angle created is sharp, that is:
d = (ρ-ρ 0 , v) = p - (ρ 0 , v)> 0.
As a result, it turns out that in the first case (ρ 0 , v)> p, in the second (ρ 0 , v) <p.
Tangent plane and its equation
A plane tangent to the surface at the point of tangency º is a plane containing all possible tangents to the curves drawn through this point on the surface.
With this kind of surface equation F (x, y, z) = 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) (zzº) = 0.
If you set the surface in explicit form z = f (x, y), then the tangent plane will be described by the equation:
zzº = f (º, º) (- º) + f (º, º) (-º).
The intersection of two planes
In three-dimensional space , the coordinate system (rectangular) Oxyz is located, two planes P ′ and P ″ are given, which intersect and do not coincide. Since any plane located in a rectangular coordinate system is determined by the general equation, we assume that P ′ and P ″ are given by the equations A′x + B′u + C′z + D ′ = 0 and A ″ x + B ″ y + C ″ z + D ″ = 0. In this case, we have the normal n ′ (A ′, B ′, C ′) of the plane P ′ and the normal n ″ (A ″, B ″, C ″) of the plane P ″. Since our planes are not parallel and do not coincide, these vectors are not collinear. Using the language of mathematics, we can write down 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 which case a = P ′ ∩ P ″.
a is a line consisting of the set of all points of the (common) planes P ′ and P ″. This means that the coordinates of any point on 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. Therefore, the coordinates of the point will be a particular solution to the following system of equations:
As a result, it turns out that the solution (general) of this system of equations will determine the coordinates of each of the points of the line, which will be the intersection point of P ′ and P ″, and determine the line a in the coordinate system Oxyz (rectangular) in space.