Equation of the plane: how to compose? Types of plane equations

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 different types. Also, if certain conditions are met, the planes can be parallel, perpendicular, intersecting, etc. We'll talk about this in this article. We will learn how to make a general equation of the plane and not only.

The normal form of equation

Suppose there is a space R ₃ that has a rectangular coordinate system XYZ. Define the vector α, which will be released from the initial point O. Through the end of the vector α draw the plane Π, which will be perpendicular to it.

We denote by Π an arbitrary point Q = (x, y, z). We will write the radius vector of the point Q by the letter p. In this case, the length of the vector α equals p = IαI and Ʋ = (cosα, cosβ, cosγ).

It is a unit vector that is directed to the side, like 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 some point QεP onto the vector Ʋ is a constant that is equal to p: (p, Ʋ) = p (p≥0).

This equation makes sense when p = 0. The only 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 Π, despite its direction, which means that the vector Ʋ is defined with Accuracy to the sign. The previous equation is the equation of our plane II, expressed in vector form. But in the coordinates of his look like this:

P is greater than or equal to 0. We have found the equation of a plane in space in the normal form.

The general equation

If the equation in the coordinates is multiplied by any number that is not equal to zero, we obtain an equation equivalent to a given one, which determines the same plane. It will look like this:

Here A, B, C are numbers that are simultaneously nonzero. This equation is referred to as the equation of a general plane.

Equations of planes. Special cases

The equation in general form can be modified under the presence of additional conditions. Let's consider some of them.

Suppose 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: Boo + Cz + D = 0.

Similarly, the form of the equation will change under the following conditions:

• First, if B = 0, then the equation will change to Ax + Cz + D = 0, which will be evidence of parallelism to the Oy axis.
• Second, if C = 0, then the equation is transformed into Ax + Boo + D = 0, which will speak of parallelism to the given axis Oz.
• Third, if D = 0, the equation will look like Ax + Boo + Cz = 0, which means that the plane intersects O (the origin).
• Fourthly, 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 Boo + D = 0, that is, it will report the parallelism to Oxz.

Type of equation in segments

In the case when the numbers A, B, C, D are different from zero, the form of equation (0) can be as follows:

X / a + y / b + z / c = 1,

In which a = -D / A, b = -D / B, c = -D / C.

As a result, we obtain the equation of the plane in the segments. 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, c).

Taking into account the equation x / a + y / b + z / c = 1, it is not difficult to visually represent the placement of the plane with respect to a given coordinate system.

Coordinates of the normal vector

The normal vector n to the plane Π 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 the given plane.

Using the equation in segments, which has the form x / a + y / b + z / c = 1, as with the general equation, we can write the coordinates of any normal vector of the given plane: (1 / a + 1 / b + 1 / from).

It is worth noting that a normal vector helps to solve a variety of tasks. The most common problems include the problem of proving the perpendicularity or parallelism of planes, the problem of finding angles between planes or the angles between planes and lines.

The form of the equation of the plane according to the coordinates of the point and the normal vector

A nonzero vector n perpendicular to a given plane is called the normal (normal) for a given plane.

Suppose that in the coordinate space (rectangular coordinate system) Oxyz are given:

• Point Mₒ with coordinates (xₒ, yₒ, z);
• The zero vector is n = A * i + B * j + C * k.

It is necessary to compose the equation of the plane, which will pass through the point Mₒ perpendicular to the normal n.

In the 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 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. Let us write down the orthogonality condition by means of the scalar product:

[MₒM, n] = 0.

Since MₒM = r-rₒ, the vector equation of the plane will look like this:

[R - rₒ, n] = 0.

This equation can have another form. To do this, we use the properties of the scalar product, 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 obtain the coordinate form of the record of the vector equation of our plane [r - rₒ, n] = 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 the equation of a plane passing through a point perpendicular to the normal n:

A * (x - xₒ) + B * (y - yₒ) C * (z-zₒ) = 0.

The form of the equation of the plane according to the coordinates of two points and a vector, the collinear plane

We define two arbitrary points M '(x', y ', z') and M "(x", y ", z"), and also the vector a (a ', a ", a).

Now we can compose the equation of the given plane, which will pass through the available points M 'and M ", and also any point M with the coordinates (x, y, z) parallel to the given vector a.

In addition, the vectors M'M = {x-x '; y-y'; zz '} and M "M = {x" -x'; y "-y '; z" -z'} must be coplanar with the vector A = (a ', a ", a), and this means that (M'M, M" M, a) = 0.

So, our equation of a plane in space will look like this:

The form 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 the same line. It is necessary to write the equation of the plane passing through the given three points. The theory of geometry asserts that such a plane does exist, but only it is unique and unrepeatable. Since this plane intersects the point (x ', y', z '), the form of its equation will be as follows:

Here A, B, C are both nonzero. Also, the given plane intersects two more points: (x ", y", z ") and (x ‴, y ‴, z ‴). In this connection, such conditions must be fulfilled:

Now we can form 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). Taking into account equation (1) and the system from equations (2) and (3), the system of equations indicated 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, this is the equation of the plane. After 3 points, it goes exactly, and it's easy to check. To do this, we need to expand our determinant by the elements in the first row. From the existing properties of the determinant it follows 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 two-sided angle between the planes

The two-sided corner represents a spatial geometric figure formed by two half-planes that emanate from one straight line. In other words, this is part of the space that is confined to these half-planes.

Suppose we have two planes with the following equations:

We know that the vectors N = (A, B, C) and N¹ = (А¹, В¹, С¹) are perpendicular according to the given planes. In this connection, the angle φ between the vectors N and N¹ is equal to the angle (two-sided) that lies between these planes. The scalar product has the form:

NN¹ = | N || N¹ | cos φ,

Precisely because

Cosφ = NN¹ / | N || N¹ | = (АА¹ + ВВ¹ + СС¹) / ((√ (А² + В² + С²)) * (√ (А¹) ² + (В¹) ² + (С¹) ²)).

It suffices to take into account that 0≤φ≤π.

In fact, two planes that intersect form two angles (two-sided): 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 sign, that is cos φ ₁ = -cos φ ₂ . If we replace A, B and C by the numbers -A, -B and -C, respectively, in equation (0), then the equation that we get will determine this same plane, the only one, φ in the equation cos φ = NN ¹ / | N || N ¹ | Will be replaced by π-φ.

The equation of the perpendicular plane

Perpendicular are the 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 + Boo + Cz + D = 0 and A¹x + Bуy + Czz + D = 0. We can say that they will be perpendicular if cosφ = 0. This means that NN¹ = AA + + BB¹ + CC¹ = 0.

Equation of a parallel plane

Parallel are two planes that do not contain common points.

The condition of parallelism of 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:

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 + Boo + Cz + D = 0 and A¹x + Bуy + Czz + D¹ = 0 describe one plane.

Distance to the plane from the point

Suppose we have a plane Π, which is given by equation (0). It is necessary to find before it the distance from the point with the coordinates (xₒ, yₒ, zₒ) = Q. To do this, we need to reduce the equation of the plane Π to the normal form:

(Ρ, v) = p (p≥0).

In this case, ρ (x, y, z) is the radius vector of our point Q located on II, 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 of a.

The difference ρ - ρ of the radius vector of any point Q = (x, y, z) belonging to Π, and also the radius vector of the given point Q ₀ = (xₒ, yₒ, zₒ) is a vector whose absolute projection on V is equal to the distance d, which must be found from Q ₀ = (xₒ, yₒ, zₒ) to Π:

D = | (ρ-ρ ⁰ , v) |, but

(Ρ-ρ ⁰ , v) = (ρ, v) - (ρ ⁰ , v) = ρ - (ρ ⁰ , v).

So it turns out,

D = | (ρ ⁰ , v) -p |.

Now we see that in order to calculate the distance d from Q ₀ to the plane II, we must use the normal form of the equation of the plane, then transfer it to the left-hand side of p, and substitute (xp, yp, zp) instead of x, y, z.

Thus, we 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 II, like the origin, then between the vector ρ-ρ ⁰ and v there is an obtuse angle, therefore:

D = - (ρ-ρ ⁰ , v) = (ρ ⁰ , v) -p> 0.

In the case where the point Q ₀ together with the origin of coordinates is located on the same side of II, then the created angle is acute, that is:

D = (ρ-ρ ⁰ , v) = ρ - (ρ ⁰ , v)> 0.

As a result, it turns out that in the first case (ρ ⁰ , v)> p, in the second case (ρ ⁰ , v)

The tangent plane and its equation

The plane tangent to the surface at the point of tangency M0 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) = 0, the equation of the tangent plane at the tangent point M0 (x, y, z0) will look like this:

Fx ( _x °, yo, z0) (x - x0) + Fx (x0, y0, z0) (y - y0) + Fx (x0, y0, z0) (z-z0) = 0.

If we define the surface in the explicit form z = f (x, y), then the tangent plane will be described by the equation:

Z - z0 = f (x0, y0) (x - x0) + f (x0, y0) (y - y0).

Intersection of two planes

In three-dimensional space the coordinate system (rectangular) Oxyz is located, two planes П 'and П "are given, which intersect and do not coincide. Since any plane in a rectangular coordinate system is defined by a general equation, we assume that Π 'and Π "are given by the equations A'x + B'y + C'z + D' = 0 and A" x + B "y + With "z + D" = 0. In this case we have the normal n '(A', B ', C') of the plane II 'and the normal n "(A", B ", C") of the plane II ". 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 П 'and П "be denoted by a, in which case a = П' ∩ П".

A is a line consisting of the set of all points of (common) planes II 'and II ". 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. Hence, the coordinates of the point will be a particular solution of the following system of equations:

As a result, it turns out that the solution (common) 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 P 'and P ", and determine the straight line a in the coordinate system Oxyz (rectangular) in space.