In the space plane can be defined in different ways (by one point and a vector and the vector of two points, three points, etc.).It is in this equation of the plane may have various kinds.Also, under certain conditions the plane can be parallel, perpendicular, intersecting, etc.On this and talk in this article.We will learn to make the overall equation of the plane and not only.
Normal equation
Suppose there is a space R3, which has a rectangular coordinate system XYZ.We define the vector α, which will be released from the initial point A. Through the end of the vector α draw the plane P, which is perpendicular to it.
Let P on an arbitrary point Q = (x, y, z).The radius vector of point Q sign the letter p.The length of the vector α is equal to p = IαI and Ʋ = (cosα, cosβ, cosγ).
It is a unit vector, which is directed to the side, as well as vector α.α, β, and γ - is the angle formed between the vector Ʋ and positive directions of the axes of the space x, y, z, respectively.The projection of a point on the vector Ʋ QεP is a constant, which is equal to p (p, Ʋ) = p (r≥0).
The above equation makes sense, when p = 0.The only plane P in this case will intersect point D (α = 0), which is the origin, and unit vector Ʋ, released from the point O will be perpendicular to P, despite its direction, which means that the vector Ʋ determinedup to sign.Previous equation is our plane II, expressed in vector form.But in the coordinates of its kind to be so:
P is greater than or equal to 0. We have found the equation of the plane in space in a normal way.
general equation
If the equation in the coordinates multiply any number that is not equal to zero, we obtain the equation equivalent to this that defines the very plane.It will have a view:
Here A, B, C - is the number at the same time different from zero.This equation is referred to as the plane equation of the general form.
equation of the plane.Particular cases
equation in general form can be modified with additional conditions.Consider some of them.
assume that the coefficient A is equal to 0. This means that the plane is parallel to a given axis Ox.In this case, change the form of the equation: Vu + Cz + D = 0.
similar form of the equation will change and under the following conditions:
- First, when B = 0, then the equation changes to Ax + Cz + D = 0 that would indicate parallel to the y-axis.
- Secondly, if C = 0, the equation is transformed into Ax + By + D = 0, there will be talk about parallel to the predetermined axis Oz.
- Third, when D = 0, the equation would look like Ax + By + Cz = 0, which would mean that the plane intersects O (the origin).
- Fourth, if A = B = 0, then the equation changes to Cz + D = 0, which will prove parallel to Oxy.
- Fifth, if B = C = 0, the equation becomes Ax + D = 0, which means that the plane is parallel to Oyz.
- Sixth, if A = C = 0, the equation takes the form Vu + D = 0, then there will be parallel to the report Oxz.
type equations in sections of
In the case where the number of A, B, C, D are different from zero, the form of equation (0) may be as follows:
x / a + y / b + z / a= 1,
wherein a = -D / A, b = -D / B, c = -D / C.
Get a result equation of the plane in pieces.It should be noted that this plane will intersect the axis Ox at coordinates (a, 0,0), Dy - (0, b, 0) and Oz - (0,0, s).
In view of the equation x / a + y / b + z / c = 1, it is easy to visualize the placement of the plane relative to a given coordinate system.
coordinates of the normal vector
normal vector n to the plane P has coordinates, which are the coefficients of the general equation of the plane, ie n (A, B, C).
In order to determine the coordinates of the normal n, is enough to know the general equation of a given plane.
When using equations in segments, which has the form x / a + y / b + z / c = 1, as when using the general equation can be written coordinates of any normal vector a given plane: (1 / a + 1 / b +1 / s).
worth noting that the normal vector helps to solve various problems.The most common are the problems, is a proof of perpendicular or parallel planes, the task of finding the angles between the planes or angles between planes and lines.
view plane equation according to the coordinates of the point and the normal vector
nonzero vector n, perpendicular to a given plane, called normal (normal) for a given plane.
assume that the coordinate space (a rectangular coordinate system) Oxyz asked:
- Mₒ point with coordinates (hₒ, uₒ, zₒ);
- zero vector n = A * i + j + B C * * k.
necessary to make the equation of the plane that passes through the point perpendicular to the normal Mₒ n.In the space
choose any arbitrary point and let her M (x y, z).Let the radius vector of any point M (x, y, z) is r = x * i + y * j + z * k, and the radius vector of the point Mₒ (hₒ, uₒ, zₒ) - rₒ = hₒ * i + uₒ* j + zₒ * k.The point M belongs to a given plane, if the vector is perpendicular to the vector MₒM n.We write the orthogonality condition by means of the scalar product:
[MₒM, n] = 0.
Since MₒM = r-rₒ, vector equation of the plane will look like this:
[r - rₒ, n] = 0.
This equation may have a different shape.For this purpose, the properties of the scalar product, and transformed the left side of the equation.[r - rₒ, n] = [r, n] - [rₒ, n].If [rₒ, n] denoted as s, we obtain the following equation: [r, n] - c = 0 or [r, n] = s, which expresses the consistency of the projections on the normal vector of the radius-vectors of the given points that belong to the plane.
Now you can get the kind of recording coordinate our plane vector equation [r - rₒ, n] = 0. Since r-rₒ = (x-hₒ) * i + (y-uₒ) * j + (z-zₒ) * kand n = A * i + j + B C * * k, we have:
turns out, is formed in our equation of the plane passing through the point perpendicular to the normal n:
A * (x hₒ) + B *(uₒ y) S * (z-zₒ) = 0.
Type of plane equation according to the coordinates of two points and a vector collinear plane
Define two points M '(x', y ', z') and M '(x ", y", z "), as well as vector a(A ', A "and ‴).
Now we can equate a given plane, which will take place through the existing points M 'and M ", as well as any point M with coordinates (x, y, z) parallel to a given vector.
This M'M vectors {x, x ', y, y'; zz '} and M "M = {x" -x', y 'y'; z "-z '} should be coplanarvector a = (a ', a ", a ‴), and that means (M'M, M' M, a) = 0.
So our equation of a plane in space would look like this:
type equation plane intersecting the three points
Suppose we have three points (x ', y', z '), (x', y", z"), (x ‴ have ‴, z ‴), which do not belong to the same line.It is necessary to write the equation of the plane passing through the specified three points.The theory of geometry argues that this kind of plane does exist, it's just one and only.Since this plane intersects the point (x ', y', z '), the form of its equation is as follows:
Here A, B, and C are different from zero at the same time.Also given plane intersects the two points (x ', y', z ') and (x ‴ have ‴, z ‴).In this connection should be carried out this kind of conditions:
Now we can create a uniform system of equations (linear) with unknowns u, v, w:
In our case, x, y, or z appears arbitrary point that satisfiesEquation (1).Considering equation (1) and a system of equations (2) and (3), a system of equations shown in the figure above, the vector satisfies N (A, B, C) which is nontrivial.That's because the determinant of the system is zero.
Equation (1), which we've got, this is the equation of the plane.After 3 point she really goes, and it's easy to check.To do this, we decompose the determinant of the elements located in the first row.Of the existing properties of the determinant it implies that our plane at the same time three crosses initially given points (x ', y', z '), (x', y ', z'), (x ‴ have ‴, z ‴).So we decided to put before us.
dihedral angle between the planes
dihedral angle is a spatial geometric shape formed by two half-planes that come from the same line.In other words, this part of the space, which is limited to the half-plane.
Suppose we have two planes with the following equations:
We know that the vectors N = (A, B, C) and N¹ = (A¹, H¹, S¹) according to the set perpendicular planes.In this regard, the angle φ between the vectors N and N¹ equal angle (dihedral), which is located between these planes.The scalar product is given by:
NN¹ = | N || N¹ | cos φ,
precisely because
cosφ = NN¹ / | N || N¹ | = (+ AA¹ VV¹ SS¹ +) / ((√ (A² + V²s² +)) * (√ (A¹) ² + (H¹) ² + (S¹) ²)).
is enough to consider that 0≤φ≤π.
actually two planes that intersect to form two angles (dihedral): φ1 and φ2.The amount is equal to their π (φ1 + φ2 = π).As for their cosines, their absolute values are equal, but they are different signs, that is, cos φ1 = -cos φ2.If in the equation (0) is replaced by A, B and C of -A, -B and -C respectively, the equation, we obtain, will determine the same plane, only the angle φ in equation cos φ = NN1 / | N|| N1 | will be replaced by π-φ.
equation perpendicular to the plane perpendicular to
called plane, between which the angle is 90 degrees.Using the material presented above, we can find the equation of a plane perpendicular to the other.Suppose we have two planes: Ax + By + Cz + D = 0 and A¹h + + S¹z V¹u + D = 0.We can say that they are perpendicular if cosφ = 0.This means that AA¹ NN¹ = + + VV¹ SS¹ = 0.
equation parallel plane
Parallel called two planes that do not contain common points.
condition of parallel planes (their equations are the same as in the previous paragraph) is that the vectors N and N¹, which to them perpendicular, collinear.This means that the following conditions of proportionality:
A / A¹ = V / H¹ = C / S¹.
If the conditions of proportionality are extended - A / A¹ = V / H¹ = C / S¹ = DD¹,
this indicates that the data plane of the same.This means that the equation Ax + By + Cz + D = 0 and + A¹h V¹u S¹z + + D¹ = 0 describe a single plane.
distance to the plane from the point
Suppose we have a plane P, which is given by Equation (0).It is necessary to find her distance from the point with coordinates (hₒ, uₒ, zₒ) = Qₒ.To do this, you need to bring the equation of the plane P in the normal form:
(ρ, v) = p (r≥0).
In this case, ρ (x, y, z) is the radius vector of our point Q, located on n, P - is the perpendicular distance P which has been discharged from the zero point, v - is the unit vector, which is located in the direction of a.
difference ρ-ρº radius vector of a point Q = (x, y, z), owned by P and the radius vector of a given point Q0 = (hₒ, uₒ, zₒ) is such a vector, the absolute valuewhose projections by v equals the distance d, which is necessary to find from Q0 = (hₒ, uₒ, zₒ) to P:
D = | (ρ-ρ0, v) |, but
(ρ-ρ0, v) = (ρ, v) - (ρ0, v) = p (ρ0, v).
It turns out,
d = | (ρ0, v) p |.
now seen to calculate the distance d from Q0 to the plane P, you must use the normal form of the equation plane, the shift to the left of the river, and the last place of x, y, z substitute (hₒ, uₒ, zₒ).
Thus, we find the absolute value of the resulting expression that is sought d.
Using the language settings, we obtain the obvious:
d = | + Ahₒ Vuₒ + Czₒ | / √ (A² + V² + s²).
If a given point Q0 is on the other side of the plane P as the origin, between the vector ρ-ρ0 and v is an obtuse angle, thus:
d = - (ρ-ρ0, v) = (ρ0, v) -p & gt; 0.
In the case when the point Q0 together with the origin located on the same side of the U, the generated angle is acute, that is:
d = (ρ-ρ0, v) = p - (ρ0, v) & gt;0.
The result is that in the first case (ρ0, v) & gt; p, the second (ρ0, v) & lt; p.
tangent plane and its equation
As for the plane to the surface at the point of contact Mº - a plane containing all possible tangent to the curve drawn through that point on the surface.
In this type of equation of the surface F (x, y, z) = 0 equation of the tangent plane at the tangent point Mº (hº, uº, zº) would look like this:
Fx (hº, uº, zº) (x hº)+ Fx (hº, uº, zº) (uº y) + Fx (hº, uº, zº) (z-zº) = 0.
If you specify explicitly the surface z = f (x, y), the tangent plane is described by the equation:
z-zº = f (hº, uº) (hº x) + f (hº, uº) (y- uº).
intersection of two planes
in three-dimensional space is a coordinate system (rectangular) Oxyz, given two planes P 'and P ", which overlap and are not the same.Since any plane, which is in a rectangular coordinate system is defined by the general equation, we assume that n 'and n "are given by equations A'x + + V'u S'z + D' = 0 and A" x + B "y +With "D + z" = 0.In this case we have normal n '(A', B ', C') of the plane P 'and the normal n' (A ', B', C ') of the plane P ".As our plane are not parallel and do not coincide, these vectors are not collinear.Using the language of mathematics, we have this condition can be written as: n '≠ n "↔ (A', B ', C') ≠ (λ * A", λ * In ", λ * C"), λεR.Let the straight line which lies at the intersection P 'and P ", will be denoted by the letter a, in this case a = n' ∩ P".
a - this is a direct, consisting of a set of points (overall) planes P 'and P ".This means that the coordinates of any point belonging to the line and must simultaneously satisfy the equation A'x + + V'u S'z + D '= 0 and A "x + B" y + C "z + D" = 0.Then, the coordinates of the point will be a particular solution of the following equations:
The result is that the decision (General) of the system of equations will determine the coordinates of each point of the line, which will be the point of intersection P 'and P ", and to determine the direct andin a coordinate system Oxyz (rectangular) space.
