Let us consider a certain plane π and an arbitrary point M 0 in space. Let's choose for the plane unit normal vector n with the beginning at some point M 1 ∈ π, and let p(M 0 ,π) be the distance from the point M 0 to the plane π. Then (Fig. 5.5)
р(М 0 ,π) = | pr n M 1 M 0 | = |nM 1 M 0 |, (5.8)
since |n| = 1.
If the π plane is given in rectangular coordinate system with its general equation Ax + By + Cz + D = 0, then its normal vector is the vector with coordinates (A; B; C) and we can choose
Let (x 0 ; y 0 ; z 0) and (x 1 ; y 1 ; z 1) be the coordinates of the points M 0 and M 1 . Then the equality Ax 1 + By 1 + Cz 1 + D = 0 holds, since the point M 1 belongs to the plane, and the coordinates of the vector M 1 M 0 can be found: M 1 M 0 = (x 0 -x 1; y 0 -y 1 ; z 0 -z 1 ). Recording scalar product nM 1 M 0 in coordinate form and transforming (5.8), we obtain
since Ax 1 + By 1 + Cz 1 = - D. So, to calculate the distance from a point to a plane, you need to substitute the coordinates of the point into the general equation of the plane, and then divide the absolute value of the result by a normalizing factor equal to the length of the corresponding normal vector.
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Finding the distance from a point to a plane is a common problem that arises when solving various problems of analytical geometry; for example, this problem can be reduced to finding the distance between two intersecting straight lines or between a straight line and a plane parallel to it.
Consider the plane $β$ and a point $M_0$ with coordinates $(x_0;y_0; z_0)$ that does not belong to the plane $β$.
Definition 1
The shortest distance between a point and a plane will be the perpendicular drawn from the point $M_0$ to the plane $β$.
Figure 1. Distance from a point to a plane. Avtor24 - online exchange of student works
Below we discuss how to find the distance from a point to a plane using the coordinate method.
Derivation of the formula for the coordinate method of finding the distance from a point to a plane in space
A perpendicular from point $M_0$ intersecting the plane $β$ at point $M_1$ with coordinates $(x_1;y_1; z_1)$ lies on a straight line whose direction vector is the normal vector of the plane $β$. In this case, the length of the unit vector $n$ is equal to one. Accordingly, the distance from $β$ to point $M_0$ will be:
$ρ= |\vec(n) \cdot \vec(M_1M_0)|\left(1\right)$, where $\vec(M_1M_0)$ is the normal vector of the $β$ plane, and $\vec(n)$ is the unit normal vector of the plane under consideration.
In the case when the equation of the plane is given in the general form $Ax+ By + Cz + D=0$, the coordinates of the normal vector of the plane are the coefficients of the equation $\(A;B;C\)$, and the unit normal vector in this case has the coordinates , calculated using the following equation:
$\vec(n)= \frac(\(A;B;C\))(\sqrt(A^2 + B^2 + C^2))\left(2\right)$.
Now we can find the coordinates of the normal vector $\vec(M_1M_0)$:
$\vec(M_0M_1)= \(x_0 – x_1;y_0-y_1;z_0-z_1\)\left(3\right)$.
We also express the coefficient $D$ using the coordinates of a point lying in the $β$ plane:
$D= Ax_1+By_1+Cz_1$
The coordinates of the unit normal vector from equality $(2)$ can be substituted into the equation of the $β$ plane, then we have:
$ρ= \frac(|A(x_0 -x_1) + B(y_0-y_1)+C(z_0-z_1)|)(\sqrt(A^2+B^2+C^2))= \frac( |Ax_0+ By_0 + Cz_0-(Ax_1+By_1+Cz_1)|)(\sqrt(A^2+B^2+C^2)) = \frac(Ax_0+ By_0 + Cz_0 + D)(\sqrt(A^2 +B^2+C^2))\left(4\right)$
Equality $(4)$ is a formula for finding the distance from a point to a plane in space.
General algorithm for finding the distance from point $M_0$ to a plane
- If the equation of the plane is not given in general form, first you need to reduce it to general form.
- After this, it is necessary to express from the general equation of the plane the normal vector of a given plane through the point $M_0$ and a point belonging to a given plane, for this we need to use the equality $(3)$.
- The next stage is searching for the coordinates of the unit normal vector of the plane using the formula $(2)$.
- Finally, you can begin to find the distance from the point to the plane, this is done by calculating the scalar product of the vectors $\vec(n)$ and $\vec(M_1M_0)$.
This article talks about determining the distance from a point to a plane. Let's analyze it using the coordinate method, which will allow us to find the distance from a given point in three-dimensional space. To reinforce this, let’s look at examples of several tasks.
The distance from a point to a plane is found using the known distance from a point to a point, where one of them is given, and the other is a projection onto a given plane.
When a point M 1 with a plane χ is specified in space, then a straight line perpendicular to the plane can be drawn through the point. H 1 is their common point of intersection. From this we obtain that the segment M 1 H 1 is a perpendicular drawn from point M 1 to the plane χ, where point H 1 is the base of the perpendicular.
Definition 1
The distance from a given point to the base of a perpendicular drawn from a given point to a given plane is called.
The definition can be written in different formulations.
Definition 2
Distance from point to plane is the length of the perpendicular drawn from a given point to a given plane.
The distance from point M 1 to the χ plane is determined as follows: the distance from point M 1 to the χ plane will be the smallest from a given point to any point on the plane. If point H 2 is located in the χ plane and is not equal to point H 2, then we obtain a right triangle of the form M 2 H 1 H 2 , which is rectangular, where there is a leg M 2 H 1, M 2 H 2 – hypotenuse. This means that it follows that M 1 H 1< M 1 H 2 . Тогда отрезок М 2 H 1 is considered inclined, which is drawn from point M 1 to the plane χ. We have that the perpendicular drawn from a given point to the plane is less than the inclined one drawn from the point to the given plane. Let's look at this case in the figure below.
Distance from a point to a plane - theory, examples, solutions
There are a number of geometric problems whose solutions must contain the distance from a point to a plane. There may be different ways to identify this. To resolve, use the Pythagorean theorem or similarity of triangles. When, according to the condition, it is necessary to calculate the distance from a point to a plane, given in a rectangular coordinate system of three-dimensional space, it is solved by the coordinate method. This paragraph discusses this method.
According to the conditions of the problem, we have that a point in three-dimensional space with coordinates M 1 (x 1, y 1, z 1) with a plane χ is given; it is necessary to determine the distance from M 1 to the plane χ. Several solution methods are used to solve this problem.
First way
This method is based on finding the distance from a point to a plane using the coordinates of point H 1, which are the base of the perpendicular from point M 1 to the plane χ. Next, you need to calculate the distance between M 1 and H 1.
To solve the problem in the second way, use the normal equation of a given plane.
Second way
By condition, we have that H 1 is the base of the perpendicular, which was lowered from point M 1 to the plane χ. Then we determine the coordinates (x 2, y 2, z 2) of point H 1. The required distance from M 1 to the χ plane is found by the formula M 1 H 1 = (x 2 - x 1) 2 + (y 2 - y 1) 2 + (z 2 - z 1) 2, where M 1 (x 1, y 1, z 1) and H 1 (x 2, y 2, z 2). To solve, you need to know the coordinates of point H 1.
We have that H 1 is the point of intersection of the χ plane with the line a, which passes through the point M 1 located perpendicular to the χ plane. It follows that it is necessary to compile an equation for a straight line passing through a given point perpendicular to a given plane. It is then that we will be able to determine the coordinates of point H 1. It is necessary to calculate the coordinates of the point of intersection of the line and the plane.
Algorithm for finding the distance from a point with coordinates M 1 (x 1, y 1, z 1) to the χ plane:
Definition 3
- draw up an equation of straight line a passing through point M 1 and at the same time
- perpendicular to the χ plane;
- find and calculate the coordinates (x 2 , y 2 , z 2) of point H 1, which are points
- intersection of line a with plane χ;
- calculate the distance from M 1 to χ using the formula M 1 H 1 = (x 2 - x 1) 2 + (y 2 - y 1) 2 + z 2 - z 1 2.
Third way
In a given rectangular coordinate system O x y z there is a plane χ, then we obtain a normal equation of the plane of the form cos α · x + cos β · y + cos γ · z - p = 0. From here we obtain that the distance M 1 H 1 with the point M 1 (x 1 , y 1 , z 1) drawn to the plane χ, calculated by the formula M 1 H 1 = cos α x + cos β y + cos γ z - p . This formula is valid, since it was established thanks to the theorem.
Theorem
If a point M 1 (x 1, y 1, z 1) is given in three-dimensional space, having a normal equation of the plane χ of the form cos α x + cos β y + cos γ z - p = 0, then calculating the distance from the point to plane M 1 H 1 is obtained from the formula M 1 H 1 = cos α · x + cos β · y + cos γ · z - p, since x = x 1, y = y 1, z = z 1.
Proof
The proof of the theorem comes down to finding the distance from a point to a line. From here we get that the distance from M 1 to the χ plane is the modulus of the difference between the numerical projection of the radius vector M 1 with the distance from the origin to the χ plane. Then we get the expression M 1 H 1 = n p n → O M → - p. The normal vector of the plane χ has the form n → = cos α, cos β, cos γ, and its length is equal to one, n p n → O M → is the numerical projection of the vector O M → = (x 1, y 1, z 1) in the direction determined by the vector n → .
Let's apply the formula for calculating scalar vectors. Then we obtain an expression for finding a vector of the form n → , O M → = n → · n p n → O M → = 1 · n p n → O M → = n p n → O M → , since n → = cos α , cos β , cos γ · z and O M → = (x 1 , y 1 , z 1) . The coordinate form of writing will take the form n → , O M → = cos α · x 1 + cos β · y 1 + cos γ · z 1 , then M 1 H 1 = n p n → O M → - p = cos α · x 1 + cos β · y 1 + cos γ · z 1 - p . The theorem has been proven.
From here we get that the distance from the point M 1 (x 1, y 1, z 1) to the plane χ is calculated by substituting cos α · x + cos β · y + cos γ · z - p = 0 into the left side of the normal equation of the plane instead of x, y, z coordinates x 1, y 1 and z 1, relating to point M 1, taking the absolute value of the obtained value.
Let's look at examples of finding the distance from a point with coordinates to a given plane.
Example 1
Calculate the distance from the point with coordinates M 1 (5, - 3, 10) to the plane 2 x - y + 5 z - 3 = 0.
Solution
Let's solve the problem in two ways.
The first method starts with calculating the direction vector of the line a. By condition, we have that the given equation 2 x - y + 5 z - 3 = 0 is a general plane equation, and n → = (2, - 1, 5) is the normal vector of the given plane. It is used as a direction vector of a straight line a, which is perpendicular to a given plane. It is necessary to write down the canonical equation of a line in space passing through M 1 (5, - 3, 10) with a direction vector with coordinates 2, - 1, 5.
The equation will become x - 5 2 = y - (- 3) - 1 = z - 10 5 ⇔ x - 5 2 = y + 3 - 1 = z - 10 5.
Intersection points must be determined. To do this, gently combine the equations into a system to move from the canonical to the equations of two intersecting lines. Let's take this point as H 1. We get that
x - 5 2 = y + 3 - 1 = z - 10 5 ⇔ - 1 · (x - 5) = 2 · (y + 3) 5 · (x - 5) = 2 · (z - 10) 5 · ( y + 3) = - 1 · (z - 10) ⇔ ⇔ x + 2 y + 1 = 0 5 x - 2 z - 5 = 0 5 y + z + 5 = 0 ⇔ x + 2 y + 1 = 0 5 x - 2 z - 5 = 0
After which you need to enable the system
x + 2 y + 1 = 0 5 x - 2 z - 5 = 0 2 x - y + 5 z - 3 = 0 ⇔ x + 2 y = 1 5 x - 2 z = 5 2 x - y + 5 z = 3
Let us turn to the Gaussian system solution rule:
1 2 0 - 1 5 0 - 2 5 2 - 1 5 3 ~ 1 2 0 - 1 0 - 10 - 2 10 0 - 5 5 5 ~ 1 2 0 - 1 0 - 10 - 2 10 0 0 6 0 ⇒ ⇒ z = 0 6 = 0 , y = - 1 10 10 + 2 z = - 1 , x = - 1 - 2 y = 1
We get that H 1 (1, - 1, 0).
We calculate the distance from a given point to the plane. We take points M 1 (5, - 3, 10) and H 1 (1, - 1, 0) and get
M 1 H 1 = (1 - 5) 2 + (- 1 - (- 3)) 2 + (0 - 10) 2 = 2 30
The second solution is to first bring the given equation 2 x - y + 5 z - 3 = 0 to normal form. We determine the normalizing factor and get 1 2 2 + (- 1) 2 + 5 2 = 1 30. From here we derive the equation of the plane 2 30 · x - 1 30 · y + 5 30 · z - 3 30 = 0. The left side of the equation is calculated by substituting x = 5, y = - 3, z = 10, and you need to take the distance from M 1 (5, - 3, 10) to 2 x - y + 5 z - 3 = 0 modulo. We get the expression:
M 1 H 1 = 2 30 5 - 1 30 - 3 + 5 30 10 - 3 30 = 60 30 = 2 30
Answer: 2 30.
When the χ plane is specified by one of the methods in the section on methods for specifying a plane, then you first need to obtain the equation of the χ plane and calculate the required distance using any method.
Example 2
In three-dimensional space, points with coordinates M 1 (5, - 3, 10), A (0, 2, 1), B (2, 6, 1), C (4, 0, - 1) are specified. Calculate the distance from M 1 to plane A B C.
Solution
First you need to write down the equation of the plane passing through the given three points with coordinates M 1 (5, - 3, 10), A (0, 2, 1), B (2, 6, 1), C (4, 0, - 1) .
x - 0 y - 2 z - 1 2 - 0 6 - 2 1 - 1 4 - 0 0 - 2 - 1 - 1 = 0 ⇔ x y - 2 z - 1 2 4 0 4 - 2 - 2 = 0 ⇔ ⇔ - 8 x + 4 y - 20 z + 12 = 0 ⇔ 2 x - y + 5 z - 3 = 0
It follows that the problem has a solution similar to the previous one. This means that the distance from point M 1 to plane A B C has a value of 2 30.
Answer: 2 30.
Finding the distance from a given point on a plane or to a plane to which they are parallel is more convenient by applying the formula M 1 H 1 = cos α · x 1 + cos β · y 1 + cos γ · z 1 - p. From this we obtain that the normal equations of planes are obtained in several steps.
Example 3
Find the distance from a given point with coordinates M 1 (- 3, 2, - 7) to the coordinate plane O x y z and the plane given by the equation 2 y - 5 = 0.
Solution
The coordinate plane O y z corresponds to an equation of the form x = 0. For the O y z plane it is normal. Therefore, it is necessary to substitute the values x = - 3 into the left side of the expression and take the absolute value of the distance from the point with coordinates M 1 (- 3, 2, - 7) to the plane. We get a value equal to - 3 = 3.
After the transformation, the normal equation of the plane 2 y - 5 = 0 will take the form y - 5 2 = 0. Then you can find the required distance from the point with coordinates M 1 (- 3, 2, - 7) to the plane 2 y - 5 = 0. Substituting and calculating, we get 2 - 5 2 = 5 2 - 2.
Answer: The required distance from M 1 (- 3, 2, - 7) to O y z has a value of 3, and to 2 y - 5 = 0 has a value of 5 2 - 2.
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Determining the distance between: 1 - point and plane; 2 - straight and flat; 3 - planes; 4 - crossing straight lines are considered together, since the solution algorithm for all these problems is essentially the same and consists of geometric constructions that need to be performed to determine the distance between a given point A and plane α. If there is any difference, it consists only in the fact that in cases 2 and 3, before starting to solve the problem, you should mark an arbitrary point A on the straight line m (case 2) or plane β (case 3). distances between intersecting straight lines, we first enclose them in parallel planes α and β and then determine the distance between these planes.
Let us consider each of the noted cases of problem solving.
1. Determining the distance between a point and a plane.
The distance from a point to a plane is determined by the length of a perpendicular segment drawn from a point to the plane.
Therefore, the solution to this problem consists of sequentially performing the following graphical operations:
1) from point A we lower the perpendicular to the plane α (Fig. 269);
2) find the point M of intersection of this perpendicular with the plane M = a ∩ α;
3) determine the length of the segment.
If the plane α is in general position, then in order to lower a perpendicular onto this plane, it is necessary to first determine the direction of the horizontal and frontal projections of this plane. Finding the meeting point of this perpendicular with the plane also requires additional geometric constructions.
The solution to the problem is simplified if the plane α occupies a particular position relative to the projection planes. In this case, both the projection of the perpendicular and the finding of the point of its meeting with the plane are carried out without any additional auxiliary constructions.
EXAMPLE 1. Determine the distance from point A to the frontally projecting plane α (Fig. 270).
SOLUTION. Through A" we draw the horizontal projection of the perpendicular l" ⊥ h 0α, and through A" - its frontal projection l" ⊥ f 0α. We mark the point M" = l" ∩ f 0α . Since AM || π 2, then [A" M"] == |AM| = d.
From the example considered, it is clear how simply the problem is solved when the plane occupies a projecting position. Therefore, if a general position plane is specified in the source data, then before proceeding with the solution, the plane should be moved to a position perpendicular to any projection plane.
EXAMPLE 2. Determine the distance from point K to the plane specified by ΔАВС (Fig. 271).
1. We transfer the plane ΔАВС to the projecting position *. To do this, we move from the system xπ 2 /π 1 to x 1 π 3 /π 1: the direction of the new x 1 axis is chosen perpendicular to the horizontal projection of the horizontal plane of the triangle.
2. Project ΔABC onto a new plane π 3 (the ΔABC plane is projected onto π 3, in [ C " 1 B " 1 ]).
3. Project point K onto the same plane (K" → K" 1).
4. Through the point K" 1 we draw (K" 1 M" 1)⊥ the segment [C" 1 B" 1]. The required distance d = |K" 1 M" 1 |
The solution to the problem is simplified if the plane is defined by traces, since there is no need to draw projections of level lines.
EXAMPLE 3. Determine the distance from point K to the plane α, specified by the tracks (Fig. 272).
* The most rational way to transfer the triangle plane to the projecting position is to replace the projection planes, since in this case it is enough to construct only one auxiliary projection.
SOLUTION. We replace the plane π 1 with the plane π 3, for this we draw a new axis x 1 ⊥ f 0α. On h 0α we mark an arbitrary point 1" and determine its new horizontal projection on the plane π 3 (1" 1). Through the points X α 1 (X α 1 = h 0α 1 ∩ x 1) and 1" 1 we draw h 0α 1. We determine the new horizontal projection of the point K → K" 1. From point K" 1 we lower the perpendicular to h 0α 1 and mark the point of its intersection with h 0α 1 - M" 1. The length of the segment K" 1 M" 1 will indicate the required distance.
2. Determining the distance between a straight line and a plane.
The distance between a line and a plane is determined by the length of a perpendicular segment dropped from an arbitrary point on the line to the plane (see Fig. 248).
Therefore, the solution to the problem of determining the distance between straight line m and plane α is no different from the examples discussed in paragraph 1 for determining the distance between a point and a plane (see Fig. 270 ... 272). As a point, you can take any point belonging to line m.
3. Determination of the distance between planes.
The distance between the planes is determined by the size of the perpendicular segment dropped from a point taken on one plane to another plane.
From this definition it follows that the algorithm for solving the problem of finding the distance between planes α and β differs from a similar algorithm for solving the problem of determining the distance between line m and plane α only in that line m must belong to plane α, i.e., in order to determine the distance between planes α and β follows:
1) take a straight line m in the α plane;
2) select an arbitrary point A on line m;
3) from point A, lower the perpendicular l to the plane β;
4) determine point M - the meeting point of the perpendicular l with the plane β;
5) determine the size of the segment.
In practice, it is advisable to use a different solution algorithm, which will differ from the one given only in that, before proceeding with the first step, the planes should be transferred to the projection position.
Including this additional operation in the algorithm simplifies the execution of all other points without exception, which ultimately leads to a simpler solution.
EXAMPLE 1. Determine the distance between planes α and β (Fig. 273).
SOLUTION. We move from the system xπ 2 /π 1 to x 1 π 1 /π 3. With respect to the new plane π 3, the planes α and β occupy a projecting position, therefore the distance between the new frontal traces f 0α 1 and f 0β 1 is the desired one.
In engineering practice, it is often necessary to solve the problem of constructing a plane parallel to a given plane and removed from it at a given distance. Example 2 below illustrates the solution to such a problem.
EXAMPLE 2. It is required to construct projections of a plane β parallel to a given plane α (m || n), if it is known that the distance between them is d (Fig. 274).
1. In the α plane we draw arbitrary horizontal lines h (1, 3) and front lines f (1,2).
2. From point 1 we restore the perpendicular l to the plane α(l" ⊥ h", l" ⊥ f").
3. On the perpendicular l we mark an arbitrary point A.
4. Determine the length of the segment - (the position indicates on the diagram the metrically undistorted direction of the straight line l).
5. Lay out the segment = d on the straight line (1"A 0) from point 1".
6. Mark on the projections l" and l" points B" and B", corresponding to point B 0.
7. Through point B we draw the plane β (h 1 ∩ f 1). To β || α, it is necessary to comply with the condition h 1 || h and f 1 || f.
4. Determining the distance between intersecting lines.
The distance between intersecting lines is determined by the length of the perpendicular enclosed between the parallel planes to which the intersecting lines belong.
In order to draw mutually parallel planes α and β through intersecting straight lines m and f, it is sufficient to draw through point A (A ∈ m) a straight line p parallel to straight line f, and through point B (B ∈ f) a straight line k parallel to straight m . The intersecting lines m and p, f and k define the mutually parallel planes α and β (see Fig. 248, e). The distance between the planes α and β is equal to the required distance between the crossing lines m and f.
Another way can be proposed for determining the distance between intersecting lines, which consists in the fact that, using some method of transforming orthogonal projections, one of the intersecting lines is transferred to the projecting position. In this case, one projection of the line degenerates into a point. The distance between the new projections of crossing lines (point A" 2 and segment C" 2 D" 2) is the required one.
In Fig. 275 shows a solution to the problem of determining the distance between crossing lines a and b, given segments [AB] and [CD]. The solution is performed in the following sequence:
1. Transfer one of the crossing lines (a) to a position parallel to the plane π 3; To do this, move from the system of projection planes xπ 2 /π 1 to the new x 1 π 1 /π 3, the x 1 axis is parallel to the horizontal projection of straight line a. Determine a" 1 [A" 1 B" 1 ] and b" 1.
2. By replacing the plane π 1 with the plane π 4, we translate the straight line
and to position a" 2, perpendicular to the plane π 4 (the new x 2 axis is drawn perpendicular to a" 1).
3. Construct a new horizontal projection of straight line b" 2 - [ C" 2 D" 2 ].
4. The distance from point A" 2 to straight line C" 2 D" 2 (segment (A" 2 M" 2 ] (is the required one.
It should be borne in mind that the transfer of one of the crossing lines to the projecting position is nothing more than the transfer of the planes of parallelism, in which the lines a and b can be enclosed, also to the projecting position.
In fact, by moving line a to a position perpendicular to the plane π 4, we ensure that any plane containing line a is perpendicular to the plane π 4, including the plane α defined by lines a and m (a ∩ m, m || b ). If we now draw a line n, parallel to a and intersecting line b, then we obtain the plane β, which is the second plane of parallelism, which contains the intersecting lines a and b. Since β || α, then β ⊥ π 4 .
