Determining the distance between a point and a plane, a line and a plane, between planes and intersecting lines. Determining the distance from a point to a plane
Let's consider the algorithm for solving problem No. 3.
1. From a given point P, draw a perpendicular t to the plane α (plane α is the plane of the figure constructed in problem No. 1); (·)PÎt; t ^ α (see example 5.1).
2. Determine the point of intersection (point T) of the perpendicular with the plane α; t ∩ α = (·) T (see example 5.2).
3. Determine the actual value │PT│ of the distance from point P to the plane (see example 5.3).
Let us consider in more detail each point of the above algorithm using the following examples.
Example 5.1. From point P, draw a perpendicular t to the plane α, defined by three points α (ABC), (Fig. 5.1).
From the theorem on the perpendicularity of a line and a plane, it is known that if a line t ^ α, then on the diagram its horizontal projection t 1 is perpendicular to the projection of the horizontal plane of the same name, that is, t 1 ^ h 1, and its frontal projection t 2 is perpendicular to the frontal projection of the same name, then there is t 2 ^ f 2 . Therefore, solving the problem must begin by constructing horizontal and frontal plane α, if they are not included in the given plane. In this case, it is necessary to remember that the construction of any horizontal must begin with a frontal projection, since the frontal projection h 2 of the horizontal h is always parallel to the OX axis (h 2 ││OX). And the construction of any frontal begins with a horizontal projection f 1 of the frontal f, which should be parallel to the OX axis (f 1 ││OX). So, in Fig. 5.1, through point C the horizontal line C-1 is drawn (C 2 -1 2; C 1 -1 1), and through point A the frontal line A-2 is drawn (A 1 -2 1; A 2 -2 2). The frontal projection t 2 of the desired perpendicular t passes through the point P 2 perpendicular to A 2 -2 2, and the horizontal projection t 1 passes through the point P 1 perpendicular to C 1 -1 1.
Example 5.2. Determine the point of intersection of the perpendicular t with the plane α (that is, determine the base of the perpendicular).
Let the plane α be defined by two intersecting lines α (h ∩ f). The straight line t is perpendicular to the plane α, since t 1 ^ f 1, and
t 2 ^ f 2 . In order to find the base of a perpendicular, it is necessary to carry out the following constructions:
1. tÎb (b – auxiliary projection plane). If b is a horizontally projecting plane, then its degenerate horizontal projection (horizontal trace b 1) coincides with the horizontal projection t 1 of straight line t, that is, b 1 ≡t 1. If b is a frontally projecting plane, then its degenerate frontal projection (frontal trace b 2) coincides with the frontal projection t 2 of straight line t, that is, b 2 ≡ t 2. IN in this example a frontal projection plane was used (see Fig. 5.2).
2. α ∩ b = 1-2 – line of intersection of two planes;
3. determine the point T - the base of the perpendicular; (·)T= t ∩ 1-2.
Example 5.3. Determine the distance from point P to the plane.
The distance from point P to the plane is determined by the length of the perpendicular segment PT. The straight line PT occupies a general position in space, therefore the order of determination natural size section, see pages 7, 8 (Fig. 3.4 and 3.5).
Diagram solution of problem No. 3 by determining the distance from point P to a flat figure, namely to the plane of a square constructed according to given conditions*, is shown in Fig. 5.3. It should be recalled that the projections of point P must be constructed according to the given coordinates (see the version of your assignment).
6. TASK OPTIONS AND EXAMPLE OF WORK PERFORMANCE
The conditions of the tasks and the coordinates of the points are given in Table 6.1.
TASK OPTIONS 148
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St. Petersburg State Marine Technical University
Department of Computer Graphics and Information Support
LESSON 4
PRACTICAL TASK No. 4
Plane.
Determining the distance from a point to a plane.
1. Determining the distance from a point to the projecting plane.
In order to find the actual distance from a point to a plane, you need to:
· from a point, lower a perpendicular to a plane;
· find the point of intersection of the drawn perpendicular with the plane;
· determine the actual size of a segment, the beginning of which is the given point, and the end is the found intersection point.
A plane can occupy space general And private position. Under private refers to the position at which the plane perpendicular to the projection plane - such a plane is called projecting. The main feature of the projecting position: a plane is perpendicular to the projection plane if it passes through the projecting line. In this case, one of the projections of the plane is a straight line - it is called following the plane.
If the plane is projecting, then it is easy to determine the actual distance from the point to the plane. Let's show this using the example of determining the distance from a point IN to the frontally projecting plane specified next Q2 on the plane P2(Fig. 1).
Plane Q is perpendicular to the frontal plane of projections, therefore, any line perpendicular to it will be parallel to the plane P2. And then a right angle to the plane P2 will be projected without distortion, and it is possible from the point B2 draw perpendicular to the trace Q2 . Segment VK is in a particular position in which the frontal projection V2K2 equal to the true value of the desired distance.
Fig.1. Determining the distance from a point to the projecting plane.
2. Determination of the distance from a point to a general plane.
If the plane occupies a general position, then it is necessary to transfer it to the projecting position. To do this, a straight line of a particular position is drawn in it (parallel to one of the projection planes), which can be transferred to the projecting position using one drawing transformation.
Straight line parallel to the plane P1, is called the horizontal plane and is denoted by the letter h. Straight line parallel to the frontal plane of projections P2, is called the frontal of the plane and is denoted by the letter f.Lines h And f are called main lines of the plane. The solution to the problem is shown in the following example (Fig. 2).
Initial condition: triangle ABC defines the plane. M- a point outside the plane. A given plane occupies a general position. To move it to the projecting position, perform the following steps. Enable mode ORTO (ORTHO), use command Segment (Line) – draw any horizontal line intersecting the frontal projection of the triangle А2В2С2 at two points. The projection of the horizontal line passing through these points is indicated h2 . Next, a horizontal projection is constructed h1 .
Main line h can be transformed into a projecting position in which the given plane also becomes projecting. To do this, it is necessary to rotate the horizontal projections of all points (auxiliary quadrilateral ABCM) to a new position at which the line h1 will take vertical position, perpendicular to the axis X. It is convenient to perform these constructions using plane-parallel transfer (a copy of the projection is placed on a free space on the screen).
As a result, the new frontal projection of the plane will look like a straight line (plane trace) A2*B2*. Now from the point M2* you can draw a perpendicular to the trace of the plane. New frontal projection M2*K2* = MK those. is the required distance from the point M to a given plane ABC.
Next, it is necessary to construct distance projections in the initial condition. To do this from the point M1 draw a segment perpendicular to the line h1 , and on it should be postponed from the point M1 a segment equal in size M1*K1*. To construct a frontal projection of a point K2 from point K1 a vertical communication line is drawn, and from the point K2* horizontal. The result of the constructions is shown in Fig. 2.
TASK No. 4. Find the true distance from a point M to the plane defined by the triangle ABC. Give the answer in mm. (Table 1)
Table 1
|
Option |
Point A |
Point B |
||||
|
Option |
Point C |
Point M |
||||
Checking and testing completed TASK No. 4.
Instructions
To find the distance from points to plane using descriptive methods: select on plane arbitrary point; draw two straight lines through it (lying in this plane); restore perpendicular to plane passing through this point (construct a line perpendicular to both intersecting lines at the same time); draw a straight line parallel to the constructed perpendicular through a given point; find the distance between the point of intersection of this line with the plane and the given point.
If the position points given by its three-dimensional coordinates, and the position plane – linear equation, then to find the distance from plane to points, use the methods of analytical geometry: indicate the coordinates points through x, y, z, respectively (x – abscissa, y – ordinate, z – applicate); denote by A, B, C, D the equations plane(A – parameter at abscissa, B – at , C – at applicate, D – free term); calculate the distance from points to plane according to the formula:s = | (Ax+By+Cz+D)/√(A²+B²+C²) |,where s is the distance between the point and the plane,|| - absolute value (or module).
Example. Find the distance between point A with coordinates (2, 3, -1) and the plane given by the equation: 7x-6y-6z+20=0. Solution. From the conditions it follows that: x=2,y=3,z =-1,A=7,B=-6,C=-6,D=20. Substitute these values into the above. You get: s = | (7*2+(-6)*3+(-6)*(-1)+20)/√(7²+(-6)²+(-6)²) | = | (14-18+6+20)/11 | = 2.Answer: Distance from points to plane equals 2 (arbitrary units).
Tip 2: How to determine the distance from a point to a plane
Determining the distance from points to plane- one of the common tasks of school planimetry. As is known, the smallest distance from points to plane there will be a perpendicular drawn from this points to this plane. Therefore, the length of this perpendicular is taken as the distance from points to plane.
You will need
- plane equation
Instructions
Let the first of the parallel f1 be given by the equation y=kx+b1. Translating the expression into general view, you get kx-y+b1=0, that is, A=k, B=-1. The normal to it will be n=(k, -1).
Now follows an arbitrary abscissa of the point x1 on f1. Then its ordinate is y1=kx1+b1.
Let the equation of the second of the parallel lines f2 be of the form:
y=kx+b2 (1),
where k is the same for both lines, due to their parallelism.
Next, you need to create the canonical equation of a line perpendicular to both f2 and f1, containing the point M (x1, y1). In this case, it is assumed that x0=x1, y0=y1, S=(k, -1). As a result, you should get the following equality:
(x-x1)/k =(y-kx1-b1)/(-1) (2).
Having solved the system of equations consisting of expressions (1) and (2), you will find the second point that determines the required distance between the parallel ones N(x2, y2). The required distance itself will be equal to d=|MN|=((x2-x1)^2+(y2-y1)^2)^1/2.
Example. Let the equations of given parallel lines on the plane f1 – y=2x +1 (1);
f2 – y=2x+5 (2). Take an arbitrary point x1=1 on f1. Then y1=3. The first point will thus have coordinates M (1,3). General perpendicular equation (3):
(x-1)/2 = -y+3 or y=-(1/2)x+5/2.
Substituting this y value into (1), you get:
-(1/2)x+5/2=2x+5, (5/2)x=-5/2, x2=-1, y2=-(1/2)(-1) +5/2= 3.
The second base of the perpendicular is at the point with coordinates N (-1, 3). The distance between parallel lines will be:
d=|MN|=((3-1)^2+(3+1)^2)^1/2=(4+16)^1/2=4.47.
Sources:
- Development of athletics in Russia
Top of any flat or volumetric geometric figure uniquely determined by its coordinates in space. In the same way, any arbitrary point in the same coordinate system can be uniquely determined, and this makes it possible to calculate the distance between this arbitrary point and the vertex of the figure.
You will need
- - paper;
- - pen or pencil;
- - calculator.
Instructions
Reduce the problem to finding the length of a segment between two points, if the coordinates of the point specified in the problem and the vertices of the geometric figure are known. This length can be calculated using the Pythagorean theorem in relation to the projections of a segment on the coordinate axis - it will be equal to square root from the sum of the squares of the lengths of all projections. For example, let point A(X₁;Y₁;Z₁) and vertex C of any geometric figure with coordinates (X₂;Y₂;Z₂) be given in a three-dimensional coordinate system. Then the lengths of the projections of the segment between them onto the coordinate axes can be as X₁-X₂, Y₁-Y₂ and Z₁-Z₂, and the length of the segment as √((X₁-X₂)²+(Y₁-Y₂)²+(Z₁-Z₂)² ). For example, if the coordinates of the point are A(5;9;1), and the vertices are C(7;8;10), then the distance between them will be equal to √((5-7)²+(9-8)²+(1- 10)²) = √(-2²+1²+(-9)²) = √(4+1+81) = √86 ≈ 9.274.
First calculate the coordinates of the vertex if they are not explicitly presented in the problem conditions. The specific method depends on the type of figure and known additional parameters. For example, if the three-dimensional coordinates of three vertices A(X₁;Y₁;Z₁), B(X₂;Y₂;Z₂) and C(X₃;Y₃;Z₃) are known, then the coordinates of its fourth vertex (opposite to vertex B) will be (X₃+X₂ -X₁;Y₃+Y₂-Y₁; Z₃+Z₂-Z₁). After determining the coordinates of the missing vertex, calculating the distance between it and an arbitrary point will again be reduced to determining the length of the segment between these two points in a given coordinate system - do this in the same way as was described in the previous step. For example, for the vertex of the parallelogram described in this step and point E with coordinates (X₄;Y₄;Z₄), the formula for calculating the distance from the previous step can be as follows: √((X₃+X₂-X₁-X₄)²+(Y₃+Y₂-Y₁- Y₄)²+(Z₃+Z₂-Z₁-Z₄)²).
For practical calculations you can use, for example, the built-in search engine Google system. So, to calculate the value using the formula obtained in the previous step, for points with coordinates A(7;5;2), B(4;11;3), C(15;2;0), E(7;9; 2), enter the following search query: sqrt((15+4-7-7)^2+(2+11-5-9)^2+(0+3-2-2)^2). The search engine will calculate and display the result of the calculation (5.19615242).
Video on the topic
Recovery perpendicular To plane is one of the important problems in geometry; it underlies many theorems and proofs. To construct a line perpendicular plane, you need to perform several steps sequentially.
You will need
- - given plane;
- - the point from which you want to draw a perpendicular;
- - compass;
- - ruler;
- - pencil.
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 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 contained 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 a plane β, which is the second plane of parallelism, which contains the intersecting lines a and b. Since β || α, then β ⊥ π 4 .
