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 surface 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 AT 2 draw perpendicular to the trace Q2 . Line segment VC is in a particular position in which the frontal projection V2K2 equal to the true value of the required 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 Line 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 occupy a 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
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Option |
Point A |
Point B |
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Option |
Point C |
Point M |
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Checking and passing completed TASK No. 4.
Let's consider the algorithm for solving problem No. 3.
1. From a given point P, draw a perpendicular t to 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 this example, a front-projection plane is 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, for the procedure for determining the natural value of a segment, 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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