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Methods for setting a rectangular coordinate system As is known, the rectangular coordinate system on the plane can be set in three ways: the 1st method is fixed by the location of the system center - T., the OX axis is performed and its positive direction is indicated, the OY axis is perpendicular to the OX axis, according to the system type (right Or the left) indicates the positive direction of the OY axis, the coordinate scale is set along the axes. In the presence of coordinate axes, to determine the coordinates of any point C, you must first omit the perpendicular from this point to the coordinate axes and then measure the length of these perpendicular; The length of the perpendicular to the OX axis is equal to the y coordinate, the length of the perpendicular to the axis Oy coordinate x point (Fig. 1). In addition to the Xoy system, you can use the X "O" Y "system, which is obtained from the Xoy system by transferring the start of the coordinates to the point O" (XO "\u003d DX, YO" \u003d DY) and rotation of the coordinate axes clockwise at an angle b. The transition from Xoy to X "O" Y "is performed by formulas: Formulas are used for reverse transition: 
The first method is generally accepted; In geodesy, this method sets the zonal system of the rectangular coordinates of Gauss. On topographic maps and plans, the system of rectangular coordinates Gauss is set in the second way. On the area, the system of rectangular coordinates is given by the third way; You can always find several geodesic items with known coordinates and determine the position of new points relative to these items, performing any measurements. Three elementary measurements On the plane you can measure the angles and distances. The angle is fixed with three points: one point is the top of the corner, and the two other points fix the directions of the 1st and 2nd sides of the angle. In the simplest case, at least one point of three does not have coordinates, it is, is determined; In the general case, one point, two points or all three can be determined. The distance is fixed with two points, and in general, one point or both can be determined. This section discusses the simplest case when the measurement of the angle or distance is performed to determine the coordinates of one point. Since, when measuring the angle, the defined point can be located either at the top of the angle, or on one of its parties, then from our point of view, there are three different dimensions on the plane, which are called elementary. The angle in paragraph A is measured with the known coordinates x4, y4 between the direction with a known directing corner of the BAB and the direction to the determined point P (Fig. 2).  Directional angle of AP directions is obtained by the formula For a straight line AP, called the position of the point P position, you can write an equation in the Xoy system: In this equation X and Y, the coordinates of any point of direct, including points p, but to find the two coordinates of the point p of one such equation are not enough. The distance S is measured from clause A with known coordinates Xa, Ya to the determined point P. From the course of the geometry it is known that the point P is on the circle of the radius s, spent around the point A, and called the position line P (Fig. 3). The circle equation is: In this equation X and Y, the coordinates of any point of the circle, including points P, but to find the two coordinates of the point of one such equation are not enough.  The angle in the defined point P is measured between the directions of two points with known coordinates; This measurement is viewed in section 8. The x and y points of the point p can be found from the joint solution of the two equations, therefore, taking any combination of three dimensions of two, we obtain the simplest ways to determine the coordinates of the point-called geodetic serifs: two equations of type (2.4) - direct angular sneak, two equations type (2.5) - linear serif, one equation of type (2.4) and one equation of type (2.5) Polar sneak, two measurements of the angles at the determined point - the inverse angular sector. The remaining combinations of measurements are called combined serifs. Each of the three elementary measurements is an invariant with respect to the coordinate systems, which allows you to solve serifs on different drawings, determining the position of the point P relative to fixed points A and B graphically. Analytical method of resolution of serifs is the calculation of the coordinates of the determined point. It can be performed through the solution of the system of two equations corresponding to the measurements performed, or through a triangle solution, the vertices of which are two source points and the defined point (this method for brevity is called a triangle method). In any geodesic construct, it is customary to allocate three types of data: the initial data (the coordinates of the initial items, directive angles of the initial directions, etc.); These data are often taken conditionally unmistakable, measured elements; Each measured element is usually accompanied by an average quadratic measurement error, unknown (or defined) elements; These elements are subject to finding a specially developed algorithm, and their values \u200b\u200bare obtained with some error depending on measurement errors and geometry of this construction. Polar sister The source data in the source data are the coordinates of paragraph A and the directions of the AB direction AB (or the coordinates of paragraph B), the measured elements are the horizontal angle B (the average quadratic error of measuring the MB angle) and the distance S (relative error of its measurement MS / S \u003d 1 / T ), unknown elements - coordinates x, y point P (Fig. 4). Source data: Xa, Ya, BAB Measured Elements: B, S Unknown Elements: X, Y  Graphic solution. From the direction of AB, to postpone the corner of the angle in and spend the direct line AQ, then around the point A carry out an arc of the circumference with the radius s on the scale of the drawing (plan or card); The intersection point of the straight line and arc is a desired point P. Analytical solution. Directional Angle B Line AR is: We write the equations of the straight line AP formula (4) and the circle of the radius S around the paragraph A - formula (5):  To find the x and y coordinates of the point P, these two equations should be solved together as the system. We substitute the value (y - ya) from the first equation to second and let out for braces (X - XA) 2: (X - Xa) 2 * (1 + TG2 b) \u003d S2. Expression (1 + TG2B) Replace 1 / COS2B and get: (X - XA) 2 \u003d S2 * COS2B, where X is Xa \u003d S * COSB. Substitute this value in the first equation (6) and we get: Y - ya \u003d s * sinb. The difference of coordinates (X - XA) and (Y - Ya) is customary to call increments and denote DX and DY. Thus, the polar sneak is unambiguously solved by the formulas:  coordinate triangulation trilateration Direct geodesic task on the plane In geodesy there are two standard tasks: a direct geodesque task on the plane and the reverse geodesic task on the plane. The direct geodesic task is the calculation of the coordinates x2, y2 of the second item, if the coordinates X1, Y1 of the first item, the directive angle b and the length s line connecting these items. The direct geodesic task is part of the polar serif, and the formulas for its solution are taken from the formula (7): Reverse geodesic task on the plane The reverse geodesic problem is the calculation of the directorate angle b and the length s line connecting two points with known coordinates x1, y1 and x2, y2 (Fig. 5).  We construct on the segment 1-2 as a rectangular triangle with customers parallel to the axes of coordinates on the hypotenuse. In this triangle of the hypotenuse is S, the katenets are equal to the increments of the coordinates of points 1 and 2 (dx \u003d x2 - x1, document \u003d y2 - y1), and one of the sharp corners is equal to Rumba R line 1-2. If d x 00 and d y 00, then we solve the triangle according to the famous formulas:   For this pattern, the direction of line 1-2 is in the second quarter, therefore, on the basis of (22) we find: The overall procedure for finding the Directorate Angle of Line 1-2 includes two operations: Determination of a quarter number in terms of coordinate increments D\u003e X and DY, calculation B according to communication formulas (22) in accordance with the quarter number. Control of the correctness of calculations is the implementation of equality:  If dx \u003d 0.0, then s \u003d ідyі; and b \u003d 90o 00 "00" with DA\u003e 0, b \u003d 270O 00 "00" with di< 0. If d \u003d 0.0, then s \u003d ідx_ and b \u003d 0o 00 "00" at dx\u003e 0, b \u003d 180O 00 "00" at dx< 0. To solve the inverse problem in automatic mode (in computer programs), a different algorithm is used that does not contain angle tangent and eliminating possible division by zero:  if d \u003d\u003e 0o, then b \u003d a, if DY.< 0o, то б = 360o - a. Straight angular sister First, consider the so-called general case of a direct angular sector, when the angles of B1 and B2 are measured at two points with known coordinates, each from its direction with a known directional angle (Fig. 6).  Source data: XA, Ya, BAC, Measured elements: in 1, B2 Unknown Elements: X, Y If BAC and BBD are not specified, it is necessary to solve the reverse geodetic problem first between points a and c and then between points B and D. Graphic solution. From the direction of AC, to postpone the angle B1 using the transporter and spend the direct line AP; From the direction of the BD to postpone the angle of B2 and spend the direct line of BP; The intersection point of these direct is the desired point P. Analytical solution. Let us give an algorithm of an option that corresponds to a common occasion of serifs: calculate Directional Angles AP and BP Lines write two equations of direct lines for the line A y - ya \u003d TGB1 * (X - XA), for the BP Y - YB \u003d TGB2 * (X - XB) (2.16) solve the system of two equations and calculate the unknown coordinates X and Y:  A special case of a direct angular sector is considered when the angles B1 and B2 are measured from the directions of AB and BA, with the angle of B1 - the right, and the angle of B2 is the left (in the general case of a sector of both corners - left) - Fig. 7.  The solution of the straight corner of the triangle method corresponds to a private sex key. The procedure for the solution will be this: to solve the inverse problem between paragraphs A and B and obtain the directory angle of the BAB and the length B of the line AB, calculate the angle g at the top of P, called the sefing angle, using the sinus theorem for the APB triangle:  calculate the lengths of the parties ap (S1) and BP (S2), calculate the directory angles B1 and B2:  solve a direct task from point A to point P and for control - from paragraph B to P. P. To calculate the coordinates X and Y in the particular case of a direct angular sector, you can use Jung's formulas:  From the general case, the direct corner serf is not difficult to move to a private case; To do this, first need to solve the reverse geodetic problem between paragraphs A and B and get the directorate angle of the BAB of the AB line and then calculate the angles in the APB triangle at the vertices A and B BAP \u003d BAB - (BAC + B1) and ABP \u003d (BBD + B2) - BBA. For machine account, all the considered ways to solve direct corner serfs for various reasons are inconvenient. One of the possible algorithms for solving a general occasion of a serif on a computer provides for the following actions: Calculation of directory angles B1 and B2, Introduction of the local coordinate system X "O" Y "with the beginning in paragraph A and with the O" X "axis directed along the AP line, and Recalculation of the coordinates of points A and B and the directing angles B1 and B2 from the Xoy system to the X "O" Y "system (Fig. 8): X "a \u003d 0, y" a \u003d 0, (24), recording the equations of AP and BP lines in the X "O" Y "system:   and the joint solution of these equations:  translation of X "and Y" coordinates from the X "O" Y "system in the Xoy system:  Since Ctgb2 "\u003d - CTGG and the seed angle of g is always greater than 0o, then the solution (27) always exists. Linear stitch From paragraph A with the known coordinates Xa, Ya, the distance S1 was measured to the specified point P, and from paragraph B with known coordinates XB, YB measured the distance S2 to point P. Graphic solution. We carry out around the point A a circle with a radius S1 (on the scale of the drawing), and around paragraph B - the circumference with the radius S2; The crossing point of the circles is the desired point; The task has two solutions, since two circles intersect at two points (Fig. 9). Original data: XA, Ya, XB, YB, Measured elements: S1, S2, Unknown Elements: X, Y. Analytical solution. Consider two analytical solutions algorithm, one - for a manual account (according to a triangle method) and one for a machine account.  The manual account algorithm consists of the following actions: the resolution of the reverse geodesic problem between points a and b and obtaining the directorate angle of BAB and the length B line AB, the calculation in the ABP triangle of the angles B1 and B2 by the cosine theorem:  calculation of the corner of the serf calculation of directory angles of the parties of the AP and BP: point P to the right of the AB line  point P to the left of the line  solution of direct geodesic problems from clause A to paragraph P and from paragraph B to paragraph P: 1 solution  2nd solution  The results of both solutions should coincide. The linear serif algorithm consists of the following actions: the solution of the reverse geodesic problem between points a and b and obtaining the directorate angle of the BAB and the length B line AB, the introduction of the local coordinate system x "o" y "with the beginning at point A and axis o" x ", directed along the AB line, and recalculate the coordinates of points A and B from the XOY system to the X" O "Y" system:  record of the circle equations in the X "O" Y "system:  and the joint solution of these equations, which involves disclosing brackets in the second equation and subtracting the second equation from the first:   If the desired point is located on the left of the line AB, then the "-" sign is taken in formula (39), if on the right, then "+". Recalculation of the coordinates x "and y" points P from the X "O" Y system "in the Xoy system according to formulas (2):  Reverse angular sister Elementary measurements include an angle measurement in the defined point P between the directions by two points A and B with the known coordinates Xa, Ya and Xb, Yb (Fig. 10). However, this measurement turns out to be theoretically quite complicated, so consider it separately. We will conduct a circle through three points A, B and P. From the school year of the geometry, it is known that an angle with a vertex on the circle is measured by half an arc to which it relies. The central angle, based on the same arc, is measured by the entire arc, therefore, it will be equal to 2V (Fig. 10).  The distance B between items A and B is considered to be known, and from a rectangular FCB triangle can be found Radius R circumference: The circle equation is: where XC and YC are the coordinates of the center of the circle. They can be calculated by deciding either a direct angular or linear serif from paragraphs A and B to a point C. in equation (42) x and y - the coordinates of any point of the circle, including points P, but for finding two coordinates of the point P one Such an equation is not enough. The reverse angle point is the method of determining the coordinate points P along two corners of B1 and B2, measured at the determined point P between directions by three points with known coordinates A, B, C (Fig. 11). Graphic solution. We give the method of marsh graphic solutions to the inverse corner. On a sheet of transparent paper (tracing), it is necessary to construct the angles of B1 and B2 with the total vertex P; Then impose a traction to the drawing and, moving it, to ensure that the direction of the carts on the tank passed through the points A, B, C in the drawing; Pass the point P with a carting on the drawing. Source data: XA, Ya, XB, Measured elements: B1, B2. Unknown Elements: X, Y.  Analytical solution. An analytical solution of the inverse angular sector provides its decomposition on simpler tasks, for example, on 2 direct angular serifs and one linear, or on 3 linear serifs, etc. More than 10 methods of analytical solution are known, but we will consider only one - through a consistent solution of three linear serifs. Suppose that the position of the point P is known, and we will carry out two circles: one radius R1 through points a, b and p and another radius R2 through points B, C and P (Fig. 11). The radii of these circles will be obtained by formula (41):  If the coordinates of the centers of circles - points O1 and O2 will be known, then the coordinates of the point P can be determined by the linear serif formulas: from the point O1 to the distance R1 and from the O2 point - distance R2. The coordinates of the center O1 can be found according to linear serif formulas from points A and B from distances R1, and from two solutions you need to take something that corresponds to the value of the angle B1: if B1<90o, то точка O1 находится справа от линии AB, если в1>90O, then the point O1 is on the left of the AB line. The coordinates of the center O2 are on linear serif formulas from points B and C from distances R2, and one solution from two possible is selected by the same rule: if B2<90o, то точка O2 находится справа от линии BC, если в2>90O, then the O2 point is located on the left of the BC line. The task has no solution if all four points a, b, c and p are on the same circle, since both circles are merged into one, and there are no points of intersection. Combined serfs In the methods considered methods of resolution, the number of measurements was taken theoretically minimal (two dimensions), which ensures the result. In practice, to find the coordinates X and Y of one point, as a rule, do not perform two, but three or more measurements of distances and angles, and these measurements are performed both at the source points and on the determined; Such serifs are called combined. It is clear that in this case the possibility of measuring control appears, and, in addition, the accuracy of the problem of problem increases. Each measurement entered into the task is over theoretically minimal amount is called redundant; It generates one additional solution. Geodesic serifs without excess measurements are called single, but serifs with excess measurements - multiple. In the presence of excessive measurements, the calculation of unknowns is performed by adjusting. Algorithms of strict equalization of multiple serifs are used with an automated account on a computer; For manual account use simplified methods of adjustment. The simplified method of adjusting any multiple serif (N of measurements) provides for the formation and solution of all possible options for independent single sections (their number is N-1), and then calculating the average values \u200b\u200bof the point coordinate from all the results obtained if they differ on the permissible value. Point position error In a one-dimensional space (on line), the point position is fixed by the value of one coordinate X, and the position error MP point is equal to the average square error MX of this coordinate. The true position of the point may be in the interval (X - T * MX) - (X + T * MX), then-there, in both sides of the X value; In practice, the coefficient T is usually specified equal to 2.0 or 2.50. In a two-dimensional space (on the surface), the position of the point is fixed by the values \u200b\u200bof the two coordinates, and the point position error should be set by two values: direction and error of the position in this direction. The geometric shape, inside which the true position of the point is located, may have a different form; In the particular case, when the point position error in all directions is the same, the radius circle R \u003d MP is obtained. The position of the point for two dimensions is obtained in the intersection of two position lines. For the measured distance s line of the position, the radius of the radius s with the center in the initial paragraph A (Fig. 2.12a); For the measured angle in the vertex in the starting point A, a straight line conducted at an angle to the original AB line (Fig. 2.12b). Due to the measurement errors, you must enter the concept of the "position band". For a distance s, measured with an average quadratic error MS is a circular belt (ring) 2 * MS width between two circles with radii (S - MS) and (S + MS); For an angle in the measured MB error, is a narrow triangle with a vertex at a point A and an angle at the top of 2 * MB. The position line is the axis of the position strip symmetry (Fig. 12).  Fig. 12. Position line and "position band" point P: a) for the measured distance, b) for the measured angle. We introduce the concept of "measurement error vector" and denote it through V. For the measured distance, the VS vector is directed along the AP line (directly or back) and has a module VS \u003d MS; For the measured angle, the VV vector is directed perpendicular to the AP (left or right from it) and has a HB module \u003d S * MB / C, where S \u003d A * P. Point P, being at the intersection of two position lines, is the center of the 4-square position formed in the intersection of two position bands (Fig. 13).  Fig. 13. 4. "Caller position: a) in linear serif, b) in a straight corner spot, This elementary 4-square can be considered a parallelogram, since within it the circumference arches can be replaced with tangential sections, and the divergers of the angle are segments of direct, parallel position lines. The distances from the point p to the boundaries of the 4-square meal, which indicates the difference in the position of the point P point in different directions. The position lines divide the 4-coal position of the positions of 4 equal parts, which we call the parallelograms of errors with angles at the vertices r and (180o - g), where g (180o - g) is the angle between the error vectors V1 and V2. Since the heights of parallelograms of errors are numerically equal to the modules of the N1 and H2 vectors, then the side of the parallelograms are obtained by known formulas:  According to the famous sides, the parallelogram of errors and the corner between them g (180o - d) can be calculated the length of both of its diagonals: short - d1 and long - d2:  Thus, the point position error in six directions (Fig. 14) is expressed by simple formulas; For all other directions of formulas will be more complex. For the generalized characteristic of the accuracy of the point of determining point P, you need to have some average point of the point position error, which can be calculated: as the radius of the circle R, the area of \u200b\u200bwhich (P * R2) is equal to the area of \u200b\u200bthe Par Experience Position Position P (4 * A * B * Sing),  as a position error in the "the weakest direction", which coincides with the direction of long diagonal: as the average quadratic from the long and short diagonals parallelogram of errors:  In practice, the third option is used more often, in which formulas are easily obtained to assess the accuracy of any single spot: polar sex (Fig. 4):  straight angular stitching (Fig. 6, 7):  linear serunel (Fig. 9):  reverse angular seed (Fig. 11). In this sector, the right part of the point error formula P should contain three terms: error of linear serif point O1 from the source points A and B (MO1), the error of the linear serif point O2 from the source points B and C (MO2), the error of the linear serif point P from the points O1 and O2 (MP),  The seed angle of g depends on the relative position of BC and BA lines and the angles of B1 and B2; For fig. 11 This angle is calculated by the formula: For many practices, it is sufficient to assume that the true position of the point P is inside the radius circle of the MP radius with the center at the point P. In strict theory, the considered criterion is called a radial error. In addition, more complex criteria are also used in this theory, such as "ELLIPS errors" (curve of the 2nd order), the "ELLIPS" (4th order curve) and others. With the amount of measurements n\u003e 2 (multiple serifs), the point P is obtained in the intersection of n position lines corresponding to the equalized measurement values; Position bands, intersecting, form 2 * N-square. The largest position error P point will be determined by the distance from point P to the vertex of this polygon remote from it. From the figure 14-b, the role of the third measurement in reducing the position error P is clearly understood; By the way, in this figure the second dimension practically does not affect the value of the point position error. Rectangular coordinate system To determine the concept of coordinates of points, we need to introduce the coordinate system in which we will determine its coordinates. The same point in different coordinate systems can have different coordinates. Here we will consider a rectangular coordinate system in space. Take in the space a point $ o $ and we introduce the coordinates of $ (0.0.0) $. We call it the beginning of the coordinate system. We will conduct three mutually perpendicular axes $ ox $, $ oy $ and $ oz $, as in Figure 1. These axes will be called the abscissa axes, ordents and the applicat, respectively. It remains only to introduce the scale on the axes (single segment) - the rectangular coordinate system in the space is ready (Fig. 1) Figure 1. Rectangular coordinate system in space. Author24 - Student Internet Exchange Coordinates of the pointNow we will analyze how to determine in such a coordinate system of any point. Take an arbitrary point $ M $ (Fig. 2). We construct a rectangular parallelepiped on the coordinate axes, so that the points $ o $ and $ m $ opposite peaks (Fig. 3). Figure 3. Building a rectangular parallelepiped. Author24 - Student Internet Exchange Then the $ M $ point will have coordinates $ (x, y, z) $, where $ x $ is a value on the numeric axis $ OX $, $ y $ - the value on the numeric axis $ oy $, and $ z $ $ oz $ numeric axis. Example 1. It is necessary to find a solution to the following task: write the coordinates of the vertices of the parallelepiped shown in Figure 4. Decision. Point $ o $ Beginning of coordinates, therefore, $ O \u003d (0,0,0) $. Points $ q $, $ n $ and $ r $ lie on the $ ox $, $ oz $ and $ oy $, respectively, then $ Q \u003d (2,0,0) $, $ n \u003d (0,0,1.5) $, $ R \u003d (0,2.5.0) $ Points $ s $, $ l $ and $ m $ lie in the planes $ oxz $, $ oxy $ and $ oyz $, respectively, then $ S \u003d (2,0,1.5) $, $ L \u003d (2,2,5,0) $, $ r \u003d (0,2.5,1.5) $ Point $ p $ has coordinates $ p \u003d (2,2.5,5.5) $ The coordinates of the vector at two points and the formulaTo find out how to find a vector by the coordinates of two points, it is necessary to consider the coordinate system introduced by us. In it from the $ o $ point in the direction of the $ ox $ axis, we postpone the $ \\ overline (I) $ vector, in the direction of the $ oy $ axis - single vector $ \\ overline (j) $, and the unit vector $ \\ overline (k) $ you need to direct the $ OZ $ axis. In order to introduce the concept of the vector coordinates, we introduce the following theorem (here we will not consider its proof). Theorem 1. An arbitrary vector in space can be decomposed in three any vectors that are not lying in the same plane, and the coefficients in such decomposition will be singlely defined. Mathematically, it looks like this: $ \\ Overline (Δ) \u003d M \\ Overline (α) + N \\ Overline (β) + L \\ OVERLINE (γ) $ Since the $ \\ overline (I) vectors $ \\ overline (j) $ and $ \\ overline (k) $ are built on the coordinate axes of the rectangular coordinate system, they obviously will not belong to the same plane. It means any vector $ \\ overline (δ) $ in this coordinate system, by Theorem 1, can take the following type $ \\ Overline (Δ) \u003d M \\ Overline (i) + N \\ Overline (j) + L \\ Overline (k) $ (1) where $ n, m, l∈R $. Definition 1. Three vector $ \\ overline (i) $, $ \\ overline (j) $ and $ \\ overline (k) $ will be referred to by coordinate vectors. Definition 2. The coefficients before the $ \\ overline (i) $, $ \\ overline (j) $ and $ \\ overline (k) $ in decomposition (1) will be called the coordinates of this vector in the coordinate system specified by us, that is $ \\ overline (δ) \u003d (m, n, l) $ Linear operations over vectorsTheorem 2. Amount Theorem: The coordinates of the sum of any number of vectors are determined by the sum of their respective coordinates. Evidence. We will prove this theorem for 2 vectors. For 3 and more vectors, proof is built in a similar way. Let $ \\ overline (α) \u003d (α_1, α_2, α_3) $, $ \\ overline (β) \u003d (β_1, β_2, β_3) $. These vector can be written as follows. $ \\ overline (α) \u003d α_1 \\ Overline (i) + α_2 \\ Overline (j) + α_3 \\ overline (k) $, $ \\ overline (β) \u003d β_1 \\ overline (i) + β_2 \\ overline (j) + β_3 \\ overline (k) $ Coordinatescorrect and linear values \u200b\u200b(numbers) are called, determining the position of the point on any surface or in space. The topography uses such coordinate systems that allow you to easily and unambiguously determine the position of the earth's surfaces both by the results of direct measurements on the ground and with the help of cards. Such systems include geographic, flat rectangular, polar and bipolar coordinates. Geographical coordinates (Fig. 1) - angular values: latitude (J) and longitude (L), determining the position of the object on the earth's surface relative to the origin of the coordinates - the intersection points of the initial (Greenwich) meridian with the equator. On the map, the geographic grid is indicated by the scale on all sides of the map frame. The Western and Eastern side of the framework are meridians, and North and South - parallels. In the corners of the card sheet, the geographical coordinates of the points of intersection of the framework of the frame. Fig. 1. System of geographical coordinates on the earth's surface In the system of geographic coordinates, the position of any point of the earth's surface relative to the start of coordinate is determined in the angular measure. For the beginning of us and in most other states adopted the intersection point of the initial (Greenwich) meridian with the equator. Being thus united for the whole of our planet, the geographic coordinate system is convenient for solving problems to determine the mutual position of objects located at considerable distances from each other. Therefore, in a military case, this system is mainly used to conduct settlements related to the use of long-range combat, such as ballistic missiles, aviation, etc. Flat rectangular coordinates(Fig. 2) - linear values \u200b\u200bthat determine the position of the object on the plane relative to the received origin - the intersection of two mutually perpendicular direct (coordinate axes x and y). In the topography, each 6-degree zone has its own system of rectangular coordinates. X axis - axial meridian zone, axis Y - equator, and the point of intersection of the axial meridian with the equator - the beginning of the coordinates.
The system of flat rectangular coordinates is zonal; It is installed for each six-genera zone, to which the surface of the Earth is divided into the image of it in the Gaussian projection, and is intended to indicate the position of the images of the earth's surface points on the plane (map) in this projection. The beginning of the coordinates in the zone is the point of intersection of the axial meridian with the equator, relative to which it is determined in the linear the position of all other points of the zone. The origin of the zone coordinates and its coordinate axes occupy a strictly defined position on the earth's surface. Therefore, the system of flat rectangular coordinates of each zone is associated with both the coordinate systems of all other zones and with the geographic coordinate system. The use of linear values \u200b\u200bto determine the position of the points makes a system of flat rectangular coordinates very convenient for making calculations both when working on the ground and on the map. Therefore, in the troops this system finds the widest application. The rectangular coordinates indicate the position of the points of the terrain, their combat orders and goals, with their help determine the mutual position of objects within the same coordinate zone or on adjacent sections of two zones. Polar and Bipolar Coordinate Systems are local systems. In military practice, they are used to determine the position of one points relative to others at relatively small areas of the area, for example, with targeting, location of landmarks and goals, the preparation of locality schemes, etc. These systems can be associated with systems of rectangular and geographic coordinates. 2. Determination of geographical coordinates and application of objects on the known coordinatesThe geographical coordinates of the point on the map are determined from the nearest parallels and meridian, the latitude and longitude of which are known. The frame of the topographic map is broken for a minute, which are separated by division points for 10 seconds in each. On the side sides of the frame marked latitudes, and on the north and southern - longitude.
Using the minute frame of the card, you can: 1 . Determine the geographic coordinates of any point on the map. For example, the coordinates of points A (Fig. 3). To do this, you need to measure the shortest distance from the point A to the southern map frame, then apply the meter to the Western frame and determine the number of minutes and seconds in the measured segment, fold the resulting (measured) minutes and seconds (0 "27") With the breadth of the south-western corner of the frame - 54 ° 30. Latitude Points on the map will be equal to: 54 ° 30 "+0" 27 "\u003d 54 ° 30" 27 ". Longitude Determined similarly. The shortest distance from the point A to the western card frame is measured using a circulator with a circulator, a circular meter is applied to the southern frame, determine the number of minutes and seconds in the measured segment (2 "35") fold the obtained (measured) value with the longitude of the south-west corner Frames - 45 ° 00 ". Longitude Points on the map will be equal to: 45 ° 00 "+2" 35 "\u003d 45 ° 02" 35 " 2. Apply any point on the card according to the specified geographic coordinates. For example, the point B of the latitude: 54 ° 31 "08", longitude of 45 ° 01 "41". To apply a dot point in the longitude, it is necessary to carry out true meridian through this point, for which it is to connect the same number of minutes along the North and Southern Frame; To apply a point on a point in latitude, it is necessary to carry out parallel through this point, for which it is to connect the same number of minutes to the Western and Eastern Frame. The intersection of two straight lines will determine the location of the point B. 3. Rectangular coordinate grid on topographic maps and its digitization. Additional mesh at the joint of the coordinate zonesThe coordinate grid on the map is a grid of squares formed by lines parallel to the coordinate axes of the zone. Mesh lines are carried out through an integer number of kilometers. Therefore, the coordinate grid is also called a kilometer grid, and its lines kilometer. On the map 1: 25000 lines forming the coordinate grid were carried out after 4 cm, that is, 1 km on the ground, and on the cards 1: 50000-1: 200000 after 2 cm (1.2 and 4 km on the ground, respectively). On the 1: 500000 card, only the outputs of the coordinate grid lines on the inner frame of each sheet after 2 cm (10 km on the ground) are applied. If necessary, on these outputs, the coordinate lines can be applied to the card. On topographic maps, the values \u200b\u200bof the abscissa and the ordents of the coordinate lines (Fig. 2) sign on the outputs of the lines behind the inner frame of the sheet and nine places on each sheet of the card. The complete values \u200b\u200bof the abscissa and ordents in kilometers are signed near the corners of the coordinate lines map close to the corners and near the nearest to the northwestern angle of crossing the coordinate lines. The remaining coordinate lines are signed by shorter two digits (tens and units of kilometers). Signatures near the horizontal lines of the coordinate grid correspond to the distances from the ordinate axis in kilometers. The signatures near the vertical lines denote the zone number (one or two first digits) and the distance in kilometers (always three digits) from the start of coordinates, conditionally transferred to the west of the axial meridian zone 500 km. For example, the signature 6740 means: 6 - the zone number, 740 is the distance from the conditional origin of the coordinates in kilometers. On the outer frame, the outputs of the coordinate lines are given ( additional grid) Related zone coordinate systems. 4. Determination of rectangular coordinates of points. Application on the card points according to their coordinatesBy coordinate grid using a circulation (ruler), you can: 1. Determine the rectangular point coordinates on the map. For example, points in (Fig. 2). To do this, it is necessary:
Definition of ordinate y produce similarly:
2. Apply the target on the specified coordinates. For example, the point r in coordinates: x \u003d 6658725 y \u003d 7362360. To do this, it is necessary:
5. Accuracy of determining coordinates on maps of various scalesThe accuracy of determining geographic coordinates on cards 1: 25000-1: 200000 is about 2 and 10 "", respectively. The accuracy of determination on the map of the rectangular coordinates of the points is limited not only by its scale, but also the values \u200b\u200bof the errors allowed when shooting or drawing up a map and applying different points and objects of the terrain The most accurate (with an error not exceeding 0.2 mm), geodesic items are applied to the card. The most sharply distinguished on the terrain and visible issued objects that have landmarks (individual bell tower, factory pipes, tower-type construction). Therefore, the coordinates of such points can be determined by about the same accuracy with which they are applied to the card, i.e., for the map of scale 1: 25000 - with an accuracy - 5-7 m, for the scale of scale 1: 50000 - with accuracy - 10- 15 m, for map scale 1: 100000 - with accuracy - 20-30 m. The remaining benchmarks and points of the contours are applied to the card, and, therefore, and are determined by it with an error of up to 0.5 mm, and the points relating to the contours related to the terrain (for example, the swamps circuit), with an error of up to 1 mm. 6. Determination of the position of objects (points) in the systems of polar and bipolar coordinates, applying to the map of objects in direction and distance, on two corners or two distancesSystem flat polar coordinates (Fig. 3, a) consists of a point about - the origin of the coordinates, or poles, and the initial direction or called polar axis.
The position of the point m on the ground or on the map in this system is determined by two coordinates: the angle of the position θ, which is measured along the clockwise arrow from the polar axis to the direction to the determined point M (from 0 to 360 °), and the abs \u003d d. Depending on the problem of the pole, the observation item, the firing position, the initial point of movement, etc., is taken, and for the polar axis - geographical (true) meridian, magnetic meridian (direction of the compass magnetic arrow) or direction on any landmark . These coordinates can serve either two angles of the position determining the directions from points A and B to the desired point M, or the distance d1 \u003d s and d2 \u003d vm to it. Corners of the position at the same time, as shown in Fig. 1, B, measured at points A and B or on the direction of the basis (i.e., angle A \u003d you and angle B \u003d AVM) or from other directions passing through points A and B and received for the initial. For example, in the second case, the location point M is determined by the angle of position θ1 and θ2, measured from the direction of magnetic meridians. System flat bipolar (bipolar) coordinates (Fig. 3, b) consists of two poles A and B and the total axis of the AB, called the basis or base of serif. The position of any point M with respect to two data on the map (locacy) points A and B is determined by the coordinates that are measured on the map or on the ground. Application of a detected object on a map This is one of the most important moments in the detection of the object. From how exactly the object (target) will be applied to the card, the accuracy of the determination of its coordinates depends. Finding an object (goal), you must first determine exactly on various features, which is detected. Then, without stopping the observation of the object and not detecting itself, apply an object on the card. To apply an object on the map there are several ways. Eyemervan: The object is applied to the card, if it is near the famous reference point. In direction and distance: To do this, you need to orient the map, find the point of your standing on it, hang on the map direction to the detected object and read the line to the object from the point of its standing, then determine the distance to the object, measuring this distance on the map and commensively with the map scale.
If it is thus graphically impossible to solve the task (interferes with the enemy, bad visibility, etc.), then you need to accurately measure the azimuth to the object, then translate it into the directory angle and read on the map from the point of standing on which to postpone the distance to the object. To get a directory angle, it is necessary to add magnetic declination of this card to magnetic azimuth (direction amendment). Direct sneak. This method is caused by an object on a card from 2-to-3-points, from which you can monitor it. To do this, from each selected point is watched on an oriented map direction to the object, then the intersection of direct lines determines the location of the object. 7. Methods of target designation on the map: in graphic coordinates, flat rectangular coordinates (full and abbreviated), in squares of kilometer grid (up to a whole square, up to 1/4, to 1/9 square), from the reference point, from the conditional line, in azimuth and range of goals in the system of bipolar coordinatesThe ability to quickly and correctly indicate goals, guidelines and other objects on the ground is important for managing divisions and fire in battle or for organizing battle. Best Description B. geographic coordinates It is very rare and only in cases where the targets are removed from the specified point on the map at a considerable distance expressing in dozens or hundreds of kilometers. At the same time, geographical coordinates are determined by the map, as described in the question number 2 of this classes. The location of the target (object) indicate the latitude and longitude, for example, height 245.2 (40 ° 8 "40" s. Sh., 65 ° 31 "00" c. D.). To the eastern (Western), the northern (southern) side of the topographic framework is applied by the circulation of the marking of the target position in latitude and longitude. From these marks to the depth of the topographic card, perpendicular to their intersection are lowered (apply commander rules, standard sheets of paper). Point of intersection of perpendiculars and is the position of the target on the map. For approximate target designation rectangular coordinates It is enough to indicate the square of the mesh square on the map, in which the object is located. The square is always indicated by numbers of kilometer lines whose intersection is formed by the South-West (bottom left) angle. When specifying the square of the card, the rules are followed: first they call two digits signed by the horizontal line (at the west side), that is, the "X" coordinate, and then two digits at the vertical line (the southern face of the sheet), that is, the "Y" coordinate. At the same time, "X" and "Y" do not talk. For example, tanks are enemy. When transmitting the report on the radio telephone number, the square number is pronounced: "Eighty-eight zero two." If the position of the point (object) must be determined more accurately, then use full or abbreviated coordinates. Work with the full coordinates. For example, it is required to determine the coordinates of the road pointer in the square 8803 on the scale map of 1: 50,000. First, they determine the distance from the lower horizontal side of the square to the road indicator (for example, 600 m on the ground). In the same way, the distance from the left vertical side of the square is measured (for example, 500 m). Now by digitizing kilometer lines, we determine the total coordinates of the object. The horizontal line has a signature 5988 (x), adding the distance from this line to the road pointer, we get: x \u003d 5988600. Similarly, we also determine the vertical line and get 2403500. The total coordinates of the road pointer are as follows: x \u003d 5988600 m, y \u003d 2403500 m. Abbreviated coordinates Accordingly, they will be equal: x \u003d 88600 m, y \u003d 03500 m. If you want to clarify the position of the target in the square, then apply the target designation by the letter or digital method inside the square of the kilometer grid. For targeting lettering Inside the square of the kilometer mesh, the square is conditionally divided into 4 parts, each part is assigned a capital letter of the Russian alphabet. Second way - Digital fashion target design inside a kilometer square (target designation snail ). This method got its name on the location of the conditional digital squares inside the square of the kilometer grid. They are arranged on the spiral, while the square is divided into 9 parts. In case of targeting in these cases, the square is called the target, and add a letter or a figure specifying the position of the target inside the square. For example, height 51.8 (5863-A) or high-voltage support (5762-2) (see Fig. 2). The target designation is the most simple and common way to design. In this case, the method of targeting is first called the nearest landmark, then the value of the angle between the direction on the benchmark and the direction on the target in the division of the negotiation (measured by binoculars) and removal to the target in meters. For example: "The benchmark is the second, right forty, further two, from a separate bush - a machine gun." Keepage from the conditional line Usually applied in motion on combat machines. In this case, the method on the map is chosen in the direction of action two points and connect them to the straight line, relative to which the target designation will be conducted. This line is denoted by letters, divided into centimeter divisions and numb them from scratch. This construction is made on maps of both the transmitting and receiving target designation. The target designation from the conditional line is usually applied in motion on combat vehicles. In this case, the method on the map is chosen in the direction of action two points and connect their straight line (Fig. 5), with respect to which it will begin target designation. This line is denoted by letters, divided into centimeter divisions and numb them from scratch.
This construction is made on maps of both the transmitting and receiving target designation. The position of the target relative to the conventional line is determined by two coordinates: a segment from the starting point to the base of the perpendicular, lowered from the point of location of the target to the conditional line, and the perpendicular segment from the conditional line to the target. When appreciation, the conditional name of the line is called, then the number of centimeters and millimeters consisting in the first segment, and, finally, the direction (left or right) and the length of the second segment. For example: "Direct AC, five, seven; Right zero, six - np ". The target designation from the conditional line can be issued by specifying the direction to the target at an angle of the conditional line and the distance to the target, for example: "Direct AC, Rights 3-40, a thousand two hundred - machine gun." Keepage in azimuth and range to goal. Azimuth directions on the target are determined using a compass in degrees, and the range to it - with the help of a surveillance device or eye in meters. For example: "Azimut thirty-five, a period of six hundred - tank in the trench." This method is most often used on the ground, where there are few landmarks. 8. Solution of tasksDetermination of the coordinates of the areas of the area (objects) and the target designation on the card is practically implemented on the training cards in advance prepared points (inflicted objects). Each learned definition of geographical and rectangular coordinates (inflicts objects according to known coordinates). Ways to designation on the card are practiced: in flat rectangular coordinates (full and abbreviated), in squares of a kilometer grid (up to a whole square, up to 1/4, to 1/9 square), from the reference point, in azimuth and target range. In this article, we will start a discussion of one "chopstice sticks", which will allow you to reduce many geometry tasks for simple arithmetic. This "stick" can significantly ease your life especially in the case when you are insecably feel in the construction of spatial figures, sections, etc. All this requires a certain imagination and practical skills. Method, which we will begin to consider here, will allow you to almost completely abstract from all kinds of geometric constructions and reasoning. The method is called "Coordinate method". In this article, we will consider the following questions:
I think you have already guessed why the coordinate method is so called? True, it received such a name, as it operates not with geometric objects, but with their numerical characteristics (coordinates). And the conversion itself, which allows you to move from geometry to algebra, is to introduce the coordinate system. If the source figure was flat, then the coordinates are two-dimensional, and if the formation figure, then the coordinates are three-dimensional. In this article we will consider only a two-dimensional case. And the main purpose of the article is to teach you to use some basic techniques of the coordinate method (they are sometimes useful in solving problems in planimetry in the USE part B). The following two sections on this subject are devoted to discussing the same methods for solving problems of Tasks C2 (task for stereometry). Why would it be logical to start discussing the method of coordinates? Probably, with the concept of the coordinate system. Remember when you with her first encountered. It seems to me that in grade 7, when you learned about the existence of a linear function, for example. Let me remind you, you built it at points. Do you remember? You chose an arbitrary number, substituted it in the formula and calculated in this way. For example, if, then, if, then, etc., what did you get in the end? And I received a point with the coordinates: and. Next, you painted the "cross" (coordinate system), chose a scale on it (how many cells you will have a single segment) and noted on it the points received, which then combined the straight line, the resulting line and there is a function graph. There are several moments that should be explained to you a little more: 1. A single segment you choose for reasons convenience, so that everything is beautiful and compactly fit in the picture 2. It is accepted that the axis goes left to right, and the axis is to the bottom 3. They intersect at right angles, and the point of their intersection is called the beginning of the coordinates. It is indicated by the letter. 4. In the recording of the coordinates of the point, for example, on the left in brackets there is a point coordinate along the axis, and on the right, along the axis. In particular, simply means that the point 5. In order to set any point on the coordinate axis, it is required to specify its coordinates (2 numbers) 6. For any point lying on the axis, 7. For any point lying on the axis, 8. The axis is called the abscissa axis 9. The axis is called the ordinate axis Now let's make the next step with you: We note two points. Connect these two points with a segment. And put the arrow as if we spend a segment from point to point: That is, we will make our segment directed!
Remember how else the directed segment is called? True, it is called a vector! Thus, if we connect a point with a point, moreover, we will have a point A, and the end - point B, then we get a vector. Did you make this building too in grade 8, remember? It turns out that vectors, like the points, can be denoted by two numbers: these numbers are called vector coordinates. Question: Do you think it is enough for us to know the coordinates of the beginning and end of the vector to find its coordinates? It turns out that yes! And it is done very simple: Thus, since the dot vector is the beginning, and the end, the vector has the following coordinates: For example, if, the coordinates of the vector Now let's do on the contrary, we will find the coordinates of the vector. What should we change for this? Yes, you need to swap the beginning and end: now the beginning of the vector will be at the point, and the end is at the point. Then: Look carefully, what is the difference between the vectors and? Their only difference is signs in coordinates. They are opposite. This fact is accepted to record like this: Sometimes, if it is not specifically stipulated, which point is the beginning of the vector, and how end, the vectors are denoted by two capital letters, but one line, for example:, etc. Now a little penate And find the coordinates of the following vectors: Check: And now deciding a problem a little more complicated: A century with on-cha-scrap at the point has a co-or-di-on-you. NAI-DITE ABS CISS DOP. All the same is pretty prose: Let the coordinates of the point. Then I am a system to determine what the coordinates of the vector. Then the point has coordinates. We are interested in the abscissa. Then Answer: What else can you do with vectors? Yes, almost all the same as with ordinary numbers (unless you can not divide, it is possible to multiply in two ways, one of which we will discuss here a little later)
All these operations have a completely visual geometric representation. For example, a rule of a triangle (or a parallelogram) for addition and subtraction:
The vector is stretched or compressed or changes the direction when multiplying or dividing by:
However, here we will be interested in the question of what is happening with the coordinates. 1. When adding (subtracting) of two vectors, we fold (deduct) alternately their coordinates. I.e: 2. When multiplying (division) of the vector by number, all its coordinates are multiplied (divided) to this number: For example: · Nay-die the sum of the co-or-di-nat's eyelid.
Let's first find the coordinates of each of the vectors. Both of them have the same start - the point of the origin. They have different ends. Then. Now we calculate the coordinates of the vector then the sum of the coordinates of the resulting vector is equal. Answer: Now the following task is: · Find the sum of the coordinates of the vector
Check: Let's consider now the following task: we have two points on the coordinate plane. How to find the distance between them? Let the first point be, and the second. Denote the distance between them through. Let's make the following drawing for clarity:
What I've done? First, I, first, connected points and, and also spent the line from the point, parallel to the axis, and spent the line from the point parallel to the axis. Have they crossed at the point by forming a wonderful figure? What is it wonderful? Yes, we are almost just knowing about the rectangular triangle. Well, the theorem of Pythagora - for sure. The desired segment is the hypotenuse of this triangle, and the segments are kartettes. What are the coordinates of the point? Yes, it is easy to find them in the picture: since the segments are parallel to the axes and, accordingly, their lengths are easy to find: if you designate the length of the segments, respectively, then Now we use the Pythagorean theorem. We know the length of the cathets, we will find hypotenuse: Thus, the distance between two points is the root of the sum of the squares of differences from the coordinates. Or - the distance between two points is the length of the segment, which connects them. It is easy to notice that the distance between points does not depend on the direction. Then: From here we make three outputs: Let's take a little exercise in the calculation of the distance between the two points: For example, if, then the distance is between and equal Or let's go differently: we find the coordinates of the vector And find the length of the vector: As you can see, the same thing! Now practice a little: Task: find the distance between the specified points: Check: Here is another pair of tasks on the same formula, though they sound a little different: 1. Nay-di KVAD-RAT of the length of the eyelid.
2. NAY-Di KVAD-RAT of the length of the eyelid-ra
I think so, did you easily manage with them? Check: 1. And this is at the attentiveness) we have already found the coordinates of the vectors and earlier :. Then the vector has coordinates. The square of its length will be equal to: 2. Find the vector coordinates Then the square of its length is equal Nothing difficult, right? Ordinary arithmetic, no more. The following tasks cannot be classified unambiguously, they are more like a general erudition and to draw simple pictures to the ability. 1. Nay-di sinus angle of the corner of the na-klo-on from-cut, co-unit-in-y-th point, with an abscissa axis.
How will we come here? It is necessary to find the sine angle between and axis. And where do we know how to look for sinus? True, in a rectangular triangle. So what do we need to do? Build this triangle!
Since the coordinates of the point and, then the segment is equal, and the segment. We need to find a sine corner. I will remind you that sinus is the attitude of the opposite catech for the hypotenuse, then What do we have to do? Find hypotenuse. You can do it in two ways: according to the Pythagore Theorem (katenets are known!) Or by the distance formula between the two points (actually the same thing as the first way!). I will go secondly: Answer: The next task will seem to you even easier. She is on the coordinates of the point. Task 2. From the point of Oposchn Pen-Pen-Di-Liar on the ABS axis. NAI-DITE ABS CIS-SU OS-NO-VIA-PEN-DI-KU-LA-RA. Let's make a drawing:
The base of the perpendicular is the point in which it crosses the abscissa axis (axis) is the point. Figure shows that it has coordinates :. We are interested in the abscissa - that is, the "ounce" component. It is equal. Answer: . Task 3. Under the conditions of the previous task, find the amount of distance from the point to the coordinate axes. The task is generally elementary if you know what is the distance from the point to the axes. You know? I hope, but still remind you of: So, on my drawing, located just above, I have already portrayed one such perpendicular? What is the axis? To the axis. And what is the length of his length then? It is equal. Now I have a perpendicular to the axis and find it length. It will be equal, right? Then their amount is equal. Answer: . Task 4. In terms of problems 2, find the order of the point, the symmetrical point relative to the abscissa axis. I think you are intuitively clear what symmetry is? Very many objects it possess: many buildings, tables, airplanes, many geometric shapes: ball, cylinder, square, rhombus, etc. Roughly speaking, symmetry can be understood as: the figure consists of two (or more) the same halves. Such symmetry is called axial. What then is the axis? This is the same line in which the figure can, relatively speaking, "cut" on the same halves (in this picture the axis of symmetry is straight):
Now let's go back to our task. We know that we are looking for a point, symmetric about the axis. Then this axis is the symmetry axis. So we need to mention such a point so that the axis can cut the segment into two equal parts. Try yourself to celebrate such a point. And now compare with my decision:
Did you do the same? Okay! At the found point we are interested in ordinary. It is equal Answer: And now tell me, thinking seconds, what will the abscissa point, the symmetrical point A relative to the ordinate axis? What is your answer? Correct answer: . In the general case, the rule can be written like this: The point, symmetric point relative to the abscissa axis, has coordinates: Point, symmetrical point relative to the ordinate axis, has coordinates: Well, now quite terrible a task: Find the coordinates of the point, symmetric point, relative to the start of the coordinates. At first, think about yourself, and then look at my drawing!
Answer: Now Pollogram problem: Task 5: Points of Java-Way-Sia Ver-Shi-na Parale-le-Lo Gram Ma. Nay-die or-di-on-point.
You can solve this problem in two ways: the logic and method of coordinates. I first apply the coordinate method, and then I will tell you how to solve otherwise. It is clear that the abscissa point is equal. (It lies on a perpendicular conducted from the point to the abscissa axis). We need to find ordinate. We use the fact that our figure is a parallelogram, it means that. Find the length of the segment using the distance formula between the two points: Lower the perpendicular connecting the point with the axis. The intersection point will indicate the letter.
The length of the segment is equal. (Find the task itself, where we discussed this moment), then we find the length of the segment on the Pythagora theorem: The length of the segment - exactly coincides with its ordinate. Answer: . Another solution (I'll just give a picture that illustrates it)
Solution: 1. Conduct 2. Find the coordinates of the point and length 3. Prove that. One more cut length problem: Points of Java-lyube-Sia Ver-Shi-on-Mi Tre-Coal-Ni. Nai di the length of its medium line, Parale-Lelle.
Do you remember what is the middle line of the triangle? Then for you this task is elementary. If you do not remember, I will remind you: the middle line of the triangle is a line that connects the mid-opposite sides. It is parallel to the base and is equal to half a half. The base is a segment. Its length we had to look earlier, it is equal. Then the length of the middle line is half smaller and equal. Answer: . Comment: This task can be solved in another way to which we turn a little later. In the meantime, now you have a few tasks, take off on them, they are completely simple, but help "fill the hand", on the use of the coordinate method! 1. Points of Java-La-Sia Ver-Shi-on-Tour-Pennation. Nai ds the length of its environment line.
2. Points and Java-Wa-Sia Ver-Shi-na Parale-le-Lo Gram Ma. Nay-die or-di-on-point.
3. Nay-di length from cut-ka, co-unit-ny-y-th point and
4. Nai di-tures of the Krai-shan-like f-gu-ry on the co-or-di-nu flat-co-po.
5. Surrounding with a price-troom in the on-cha-le co-or-di-nat Pro-Ho-dit through the point. Nay-di her ra di-musty.
6. NAY-DI-DI-DI-SCHIE-NO-POCI, OPI-SAN-NOE ROD-MO-COMPUT-NI-KA, VER-ShI-RO-RO -Di-on-you co-from-vet
Solutions: 1. It is known that the middle line of the trapezium is equal to half the base. The base is equal, and the base. Then Answer: 2. The easiest way to solve this task is: notice that (the rule of the parallelogram). Calculate the coordinates of the vectors and is not possible :. In addition, the coordinate vectors are folded. Then has coordinates. The same coordinates also have a point, since the beginning of the vector is a point with coordinates. We are interested in ordinary. It is equal. Answer: 3. We act right away by the distance formula between the two points: Answer: 4. Look at the picture and say, between which two figures "clamped" the shaded region? It is clamped between two squares. Then the area of \u200b\u200bthe desired figure is equal to the square of a large square minus the square is small. The side of a small square is a segment connecting points and its length is equal Then the small square square is equal Similarly, with a big square: his side is a segment connecting points and its length is equal Then the large square square is equal Place the desired figure will find by the formula: Answer: 5. If the circle has the origin as a center and passes through the point, its radius will exactly equal to the length of the segment (make a drawing and you will understand why it is obvious). Find the length of this segment: Answer: 6. It is known that the radius of the circumference described near the rectangle is equal to half of its diagonal. We will find the length of any of two diagonals (after all, in a rectangle they are equal!) Answer: Well, you coped with everything? It was not very difficult to figure out, because so? The rule here is one thing - to be able to make a visual picture and simply "count" from it all the data. We left quite a bit. There are still literally two points that I would like to discuss. Let's try to decide this is such a simple task. Let two points and. Find the coordinates of the middle of the segment. The solution to this task is the following: Let the point - the search for the middle, then coordinates: I.e: the coordinates of the middle of the segment \u003d the arithmetic average of the corresponding coordinates of the ends of the segment. This rule is very simple and as a rule does not cause difficulties in students. Let's see what tasks and how it is used: 1. Nay-di or-di-on-tu se-di-di from-cut, co-unit-ny-yu-th point and
2. Points of Java-lyube-Sia Ver-Shi-na-mi-twh-coal-ni-ka. Nay-di or-di-on-ta dots of his di-go-on-lei.
3. Nai-di ABS-SU-SU price-tra surroundings of the neighborhood, opi-san-san near the right-Mo-Ni-ka, the Ver-Shi-RO Co-or-di-on-you co-ot-vet.
Solutions: 1. The first task is just a classic. We act immediately by definition of the middle of the segment. It has coordinates. Ordinate is equal. Answer: 2. It is easy to see that this quadrilateral is a parallelogram (even rhombus!). You yourself can prove it yourself, the calculation of the length of the parties and comparing them between themselves. What do I know about parallelograms? His diagonal point of intersection is divided in half! Yeah! So the point of intersection of diagonals is what? This is a middle of any of the diagonals! Select, in particular, diagonal. Then the point has the coordinates of the ordinate point equal. Answer: 3. What is the coincidence of the center described near the Circle Rectangle? It coincides with the intersection point of his diagonals. And what do you know about the diagonal of the rectangle? They are equal and the intersection point is divided by half. The task was drove to the previous one. I will take, for example, diagonal. Then if the center of the described circle, then the middle. Looking for coordinates: Absissal is equal. Answer: Now practice a little alone, I will only give answers to each task so that you can check yourself. 1. NAY-DI-TE-DI-SCHIE-NO-EI, opi-san about Tre-coal-ni-ka, Ver-Shi-GO-RO have co-or-di -no misters
2. NAY-DI-TE-DI-OU-TU-TUR DISCIEMBER-NOES, OPI-SAN-NOE PROTAGE-NI-KA, VER-Shi-GO-RO have coordinates 3. Ka-Ko-go-di-u-sa must-on being surrounded with a price-triple at point so that she can sa-las axis ABS?
4. Na-di or-di-on-ta dots of the axis of the axis and from-cut, the co-unit-yu-th point and
Answers: Everything succeeded? I really hope for it! Now - the last jerk. Now be especially attentive. The material that I will now explain now is directly related not only to simple tasks on the coordinate method from b of the part, but also occurs everywhere in the task C2. Which of my promises I have not yet restrained? Remember what operations on vectors I promised to enter and what ultimately introduced? I did not forget exactly? Forgot! I forgot to explain what the multiplication of vectors means. There are two ways to multiply vector on the vector. Depending on the selected method, we will have objects of different nature: Vector product is performed pretty cunning. How to do it and why it is necessary, we will discuss in the next article. And in this we will focus on the scalar product. There are already two ways to allow us to calculate: As you guessed, the result should be the same! So, let's first consider the first way: Scalar product through coordinates Find: - generally accepted indication of a scalar product Formula for calculation Next: That is, the scalar product \u003d the amount of the works of the coordinates of the vectors! Example: Nai di
Decision: We will find the coordinates of each of the vectors: Calculate the scalar product by the formula: Answer: See, absolutely nothing complicated! Well, now try myself: · Nay-di SKA-LAR-NEE pro-from-ve-de -ity of events and
Cope? Maybe I noticed the trick small? Let's check: The coordinates of the vectors as in the past task! Answer:. In addition to the coordinate, there is another way to calculate a scalar product, namely, through the lengths of vectors and cosine angle between them: Indicates the angle between the vectors and. That is, the scalar product is equal to the product of the lengths of the vectors on the cosine of the corner between them. Why do we have this second formula if we have the first one that is much easier, there are at least no cosine in it. And it is necessary for the fact that from the first and second formula we can withdraw how to find the angle between vectors! Let then remember the formula for the length of the vector! Then if I substitute this data in the formula of the scalar product, then I will get: But on the other side: So what did I get to you? We now have a formula that allows you to calculate the angle between two vectors! Sometimes it is also written for brevity as follows: That is, the algorithm for calculating the angle between the vectors is as follows:
Let's practice in the examples: 1. Nay-di the corner between the eyelid-ray and. Give the answer in gra-du-sac.
2. Under the conditions of the previous task, find the cosine between the vectors. We will do this: the first task I will help you decide, and try to do the second yourself! I agree? Then begin! 1. These vector are our old familiar. We already considered their scalar work and it was equal. They have such coordinates :, Then we find their lengths: Then we are looking for a cosine between vectors: Kosinus which corner is equal? This is the angle. Answer: Well, now I myself solve the second task, and then compare! I will give only a very brief solution: 2. It has coordinates, has coordinates. Let - the angle between the vectors and, then Answer: It should be noted that the tasks are directly in the vector and the coordinate method in the part b of the examination work is rather rare. However, the overwhelming majority of C2 tasks can be easily solved by resorting to the introduction of the coordinate system. So you can consider this article by the foundation, on the basis of which we will make enough tricky construction, which will be needed to solve complex tasks. Coordinates and vectors. Middle rovingWe continue to study the coordinate method. In the last part, we brought a number of important formulas that allow:
Of course, the entire coordinate method does not fit into these 6 points. It underlies such science as an analytical geometry with which you have to get to know the university. I just want to build a foundation that will allow you to solve problems in a single state. Exam. With the tasks of the part B we figured out in now it's time to go to a qualitatively new level! This article will be devoted to the method of solving those C2 tasks, in which it will be reasonable to move to the coordinate method. This rationality is determined by the fact that the task is required to find and which figure is given. So, I would apply the coordinate method if you are issued:
If the figure in the condition of the problem is the body of rotation (ball, cylinder, cone ...) Suitable figures for the coordinate method are:
Also in my experience it is impractical to use the coordinate method for:
However, it should be immediately noted that three "unprofitable" for the method of coordinate situation in practice is quite rare. In most tasks, he can become your Savior, especially if you are not very strong in three-dimensional buildings (which sometimes are quite intricate). What are all the above figures? They are no longer flat, such as, for example, a square, triangle, circle, and bulk! Accordingly, we need to consider not a two-dimensional, but a three-dimensional coordinate system. It is built fairly easy: just besides the axis of the abscissa and ordinates, we introduce another axis, the appliquet axis. The figure schematically shows their mutual location:
All of them are mutually perpendicular, intersect at one point, which we will call the start of coordinates. The axis of the abscissa, as before, we denote the axis of the ordinate - and the introduced axis of the application -. If earlier each point on the plane was characterized by two numbers - abscissa and ordinary, then each point in space is already described by three numbers - abscissa, ordinate, applikate. For example:
Accordingly, the abscissa of the point is equal, ordinate -, and the applicatis -. Sometimes the abscissa point is also called the projection of the point on the abscissa axis, the ordinate - the projection of the point on the axis of the ordinate, and the application - the projection of the point on the appliquet axis. Accordingly, if the point is set, the point with the coordinates: call the projection point to the plane call the projection point to the plane Natural question arises: Are all formulas derived for a two-dimensional case in space? The answer is affirmative, they are fair and have the same look. For a small detail. I think you have already guessed yourself for which one. In all formulas, we must add another member responsible for the appliquet axis. Namely. 1. If two points are set: then:
2. If two versions are given: and then:
However, the space is not so simple. As you understand, adding another coordinate makes a significant variety in the spectrum of figures, "living" in this space. And for further narration, I need to introduce some, roughly speaking, "generalization" straight. This "generalization" will be plane. What do you know about the plane? Try to answer the question, and what is a plane? It is very difficult to say. However, we all intuitively imagine how it looks like:
Roughly speaking, this is a non-infinite "leaf", covered in space. "Infinity" should be understood that the plane applies to all directions, that is, its area is equal to infinity. However, this explanation "on the fingers" does not give the slightest idea of \u200b\u200bthe structure of the plane. And it will be interested in it. Let's remember one of the main axes of geometry:
Or its analogue in space:
Of course, you remember how to remove the equation directly in two predetermined points: if the first point has coordinates: and the second, then the direct equation will be as follows: That you passed in the 7th grade. In the space, the direct equation looks like this: let us give two points with coordinates :, the equation is straight, through them passing, has the appearance: For example, through points, the straight line passes: How should it be understood? This should be understood as: the point lies on the line if its coordinates satisfy the following system: We will not really be interested in the equation straight, but we need to pay attention to the very important concept of the direct vectors. - Any nonzero vector lying on this direct or parallel to it.
For example, both vectors are guide vectors direct. Let the point lying on the line, and its guide vector. Then the equation direct can be written in the following form: I repeat once again, I will not be very interested in the equation straight, but I really need you to remember what the guide vector is! Again: this is any nonsense vector lying on a straight line, or parallel to it. Display the equation of the plane for three specified points Not so trivially, and usually this question is not considered aware of high school. And in vain! This technique is vital when we resort to the coordinate method for solving complex tasks. However, I assume that you are full of desire to learn something new? Moreover, you can hit your teacher at the university when it turns out that you already know how you are already with a technique that is usually studied in the course of analytical geometry. So, proceed. The plane equation is not too different from the direct equation on the plane, namely it looks: some numbers (not all equal zero), and variables, for example: etc. As you can see, the equation of the plane is not very different from the equation of the straight line (linear function). However, remember that we have argued with you? We said that if we have three points that are not lying on one straight line, the equation of the plane is definitely restored by them. But how? I will try to explain to you. Since the plane equation is: And the points belong to this plane, then when substituting the coordinates of each point to the plane equation, we must receive a true identity: Thus, it becomes necessary to solve three equations already with unknown! Dilemma! However, it can always be assumed that (for this you need to divide on). Thus, we obtain three equations with three unknowns: However, we will not solve such a system, and we will divert the mysterious expression that follows from it: The equation of the plane passing through three setpoints\\ [\\ left | (\\ begin (array) (* (20) (C)) (x - (x_0)) & ((x_1) - (x_0)) & ((x_2) - (x_0)) \\\\ (y - (y_0) ) & ((y_1) - (y_0)) & ((y_2) - (y_0)) \\\\ (z - (z_0)) & ((z_1) - (z_0)) & ((z_2) - (z_0)) \\ END (Array)) \\ Right | \u003d 0 \\] Stop! What else is what? Some very unusual module! However, the object that you see in front of yourself has nothing to do with the module. This object is called the third order determinant. From now on, in the future, when you are dealing with the coordinate method on the plane, you will very often meet these identifies. What is the third order determinant? Oddly enough, it is just a number. It remains to understand what specifically the number we will compare with the determinant. Let's first write the third order determinant in a more general form: Where are some numbers. And under the first index, we understand the line number, and under the index - the number of the column. For example, it means that this number is at the intersection of the second line and the third column. Let's raise the following question: How exactly will we calculate such a determinant? That is, what specific number will we compare him? For a third-order determinant, there is a heuristic (visual) triangle rule it looks like this:
If you write all these numbers, then we will get the following expression: Nevertheless, remembering the method of calculation in this form is not necessary, it is enough in the head just to keep triangles and the idea itself, which is what it makes up and what is then deducted from something). Let's illustrate the triangle method on the example: 1. Calculate the determinant: Let's deal with what we fold, and what - we subtract: The components that go with the "plus": This is the main diagonal: the product of the elements is equal The first triangle, "perpendicular main diagonal: the product of elements is equal The second triangle, "perpendicular main diagonal: the product of the elements is equal We fold three numbers: The components that go with the "minus" This is a side diagonal: the product of the elements is equal The first triangle, "perpendicular to the side diagonal: the product of the elements is equal The second triangle, "perpendicular to the side diagonal: the product of the elements is equal We fold three numbers: All that remains to do is to deduct from the sum of the terms "with a plus" amount of the terms "with a minus": In this way, As you can see, nothing complicated and supernatural in the calculation of the third order determinants is not. It is just important to remember about the triangles and not allow arithmetic errors. Now try to calculate yourself: Check:
Here is another couple of determinants, they calculated their meanings on their own and compare with the answers: Answers: Well, all the coincided? Great, then you can move on! If there are difficulties, then the Council is mine: there is a bunch of software for calculating the determinant online. All you need is to come up with your identifier, calculate it yourself, and then compare with what the program will consider. And so as long as the results do not start the coincidence. I'm sure this moment will not wait long to wait! Now let's go back to the determinant who I wrote when he spoke about the equation of the plane passing through three setpoints: All you need is to calculate its value directly (by the method of triangles) and equate the result to zero. Naturally, because - variables, then you will get some expression, depending on them. It is this expression that will be the equation of a plane passing through three set points that are not lying on one straight line! Let's illustrate the above-mentioned example: 1. Construct the equation of the plane passing through points We write the determinant for these three points: Simplify: Now we calculate it directly according to the rule of triangles: \\ [(\\ left | (\\ begin (array) (* (20) (C)) (x + 3) & 2 & 6 \\\\ (y - 2) & 0 & 1 \\ (Z + 1) & 5 & 0 \\ END (Array)) \\ \\ CDOT 5 \\ CDOT 6 -) \\] Thus, the equation of the plane passing through the points has the form: Now try to solve one task on your own, and then we will discuss it: 2. Find the equation of the plane passing through the points Well, let's now discuss the decision: We make a determinant: And calculate its value: Then the equation of the plane is: Or, shorting on, we get: Now two tasks for self-control:
Answers: All coincided? Again, if there are certain difficulties, then my advice is: you take three points from my head (with a great degree of probability they will not lie on one straight), build a plane on them. And then check yourself online. For example, on the site: However, with the help of determinants, we will build not only the equation of the plane. Remember, I told you that the vectors defined not only the scalar product. There is still a vector, as well as a mixed work. And if the scalar product of two vectors and there will be a number, then the vector product of two vectors and will be vector, and this vector will be perpendicular to the specified:
Moreover, its module will be equal to the area of \u200b\u200bthe parallelogram, which is preceded by vectors and. This vector will need to calculate the distance from the point to direct. How do we consider the vector product of vectors and, if their coordinates are set? The third order determinant comes to the rescue. However, before I proceed to the algorithm for calculating the vector art, I have to make a small lyrical retreat. This retreat concerns basic vectors. Schematically, they are depicted in the picture:
What do you think, why are they called basic? The fact is that : Or in the picture:
The justice of this formula is obvious, because: Vector artNow I can proceed to the introduction of a vector work: The vector product of two vectors is called the vector that is calculated by the following rule: Now let's give a few examples of the vector art calculation: Example 1: Find vector vectors: Solution: I make up the determinant: And calculate it: Now from writing through basic vectors, I will return to the usual recording of the vector: In this way: Now try. Ready? Check: And traditionally two tasks for control:
Answers: Mixed product of three vectorsThe last design that will need me is a mixed product of three vectors. It, as well as scalar, is a number. There are two ways to calculate it. - Through the determinant, - through a mixed work. Namely, let us have three versions: Then the mixed product of three vectors, denoted by can be calculated as: 1. - That is, a mixed product is a scalar product of a vector on a vector product of two other vectors. For example, a mixed product of three vectors is: Independently try to calculate it through a vector product and make sure that the results will match! And again - two examples for self solutions: Answers: Select the coordinate systemWell, now we have all the necessary foundation of knowledge to solve complex stereometric tasks on geometry. However, before proceeding directly to the examples and algorithms of their decision, I believe that it will be useful to stop still on what question: how exactly select the coordinate system for a particular figure. After all, it is the choice of the mutual location of the coordinate system and the figure in space will ultimately determine how cumbersome will be calculations. I remind you that in this section we consider the following figures:
For rectangular parallelepipeda or cube, I recommend that you build:
That is, I will put an "into the angle". Cube and parallelepiped are very good figures. For them, you can always easily find the coordinates of his vertices. For example, if (as shown in Figure)
the coordinates of the vertices are as follows: To remember it, of course, no need to remember how it is better to have a cube or rectangular parallelepiped - preferably. Direct prismPrism is a more harmful figure. Its located in space can be different. However, I seem most acceptable to me: Triangular Prism:
That is, one of the sides of the triangle we fully put on the axis, and one of the vertices coincides with the start of coordinates. Hexagonal prism:
That is, one of the vertices coincides with the start of coordinates, and one of the parties lies on the axis. Quadrangular and hexagonal pyramid:
Situation, similar to Cuba: Two sides of the base we combine with the coordinate axes, one of the vertices we combine with the beginning of the coordinates. The only minor complexity will calculate the coordinates of the point. For hexagonal pyramid - similarly as for a hexagonal prism. The main task is again in the search for the coordinates of the vertex.
Tetrahedron (triangular pyramid)
The situation is very similar to the one that I led for a triangular prism: one peak coincides with the beginning of the coordinates, one side lies on the coordinate axis. Well, now we are finally close to getting close to solving problems. From what I said at the very beginning of the article, you could make this conclusion: most C2 tasks are divided into 2 categories: challenges at the angle and tasks per distance. At first, we will consider the tasks of finding the angle. They, in turn, are divided into the following categories (as complexity increases): Tasks for searching corners
Let's consider these tasks consistently: let's start by finding the angle between two straight. Well, remember, and did we decide with you similar examples before? I remember, because we had something like that ... We were looking for the angle between two vectors. I will remind you if two versions are given: and, the corner between them is from the ratio: Now we have a goal - finding an angle between two straight. Let's turn to the "flat picture":
How much did the corners do with the intersection of two straight lines? Already pieces. The truth is not equal from them only two, others are vertical to them (and therefore they coincide with them). So what kind of angle should we be considered an angle between two straight: or? Here is a rule: the angle between two direct always no more than degrees. That is, from two angles, we will always choose an angle with the smallest degree. That is, in this picture, the angle between two straight is equal. To do not bother with a search for the smallest of two angles, sly mathematics offered to use the module. Thus, the angle between two direct is determined by the formula: You, like a careful reader, had to arouse the question: And where, in fact, we will take these the most numbers that we need to calculate the cosine of the corner? Answer: We will take them from direct vectors! Thus, the algorithm for finding an angle between two straight lines is as follows:
Or in more detail:
Well, now it's time to move to the tasks: the solution of the first two I will demonstrate in detail, I will introduce another decision in short form, and I will only give answers to the last two tasks, you should spend all the calculations to them. Tasks: 1. In Pra-Ville-Nome Tet-Ra-Ed-Rea Nai di, the corner between you-co-Tet-Ra-ED-RA and the mea-di-bo-ko-coordinates. 2. In the Pra-Ville-Neu-coal Pi-Ra-Mi-de Stro-Ros, OS-Na-Viya are equal, and the bouquet of ribs are equal, Nay-di the corner between the straight and. 3. The lengths of all the ribs of Pra-Ville Che-you-Rah-Coal Pi-Ra-Mi are equal to each other. Nai-di the angle between the straighties and if from-re-zok - the co-alone given Pi-Ra-Mi-dwi, the point is se-re-di-on her bouquet rib 4. On the edge of the cube from-me-per point so that the nai-di the corner between the straight and 5. Point - se-re-di-on edges Cuba Nai-di the corner between straight and. I misunderstood the tasks in this order. While you have not had time to start navigate in the coordinate method, I myself disassemble the most "problem" figures, and you will give you to deal with the simplest cube! Gradually, you have to learn to work with all the figures, the complexity of the tasks I will increase from the topic to the topic. We proceed to solving problems: 1. Draw a tetrahedron, put it in the coordinate system as I designed earlier. Since the tetrahed is correct - then all its faces (including base) - the right triangles. Since we are not given the length of the side, then I can take it equal. I think you understand that the angle will not really depend on how our tetrahedron will "stretch"?. Also spend in Tetrahedra height and median. Along the way, I paint its base (it will also come in handy).
I need to find the angle between and. What do we know? We know only the point coordinate. So, it is necessary to find more coordinates of points. Now we think: the point is the point of intersection of heights (or bisetriss or median) of the triangle. And the point is a raised point. The point is the middle of the segment. Then we definitively need to find: the coordinates of the points :. Let's start with the simplest: the coordinates of the point. Look at Figure: It is clear that the point of the point is zero (the point lies on the plane). Her ordinate is equal to (since - the median). It is harder to find it abscissa. However, it is easily done on the basis of the Pythagora theorem: Consider a triangle. Its hypotenuse is equal, and one of the cathets is equal then: Finally we have :. Now we find the coordinates of the point. It is clear that her appliquet is again zero, and its ordinate is the same as the point, that is. Find her abscissa. This is done trivially, if you remember that the heights of the equilateral triangle of the intersection point are divided in proportion, counting from the top. Since:, then the desired abscissa point equal to the length of the segment is equal to :. Thus, the coordinates of the point are equal: Find the coordinates of the point. It is clear that its abscissa and ordinate coincide with the abscissa and the ordinary point. And the applicature is equal to the length of the segment. - This is one of the cathes of the triangle. Triangle hypotenuse is a cut - catat. He is looking for for reasons that I highlighted a bold: The point is the middle of the segment. Then we need to remember the formula of the coordinates of the middle of the segment: Well, all, now we can search the coordinates of the guide vectors: Well, everything is ready: we substitute all the data in the formula: In this way, Answer: You should not scare such "scary" answers: for the tasks C2 is common practice. I would rather surprised the "beautiful" answer in this part. Also, as you noted, I practically did not resort to anything, except for the Pythagoreo theorem and the property of the equilateral triangle heights. That is, to solve the stereometer task, I used the minimum of stereometry. The win in this partially "quenching" is quite bulky computing. But they are algorithm enough! 2. I will show the correct hexagonal pyramid along with the coordinate system, as well as its base:
We need to find the angle between straight and. Thus, our task is reduced to the search for the coordinates of the points :. The coordinates of the last three we will find on a small pattern, and we will find the coordinate of the vertices through the point coordinate. Works in bulk, but you need to start it! a) The coordinate: it is clear that its applicatis and ordinate are equal to zero. We find the abscissa. To do this, consider a rectangular triangle. Alas, we are known only by hypotenuse, which is equal. Watch we will try to find (for it is clear that the doublest length of the category will give us an abscissue point). How do we look for her? Let's remember that for the figure we lie at the base of the pyramid? This is the right hexagon. What does it mean? This means that he has all the parties and all the corners are equal. It would be necessary to find one such angle. Any ideas? Ideas mass, but there is a formula: The sum of the corners of the correct N-parliament is equal to . Thus, the sum of the angles of the correct hexagon is equal to degrees. Then each of the corners is equal to: We look at the picture again. It is clear that the cut - bisector angle. Then the angle is equal to degrees. Then: Then, from where. Thus, has coordinates b) Now you can easily find the point coordinate :. c) We will find the coordinates of the point. Since its abscissa coincides with the length of the segment, it is equal to. It is not very difficult to find the ordinate: if we connect the points and a point of intersection of direct designation, let's say for. (Make a simple construction itself). Then thus, the ordinate point B is equal to the sum of the lengths of the segments. Re-turn to the triangle. Then Then because then the point has coordinates d) Now we will find the coordinates of the point. Consider a rectangle and prove that thus point coordinates: e) It remains to find the coordinates of the vertices. It is clear that its abscissa and ordinate coincides with the abscissa and the ordinary point. We find the applicant. Since, then. Consider a rectangular triangle. By the condition of the problem, the lateral edge. This is the hypotenus of my triangle. Then the height of the pyramid - catat. Then the point has coordinates: Well, everything, I have the coordinates of all points of interest to me. I am looking for coordinates of direct vectors of direct: We are looking for an angle between these vectors: Answer: Again, when solving this task, I did not use any irregular techniques, except for the formula of the amount of the corners of the correct N-square, as well as the definition of cosine and the sine of the rectangular triangle. 3. Since we again are not given the length of the ribs in the pyramid, then I will consider them equal to one. Thus, since all ribs, not just side, are equal to each other, then at the base of the pyramid and the square is lying, and the side faces are the right triangles. We will show this pyramid, as well as its base on the plane, noting all the data given in the text of the task:
We are looking for the angle between and. I will make very short calculations when I search for the coordinates of points. You will need to "decipher" them: b) - the middle of the segment. Its coordinates: c) Cut length I will find on the Pythagora theorem in the triangle. I will find on the Pythagorean theorem in the triangle. Coordinates: d) - midt of segment. Its coordinates are equal e) vector coordinates f) vector coordinates g) We are looking for an angle: Cube is the simplest figure. I am sure that you will deal with her yourself. Answers to tasks 4 and 5 are as follows: Finding the angle between the straight and planeWell, the time of simple tasks is over! Now the examples will be even more difficult. To find the corner between the straight and plane, we will be as follows:
As you can see, this formula is very similar to the fact that we used to search for corners between two straight. The structure of the right part is simply the same, and we are now looking for sinus now, and not cosine, as before. Well, one opposite action was added - searching the equation of the plane. Let's not postpone in a long box solution of examples: 1. Os-no-va-ni-it is a straight procure-lap-ben-smta Raughty-na-de-re-coal nickname you-so-one prize is equal. Nay-di the corner between straight and flat-co-st 2. In the direct-Mo-Mr. Pa-Ral-les-Le-Pi-de-Die from the Nay-di-level corner between straight and flat-co - 3. In Pra-Ville, the neck-coal prize-alto all ribs are equal. Nai-di the corner between straight and flat-co-st. 4. In Pra-Ville Tre-Coal Pi-Ra-Mi-de with Os-no-Va Ni-West-Na-Di-Thief, Obra-Zo-Wan a flat-co-co-copy of Os-no-va and straight, pro-ho-son through the re-di ribs and 5. The lengths of all the ribs of pra-vil-ote four-born Pi-Ra-Mi-dy are equal to each other. Nay-di the angle between straight and flat-co-stew if the point is Ce-re-di-on-co-co-rib Pi-Ra-Mi-dy. Again, I will decide the first two tasks in detail, the third - briefly, and the last two leave you for an independent decision. In addition, you have already had to deal with triangular and quadrangular pyramids, but with prisms - so far there are no. Solutions: 1. Show a prism as well as its base. It is compatible with the coordinate system and note all the data that is given in the Terk Condition:
I apologize for some non-compliance with the proportions, but to solve the problem, it is essentially not so important. The plane is simply the "rear wall" of my prism. It is enough just to guess that the equation of such a plane is: However, it can be shown directly: Choose arbitrary three points on this plane: for example,. Make a plane equation: Exercise to you: independently calculate this determinant. Did you succeed? Then the equation of the plane is: Or simply In this way, To solve the example, I need to find the coordinates of the guide vector straight. Since the point has fallen with the beginning of coordinates, the vector coordinates simply coincide with the coordinates of the point for this we will find at the beginning of the coordinates of the point. To do this, consider a triangle. We will spend the height (it is median and bisector) from the top. Since, the ordinate point is equal. In order to find the abscissa of this point, we need to calculate the length of the segment. According to Pythagora theorem, we have: Then the point has coordinates: The point is "raised" to the point: Then the coordinates of the vector: Answer: As you can see, there is nothing fundamentally difficult in solving such tasks. In fact, the process further simplifies "straight" such a figure as Prism. Now let's move on to the following example: 2. Draw a parallelepiped, we carry out a plane and direct, as well as separately draw its lower base:
First we find the plane equation: the coordinates of the three points lying in it: (The first two coordinates are obtained by an obvious way, and the last coordinate you can easily find pictures from the point). Then constitute the equation of the plane: Calculate: We are looking for the coordinates of the guide vector: it is clear that its coordinates coincide with the coordinates of the point, is not it? How to find coordinates? This is the coordinates of the point raised along the appliquet axis per unit! . Then look for the desired angle: Answer: 3. Put the correct hexagonal pyramid, and then spend the plane and direct.
There is even a plane to draw a problem, not to mention the solution of this task, however, the coordinate method is still! It is in his versatility and is its main advantage! The plane passes through three points :. We are looking for their coordinates: one) . Himself output coordinates for the last two points. You will be useful for this solution to the challenge with a hexagonal pyramid! 2) we build the equation of the plane: We are looking for the coordinates of the vector :. (see the task with a triangular pyramid again!) 3) We are looking for an angle: Answer: As you can see, nothing is supernatural in these tasks. It is only necessary to be very careful with roots. To the last two tasks I will give only answers: How could you make sure the technique of solving tasks everywhere the same: the main task to find the coordinates of the vertices and substitute them into certain formulas. We left to consider another class of challenges for calculating the corners, namely: Calculation of angles between two planesThe algorithm solutions will be:
As you can see, the formula is very similar to the previous two, with which we searched the corners between straight and between the straight and plane. So remember this you will not be much difficulty. We immediately go to the analysis of tasks: 1. ST-RO-OS-NO-VIL-VIL-VILTER TRE-CONSIDE CONSUITE WHERE, and DI-HALL BO-KO-CO-COP is equal. Nay-di the angle between the F-Co-Stew and the F-Co-Stew Os-no-Viya prize-we. 2. In Pra-Ville-Mi-Deh-Coal Pi-Ra-Mi-de, all ribs are equal, the sinus of the corner between the F-Co-Stew and Co-Stew, Pro-Ho-Fith through the Pen-Pen-Di-Liar Pen-Di-Liar, but straight. 3. In the correct che-the-coal-coal prize of ST-Ro-us, Os-Na-Via is equal, and the bouquet of edges are equal. On the edge of from-me - to the point so that. Find the angle between the flat-ko-mi and 4. In the Pra-Willian four-born prize-ours, the OS-NA-VIA is equal, and the bou--way Ribra is equal. On the edge of from-me-a point so that the nai-di the corner between the flat-ko-mi and. 5. In Cuba, the Nau-Di Ko-Si-Nus angle between the flat-co-stey and Tasks solutions: 1. Rise the correct (at the base is an equilateral triangle) a triangular prism and note on it planes that appear in the condition of the problem:
We need to find the equations of two planes: the base equation is obtained trivial: you can make an appropriate determinant for three points, I will constitute the equation immediately: Now we will find the point equation has the coordinates of the point - as it is the median and the height of the triangle, it is easily located on the Pythagora theorem in the triangle. Then the point has coordinates: find the application point for this Consider a rectangular triangle Then we obtain these coordinates: we will compose the equation of the plane. Calculate the angle between the planes: Answer: 2. Make drawing: The most difficult thing is to understand that this is such a mysterious plane passing through the point perpendicularly. Well, the main thing is that? The main thing is attentiveness! In fact, direct is perpendicular. Straight is also perpendicular. Then the plane passing through these two straight lines will be perpendicular to the straight, and, by the way, pass through the point. This plane also passes through the top of the pyramid. Then the desired plane - and the plane is already given to us. We are looking for the coordinates of the points. The coordinate of the point will find through the point. From a small drawing it is easy to remove that the coordinates of the point will be like that: what is left now to find to find the coordinates of the peak of the pyramid? You still need to calculate its height. This is done with the help of the same Pythagore's theorem: first prove that (trivially of small triangles, forming a square at the base). Since under the condition, we have: Now everything is ready: the coordinates of the vertices: Make a plane equation: You are already special in the calculation of the determinants. Without difficult, you will get: Or otherwise (if there are both parts for the root of two) Now we find the equation of the plane: (You didn't forget how we get the equation of the plane, however? If you do not understand where this minus one came from, then come back to the definition of the equation of the plane! Just always before that it turned out that my plane belonged to the beginning of the coordinates!) Calculate the determinant: (You can notice that the equation of the plane coincided with the equation of direct passing through the points and! Think why!) Now we calculate the angle: We also need to find sinus: Answer: 3. Caverny question: What is a rectangular prism, what do you think? This is just a particularly well-known parallelepiped! Immediately do the drawing! You can even separately do not depict the basis, the benefits of it here is a bit here:
The plane, as we have already noticed before, is written in the form of equation: Now make a plane The equation is the equation of the plane: We are looking for an angle: Now answers to the last two tasks: Well, now it's time to rest a little, because we are great and you have done a huge job! Coordinates and vectors. Advanced levelIn this article we will discuss with you another class of tasks that can be solved using the coordinate method: tasks for calculating the distance. Namely, we will consider the following cases:
I ordered these tasks as their complexity increases. Most just turns out to find distance from point to planeAnd the most difficult thing is to find distance between cross-country straight. Although, of course, there is nothing impossible! Let's not postpone in a long box and immediately proceed to consider the first class of tasks: Calculating the distance from the point to the planeWhat do we need to solve this task? 1. The coordinates of the point So, as soon as we get all the necessary data, we use the formula: As we build the equation of the plane you should already be known from the previous tasks that I understood in the last part. Let's immediately proceed to tasks. The scheme is as follows: 1, 2 - I help you to decide, and quite detail, 3, 4 - Only the answer, the decision you spend and compare. Started! Tasks: 1. Dan Cube. Cube's edge length is equal. Nai di-Thief from Ce-di-di from-cut to flat-co-st 2. Dana Pra-Vil-Naya Che-Mi-Ya-Coal-Naya Pi-Mi-da Boe-co-co-rib Stro-Ro-on Os-No-Via is equal. NAY-DI-THAT STO-YAST from the point to the flat-co-comma where - se-re-di-on ribs. 3. In Pra-Ville Tre-Coal Pi-Ra-Mi-de with Os-No-Va-Ni-Ko-Co-Way Equally, and a hundred-ro-on Os-na- equal to. Nai di-Thief from the Ver-Shi-we to flat-co-st. 4. In the Pra-Ville seam-coal prize-alone, all Ribra is equal. NAY-DI-THAT STO-YAST from point to flat-co-st. Solutions: 1. Draw a cube with single edges, we build a segment and a plane, the middle of the segment we denote the letter
At first, let's start with the lung: find the coordinates of the point. Since that (remember the coordinates of the middle of the segment!) Now we compile the equation of the three-point plane \\ [\\ left | (\\ Begin (Array) (* (20) (C)) X & 0 & 1 \\\\ Y & 1 & 0 \\\\ Z & 1 & 1 \\ END (Array)) \\ Right | \u003d 0 \\] Now I can proceed to the search for the distance: 2. We begin again from the drawing, where we celebrate all the data! For the pyramid it would be useful separately to draw its base.
Even the fact that I paint as a chicken paw does not prevent us with ease to solve this task! Now it's easy to find the point coordinates As the coordinates of the point, then 2. Since the coordinates of the point A - the middle of the segment, then We also find the coordinates of two more points on the plane to make the equation of the plane and simplify it: \\ [\\ left | (\\ Left | (\\ Begin (Array) (* (20) (C)) X & 1 & (\\ FRAC (3) (2)) \\\\ Y & 0 & (\\ FRAC (3) (2)) \\\\ Z & 0 & (\\ FRAC ( (\\ SQRT 3)) (2)) \\ END (Array)) \\ Right |) \\ RIGHT | \u003d 0 \\] Since the point has coordinates:, then we calculate the distance: Answer (very rare!): Well, figured out? It seems to me that everything is also technically as in those examples that we considered with you in the previous part. So I am sure that if you have mastered the material, then you will not be difficult to solve the remaining two tasks. I will only give answers: Calculation of distance from direct to the planeIn fact, there is nothing new here. How can the straight and plane relative to each other can be located? They have all the possibilities: cross, or directly parallel to the plane. What do you think, the equal distance from a straight line to the plane with which this direct intersects? It seems to me that it is clear that the distance is zero. Uninteresting case.
The second case is cunning: the distance is already nonzero. However, since the straight parallel plane, then each point is equivalent to this plane:
In this way: This means that my task was drilled to the previous one: we are looking for the coordinates of any point on a straight line, we look for the equation of the plane, calculate the distance from the point to the plane. In fact, such tasks in the exam are extremely rare. I managed to find only one task, and the data in it were such that the coordinate method was not very applied to it! We now turn to another, a much more important class of tasks: Calculation of the distance point to directWhat do we need? 1. The coordinates of the point from which we are looking for the distance: 2. Coordinates of any point lying on the line 3. Direct vectors direct coordinates What formula? What does the denominator of this fraction mean to you and so it should be clear: it is the length of the guide vector straight. Here is a very cunning numerator! The expression means the module (length) of the vector product of vectors and how to calculate the vector work, we were studied in the previous part of the work. Update your knowledge, they will now be very useful to us! Thus, the problem solving algorithm will be the following: 1. We are looking for the coordinates of the point from which we are looking for the distance: 2. We are looking for the coordinates of any point on the line, which we are looking for the distance: 3. Build vector 4. Build the line guide vector 5. Calculate vector art 6. We are looking for the length of the resulting vector: 7. Calculate the distance: We have a lot of work, and examples will be quite complicated! So now focus all the attention! 1. Dana Pra-Vil-Naya Tre-Coal-Naya Pi-Ra-Mi-yes with Ver-Shih. One-Ro-on Os-no-Viya Pi-Ra-Mi-dya is equal, you-so-so is equal. Nai di-Thief from the se d-di-ko-co-rib to straight, where the points and - the re-di ribs and co-from- VET-DOWN. 2. The lengths of the ribs and the direct-m-coal-but-goa Parale-le-le-pi-yes are equal to co-from-vessels - but also the NAI-Di-Thief from Ver-Shi-re-direct 3. In the Pra-Ville seam-coal prize-alone, all ribs are equal to the NAI-Di-Thief Route from point to straight Solutions: 1. Make a neat drawing, which mark all the data:
We have a lot of work with you! I would first like to describe the words that we will look for and in what order: 1. The coordinates of the points and 2. Coordinates of the point 3. The coordinates of the points and 4. The coordinates of vectors and 5. Their vector art 6. Vector length 7. Vector length length 8. Distance from to Well, we have a lot of work! We are accepted for her, drossing the sleeves! 1. To find the coordinates of the pyramid height, we need to know the coordinates of the point of its applicature equal to zero, and the ordinate is equal to the abscissa of it is equal to the length of the segment since the height of the equilateral triangle, then it is divided into relationship, counting from the top, hence. Finally, they received coordinates: Coordinates of the point 2. - Middle Cut 3. - Mid Segment Mid-cut 4.Copinates Coordinates of the vector 5. Calculate vector art:
6. The length of the vector: the easiest way is to replace that the segment is the middle line of the triangle, which means it is equal to half the base. So that. 7. We consider the length of the vector work: 8. Finally, we find the distance: UV, well! Honestly, I will say: the solution to this problem with traditional methods (through construction) would be much faster. But here I all reduced to the finished algorithm! So I think that the algorithm is clear to you? Therefore, I will ask you to solve the remaining two tasks yourself. Compare answers? Again, I repeat: these tasks are easier (faster) to solve through the constructions, and not resorting to the coordinate method. I demonstrated such a solution to the decision only to show you a universal method that allows you to hold anything. " Finally, consider the last class of tasks: Calculating the distance between cross-country straightHere, the algorithm for solving tasks will be similar to the previous one. What we have: 3. Any vector connecting points first and second straight: How are we looking for the distance between straight? The formula is as follows: The numerator is a module of a mixed product (we were administered in the previous part), and the denominator is as in the previous formula (module of the vector product of the direct vectors of direct, the distance between which we are looking for). I will remind you that
then the formula for the distance can be rewritten in the form: Single determinant to share the determinant! Although, to be honest, I'm not at all joking here! This formula, in fact, is very cumbersome and leads to sufficiently complex calculations. In your place, I would resort to it only in the most extreme case! Let's try to solve several tasks using the method outlined above: 1. In Pra-Ville, the Tre-Coal Prize, all Ribr Ko-Roy are equal, the Nai di-Those Rouner of the straight and. 2. Dana-Vil-Naya Tre-Coal-Naya Prize-Ma All Rib Os-no-Via-Way is equal to se-human, pro-Ho-grandfather through the Bo-Co-Woe The edge and se-re-di is the ribs of Java-Lhaa-Xia Kvad-Ra-Tom. NAY-DI-THES ROSTOZHONE BETWEEN RIATS AND I solve the first, and relying on it, you decide the second! 1. I draw a prism and mark straight and
The coordinates of the point C: then Coordinates of the point Coordinates of the vector Coordinates of the point Coordinates of the vector Coordinates of the vector \\ [\\ left ((B, \\ Overrightarrow (A (A_1)) \\ Overrightarrow (B (B_1))) \\ Right) \u003d \\ left | (\\ Begin (Array) (* (20) (L)) (\\ Begin (Array) (* (20) (C)) 0 & 1 & 0 End (Array)) \\\\ (\\ Begin (Array) (* (20) (C)) 0 & 0 & 1 \\ END (Array)) \\\\ (\\ Begin (Array) (* (20) (C)) (\\ FRAC ((\\ SQRT 3)) (2)) & (- \\ FRAC (1) (2)) & 1 \\ End (Array)) \\ END (Array)) \\ RIGHT | \u003d \\ FRAC ((\\ SQRT 3)) (2) \\] We consider vector product between vectors and \\ [\\ Overrightarrow (A (A_1)) \\ CDOT \\ OVERRIGHTARROW (B (C_1)) \u003d \\ left | \\ Begin (Array) (L) \\ Begin (Array) (* (20) (C)) (\\ Overrightarrow i) & (\\ Overrightarrow J) & (\\ Overrightarrow K) \\ END (Array) \\\\\\ Begin (Array ) (* (20) (C)) 0 & 0 & 1 \\ END (Array) \\\\\\ Begin (Array) (* (20) (C)) (\\ FRAC ((\\ SQRT 3)) (2)) & (- \\ - \\ FRAC ((\\ SQRT 3)) (2) \\ Overrightarrow K + \\ FRAC (1) (2) \\ Overrightarrow i \\] Now we consider it a length: Answer: Now try to accurately fulfill the second task. The answer to it will be :. Coordinates and vectors. Brief description and basic formulasVector - directional cut. - Beginning of vector, - Content vector. Absolute valuevector - Cut length depicting vector. Referred to as. Coordinates of the vector: , Sum of vectors :. Making vectors: Scalar product of vectors: The scalar product of vectors is equal to the product of their absolute values \u200b\u200bon the cosine of the angle between them: Become a student YouClever, Prepare for OGE or EGE in mathematics, And also access YouClever's textbook without restrictions ... Video Tutorial "Geographical latitude and geographical longitude. Geographic coordinates "will help you get an idea of \u200b\u200bgeographic latitude and geographical longitude. The teacher will tell how to properly determine the geographical coordinates. Geographic latitude- The length of the arc in degrees from the equator to a given point. To determine the latitude of the object, you need to find the parallel on which this object is located. For example, the latitude of Moscow is 55 degrees and 45 minutes of northern latitude, it is written as follows: Moscow 55 ° 45 "S.Sh.; New York latitude - 40 ° 43" S.Sh.; Sydney - 33 ° 52 "Yu.Sh. Geographical longitude is determined by meridians. Longitude can be Western (from 0 Meridian to west to 180 meridian) and Eastern (from 0 Meridian east to 180 meridian). The value of longitudes is measured in degrees and minutes. Geographical longitude may have a value from 0 to 180 degrees. Geographic longitude- The length of the Equator arc in degrees from the initial meridian (0 degrees) to the meridian of the specified point. Greenwich Meridian is considered the initial meridian (0 degrees).
Fig. 2. Definition of longitude () To determine the longitude, you need to find the meridian on which the specified object is located. For example, the longitude of Moscow - 37 degrees and 37 minutes of Eastern longitude, it is written in this way: 37 ° 37 "VD; longitude of Mexico City - 99 ° 08" ZD
Fig. 3. Geographic latitude and geographic longitude To accurately locate the object on the surface of the Earth, it is necessary to know its geographical latitude and geographical longitude. Geographical coordinates- The values \u200b\u200bdetermining the position of the point on the earth's surface with the help of latitudes and longitude. For example, Moscow has the following geographic coordinates: 55 ° 45 "S.Sh. and 37 ° 37" V.D. The city of Beijing has the following coordinates: 39 ° 56 'S.Sh. 116 ° 24 'VD First, the value of the latitude is recorded. Sometimes it is required to find an object on the already specified coordinates, for this you must first assume, in which hemispheres is this object. Homework Paragraphs 12, 13. 1. What is a geographical latitude and longitude? Bibliography Basic 1. The initial course of geography: studies. for 6 cl. general education. institutions / etc. Gerasimova, N.P. Nezlukov. - 10th ed., Stereotype. - M.: Drop, 2010. - 176 p. 2. Geography. 6 cl.: Atlas. - 3rd ed., Stereotype. - M.: Drop, Dick, 2011. - 32 s. 3. Geography. 6 cl.: Atlas. - 4th ed., Stereotype. - M.: Drop, Dick, 2013. - 32 s. 4. Geography. 6 CL.: CONT cards. - M.: Dick, Drop, 2012. - 16 p. Encyclopedias, Dictionaries, Directory and Statistical Collections 1. Geography. Modern illustrated encyclopedia / A.P. Gorkin. - M.: Rosman-Press, 2006. - 624 p. Literature for preparing for GIA and EGE 1. Geography: initial course. Tests. Studies. Handbook for students of 6 cl. - M.: Humanit. ed. Center Vlados, 2011. - 144 p. 2. Tests. Geography. 6-10 CL.: Educational and methodical manual / A.A. Letlyagin. - M.: Agency "Agency" Krp "Olymp": "Astrel", "AST", 2001. - 284 p. Materials on the Internet 1. Federal Institute of Pedagogical Measurements (). 2. Russian Geographical Society (). |
Fig. 2. System of flat rectangular coordinates on maps
Fig. 3. Determination of the geographic coordinates of the point on the map (point A) and applying a point by geographic coordinates (point b)
Fig. 2. Determination of the rectangular coordinates of the point on the map (point B) and applying a point on the rectangular coordinates (point d)
Fig. 3. A - polar coordinates; b - bipolar coordinates
Fig. 4. Application of the goal on the card direct serif from two points.
Fig. 5. Camering from the conditional line
and
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