Each point on the planet's surface has a specific position, which corresponds to its own latitude and longitude coordinates. It is located at the intersection of the spherical arcs of the meridian, which corresponds to longitude, with the parallel, which corresponds to latitude. It is denoted by a pair of angular quantities expressed in degrees, minutes, seconds, which has the definition of a coordinate system.

Latitude and longitude are the geographic aspect of a plane or sphere translated into topographic images. To more accurately locate a point, its altitude above sea level is also taken into account, which makes it possible to find it in three-dimensional space.

Latitude and longitude

The need to find a point using latitude and longitude coordinates arises due to the duty and occupation of rescuers, geologists, military personnel, sailors, archaeologists, pilots and drivers, but it may also be necessary for tourists, travelers, seekers, and researchers.

What is latitude and how to find it

Latitude is the distance from an object to the equator line. Measured in angular units (such as degrees, degrees, minutes, seconds, etc.). Latitude on a map or globe is indicated by horizontal parallels - lines that describe a circle parallel to the equator and converge in the form of a series of tapering rings towards the poles.

Lines of latitude

Therefore, they distinguish the northern latitude - this is the entire part earth's surface north of the equator, as well as south - this is the entire part of the planet’s surface south of the equator. The equator is the zero, longest parallel.

• Parallels from the equator line to the north pole are considered to be a positive value from 0° to 90°, where 0° is the equator itself, and 90° is the top north pole. They are counted as northern latitude (N).
• Parallels extending from the equator to the side south pole, indicated by a negative value from 0° to -90°, where -90° is the location of the south pole. They are counted as southern latitude (S).
• On the globe, parallels are depicted as circles encircling the ball, which become smaller as they approach the poles.
• All points on the same parallel will be designated by the same latitude, but different longitudes.
  On maps, based on their scale, parallels have the form of horizontal, curved stripes - the smaller the scale, the straighter the parallel strip is depicted, and the larger it is, the more curved it is.

Remember! The closer to the equator a given area is located, the smaller its latitude will be.

What is longitude and how to find it

Longitude is the amount by which the position of a given area is removed relative to Greenwich, that is, the prime meridian.

Lines of longitude

Longitude is similarly characterized by measurement in angular units, only from 0° to 180° and with a prefix - eastern or western.

• The Greenwich Prime Meridian vertically encircles the globe of the Earth, passing through both poles, dividing it into the western and eastern hemispheres.
• Each of the parts located west of Greenwich (in the Western Hemisphere) will be designated west longitude (w.l.).
• Each of the parts distant from Greenwich to the east and located in the eastern hemisphere will bear the designation east longitude (E.L.).
• Finding each point along one meridian has the same longitude, but different latitude.
• Meridians are drawn on maps in the form of vertical stripes curved in the shape of an arc. The smaller the map scale, the straighter the meridian strip will be.

How to find the coordinates of a given point on the map

Often you have to find out the coordinates of a point that is located on the map in a square between the two nearest parallels and meridians. Approximate data can be obtained by eye by sequentially estimating the step in degrees between the mapped lines in the area of ​​interest, and then comparing the distance from them to the desired area. For accurate calculations you will need a pencil with a ruler, or a compass.

• For the initial data we take the designations of the parallels closest to our point with the meridian.
• Next, we look at the step between their stripes in degrees.
• Then we look at the size of their step on the map in cm.
• We measure with a ruler in cm the distance from a given point to the nearest parallel, as well as the distance between this line and the neighboring one, convert it to degrees and take into account the difference - subtracting from the larger one, or adding to the smaller one.
• This gives us the latitude.

Example! The distance between the parallels 40° and 50°, among which our area is located, is 2 cm or 20 mm, and the step between them is 10°. Accordingly, 1° is equal to 2 mm. Our point is 0.5 cm or 5 mm away from the fortieth parallel. We find the degrees to our area 5/2 = 2.5°, which must be added to the value of the nearest parallel: 40° + 2.5° = 42.5° - this is our northern latitude of the given point. In the southern hemisphere, the calculations are similar, but the result has a negative sign.

Similarly, we find longitude - if the nearest meridian is further from Greenwich, and the given point is closer, then we subtract the difference, if the meridian is closer to Greenwich, and the point is further, then we add it.

If you only have a compass at hand, then each of the segments is fixed with its tips, and the spread is transferred to the scale.

In a similar way, calculations of coordinates on the surface of the globe are carried out.

The best services for finding a place by coordinates

The easiest way to find out your location is by logging into the PC version of the service, which works directly with Google Maps. Many utilities make it easy to enter latitude and longitude in a browser. Let's look at the best of them.

Map & Directions

In addition, Maps & Directions allows you to determine the coordinates of your position on the map for free by clicking just one button. Click on “Find my coordinates”, and the service will immediately place a marker and determine the latitude, longitude to many thousandths, as well as altitude.

On the same site you can measure the distance between settlements or the area of ​​any given territory, draw a route or calculate travel time. The service will be useful for both travelers and simply curious users.

Mapcoordinates.net

A useful utility, Mapcoordinates.net, allows you to find out the coordinates of a point in any region of the world. The service is also integrated with Google Maps, but has a simplified interface, thanks to which even an untrained user can use it.

In the address bar of the utility, where it says “Search,” enter the address of the place, latitude and longitude of which you want to get. A map with coordinates will appear along with a marker at the desired location. The latitude, longitude, and altitude of the selected point will be displayed above the marker.

Unfortunately, Mapcoordinates.net is not suitable for searching for points knowing their coordinates. However, for the reverse procedure, this is a very convenient utility. The service supports many languages, including Russian.

Search by coordinates on the map through a browser using the Google Maps service

If for some reason you prefer to work not with simplified services, but directly with Google Maps, then these instructions will be useful for you. The process of searching by coordinates through Google Maps is a little more complicated than in the methods described earlier, but it can be mastered quickly and without much difficulty.

To find out the exact coordinates of a place, follow these simple instructions:

Open the service on your PC. It is important that the full mode must be turned on, and not the light mode (marked with a special lightning icon), otherwise it will not be possible to obtain information;

Click on the section of the map where the item or point you need is located, with the right mouse button;

Select the “What’s here?” option in the menu that appears;

Look at the tab that appears at the bottom of the screen. It will display latitude, longitude and altitude.

To determine a location using known geographic coordinates, a different procedure will be required:

1. Open Google Maps in full mode on your computer;

   In the search bar at the top of the screen you can enter coordinates. This can be done in the following formats: degrees, minutes and seconds; degrees and decimal minutes; decimal degrees;

Press the “Enter” key, and a special marker will appear on the map at the required location.

Most important when using Google service Maps should be specified correctly geographical coordinates. Cards only recognize a few data formats, so be sure to keep the following input rules in mind:

When entering degrees, use the special character to indicate it as "°" rather than "d";

You must use a dot rather than a comma as a separator between the integer and fractional parts, otherwise the search string will not be able to return the location;

Latitude is indicated first, then longitude. The first parameter must be written in the range from -90 to 90, the second - from -180 to 180.

Finding a special character on a PC keyboard is difficult, and in order to adhere to the required list of rules, you need to put in a lot of effort. It is much easier to use special utilities - we have listed the best of them in the section above.

Finding a place by latitude and longitude on Android OS

Often you need to find a place by coordinates far from your laptop or personal computer. Will help out mobile app Google Maps running on the Android platform. It is usually used to get directions or find out the schedule. Vehicle, however, the program is also suitable for finding the location of an item or point.

You can download the application for Android on the official page on Google Play. It is available in both Russian and English languages. After installing the program, follow the following instructions:

Open Google Maps on your device and wait for the map to appear;

Find a place that interests you. Click on it and hold until a special marker appears;

A tab will appear at the top of the screen with a search window and full coordinates of the location;

If you need to find a place by coordinates, and not vice versa, then the method is mobile device no different from its PC counterpart.

The mobile version of the service, like the one running on a PC, will allow you to study the desired location in detail, find out its exact coordinates, or vice versa, recognize the address using known data. This is a convenient way both at home and on the road.

In this article, we will begin to discuss one “magic wand” that will allow you to reduce many geometry problems to simple arithmetic. This “stick” can make your life much easier, especially when you feel unsure of constructing spatial figures, sections, etc. All this requires a certain imagination and practical skills. The method that 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:

1. Coordinate plane
2. Points and vectors on the plane
3. Constructing a vector from two points
4. Vector length (distance between two points)​
5. Coordinates of the middle of the segment
6. Dot product of vectors
7. Angle between two vectors​

I think you've already guessed why the coordinate method is called that? That's right, it got this name because it operates not with geometric objects, but with their numerical characteristics (coordinates). And the transformation itself, which allows us to move from geometry to algebra, consists in introducing a coordinate system. If the original figure was flat, then the coordinates are two-dimensional, and if the figure is three-dimensional, then the coordinates are three-dimensional. In this article we will consider only the two-dimensional case. And the main goal of the article is to teach you how to use some basic techniques of the coordinate method (they sometimes turn out to be useful when solving problems on planimetry in Part B of the Unified State Exam). The next two sections on this topic are devoted to a discussion of methods for solving problems C2 (the problem of stereometry).

Where would it be logical to start discussing the coordinate method? Probably from the concept of a coordinate system. Remember when you first encountered her. It seems to me that in 7th grade, when you learned about the existence of a linear function, for example. Let me remind you that you built it point by point. Do you remember? You chose an arbitrary number, substituted it into the formula and calculated it that way. For example, if, then, if, then, etc. What did you get in the end? And you received points with coordinates: and. Next, you drew a “cross” (coordinate system), chose a scale on it (how many cells you will have as a unit segment) and marked the points you obtained on it, which you then connected with a straight line; the resulting line is the graph of the function.

There are a few points here that should be explained to you in a little more detail:

1. You choose a single segment for reasons of convenience, so that everything fits beautifully and compactly in the drawing.

2. It is accepted that the axis goes from left to right, and the axis goes from bottom to top

3. They intersect at right angles, and the point of their intersection is called the origin. It is indicated by a letter.

4. In writing the coordinates of a point, for example, on the left in parentheses there is the coordinate of the point along the axis, and on the right, along the axis. In particular, it simply means that at the point

5. In order to specify any point on the coordinate axis, you need to indicate 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 x-axis

9. The axis is called the y-axis

Now let's take the next step: mark two points. Let's connect these two points with a segment. And we’ll put the arrow as if we were drawing a segment from point to point: that is, we’ll make our segment directed!

Remember what another directional segment is called? That's right, it's called a vector!

So if we connect dot to dot, and the beginning will be point A, and the end will be point B, then we get a vector. You also did this construction in 8th grade, remember?

It turns out that vectors, like 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 a vector to find its coordinates? It turns out that yes! And this is done very simply:

Thus, since in a vector the point is the beginning and the point is the end, the vector has the following coordinates:

For example, if, then the coordinates of the vector

Now let's do the opposite, find the coordinates of the vector. What do we need to 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 will be at the point. Then:

Look carefully, what is the difference between vectors and? Their only difference is the signs in the coordinates. They are opposites. This fact is usually written like this:

Sometimes, if it is not specifically stated which point is the beginning of the vector and which is the end, then vectors are denoted by more than two in capital letters, and one lowercase, for example: , etc.

Now a little practice yourself and find the coordinates of the following vectors:

Examination:

Now solve a slightly more difficult problem:

A vector with a beginning at a point has a co-or-di-na-you. Find the abs-cis-su points.

All the same is quite prosaic: Let be the coordinates of the point. Then

I compiled the system based on the definition of what vector coordinates are. Then the point has coordinates. We are interested in the abscissa. Then

Answer:

What else can you do with vectors? Yes, almost everything is the same as with ordinary numbers (except that you can’t divide, but you can multiply in two ways, one of which we will discuss here a little later)

1. Vectors can be added to each other
2. Vectors can be subtracted from each other
3. Vectors can be multiplied (or divided) by an arbitrary non-zero number
4. Vectors can be multiplied by each other

All these operations have a very clear geometric representation. For example, the triangle (or parallelogram) rule for addition and subtraction:

A vector stretches or contracts or changes direction when multiplied or divided by a number:

However, here we will be interested in the question of what happens to the coordinates.

1. When adding (subtracting) two vectors, we add (subtract) their coordinates element by element. That is:

2. When multiplying (dividing) a vector by a number, all its coordinates are multiplied (divided) by this number:

For example:

· Find the amount of co-or-di-nat century-to-ra.

Let's first find the coordinates of each of the vectors. They both have the same origin - the origin point. Their ends are different. Then, . Now let's calculate the coordinates of the vector. Then the sum of the coordinates of the resulting vector is equal.

Answer:

Now solve the following problem yourself:

· Find the sum of vector coordinates

We check:

Let's now consider the following problem: we have two points on coordinate plane. How to find the distance between them? Let the first point be, and the second. Let us denote the distance between them by. Let's make the following drawing for clarity:

What I've done? First of all, I connected dots and,a also from a point I drew a line parallel to the axis, and from a point I drew a line parallel to the axis. Did they intersect at a point, forming a remarkable figure? What's so special about her? Yes, you and I know almost everything about the right triangle. Well, the Pythagorean theorem for sure. The required segment is the hypotenuse of this triangle, and the segments are the legs. What are the coordinates of the point? Yes, they are easy to find from the picture: Since the segments are parallel to the axes and, respectively, their lengths are easy to find: if we denote the lengths of the segments by, respectively, then

Now let's use the Pythagorean theorem. We know the lengths of the legs, we will find the hypotenuse:

Thus, the distance between two points is the root of the sum of the squared differences from the coordinates. Or - the distance between two points is the length of the segment connecting them. It is easy to see that the distance between points does not depend on the direction. Then:

From here we draw three conclusions:

Let's practice a little bit about calculating the distance between two points:

For example, if, then the distance between and is equal to

Or let's go another way: find the coordinates of the vector

And find the length of the vector:

As you can see, it's the same thing!

Now practice a little yourself:

Task: find the distance between the indicated points:

We check:

Here are a couple more problems using the same formula, although they sound a little different:

1. Find the square of the length of the eyelid.

2. Find the square of the length of the eyelid

I think you dealt with them without difficulty? We check:

1. And this is for attentiveness) We have already found the coordinates of the vectors earlier: . Then the vector has coordinates. The square of its length will be equal to:

2. Find the coordinates of the vector

Then the square of its length is

Nothing complicated, right? Simple arithmetic, nothing more.

The following problems cannot be classified unambiguously; they are more about general erudition and the ability to draw simple pictures.

1. Find the sine of the angle from the cut, connecting the point, with the abscissa axis.

And

How are we going to proceed here? We need to find the sine of the angle between and the axis. Where can we look for sine? That's right, in a right triangle. So what do we need to do? Build this triangle!

Since the coordinates of the point are and, then the segment is equal to, and the segment. We need to find the sine of the angle. Let me remind you that sine is the ratio of the opposite side to the hypotenuse, then

What's left for us to do? Find the hypotenuse. You can do this in two ways: using the Pythagorean theorem (the legs are known!) or using the formula for the distance between two points (in fact, the same thing as the first method!). I'll go the second way:

Answer:

The next task will seem even easier to you. She is on the coordinates of the point.

Task 2. From the point the per-pen-di-ku-lyar is lowered onto the ab-ciss axis. Nai-di-te abs-cis-su os-no-va-niya per-pen-di-ku-la-ra.

Let's make a drawing:

The base of a perpendicular is the point at which it intersects the x-axis (axis), for me this is a point. The figure shows that it has coordinates: . We are interested in the abscissa - that is, the “x” component. She is equal.

Answer: .

Task 3. In the conditions of the previous problem, find the sum of the distances from the point to the coordinate axes.

The task is generally elementary if you know what the distance from a point to the axes is. You know? I hope, but still I will remind you:

So, in my drawing just above, have I already drawn one such perpendicular? Which axis is it on? To the axis. And what is its length then? She is equal. Now draw a perpendicular to the axis yourself and find its length. It will be equal, right? Then their sum is equal.

Answer: .

Task 4. In the conditions of task 2, find the ordinate of a point symmetrical to the point relative to the abscissa axis.

I think it is intuitively clear to you what symmetry is? Many objects have it: many buildings, tables, airplanes, many geometric figures: ball, cylinder, square, rhombus, etc. Roughly speaking, symmetry can be understood as follows: a figure consists of two (or more) identical halves. This symmetry is called axial symmetry. What then is an axis? This is exactly the line along which the figure can, relatively speaking, be “cut” into equal halves (in this picture the axis of symmetry is straight):

Now let's get back to our task. We know that we are looking for a point that is symmetrical about the axis. Then this axis is the axis of symmetry. This means that we need to mark a point such that the axis cuts the segment into two equal parts. Try to mark such a point yourself. Now compare with my solution:

Did it work out the same way for you? Fine! We are interested in the ordinate of the found point. It is equal

Answer:

Now tell me, after thinking for a few seconds, what will be the abscissa of a point symmetrical to point A relative to the ordinate? What is your answer? Correct answer: .

In general, the rule can be written like this:

A point symmetrical to a point relative to the abscissa axis has the coordinates:

A point symmetrical to a point relative to the ordinate axis has coordinates:

Well, now it's completely scary task: find the coordinates of a point symmetrical to the point relative to the origin. You first think for yourself, and then look at my drawing!

Answer:

Now parallelogram problem:

Task 5: The points appear ver-shi-na-mi pa-ral-le-lo-gram-ma. Find or-di-on-that point.

You can solve this problem in two ways: logic and the coordinate method. I'll use the coordinate method first, and then I'll tell you how you can solve it differently.

It is quite clear that the abscissa of the point is equal. (it lies on the perpendicular drawn from the point to the abscissa axis). We need to find the ordinate. Let's take advantage of the fact that our figure is a parallelogram, this means that. Let's find the length of the segment using the formula for the distance between two points:

We lower the perpendicular connecting the point to the axis. I will denote the intersection point with a letter.

The length of the segment is equal. (find the problem yourself where we discussed this point), then we will find the length of the segment using the Pythagorean theorem:

The length of a segment coincides exactly with its ordinate.

Answer: .

Another solution (I'll just give a picture that illustrates it)

Solution progress:

1. Conduct

2. Find the coordinates of the point and length

3. Prove that.

Another one segment length problem:

The points appear on top of the triangle. Find the length of its midline, parallel.

Do you remember what the middle line of a triangle is? Then this task is elementary for you. If you don’t remember, I’ll remind you: the middle line of a triangle is the line that connects the midpoints of opposite sides. It is parallel to the base and equal to half of it.

The base is a segment. We had to look for its length earlier, it is equal. Then the length of the middle line is half as large and equal.

Answer: .

Comment: this problem can be solved in another way, which we will turn to a little later.

In the meantime, here are a few problems for you, practice on them, they are very simple, but they help you get better at using the coordinate method!

1. The points are the top of the tra-pe-tions. Find the length of its midline.

2. Points and appearances ver-shi-na-mi pa-ral-le-lo-gram-ma. Find or-di-on-that point.

3. Find the length from the cut, connecting the point and

4. Find the area behind the colored figure on the co-ordi-nat plane.

5. A circle with a center in na-cha-le ko-or-di-nat passes through the point. Find her ra-di-us.

6. Find-di-te ra-di-us of the circle, describe-san-noy about the right-angle-no-ka, the tops of something have a co-or -di-na-you are so-responsible

Solutions:

1. It is known that the midline of a trapezoid is equal to half the sum of its bases. The base is equal, and the base. Then

Answer:

2. The easiest way to solve this problem is to note that (parallelogram rule). Calculating the coordinates of vectors is not difficult: . When adding vectors, the coordinates are added. Then has coordinates. The point also has these coordinates, since the origin of the vector is the point with the coordinates. We are interested in the ordinate. She is equal.

Answer:

3. We immediately act according to the formula for the distance between two points:

Answer:

4. Look at the picture and tell me which two figures the shaded area is “sandwiched” between? It is sandwiched between two squares. Then the area of ​​the desired figure is equal to the area of ​​the large square minus the area of ​​the small one. The side of a small square is a segment connecting the points and Its length is

Then the area of ​​the small square is

We do the same with a large square: its side is a segment connecting the points and its length is

Then the area of ​​the large square is

We find the area of ​​the desired figure using the formula:

Answer:

5. If a circle has the origin as its center and passes through a point, then its radius will be exactly equal to length segment (make a drawing and you will understand why this is obvious). Let's find the length of this segment:

Answer:

6. It is known that the radius of a circle circumscribed about a rectangle is equal to half its diagonal. Let's find the length of any of the two diagonals (after all, in a rectangle they are equal!)

Answer:

Well, did you cope with everything? It wasn't very difficult to figure it out, was it? There is only one rule here - be able to make a visual picture and simply “read” all the data from it.

We have very little left. There are literally two more points that I would like to discuss.

Let's try to solve this simple problem. Let two points and be given. Find the coordinates of the midpoint of the segment. The solution to this problem is as follows: let the point be the desired middle, then it has coordinates:

That is: coordinates of the middle of the segment = the arithmetic mean of the corresponding coordinates of the ends of the segment.

This rule is very simple and usually does not cause difficulties for students. Let's see in what problems and how it is used:

1. Find-di-te or-di-na-tu se-re-di-ny from-cut, connect-the-point and

2. The points appear to be the top of the world. Find-di-te or-di-na-tu points per-re-se-che-niya of his dia-go-na-ley.

3. Find-di-te abs-cis-su center of the circle, describe-san-noy about the rectangular-no-ka, the tops of something have co-or-di-na-you so-responsibly-but.

Solutions:

1. The first problem is simply a classic. We proceed immediately to determine the middle of the segment. It has coordinates. The ordinate is equal.

Answer:

2. It is easy to see that this quadrilateral is a parallelogram (even a rhombus!). You can prove this yourself by calculating the lengths of the sides and comparing them with each other. What do I know about parallelograms? Its diagonals are divided in half by the point of intersection! Yeah! So what is the point of intersection of the diagonals? This is the middle of any of the diagonals! I will choose, in particular, the diagonal. Then the point has coordinates The ordinate of the point is equal to.

Answer:

3. What does the center of the circle circumscribed about the rectangle coincide with? It coincides with the intersection point of its diagonals. What do you know about the diagonals of a rectangle? They are equal and the point of intersection divides them in half. The task was reduced to the previous one. Let's take, for example, the diagonal. Then if is the center of the circumcircle, then is the midpoint. I'm looking for coordinates: The abscissa is equal.

Answer:

Now practice a little on your own, I’ll just give the answers to each problem so you can test yourself.

1. Find-di-te ra-di-us of the circle, describe-san-noy about the tri-angle-no-ka, the tops of something have a co-or-di -no misters

2. Find-di-te or-di-on-that center of the circle, describe-san-noy about the triangle-no-ka, the tops of which have coordinates

3. What kind of ra-di-u-sa should there be a circle with a center at a point so that it touches the ab-ciss axis?

4. Find-di-those or-di-on-that point of re-se-ce-tion of the axis and from-cut, connect-the-point and

Answers:

Was everything successful? I really hope for it! Now - the last push. Now be especially careful. The material that I will now explain is directly related not only to simple problems on the coordinate method from Part B, but is also found everywhere in Problem C2.

Which of my promises have I not yet kept? Remember what operations on vectors I promised to introduce and which ones I ultimately introduced? Are you sure I haven't forgotten anything? Forgot! I forgot to explain what vector multiplication means.

There are two ways to multiply a vector by a vector. Depending on the chosen method, we will get objects of different natures:

The cross product is done quite cleverly. We will discuss how to do it and why it is needed in the next article. And in this one we will focus on the scalar product.

There are two ways that allow us to calculate it:

As you guessed, the result should be the same! So let's look at the first method first:

Dot product via coordinates

Find: - generally accepted notation for scalar product

The formula for calculation is as follows:

That is, the scalar product = the sum of the products of vector coordinates!

Example:

Find-di-te

Solution:

Let's find the coordinates of each of the vectors:

We calculate the scalar product using the formula:

Answer:

See, absolutely nothing complicated!

Well, now try it yourself:

· Find a scalar pro-iz-ve-de-nie of centuries and

Did you manage? Maybe you noticed a small catch? Let's check:

Vector coordinates, as in the previous problem! Answer: .

In addition to the coordinate one, there is another way to calculate the scalar product, namely, through the lengths of the vectors and the cosine of the angle between them:

Denotes the angle between the vectors and.

That is, the scalar product is equal to the product of the lengths of the vectors and the cosine of the angle between them.

Why do we need this second formula, if we have the first one, which is much simpler, at least there are no cosines in it. And it is needed so that from the first and second formulas you and I can deduce how to find the angle between vectors!

Let Then remember the formula for the length of the vector!

Then if I substitute this data into the scalar product formula, I get:

But in other way:

So what did you and I get? We now have a formula that allows us to calculate the angle between two vectors! Sometimes it is also written like this for brevity:

That is, the algorithm for calculating the angle between vectors is as follows:

1. Calculate the scalar product through coordinates
2. Find the lengths of the vectors and multiply them
3. Divide the result of point 1 by the result of point 2

Let's practice with examples:

1. Find the angle between the eyelids and. Give the answer in grad-du-sah.

2. In the conditions of the previous problem, find the cosine between the vectors

Let's do this: I'll help you solve the first problem, and try to do the second yourself! Agree? Then let's begin!

1. These vectors are our old friends. We have already calculated their scalar product and it was equal. Their coordinates are: , . Then we find their lengths:

Then we look for the cosine between the vectors:

What is the cosine of the angle? This is the corner.

Answer:

Well, now solve the second problem yourself, and then compare! I will give just a very short solution:

2. has coordinates, has coordinates.

Let be the angle between the vectors and, then

Answer:

It should be noted that the problems directly on vectors and the coordinate method in part B exam paper quite rare. However, the vast majority of C2 problems can be easily solved by introducing a coordinate system. So you can consider this article the foundation on the basis of which we will make some rather clever constructions that we will need to solve complex problems.

COORDINATES AND VECTORS. AVERAGE LEVEL

You and I continue to study the coordinate method. In the last part, we derived a number of important formulas that allow you to:

1. Find vector coordinates
2. Find the length of a vector (alternatively: the distance between two points)
3. Add and subtract vectors. Multiply them by a real number
4. Find the midpoint of a segment
5. Calculate dot product of vectors
6. Find the angle between vectors

Of course, the entire coordinate method does not fit into these 6 points. It underlies such a science as analytical geometry, which you will become familiar with at university. I just want to build a foundation that will allow you to solve problems in a single state. exam. We have dealt with the tasks of Part B. Now it’s time to move to a whole new level! This article will be devoted to a method for solving those C2 problems in which it would be reasonable to switch to the coordinate method. This reasonableness is determined by what is required to be found in the problem and what figure is given. So, I would use the coordinate method if the questions are:

1. Find the angle between two planes
2. Find the angle between a straight line and a plane
3. Find the angle between two straight lines
4. Find the distance from a point to a plane
5. Find the distance from a point to a line
6. Find the distance from a straight line to a plane
7. Find the distance between two lines

If the figure given in the problem statement is a body of rotation (ball, cylinder, cone...)

Suitable figures for the coordinate method are:

1. Rectangular parallelepiped
2. Pyramid (triangular, quadrangular, hexagonal)

Also from my experience it is inappropriate to use the coordinate method for:

1. Finding cross-sectional areas
2. Calculation of volumes of bodies

However, it should immediately be noted that the three “unfavorable” situations for the coordinate method are quite rare in practice. In most tasks, it can become your savior, especially if you are not very good at three-dimensional constructions (which can sometimes be quite intricate).

What are all the figures I listed above? They are no longer flat, like, for example, a square, a triangle, a circle, but voluminous! Accordingly, we need to consider not a two-dimensional, but a three-dimensional coordinate system. It is quite easy to construct: just in addition to the abscissa and ordinate axis, we will introduce another axis, the applicate axis. The figure schematically shows their relative position:

All of them are mutually perpendicular and intersect at one point, which we will call the origin of coordinates. As before, we will denote the abscissa axis, the ordinate axis - , and the introduced applicate axis - .

If previously each point on the plane was characterized by two numbers - the abscissa and the ordinate, then each point in space is already described by three numbers - the abscissa, the ordinate, and the applicate. For example:

Accordingly, the abscissa of a point is equal, the ordinate is , and the applicate is .

Sometimes the abscissa of a point is also called the projection of a point onto the abscissa axis, the ordinate - the projection of a point onto the ordinate axis, and the applicate - the projection of a point onto the applicate axis. Accordingly, if a point is given, then a point with coordinates:

called the projection of a point onto a plane

called the projection of a point onto a plane

A natural question arises: are all the formulas derived for the two-dimensional case valid in space? The answer is yes, they are fair and have the same appearance. For a small detail. I think you've already guessed which one it is. In all formulas we will have to add one more term responsible for the applicate axis. Namely.

1. If two points are given: , then:

• Vector coordinates:
• Distance between two points (or vector length)
• The midpoint of the segment has coordinates

2. If two vectors are given: and, then:

• Their scalar product is equal to:
• The cosine of the angle between the vectors is equal to:

However, space is not so simple. As you understand, adding one more coordinate introduces significant diversity into the spectrum of figures “living” in this space. And for further narration I will need to introduce some, roughly speaking, “generalization” of the straight line. This “generalization” will be a plane. What do you know about plane? Try to answer the question, what is a plane? It's very difficult to say. However, we all intuitively imagine what it looks like:

Roughly speaking, this is a kind of endless “sheet” stuck into space. “Infinity” should be understood that the plane extends in all directions, that is, its area is equal to infinity. However, this “hands-on” explanation does not give the slightest idea about the structure of the plane. And it is she who will be interested in us.

Let's remember one of the basic axioms of geometry:

• a straight line passes through two different points on a plane, and only one:

Or its analogue in space:

Of course, you remember how to derive the equation of a line from two given points; it’s not at all difficult: if the first point has coordinates: and the second, then the equation of the line will be as follows:

You took this in 7th grade. In space, the equation of a line looks like this: let us be given two points with coordinates: , then the equation of the line passing through them has the form:

For example, a line passes through points:

How should this be understood? This should be understood as follows: a point lies on a line if its coordinates satisfy the following system:

We will not be very interested in the equation of a line, but we need to pay attention to the very important concept of the direction vector of a line. - any non-zero vector lying on a given line or parallel to it.

For example, both vectors are direction vectors of a straight line. Let be a point lying on a line and let be its direction vector. Then the equation of the line can be written in the following form:

Once again, I won’t be very interested in the equation of a straight line, but I really need you to remember what a direction vector is! Again: this is ANY non-zero vector lying on a line or parallel to it.

Withdraw equation of a plane based on three given points is no longer so trivial, and usually this issue is not addressed in the course high school. But in vain! This technique is vital when we resort to the coordinate method to solve complex problems. However, I assume that you are eager to learn something new? Moreover, you will be able to impress your teacher at the university when it turns out that you already know how to use a technique that is usually studied in an analytical geometry course. So let's get started.

The equation of a plane is not too different from the equation of a straight line on a plane, namely, it has the form:

some numbers (not all equal to zero), but variables, for example: etc. As you can see, the equation of a plane is not very different from the equation of a straight line (linear function). However, remember what you and I argued? We said that if we have three points that do not lie on the same line, then the equation of the plane can be uniquely reconstructed from them. But how? I'll try to explain it to you.

Since the equation of the plane is:

And the points belong to this plane, then when substituting the coordinates of each point into the equation of the plane we should obtain the correct identity:

Thus, there is a need to solve three equations with unknowns! Dilemma! However, you can always assume that (to do this you need to divide by). Thus, we get three equations with three unknowns:

However, we will not solve such a system, but will write out the mysterious expression that follows from it:

Equation of a plane passing through three given points

\[\left| (\begin(array)(*(20)(c))(x - (x_0))&((x_1) - (x_0))&((x_2) - (x_0))\\(y - (y_0) )&((y_1) - (y_0))&((y_2) - (y_0))\\(z - (z_0))&((z_1) - (z_0))&((z_2) - (z_0)) \end(array)) \right| = 0\]

Stop! What is this? Some very unusual module! However, the object that you see in front of you has nothing to do with the module. This object is called a third-order determinant. From now on, when you deal with the method of coordinates on a plane, you will very often encounter these same determinants. What is a third order determinant? Oddly enough, it's just a number. It remains to understand what specific number we will compare with the determinant.

Let's first write the third-order determinant in a more general form:

Where are some numbers. Moreover, by the first index we mean the row number, and by the index we mean the column number. For example, it means that this number is at the intersection of the second row and third column. Let's pose the following question: how exactly will we calculate such a determinant? That is, what specific number will we compare to it? For the third-order determinant there is a heuristic (visual) triangle rule, it looks like this:

1. The product of the elements of the main diagonal (from the upper left corner to the lower right) the product of the elements forming the first triangle “perpendicular” to the main diagonal the product of the elements forming the second triangle “perpendicular” to the main diagonal
2. The product of the elements of the secondary diagonal (from the upper right corner to the lower left) the product of the elements forming the first triangle “perpendicular” to the secondary diagonal the product of the elements forming the second triangle “perpendicular” to the secondary diagonal
3. Then the determinant is equal to the difference between the values ​​obtained at the step and

If we write all this down in numbers, we get the following expression:

However, you don’t need to remember the method of calculation in this form; it’s enough to just keep in your head the triangles and the very idea of ​​what adds up to what and what is then subtracted from what).

Let's illustrate the triangle method with an example:

1. Calculate the determinant:

Let's figure out what we add and what we subtract:

Terms that come with a plus:

This is the main diagonal: the product of the elements is equal to

The first triangle, "perpendicular to the main diagonal: the product of the elements is equal to

Second triangle, "perpendicular to the main diagonal: the product of the elements is equal to

Add up three numbers:

Terms that come with a minus

This is a side diagonal: the product of the elements is equal to

The first triangle, “perpendicular to the secondary diagonal: the product of the elements is equal to

The second triangle, “perpendicular to the secondary diagonal: the product of the elements is equal to

Add up three numbers:

All that remains to be done is to subtract the sum of the “plus” terms from the sum of the “minus” terms:

Thus,

As you can see, there is nothing complicated or supernatural in calculating third-order determinants. It’s just important to remember about triangles and not make arithmetic errors. Now try to calculate it yourself:

We check:

1. The first triangle perpendicular to the main diagonal:
2. Second triangle perpendicular to the main diagonal:
3. Sum of terms with plus:
4. The first triangle perpendicular to the secondary diagonal:
5. Second triangle perpendicular to the side diagonal:
6. Sum of terms with minus:
7. The sum of the terms with a plus minus the sum of the terms with a minus:

Here are a couple more determinants, calculate their values ​​yourself and compare them with the answers:

Answers:

Well, did everything coincide? Great, then you can move on! If there are difficulties, then my advice is this: on the Internet there are a lot of programs for calculating the determinant online. All you need is to come up with your own determinant, calculate it yourself, and then compare it with what the program calculates. And so on until the results begin to coincide. I am sure this moment will not take long to arrive!

Now let's go back to the determinant that I wrote out when I talked about the equation of a plane passing through three given points:

All you need is to calculate its value directly (using the triangle method) and set the result to zero. Naturally, since these are variables, you will get some expression that depends on them. It is this expression that will be the equation of a plane passing through three given points that do not lie on the same straight line!

Let's illustrate this with a simple example:

1. Construct the equation of a plane passing through the points

We compile a determinant for these three points:

Let's simplify:

Now we calculate it directly using the triangle rule:

\[(\left| (\begin(array)(*(20)(c))(x + 3)&2&6\\(y - 2)&0&1\\(z + 1)&5&0\end(array)) \ right| = \left((x + 3) \right) \cdot 0 \cdot 0 + 2 \cdot 1 \cdot \left((z + 1) \right) + \left((y - 2) \right) \cdot 5 \cdot 6 - )\]

Thus, the equation of the plane passing through the points is:

Now try to solve one problem yourself, and then we will discuss it:

2. Find the equation of the plane passing through the points

Well, let's now discuss the solution:

Let's create a determinant:

And calculate its value:

Then the equation of the plane has the form:

Or, reducing by, we get:

Now two tasks for self-control:

1. Construct the equation of a plane passing through three points:

Answers:

Did everything coincide? Again, if there are certain difficulties, then my advice is this: take three points from your head (with a high degree of probability they will not lie on the same straight line), build a plane based on them. And then you check yourself online. For example, on the site:

However, with the help of determinants we will construct not only an equation of the plane. Remember, I told you that not only dot product is defined for vectors. There is also a vector product, as well as a mixed product. And if the scalar product of two vectors is a number, then the vector product of two vectors will be a vector, and this vector will be perpendicular to the given ones:

Moreover, its module will be equal to area parallelogram constructed on vectors and. This vector We will need it to calculate the distance from a point to a line. How can we calculate the vector product of vectors and, if their coordinates are given? The third-order determinant comes to our aid again. However, before I move on to the algorithm for calculating the vector product, I have to make a small digression.

This digression concerns basis vectors.

They are shown schematically in the figure:

Why do you think they are called basic? The fact is that :

Or in the picture:

The validity of this formula is obvious, because:

Vector artwork

Now I can start introducing the cross product:

The vector product of two vectors is a vector, which is calculated according to the following rule:

Now let's give some examples of calculating the cross product:

Example 1: Find the cross product of vectors:

Solution: I make up a determinant:

And I calculate it:

Now from writing through basis vectors, I will return to the usual vector notation:

Thus:

Now try it.

Ready? We check:

And traditionally two tasks for control:

1. Find the vector product of the following vectors:
2. Find the vector product of the following vectors:

Answers:

Mixed product of three vectors

The last construction I'll need is the mixed product of three vectors. It, like a scalar, is a number. There are two ways to calculate it. - through a determinant, - through a mixed product.

Namely, let us be given three vectors:

Then the mixed product of three vectors, denoted by, can be calculated as:

1. - that is, the mixed product is the scalar product of a vector and the vector product of two other vectors

For example, the mixed product of three vectors is:

Try to calculate it yourself using the vector product and make sure that the results match!

And again, two examples for independent solutions:

Answers:

Selecting a coordinate system

Well, now we have all the necessary foundation of knowledge to solve complex stereometric geometry problems. However, before proceeding directly to examples and algorithms for solving them, I believe that it will be useful to dwell on the following question: how exactly choose a coordinate system for a particular figure. After all, it’s the choice relative position coordinate systems and shapes in space will ultimately determine how cumbersome the calculations will be.

Let me remind you that in this section we consider the following figures:

1. Rectangular parallelepiped
2. Straight prism (triangular, hexagonal...)
3. Pyramid (triangular, quadrangular)
4. Tetrahedron (same as triangular pyramid)

For a rectangular parallelepiped or cube, I recommend you the following construction:

That is, I will place the figure “in the corner”. The cube and parallelepiped are very good figures. For them, you can always easily find the coordinates of its vertices. For example, if (as shown in the picture)

then the coordinates of the vertices are as follows:

Of course, you don’t need to remember this, but remembering how best to position a cube or rectangular parallelepiped is advisable.

Straight prism

The prism is a more harmful figure. It can be positioned in space in different ways. However, the following option seems to me the most acceptable:

Triangular prism:

That is, we place one of the sides of the triangle entirely on the axis, and one of the vertices coincides with the origin of coordinates.

Hexagonal prism:

That is, one of the vertices coincides with the origin, and one of the sides lies on the axis.

Quadrangular and hexagonal pyramid:

The situation is similar to a cube: we align two sides of the base with the coordinate axes, and align one of the vertices with the origin of coordinates. The only slight difficulty will be to calculate the coordinates of the point.

For a hexagonal pyramid - the same as for a hexagonal prism. The main task will again be to find the coordinates of the vertex.

Tetrahedron (triangular pyramid)

The situation is very similar to the one I gave for a triangular prism: one vertex coincides with the origin, one side lies on the coordinate axis.

Well, now you and I are finally close to starting to solve problems. From what I said at the very beginning of the article, you could draw the following conclusion: most C2 problems are divided into 2 categories: angle problems and distance problems. First, we will look at the problems of finding an angle. They are in turn divided into the following categories (as complexity increases):

Problems for finding angles

1. Finding the angle between two straight lines
2. Finding the angle between two planes

Let's look at these problems sequentially: let's start by finding the angle between two straight lines. Well, remember, didn’t you and I decide? similar examples earlier? Do you remember, we already had something similar... We were looking for the angle between two vectors. Let me remind you, if two vectors are given: and, then the angle between them is found from the relation:

Now our goal is to find the angle between two straight lines. Let's look at the “flat picture”:

How many angles did we get when two straight lines intersected? Just a few things. True, only two of them are not equal, while the others are vertical to them (and therefore coincide with them). So which angle should we consider the angle between two straight lines: or? Here the rule is: the angle between two straight lines is always no more than degrees. That is, from two angles we will always choose the angle with the smallest degree measure. That is, in this picture the angle between two straight lines is equal. In order not to bother each time with finding the smallest of two angles, cunning mathematicians suggested using a modulus. Thus, the angle between two straight lines is determined by the formula:

You, as an attentive reader, should have had a question: where, exactly, do we get these very numbers that we need to calculate the cosine of an angle? Answer: we will take them from the direction vectors of the lines! Thus, the algorithm for finding the angle between two straight lines is as follows:

1. We apply formula 1.

Or in more detail:

1. We are looking for the coordinates of the direction vector of the first straight line
2. We are looking for the coordinates of the direction vector of the second straight line
3. We calculate the modulus of their scalar product
4. We are looking for the length of the first vector
5. We are looking for the length of the second vector
6. Multiply the results of point 4 by the results of point 5
7. We divide the result of point 3 by the result of point 6. We get the cosine of the angle between the lines
8. If this result allows you to accurately calculate the angle, look for it
9. Otherwise we write through arc cosine

Well, now it’s time to move on to the problems: I will demonstrate the solution to the first two in detail, I will present the solution to another one in a brief form, and to the last two problems I will only give the answers; you must carry out all the calculations for them yourself.

Tasks:

1. In the right tet-ra-ed-re, find the angle between the height of the tet-ra-ed-ra and the middle side.

2. In the right-hand six-corner pi-ra-mi-de, the hundred os-no-va-niyas are equal, and the side edges are equal, find the angle between the lines and.

3. The lengths of all the edges of the right four-coal pi-ra-mi-dy are equal to each other. Find the angle between the straight lines and if from the cut - you are with the given pi-ra-mi-dy, the point is se-re-di-on its bo-co- second ribs

4. On the edge of the cube there is a point so that Find the angle between the straight lines and

5. Point - on the edges of the cube Find the angle between the straight lines and.

It is no coincidence that I arranged the tasks in this order. While you have not yet begun to navigate the coordinate method, I will analyze the most “problematic” figures myself, and I will leave you to deal with the simplest cube! Gradually you will have to learn how to work with all the figures; I will increase the complexity of the tasks from topic to topic.

Let's start solving problems:

1. Draw a tetrahedron, place it in the coordinate system as I suggested earlier. Since the tetrahedron is regular, all its faces (including the base) are regular triangles. Since we are not given the length of the side, I can take it to be equal. I think you understand that the angle will not actually depend on how much our tetrahedron is “stretched”?. I will also draw the height and median in the tetrahedron. Along the way, I will draw its base (it will also be useful to us).

I need to find the angle between and. What do we know? We only know the coordinate of the point. This means that we need to find the coordinates of the points. Now we think: a point is the point of intersection of the altitudes (or bisectors or medians) of the triangle. And a point is a raised point. The point is the middle of the segment. Then we finally need to find: the coordinates of the points: .

Let's start with the simplest thing: the coordinates of a point. Look at the figure: It is clear that the applicate of a point is equal to zero (the point lies on the plane). Its ordinate is equal (since it is the median). It is more difficult to find its abscissa. However, this is easily done based on the Pythagorean theorem: Consider a triangle. Its hypotenuse is equal, and one of its legs is equal Then:

Finally we have: .

Now let's find the coordinates of the point. It is clear that its applicate is again equal to zero, and its ordinate is the same as that of the point, that is. Let's find its abscissa. This is done quite trivially if you remember that heights equilateral triangle the intersection point is divided in proportion, counting from the top. Since: , then the required abscissa of the point, equal to the length of the segment, is equal to: . Thus, the coordinates of the point are:

Let's find the coordinates of the point. It is clear that its abscissa and ordinate coincide with the abscissa and ordinate of the point. And the applicate is equal to the length of the segment. - this is one of the legs of the triangle. The hypotenuse of a triangle is a segment - a leg. It is sought for reasons that I have highlighted in bold:

The point is the middle of the segment. Then we need to remember the formula for the coordinates of the midpoint of the segment:

That's it, now we can look for the coordinates of the direction vectors:

Well, everything is ready: we substitute all the data into the formula:

Thus,

Answer:

You should not be scared by such “scary” answers: for C2 tasks this is common practice. I would rather be surprised by the “beautiful” answer in this part. Also, as you noticed, I practically did not resort to anything other than the Pythagorean theorem and the property of altitudes of an equilateral triangle. That is, to solve the stereometric problem, I used the very minimum of stereometry. The gain in this is partially “extinguished” by rather cumbersome calculations. But they are quite algorithmic!

2. Let us depict a regular hexagonal pyramid along with the coordinate system, as well as its base:

We need to find the angle between the lines and. Thus, our task comes down to finding the coordinates of the points: . We will find the coordinates of the last three using a small drawing, and we will find the coordinate of the vertex through the coordinate of the point. There's a lot of work to do, but we need to get started!

a) Coordinate: it is clear that its applicate and ordinate are equal to zero. Let's find the abscissa. To do this, consider a right triangle. Alas, in it we only know the hypotenuse, which is equal. We will try to find the leg (for it is clear that double the length of the leg will give us the abscissa of the point). How can we look for it? Let's remember what kind of figure we have at the base of the pyramid? This is a regular hexagon. What does it mean? This means that all sides and all angles are equal. We need to find one such angle. Any ideas? There are a lot of ideas, but there is a formula:

The sum of the angles of a regular n-gon is .

Thus, the sum of the angles of a regular hexagon is equal to degrees. Then each of the angles is equal to:

Let's look at the picture again. It is clear that the segment is the bisector of the angle. Then the angle is equal to degrees. Then:

Then where from.

Thus, has coordinates

b) Now we can easily find the coordinate of the point: .

c) Find the coordinates of the point. Since its abscissa coincides with the length of the segment, it is equal. Finding the ordinate is also not very difficult: if we connect the dots and designate the point of intersection of the straight line as, say, . (do it yourself simple construction). Then Thus, the ordinate of point B is equal to the sum of the lengths of the segments. Let's look at the triangle again. Then

Then since Then the point has coordinates

d) Now let's find the coordinates of the point. Consider the rectangle and prove that Thus, the coordinates of the point are:

e) It remains to find the coordinates of the vertex. It is clear that its abscissa and ordinate coincide with the abscissa and ordinate of the point. Let's find the applica. Since, then. Consider a right triangle. According to the conditions of the problem, a side edge. This is the hypotenuse of my triangle. Then the height of the pyramid is a leg.

Then the point has coordinates:

Well, that's it, I have the coordinates of all the points that interest me. I am looking for the coordinates of the directing vectors of straight lines:

We are looking for the angle between these vectors:

Answer:

Again, in solving this problem I did not use any sophisticated techniques other than the formula for the sum of the angles of a regular n-gon, as well as the definition of the cosine and sine of a right triangle.

3. Since we are again not given the lengths of the edges in the pyramid, I will count them equal to one. Thus, since ALL edges, and not just the side ones, are equal to each other, then at the base of the pyramid and me there is a square, and the side faces are regular triangles. Let us draw such a pyramid, as well as its base on a plane, noting all the data given in the text of the problem:

We are looking for the angle between and. I will make very brief calculations when I search for the coordinates of the points. You will need to “decipher” them:

b) - the middle of the segment. Its coordinates:

c) I will find the length of the segment using the Pythagorean theorem in a triangle. I can find it using the Pythagorean theorem in a triangle.

Coordinates:

d) - the middle of the segment. Its coordinates are

e) Vector coordinates

f) Vector coordinates

g) Looking for the angle:

A cube is the simplest figure. I'm sure you'll figure it out on your own. The answers to problems 4 and 5 are as follows:

Finding the angle between a straight line and a plane

Well, the time for simple puzzles is over! Now the examples will be even more complicated. To find the angle between a straight line and a plane, we will proceed as follows:

1. Using three points we construct an equation of the plane
   ,
   using a third order determinant.
2. Using two points, we look for the coordinates of the directing vector of the straight line:
3. We apply the formula to calculate the angle between a straight line and a plane:

As you can see, this formula is very similar to the one we used to find angles between two straight lines. The structure on the right side is simply the same, and on the left we are now looking for the sine, not the cosine as before. Well, one nasty action was added - searching for the equation of the plane.

Let's not procrastinate solution examples:

1. The main-but-va-ni-em direct prism-we are an equal-to-poor triangle. Find the angle between the straight line and the plane

2. In a rectangular par-ral-le-le-pi-pe-de from the West Find the angle between the straight line and the plane

3. In a right six-corner prism, all edges are equal. Find the angle between the straight line and the plane.

4. In the right triangular pi-ra-mi-de with the os-no-va-ni-em of the known ribs Find a corner, ob-ra-zo-van -flat in base and straight, passing through the gray ribs and

5. The lengths of all the edges of a right quadrangular pi-ra-mi-dy with a vertex are equal to each other. Find the angle between the straight line and the plane if the point is on the side of the pi-ra-mi-dy’s edge.

Again, I will solve the first two problems in detail, the third briefly, and leave the last two for you to solve on your own. Besides, you have already had to deal with triangular and quadrangular pyramids, but not yet with prisms.

Solutions:

1. Let us depict a prism, as well as its base. Let's combine it with the coordinate system and note all the data that is given in the problem statement:

I apologize for some non-compliance with the proportions, but for solving the problem this is, in fact, not so important. The plane is simply the "back wall" of my prism. It is enough to simply guess that the equation of such a plane has the form:

However, this can be shown directly:

Let's choose arbitrary three points on this plane: for example, .

Let's create the equation of the plane:

Exercise for you: calculate this determinant yourself. Did you succeed? Then the equation of the plane looks like:

Or simply

Thus,

To solve the example, I need to find the coordinates of the direction vector of the straight line. Since the point coincides with the origin of coordinates, the coordinates of the vector will simply coincide with the coordinates of the point. To do this, we first find the coordinates of the point.

To do this, consider a triangle. Let's draw the height (also known as the median and bisector) from the vertex. Since, the ordinate of the point is equal to. In order to find the abscissa of this point, we need to calculate the length of the segment. According to the Pythagorean theorem we have:

Then the point has coordinates:

A dot is a "raised" dot:

Then the vector coordinates are:

Answer:

As you can see, there is nothing fundamentally difficult when solving such problems. In fact, the process is simplified a little more by the “straightness” of a figure such as a prism. Now let's move on to the next example:

2. Draw a parallelepiped, draw a plane and a straight line in it, and also separately draw its lower base:

First, we find the equation of the plane: The coordinates of the three points lying in it:

(the first two coordinates are obtained in an obvious way, and you can easily find the last coordinate from the picture from the point). Then we compose the equation of the plane:

We calculate:

We are looking for the coordinates of the guiding vector: It is clear that its coordinates coincide with the coordinates of the point, isn’t it? How to find coordinates? These are the coordinates of the point, raised along the applicate axis by one! . Then we look for the desired angle:

Answer:

3. Draw a regular hexagonal pyramid, and then draw a plane and a straight line in it.

Here it’s even problematic to draw a plane, not to mention solving this problem, but the coordinate method doesn’t care! Its versatility is its main advantage!

The plane passes through three points: . We are looking for their coordinates:

1) . Find out the coordinates for the last two points yourself. You'll need to solve the hexagonal pyramid problem for this!

2) We construct the equation of the plane:

We are looking for the coordinates of the vector: . (See the triangular pyramid problem again!)

3) Looking for an angle:

Answer:

As you can see, there is nothing supernaturally difficult in these tasks. You just need to be very careful with the roots. I will only give answers to the last two problems:

As you can see, the technique for solving problems is the same everywhere: the main task is to find the coordinates of the vertices and substitute them into certain formulas. We still have to consider one more class of problems for calculating angles, namely:

Calculating angles between two planes

The solution algorithm will be as follows:

1. Using three points we look for the equation of the first plane:
2. Using the other three points we look for the equation of the second plane:
3. We apply the formula:

As you can see, the formula is very similar to the two previous ones, with the help of which we looked for angles between straight lines and between a straight line and a plane. So it won’t be difficult for you to remember this one. Let's move on to the analysis of the tasks:

1. The side of the base of the right triangular prism is equal, and the dia-go-nal of the side face is equal. Find the angle between the plane and the plane of the axis of the prism.

2. In the right four-corner pi-ra-mi-de, all the edges of which are equal, find the sine of the angle between the plane and the plane bone, passing through the point per-pen-di-ku-lyar-but straight.

3. In a regular four-corner prism, the sides of the base are equal, and the side edges are equal. There is a point on the edge from-me-che-on so that. Find the angle between the planes and

4. In a right quadrangular prism, the sides of the base are equal, and the side edges are equal. There is a point on the edge from the point so that Find the angle between the planes and.

5. In a cube, find the co-si-nus of the angle between the planes and

Problem solutions:

1. I draw the correct one (at the base there is an equilateral triangle) triangular prism and mark on it the planes that appear in the problem statement:

We need to find the equations of two planes: The equation of the base is trivial: you can compose the corresponding determinant using three points, but I will compose the equation right away:

Now let’s find the equation Point has coordinates Point - Since is the median and altitude of the triangle, it is easily found using the Pythagorean theorem in the triangle. Then the point has coordinates: Let's find the applicate of the point. To do this, consider a right triangle

Then we get the following coordinates: We compose the equation of the plane.

We calculate the angle between the planes:

Answer:

2. Making a drawing:

The most difficult thing is to understand what kind of mysterious plane this is, passing perpendicularly through the point. Well, the main thing is, what is it? The main thing is attentiveness! In fact, the line is perpendicular. The straight line is also perpendicular. Then the plane passing through these two lines will be perpendicular to the line, 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 has already been given to us. We are looking for the coordinates of the points.

We find the coordinate of the point through the point. From the small picture it is easy to deduce that the coordinates of the point will be as follows: What now remains to be found to find the coordinates of the top of the pyramid? You also need to calculate its height. This is done using the same Pythagorean theorem: first prove that (trivially from small triangles forming a square at the base). Since by condition, we have:

Now everything is ready: vertex coordinates:

We compose the equation of the plane:

You are already an expert in calculating determinants. Without difficulty you will receive:

Or otherwise (if we multiply both sides by the root of two)

Now let's find the equation of the plane:

(You haven’t forgotten how we get the equation of a plane, right? If you don’t understand where this minus one came from, then go back to the definition of the equation of a plane! It just always turned out before that my plane belonged to the origin of coordinates!)

We calculate the determinant:

(You may notice that the equation of the plane coincides with the equation of the line passing through the points and! Think about why!)

Now let's calculate the angle:

We need to find the sine:

Answer:

3. Tricky question: what do you think a rectangular prism is? This is just a parallelepiped that you know well! Let's make a drawing right away! You don’t even have to depict the base separately; it’s of little use here:

The plane, as we noted earlier, is written in the form of an equation:

Now let's create a plane

We immediately create the equation of the plane:

Looking for an angle:

Now the answers to the last two problems:

Well, now is the time to take a little break, because you and I are great and have done a great job!

Coordinates and vectors. Advanced level

In this article we will discuss with you another class of problems that can be solved using the coordinate method: distance calculation problems. Namely, we will consider the following cases:

1. Calculation of the distance between intersecting lines.

I have ordered these assignments in order of increasing difficulty. It turns out to be easiest to find distance from point to plane, and the most difficult thing is to find distance between crossing lines. Although, of course, nothing is impossible! Let's not procrastinate and immediately proceed to consider the first class of problems:

Calculating the distance from a point to a plane

What do we need to solve this problem?

1. Point coordinates

So, as soon as we receive all the necessary data, we apply the formula:

You should already know how we construct the equation of a plane from the previous problems that I discussed in the last part. Let's get straight to the tasks. The scheme is as follows: 1, 2 - I help you decide, and in some detail, 3, 4 - only the answer, you carry out the solution yourself and compare. Let's start!

Tasks:

1. Given a cube. The length of the edge of the cube is equal. Find the distance from the se-re-di-na from the cut to the plane

2. Given the right four-coal pi-ra-mi-yes, the side of the side is equal to the base. Find the distance from the point to the plane where - se-re-di-on the edges.

3. In the right triangular pi-ra-mi-de with the os-no-va-ni-em, the side edge is equal, and the hundred-ro-on the os-no-va- nia is equal. Find the distance from the top to the plane.

4. In a right hexagonal prism, all edges are equal. Find the distance from a point to a plane.

Solutions:

1. Draw a cube with single edges, construct a segment and a plane, denote the middle of the segment with a letter

.

First, let's start with the easy one: find the coordinates of the point. Since then (remember the coordinates of the middle of the segment!)

Now we compose the equation of the plane using three points

\[\left| (\begin(array)(*(20)(c))x&0&1\\y&1&0\\z&1&1\end(array)) \right| = 0\]

Now I can start finding the distance:

2. We start again with a drawing on which we mark all the data!

For a pyramid, it would be useful to draw its base separately.

Even the fact that I draw like a chicken with its paw will not prevent us from solving this problem with ease!

Now it's easy to find the coordinates of a point

Since the coordinates of the point, then

2. Since the coordinates of point a are the middle of the segment, then

Without any problems, we can find the coordinates of two more points on the plane. We create an equation for the plane and simplify it:

\[\left| (\left| (\begin(array)(*(20)(c))x&1&(\frac(3)(2))\\y&0&(\frac(3)(2))\\z&0&(\frac( (\sqrt 3 ))(2))\end(array)) \right|) \right| = 0\]

Since the point has coordinates: , we calculate the distance:

Answer (very rare!):

Well, did you figure it out? It seems to me that everything here is just as technical as in the examples that we looked at in the previous part. So I am sure that if you have mastered that material, then it will not be difficult for you to solve the remaining two problems. I'll just give you the answers:

Calculating the distance from a straight line to a plane

In fact, there is nothing new here. How can a straight line and a plane be positioned relative to each other? They have only one possibility: to intersect, or a straight line is parallel to the plane. What do you think is the distance from a straight line to the plane with which this straight line intersects? It seems to me that it is clear here that such a distance is equal to zero. Not an interesting case.

The second case is trickier: here the distance is already non-zero. However, since the line is parallel to the plane, then each point of the line is equidistant from this plane:

Thus:

This means that my task has been reduced to the previous one: we are looking for the coordinates of any point on a straight line, looking for the equation of the plane, and calculating the distance from the point to the plane. In fact, such tasks are extremely rare in the Unified State Examination. I managed to find only one problem, and the data in it were such that the coordinate method was not very applicable to it!

Now let's move on to another, much more important class of problems:

Calculating the distance of a point to a line

What do we need?

1. Coordinates of the point from which we are looking for the distance:

2. Coordinates of any point lying on a line

3. Coordinates of the directing vector of the straight line

What formula do we use?

What the denominator of this fraction means should be clear to you: this is the length of the directing vector of the straight line. This is a very tricky numerator! The expression means the modulus (length) of the vector product of vectors and How to calculate the vector product, we studied in the previous part of the work. Refresh your knowledge, we will need it very much now!

Thus, the algorithm for solving problems will be as follows:

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 to which we are looking for the distance:

3. Construct a vector

4. Construct a directing vector of a straight line

5. Calculate the vector product

6. We look for the length of the resulting vector:

7. Calculate the distance:

We have a lot of work to do, and the examples will be quite complex! So now focus all your attention!

1. Given a right triangular pi-ra-mi-da with a top. The hundred-ro-on the basis of the pi-ra-mi-dy is equal, you are equal. Find the distance from the gray edge to the straight line, where the points and are the gray edges and from veterinary.

2. The lengths of the ribs and the straight-angle-no-go par-ral-le-le-pi-pe-da are equal accordingly and Find the distance from the top to the straight line

3. In a right hexagonal prism, all edges are equal, find the distance from a point to a straight line

Solutions:

1. We make a neat drawing on which we mark all the data:

We have a lot of work to do! First, I would like to describe in words what we will look for and in what order:

1. Coordinates of points and

2. Point coordinates

3. Coordinates of points and

4. Coordinates of vectors and

5. Their cross product

6. Vector length

7. Length of the vector product

8. Distance from to

Well, we have a lot of work ahead of us! Let's get to it with our sleeves rolled up!

1. To find the coordinates of the height of the pyramid, we need to know the coordinates of the point. Its applicate is zero, and its ordinate is equal to its abscissa is equal to the length of the segment. Since is the height of an equilateral triangle, it is divided in the ratio, counting from the vertex, from here. Finally, we got the coordinates:

Point coordinates

2. - middle of the segment

3. - middle of the segment

Midpoint of the segment

4.Coordinates

Vector coordinates

5. Calculate the vector product:

6. Vector length: the easiest way to replace is that the segment is the midline of the triangle, which means it is equal to half the base. So.

7. Calculate the length of the vector product:

8. Finally, we find the distance:

Ugh, that's it! I'll tell you honestly: the solution to this problem is traditional methods(via construction), it would be much faster. But here I reduced everything to a ready-made algorithm! I think the solution algorithm is clear to you? Therefore, I will ask you to solve the remaining two problems yourself. Let's compare the answers?

Again, I repeat: it is easier (faster) to solve these problems through constructions, rather than resorting to coordinate method. I demonstrated this method of solution only to show you a universal method that allows you to “not finish building anything.”

Finally, consider the last class of problems:

Calculating the distance between intersecting lines

Here the algorithm for solving problems will be similar to the previous one. What we have:

3. Any vector connecting the points of the first and second line:

How do we find the distance between lines?

The formula is as follows:

The numerator is the modulus mixed product(we introduced it in the previous part), and the denominator is as in the previous formula (the modulus of the vector product of the directing vectors of the straight lines, the distance between which we are looking for).

I'll remind you that

Then the formula for the distance can be rewritten as:

This is a determinant divided by a determinant! Although, to be honest, I have no time for jokes here! This formula, in fact, is very cumbersome and leads to quite complex calculations. If I were you, I would resort to it only as a last resort!

Let's try to solve a few problems using the above method:

1. In a right triangular prism, all the edges of which are equal, find the distance between the straight lines and.

2. Given a right triangular prism, all the edges of the base are equal to the section passing through the body rib and se-re-di-well ribs are a square. Find the distance between the straight lines and

I decide the first, and based on it, you decide the second!

1. I draw a prism and mark straight lines and

Coordinates of point C: then

Point coordinates

Vector coordinates

Point coordinates

Vector coordinates

Vector coordinates

\[\left((B,\overrightarrow (A(A_1)) \overrightarrow (B(C_1)) ) \right) = \left| (\begin(array)(*(20)(l))(\begin(array)(*(20)(c))0&1&0\end(array))\\(\begin(array)(*(20) (c))0&0&1\end(array))\\(\begin(array)(*(20)(c))(\frac((\sqrt 3 ))(2))&( - \frac(1) (2))&1\end(array))\end(array)) \right| = \frac((\sqrt 3 ))(2)\]

We calculate the vector product between vectors and

\[\overrightarrow (A(A_1)) \cdot \overrightarrow (B(C_1)) = \left| \begin(array)(l)\begin(array)(*(20)(c))(\overrightarrow i )&(\overrightarrow j )&(\overrightarrow k )\end(array)\\\begin(array )(*(20)(c))0&0&1\end(array)\\\begin(array)(*(20)(c))(\frac((\sqrt 3 ))(2))&( - \ frac(1)(2))&1\end(array)\end(array) \right| - \frac((\sqrt 3 ))(2)\overrightarrow k + \frac(1)(2)\overrightarrow i \]

Now we calculate its length:

Answer:

Now try to complete the second task carefully. The answer to it will be: .

Coordinates and vectors. Brief description and basic formulas

A vector is a directed segment. - the beginning of the vector, - the end of the vector.
A vector is denoted by or.

Absolute value vector - the length of the segment representing the vector. Denoted as.

Vector coordinates:

,
where are the ends of the vector \displaystyle a .

Sum of vectors: .

Product of vectors:

Dot product of vectors:

The scalar product of vectors is equal to their product absolute values by the cosine of the angle between them:

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Methods for specifying a rectangular coordinate system

As is known, a system of rectangular coordinates on a plane can be specified in three ways: the 1st method fixes the location of the center of the system - i.e. O, draws the OX axis and indicates its positive direction, draws the OY axis perpendicular to the OX axis, in accordance with the type of system (right or left) the positive direction of the OY axis is indicated, the coordinate scale along the axes is set.

If there are coordinate axes, to determine the coordinates of any point C, you must first lower perpendiculars from this point to the coordinate axes and then measure the length of these perpendiculars; the length of the perpendicular to the OX axis is equal to the Y coordinate, the length of the perpendicular to the OY axis is equal to the X coordinate of the 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 moving the origin of coordinates to point O" (Xo"=дx, Yo"=дy) and rotating the coordinate axes clockwise by angle b.

The transition from XOY to X"O"Y" is performed using the formulas:

For the reverse transition, the following formulas are used:

• 2nd method: two mutually perpendicular systems of parallel lines are drawn; the distances between the lines are the same, these lines are considered to be parallel to the coordinate axes, and each line is labeled with the value of the corresponding coordinate (a coordinate grid is obtained).
• The 3rd method indicates the numerical values ​​of the coordinates of two fixed points.

The first method is generally accepted; in geodesy, this method defines a zonal system of rectangular Gaussian coordinates.

On topographic maps and plans, the Gaussian rectangular coordinate system is specified in the second way.

On the ground, a system of rectangular coordinates is specified in the third way; You can always find several geodetic points with known coordinates and determine the position of new points relative to these points by performing any measurements.

Three Elementary Dimensions

You can measure angles and distances on a plane.

An angle is fixed by three points: one point is the vertex of the angle, and the other two points fix the directions of the 1st and 2nd sides of the angle. In the simplest case, at least one point out of three has no coordinates, that is, it is definable; in general, one point, two points, or all three can be determined.

The distance is fixed by two points, and in general, one point or both can be determined.

This section discusses the simplest case, when measuring an angle or distance is performed to determine the coordinates of a single point. Since when measuring an angle, the point being determined can be located either at the vertex of the angle or on one of its sides, then from our point of view on the plane there are three different measurements, which we will call elementary.

The angle b is measured at point A with known coordinates X4, Y4 between the direction with a known directional angle bAB and the direction to the determined point P (Fig. 2).

The directional angle b of the direction AP is obtained by the formula

For a straight line AP, called the position line of point P, we can write an equation in the XOY system:

In this equation, X and Y are the coordinates of any point on the line, including point P, but to find two coordinates of point P, one such equation is not enough.

The distance S is measured from point A with known coordinates XA, YA to the determined point P. From the geometry course it is known that point P is located on a circle of radius S drawn around point A, and called the position line of point P (Fig. 3). The equation of a circle is:

In this equation, X and Y are the coordinates of any point on the circle, including point P, but to find two coordinates of a point, one such equation is not enough.

The angle b is measured at a determined point P between the directions of two points with known coordinates; this measurement is discussed in Section 8.

The X and Y coordinates of point P can be found from a joint solution of two equations, therefore, taking any combination of three dimensions by two, we obtain the simplest methods for determining the coordinates of a point, called geodetic intersections: two equations of type (2.4) - a straight angular intersection, two equations of the type (2.5) - linear intersection, one equation of type (2.4) and one equation of type (2.5) polar intersection, two measurements of angles at the determined point - inverse angle intersection.

The remaining combinations of measurements are called combined notches.

Each of the three elementary dimensions is invariant with respect to coordinate systems, which makes it possible to solve serifs in various drawings by determining the position of point P relative to fixed points A and B graphically.

An analytical way to solve intersections is to calculate the coordinates of the point being determined. It can be performed through solving a system of two equations corresponding to the measurements performed, or through solving a triangle, the vertices of which are two starting points and a determined point (for brevity, we will call this method the triangle method).

In any geodetic construction, it is customary to distinguish three types of data: initial data (coordinates of initial points, directional angles of initial directions, etc.); these data are often assumed to be conditionally error-free, measured elements; each measured element is usually accompanied by a value of the mean square error of measurement, unknown (or determined) elements; these elements must be found using a specially developed algorithm, and their values ​​are obtained with some error, depending on measurement errors and the geometry of the given construction.

Polar notch

In a polar intersection, the initial data are the coordinates of point A and the directional angle of direction AB (or the coordinates of point B), the measured elements are the horizontal angle b (the root mean square error of measuring the angle mв) and the distance S (the relative error of its measurement mS / S = 1 / T ), unknown elements are the X, Y coordinates of point P (Fig. 4).

Input data: XA, YA, bAB

Measured elements: V, S

Unknown elements: X, Y

Graphic solution. From direction AB, use a protractor to plot angle B and draw a straight line AQ, then draw a circular arc of radius S around point A on the scale of the drawing (plan or map); the point of intersection of the straight line and the arc is the desired point P.

Analytical solution. Directional angle b of line AP is equal to:

Let's write down the equations of a straight line AP - formula (4) and a circle of radius S around point A - formula (5):

To find the X and Y coordinates of point P, you need to solve these two equations together as a system. Let's substitute the value (Y - YA) from the first equation into the second and put (X - XA) 2 out of brackets:

(X - XA) 2 * (1 + tan2 b)= S2.

We replace the expression (1 + tan2b) with 1 / Cos2b and get:

(X - XA) 2 = S2 * Cos2b, whence X - XA = S* Cosb.

Substitute this value into the first equation (6) and get:

Y - YA = S * Sinb.

The differences between the coordinates (X - XA) and (Y - YA) are usually called increments and denoted DX and DY.

Thus, the polar notch is uniquely solved using the formulas:

coordinate triangulation trilateration

Direct geodetic problem on a plane

There are two standard problems in geodesy: the direct geodetic problem on a plane and the inverse geodetic problem on a plane.

A direct geodetic problem is the calculation of the coordinates X2, Y2 of the second point, if the coordinates X1, Y1 of the first point, the directional angle b and the length S of the line connecting these points are known. The direct geodetic problem is part of a polar intersection, and the formulas for solving it are taken from the set of formulas (7):

Inverse geodetic problem on a plane

The inverse geodetic problem is the calculation of the directional angle b and the length S of the line connecting two points with known coordinates X1, Y1 and X2, Y2 (Fig. 5).

Let us construct on segment 1-2, as on the hypotenuse, a right triangle with legs parallel to the coordinate axes. In this triangle, the hypotenuse is equal to S, the legs are equal to the increments of the coordinates of points 1 and 2 (ДX = X2 - X1, ДY = Y2 - Y1), and one of the acute angles is equal to the point r of line 1-2.

If D X 00 and D Y 00, then we solve the triangle using the known formulas:

For this figure, the direction of line 1-2 is in the second quarter, so based on (22) we find:

The general procedure for finding the directional angle of line 1-2 includes two operations: determining the quarter number by the signs of increments of coordinates D>X and DY, calculating b using connection formulas (22) in accordance with the quarter number.

Control of the correctness of calculations is the fulfillment of the equality:

If DX = 0.0, then S = iДYі;

and b = 90o 00 "00" for DY > 0,

b = 270o 00" 00" at DY< 0.

If DY = 0.0, then S = iДXi

and b = 0o 00 "00" for DX > 0,

b = 180o 00 "00" at DX< 0.

To solve the inverse problem automatically (in computer programs), another algorithm is used that does not contain the tangent of the angle and excludes possible division by zero:

if ДY => 0o, then b = a,

if DY< 0o, то б = 360o - a.

Straight corner serif

First, let's consider the so-called general case of a straight corner intersection, when angles b1 and b2 are measured at two points with known coordinates, each from its own direction with a known directional angle (Fig. 6).

Initial data: XA, YA, bAC,

Measured elements: v 1, v2

Unknown elements: X, Y

If bAC and bBD are not specified explicitly, you need to solve the inverse geodetic problem first between points A and C and then between points B and D.

Graphic solution. From the direction AC, use a protractor to make an angle b1 and draw a straight line AP; from direction BD, set aside angle b2 and draw a straight line BP; the point of intersection of these lines is the desired point P.

Analytical solution. We present the variant algorithm corresponding to the general case of a notch:

calculate directional angles of lines AP and BP

write two equations of straight lines

for the AP line Y - YA= tgb1 * (X - XA), for the BP line Y - YB= tgb2 * (X - XB) (2.16)

solve the system of two equations and calculate the unknown coordinates X and Y:

A special case of a straight corner notch is considered to be the case when angles b1 and b2 are measured from the directions AB and BA, and angle b1 is right, and angle b2 is left (in the general case of notches, both angles are left) - Fig. 7.

The solution of a straight corner intersection using the triangle method corresponds to a special case of a intersection. The procedure for solving this will be as follows: solve the inverse problem between points A and B and obtain the directional angle bAB and the length b of line AB, calculate the angle r at vertex P, called the notch angle,

using the sine theorem for triangle APB:

calculate the lengths of the sides AP (S1) and BP (S2), calculate the directional angles b1 and b2:

solve a direct problem from point A to point P and for control - from point B to point P.

To calculate the X and Y coordinates in the special case of a straight corner intersection, you can use Young’s formulas:

From general case with a straight angular serif it is easy to move on to a special case; to do this, you first need to solve the inverse geodetic problem between points A and B and obtain the directional angle bAB of line AB and then calculate the angles in the triangle APB at vertices A and B

BAP = bAB - (bAC + b1) and ABP = (bBD + b2) - bBA.

For machine calculation, all the considered methods for solving right angle intersections are inconvenient for various reasons. One of the possible algorithms for solving the general case of notching on a computer involves the following actions: calculating directional angles b1 and b2, introducing a local coordinate system X"O"Y" with the origin at point A and with the O"X axis directed along the line AP, and recalculation of the coordinates of points A and B and directional angles b1 and b2 from the XOY system to the X"O"Y" system (Fig. 8):

X"A = 0, Y"A = 0,

(24), writing the equations of lines AP and BP in the X"O"Y" system:

and joint solution of these equations:

conversion of X" and Y" coordinates from the X"O"Y" system to the XOY system:

Since Ctgb2" = - Ctgg and the notch angle r is always greater than 0°, then solution (27) always exists.

Linear serif

From point A with known coordinates XA, YA, the distance S1 to the determined point P is measured, and from point B with known coordinates XB, YB, the distance S2 to point P is measured.

Graphic solution. Let's draw a circle around point A with radius S1 (on the scale of the drawing), and around point B - a circle with radius S2; the point of intersection of the circles is the desired point; the problem has two solutions, since two circles intersect at two points (Fig. 9).

Input data: XA, YA, XB, YB,

Measured elements: S1, S2,

Unknown elements: X, Y.

Analytical solution. Let's consider two analytical solution algorithms, one for manual calculation (using the triangle method) and one for machine calculation.

The manual counting algorithm consists of the following steps:

solving the inverse geodetic problem between points A and B and obtaining the directional angle bAB and the length b of line AB, calculating angles b1 and b2 in triangle ABP using the cosine theorem:

calculating the intersection angle r

calculation of directional angles of sides AP and BP:

point P to the right of line AB

point P to the left of line AB

solving direct geodetic problems from point A to point P and from point B to point P:

1st solution

2nd solution

The results of both solutions should be the same.

The algorithm for a machine solution of a linear intersection consists of the following actions: solving the inverse geodetic problem between points A and B and obtaining the directional angle bAB and the length b of line AB, introducing a local coordinate system X"O"Y" with the origin at point A and the axis O"X ", directed along the line AB, and recalculation of the coordinates of points A and B from the XOY system to the X"O"Y" system:

writing the equations of circles in the X"O"Y" system:

and a joint solution of these equations, which involves opening the parentheses in the second equation and subtracting the second equation from the first:

If the desired point is to the left of line AB, then in formula (39) the “-” sign is taken, if to the right, then “+”.

Converting the X" and Y" coordinates of point P from the X"O"Y" system to the XOY system using formulas (2):

Reverse notch

Elementary measurements also include measuring the angle at a determined point P between the directions to two points A and B with known coordinates XA, YA and XB, YB (Fig. 10). However, this measurement turns out to be quite complex theoretically, so we will consider it separately.

Let's draw a circle through three points A, B and P. From school course Geometry knows that an angle with a vertex on a circle is measured by half the arc on which it rests. The central angle based on the same arc is measured by the entire arc, therefore it will be equal to 2c (Fig. 10).

The distance b between points A and B is assumed to be known, and from the right triangle FCB the radius R of the circle can be found:

The equation of a circle is:

where XC and YC are the coordinates of the center of the circle. They can be calculated by solving either a straight angular or linear intersection from points A and B to point C. In equation (42), X and Y are the coordinates of any point on the circle, including point P, but to find two coordinates of point P one such an equation is not enough.

Reverse angular intersection is a method of determining the coordinates of point P from two angles b1 and b2, measured at the determined point P between the directions of three points with known coordinates A, B, C (Fig. 11).

Graphic solution. Let us present Bolotov's method of graphically solving reverse corner intersection. On a sheet of transparent paper (tracing paper) you need to construct angles b1 and b2 with a common vertex P; then place the tracing paper on the drawing and, moving it, ensure that the directions of the angles on the tracing paper pass through points A, B, C in the drawing; pin point P from the tracing paper onto the drawing.

Source data: XA, YA, XB,

Measured elements: v1, v2.

Unknown elements: X, Y.

Analytical solution. The analytical solution of a reverse corner intersection involves its decomposition into simpler problems, for example, into 2 straight corner intersections and one linear, or into 3 linear intersections, etc. More than 10 methods of analytical solution are known, but we will consider only one - through the sequential solution of three linear notches.

Let's assume that the position of point P is known, and draw two circles: one with radius R1 through points A, B and P and another with radius R2 through points B, C and P (Fig. 11). We obtain the radii of these circles using formula (41):

If the coordinates of the centers of the circles - points O1 and O2 - are known, then the coordinates of point P can be determined using linear intersection formulas: from point O1 by distance R1 and from point O2 - by distance R2.

The coordinates of the center O1 can be found using the formulas for a linear intersection from points A and B along the distances R1, and from the two solutions you need to take the one that corresponds to the value of the angle in1: if in1<90o, то точка O1 находится справа от линии AB, если в1>90o, then point O1 is to the left of line AB.

The coordinates of the center O2 are found using linear intersection formulas from points B and C along distances R2, and one solution out of two possible ones is selected according to the same rule: if in2<90o, то точка O2 находится справа от линии BC, если в2>90o, then point O2 is to the left of line BC.

The problem has no solution if all four points A, B, C and P are on the same circle, since both circles merge into one, and there are no points of intersection.

Combined serifs

In the considered methods for solving serifs, the number of measurements was taken to be theoretically minimal (two measurements) to ensure obtaining the result.

In practice, to find the X and Y coordinates of one point, as a rule, not two, but three or more measurements of distances and angles are performed, and these measurements are performed both at the starting points and at the ones being determined; such serifs are called combined. It is clear that in this case it becomes possible to control measurements, and, in addition, the accuracy of solving the problem increases.

Each dimension introduced into a problem in excess of the theoretical minimum quantity is called redundant; it generates one additional solution. Geodetic intersections without redundant measurements are usually called single, and intersections with redundant measurements are called multiple.

If there are redundant measurements, the unknowns are calculated using the adjustment method. Algorithms for strict equalization of multiple intersections are used in automated computer calculations; For manual counting, simplified adjustment methods are used.

A simplified method for adjusting any multiple intersection (n measurements) involves first generating and solving all possible variants of independent single intersections (their number is n-1), and then calculating the average values ​​of the point coordinates from all the results obtained, if they differ from each other to the permissible value.

Point position error

In one-dimensional space (on a line), the position of a point is fixed by the value of one X coordinate, and the position error of the point Mp is equal to the mean square error mx of this coordinate. The true position of a point can be in the interval (X - t * mx) - (X + t * mx), that is, in both directions from the value of X; in practice, the t factor is usually set to 2.0 or 2.50.

In two-dimensional space (on a surface), the position of a point is fixed by the values ​​of two coordinates, and the position error of the point must be given by two quantities: the direction and the position error in this direction. Geometric figure, within which the true position of the point is located, may have different shapes; in the particular case when the error in the position of a point in all directions is the same, a circle of radius R = Mp is obtained.

The position of a point in two dimensions is obtained at the intersection of two position lines. For the measured distance S, the position line is a circle of radius S with the center at the starting point A (Fig. 2.12a); for the measured angle b with the vertex at the starting point A - a straight line drawn at an angle b to the starting line AB (Fig. 2.12b).

Due to measurement errors, it is necessary to introduce the concept of “position band”. For a distance S measured with a mean square error ms is a circular belt (ring) with a width of 2 * ms between two circles of radii (S - ms) and (S + ms); for angle b, measured with an error mв, it is a narrow triangle with a vertex at point A and an angle at the vertex of 2 * mв. The position line of the point is the axis of symmetry of the position strip (Fig. 12).

Rice. 12. Position line and “position strip” of point P: a) for the measured distance, b) for the measured angle.

Let us introduce the concept of “measurement error vector” and denote it by V. For the measured distance, the vector Vs is directed along the line AP (directly or reversely) and has modulus vs = ms; for the measured angle, the vector Vв is directed perpendicular to the line AP (to the left or right of it) and has a modulus nв = S * mв / s, where S = A * P.

Point P, being at the intersection of two position lines, is the center of a position 4-gon formed at the intersection of two position lines (Fig. 13).

Rice. 13.4 -position angle: a) in a linear notch, b) in a right angular notch,

This elementary 4-gon can be considered a parallelogram, since within its limits the arcs of circles can be replaced by segments of tangents, and the diverging sides of the angle by segments of straight lines parallel to the line of position. The distances from point P to the boundaries of the 4-gon are not the same, which indicates that the position errors of point P differ in different directions.

The position lines divide the position 4-gon into 4 equal parts, which we call error parallelograms with angles at the vertices r and (180o - z), where r (180o - z) is the angle between the error vectors V1 and V2. Since the heights of the parallelograms of errors are numerically equal to the modules of the vectors n1 and n2, the sides of the parallelograms are obtained according to the well-known formulas:

Using the known sides of the error parallelogram and the angle between them r (180o - r), we can calculate the length of both of its diagonals: short - d1 and long - d2:

Thus, the error in the position of a point in six directions (Fig. 14) is expressed by simple formulas; for all other directions the formulas will be more complex.

For a generalized characteristic of the accuracy of determining point P, you need to have some average value of the error in the position of point P, which can be calculated: as the radius of a circle R, the area of ​​which (p * R2) is equal to the area of ​​the parallelogram of the position of point P (4 * a * b * Sing),

as a position error in the “weakest direction”, coinciding with the direction of the long diagonal:

as the mean square of the long and short diagonals of the error parallelogram:

In practice, the third option is used more often than others, in which formulas for assessing the accuracy of any single notch are easily obtained:

polar notch (Fig. 4):

straight corner notch (Fig. 6, 7):

linear notch (Fig. 9):

reverse angular notching (Fig. 11).

In this notch, the right side of the formula for the position error of point P must contain three terms:

error of linear intersection of point O1 from initial points A and B (mO1), error of linear intersection of point O2 from initial points B and C (mO2), error of linear intersection of point P from points O1 and O2 (mP),

The notch angle r depends on the relative position of lines BC and BA and angles b1 and b2; for fig. 11 this angle is calculated by the formula:

For many practical cases, it is enough to assume that the true position of point P is inside a circle of radius MP with the center at point P. In strict theory, the considered criterion is called radial error. In addition, this theory uses more complex criteria, such as “error ellipse” (2nd order curve), “poder of error ellipse” (4th order curve), etc.

When the number of measurements is n>2 (multiple intersections), point P is obtained at the intersection of n position lines corresponding to the adjusted measurement values; the position stripes, intersecting, form a 2 * n-gon. The largest error in the position of point P will be determined by the distance from point P to the vertex of this polygon farthest from it. From Figure 14-b the role of the third dimension in reducing the error in the position of point P is clear; By the way, in this figure the second measurement has virtually no effect on the value of the point position error.

Rectangular coordinate system

To define the concept of coordinates of points, we need to introduce a 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.

Let's take a point $O$ in space and introduce coordinates $(0,0,0)$ for it. Let's call it the origin of the coordinate system. Let us draw three mutually perpendicular axes $Ox$, $Oy$ and $Oz$ through it, as in Figure 1. These axes will be called the abscissa, ordinate and applicate axes, respectively. All that remains is to enter the scale on the axes (unit segment) - the rectangular coordinate system in space is ready (Fig. 1)

Figure 1. Rectangular coordinate system in space. Author24 - online exchange of student works

Point coordinates

Now let’s look at how the coordinates of any point are determined in such a system. Let's take an arbitrary point $M$ (Fig. 2).

Let's construct a rectangular parallelepiped on the coordinate axes, so that the points $O$ and $M$ are opposite its vertices (Fig. 3).

Figure 3. Construction of a rectangular parallelepiped. Author24 - online exchange of student works

Then point $M$ will have coordinates $(X,Y,Z)$, where $X$ is the value on the number axis $Ox$, $Y$ is the value on the number axis $Oy$, and $Z$ is the value on number axis $Oz$.

Example 1

It is necessary to find a solution to the following problem: write the coordinates of the vertices of the parallelepiped shown in Figure 4.

Solution.

Point $O$ is the origin of coordinates, therefore $O=(0,0,0)$.

Points $Q$, $N$ and $R$ lie on the axes $Ox$, $Oz$ and $Oy$, respectively, which means

$Q=(2,0,0)$, $N=(0,0,1.5)$, $R=(0,2.5,0)$

Points $S$, $L$ and $M$ lie in the planes $Oxz$, $Oxy$ and $Oyz$, respectively, which means

$S=(2,0,1.5)$, $L=(2,2.5,0)$, $R=(0,2.5,1.5)$

Point $P$ has coordinates $P=(2,2.5,1.5)$

Vector coordinates based on two points and the formula for finding

To find out how to find a vector from the coordinates of two points, you need to consider the coordinate system we introduced earlier. In it, from the point $O$ in the direction of the $Ox$ axis we plot the unit vector $\overline(i)$, in the direction of the $Oy$ axis - the unit vector $\overline(j)$, and the unit vector $\overline(k) $ must be directed along the $Oz$ axis.

In order to introduce the concept of vector coordinates, we introduce the following theorem (we will not consider its proof here).

Theorem 1

An arbitrary vector in space can be expanded into any three vectors that do not lie in the same plane, and the coefficients in such an expansion will be uniquely determined.

Mathematically it looks like this:

$\overline(δ)=m\overline(α)+n\overline(β)+l\overline(γ)$

Since the vectors $\overline(i)$, $\overline(j)$ and $\overline(k)$ are constructed on the coordinate axes of a rectangular coordinate system, they obviously will not belong to the same plane. This means that any vector $\overline(δ)$ in this coordinate system, according to Theorem 1, can take the following form

$\overline(δ)=m\overline(i)+n\overline(j)+l\overline(k)$ (1)

where $n,m,l∈R$.

Definition 1

The three vectors $\overline(i)$, $\overline(j)$ and $\overline(k)$ will be called coordinate vectors.

Definition 2

The coefficients in front of the vectors $\overline(i)$, $\overline(j)$ and $\overline(k)$ in expansion (1) will be called the coordinates of this vector in the coordinate system given by us, that is

$\overline(δ)=(m,n,l)$

Linear operations on vectors

Theorem 2

Sum Theorem: The coordinates of the sum of any number of vectors are determined by the sum of their corresponding coordinates.

Proof.

We will prove this theorem for 2 vectors. For 3 or more vectors, the proof is constructed in a similar way. Let $\overline(α)=(α_1,α_2,α_3)$, $\overline(β)=(β_1,β_2 ,β_3)$.

These vectors can be written as follows

$\overline(α)=α_1\overline(i)+ α_2\overline(j)+α_3\overline(k)$, $\overline(β)=β_1\overline(i)+ β_2\overline(j)+ β_3\overline(k)$

If a certain point A is given on the coordinate plane and it is necessary to determine its coordinates, then this is done as follows. Two straight lines are drawn through point A: one parallel to the y-axis, the other - x. A straight line parallel to the y-axis intersects the x-axis (x-axis). The point of intersection of the axis and the line is the x coordinate of point A. A line parallel to the x-axis intersects the y-axis. The point of intersection of the axis and the line is the y coordinate of point A. For example, if a line parallel to y intersects the x axis at point –5, and a line parallel to x intersects the y axis at point 2.3, then the coordinates of point A are written as follows: A (–5; 2.3).

The inverse problem, when you need to draw a point at given coordinates, is solved in a similar way. Through points whose values ​​are equal to the given coordinates, lines are drawn on the x and y axes, parallel to each other: through the x coordinate - a straight line parallel to y, through the y coordinate - a straight line parallel to x. The point of intersection of these lines will be the desired point with the given coordinates. For example, given point B (–1.5; –3), you need to depict it on the coordinate plane. To do this, through the point (–1.5; 0), which lies on the x axis, draw a straight line parallel to the y axis. Through the point (0; –3) a straight line is drawn parallel to the x axis. Where these lines intersect, point B will be located (–1.5; –3).