What is the formula for finding the angle between vectors. §eleven

You already know how the coordinates of a vector are related to the coordinates of points on a plane.

A vector whose end coincides with a given point and whose origin coincides with the origin point is called radius vector ... And on the plane, the coordinates of any point are equal to the corresponding coordinates of its radius vector.

The same statement is true for space.

Definition:

Vector OM is radius vector point M of space. And the coordinates of any point in space are equal to the corresponding coordinates of its radius vector.

Let's prove, this statement.

So, we have proved that the coordinates of any point in space are equal to the corresponding coordinates of its radius vector.

Now we express the coordinates of an arbitrary vector of space AB through the coordinates of its beginning and end.

Let the coordinates of the origin A be equal to x 1, y 1 and z 1. And the coordinates of the end of B are x 2, y 2 and z 2. Let us draw the radius vectors of points A and B. It is clear that the coordinates of the vector OA are equal to the corresponding coordinates of point A. And the coordinates of the vector OB are equal to the coordinates of point B.

Using the rule for constructing the vector of the difference between two vectors, we can note that the vector AB is equal to the difference between the vectors OB and OA. Why is OB minus OA and not vice versa? Because the difference vector is directed from the end of the vector subtracted to the end of the vector of the reduced... This means that the vector OB is decreasing, and the vector OA is subtracted.

The coordinates of the vectors OA and OB are known. Then we find the coordinates of the vector AB, as the difference of the corresponding coordinates of the vectors OB and OA. We get (𝑥 2 - 𝑥 1; 𝑦 2 - 1; 𝑧 2 - 1). This is how we expressed the coordinates of the vector in terms of the coordinates of its beginning and end.

Can be done output, what each vector coordinate is equal to the difference between the corresponding coordinates of its end and beginning .

The task: by the coordinates of the points 𝐴 (2; −3; 0), 𝐵 (7; −12; 18) and 𝐶 (−8; 0; 5) determine the coordinates of the vectors 𝑂𝐴, 𝑂𝐵, 𝑂𝐶, 𝐴𝐵, 𝐵𝐶 and 𝐴𝐶, if the point 𝑂 is the origin point.

Decision:

The task: By the coordinates of the vectors 𝑂𝐴 (4; −7; 1), 𝑂𝐵 (−2; 0; 3), 𝑂𝐶 (0.5; −4; 8), 𝐴𝐷 (13; −2; 5), 𝐵𝐸 (1; - 3; 0) and 𝐶𝐹 (−9; 0; 0) determine the coordinates of points 𝐴, 𝐵, 𝐶, 𝐷, 𝐸 and 𝐹, if point 𝑂 is the origin point.

Decision:

We have considered examples of applying the relationship between the coordinates of a vector and the coordinates of its beginning and end.

Now let's talk about collinear vectors.

Do you remember if vectors A and B collinear , then one can be expressed in terms of the other by multiplying by some number k... Moreover, if, then vectors A and B co-directed , if, then vectors A and B opposite direction .

If the coordinates of the vector, then the coordinates of the vector .

Let's write down the relations of the corresponding coordinates of these vectors.

They are all equal and equal k.

So it can be noted that if the coordinates of the vectors are proportional, then these vectors are collinear .

The task: by the coordinates of the vectors, determine whether the pairs of vectors are collinear.

Decision:

The task: find the values \u200b\u200bof the variables m and n, at which these vectors will be collinear.

Decision:

You are also familiar with the concept coplanar vectors. The question arises, is it possible to determine whether they are coplanar by the coordinates of the vectors or not?

Recall that coplanar vectors are called if, when they are deposited from the same point, they lie in the same plane.

In other words, vectors are coplanar if there are equal vectors lying in the same plane .

We also found out that any two vectors are always coplanar, but three vectors can be either coplanar or non-coplanar.

In this regard, we have proved vector coplanarity.

If the vector can be decomposed into vectors and: , then vectors, and coplanar.

The converse is also true, which is called the coplanar vector property.

If vectors, and coplanar (), then the vector can be decomposed into vectors and: , and the expansion coefficients are uniquely determined.

The task: consider triples of vectors with known coordinates and find out whether they are coplanar or not.

Decision:

Outcomes:

You learned that the radius vector of a point in space is a vector whose end coincides with a given point, and its origin coincides with the origin. The coordinates of any point are equal to the corresponding coordinates of its radius vector.

We have proved that each coordinate of a vector is equal to the difference between the corresponding coordinates of its end and beginning. And also, taking advantage of the fact that any vector can be expressed in terms of a vector collinear to it, we noted that if the coordinates of the vectors are proportional, then these vectors are collinear. We also looked at examples of determining the coplanarity of vectors by their coordinates.

Let be V – n-dimensional vector space, in which two bases are given: e 1 , e 2 , …, e n - old basis, e" 1 , e" 2 , …, e" n - a new basis. An arbitrary vector a there are coordinates in each of them:

a \u003d a 1 e 1 + a 2 e 2 + ... + a n e n;

a \u003d a "1 e"1 + a" 2 e"2 +… + a" n e" n.

To establish a relationship between the columns of the coordinates a in the old and new bases, it is necessary to expand the vectors of the new basis into the vectors of the old basis:

e"1 \u003d a 11 e 1 + a 21 e 2 + ... + a n 1 e n,

e"2 \u003d a 12 e 1 + a 22 e 2 + ... + a n 2 e n,

………………………………..

e" n \u003d a 1 n e 1 + a 2 n e 2 + ... + a nn e n.

Definition 8.14. The transition matrix from the old basis to the new basis is called the matrix composed of the coordinates of the vectors of the new basis relative to the old basis, written in columns, i.e.

Matrix columns T - these are the coordinates of the basic, and hence linearly independent, vectors, therefore, these columns are linearly independent. A matrix with linearly independent columns is nondegenerate, its determinant is not equal to zero, and for the matrix T there is an inverse matrix T –1 .

Let us denote the columns of coordinates of the vector a in the old and new bases, respectively, as [ a] and [ a] ". Using the transition matrix, a link is established between [ a] and [ a]".

Theorem 8.10. Vector coordinate column a in the old basis is equal to the product of the transition matrix by the column of coordinates of the vector a in a new basis, that is [ a] = T[a]".

Consequence... Vector coordinate column a in the new basis is equal to the product of the matrix inverse to the transition matrix by the column of coordinates of the vector a in the old basis, that is [ a]" = T –1 [a].

Example 8.8. Create a matrix of transition from the basis e 1 , e 2, to the basis e" 1 , e"2, where e" 1 = 3e 1 + e 2 , e" 2 = 5e 1 + 2e 2, and find the coordinates of the vector a = 2e" 1 – 4e"2 in the old basis.

Decision... The coordinates of the new basis vectors relative to the old basis are rows (3, 1) and (5, 2), then the matrix T will take the form. Because [ a] "\u003d, then [ a] = × = .

Example 8.9. Two bases are given e 1 , e 2 - old basis, e" 1 , e"2 is a new basis, and e" 1 = 3e 1 + e 2 , e" 2 = 5e 1 + 2e 2. Find vector coordinates a = 2e 1 – e 2 in a new basis.

Decision. 1 way... By condition, the coordinates of the vector are given and in the old basis: [ a] \u003d. Find the transition matrix from the old basis e 1 , e 2 to a new basis e" 1 , e"2. Get the matrix T \u003d for it we find the inverse matrix T –1 \u003d. Then, according to the corollary to Theorem 8.10, we have [ a]" = T –1 [a] = × = .

Method 2. Because e" 1 , e"2 is a basis, then the vector and decomposes in basis vectors as follows a = k 1 e" 1 – k 2 e"2. Find the numbers k 1 and k 2 - these are the coordinates of the vector andin a new basis.

a = k 1 e" 1 – k 2 e" 2 = k 1 (3e 1 + e 2) – k 2 (5e 1 + 2e 2) =

= e 1 (3k 1 + 5k 2) + e 2 (k 1 + 2k 2) = 2e 1 – e 2 .

Since the coordinates of one and the same vector in this basis are uniquely determined, we have the system: Solving this system, we get k 1 \u003d 9 and k 2 \u003d –5, so. [ a]" = .

In this article, we will begin a discussion of 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 insecure about building spatial figures, sections, etc. All this requires a certain imagination and practical skills. The method, which we will begin to consider here, will allow you to almost completely abstract from all kinds of geometric constructions and reasoning. The method is called "Coordinate method"... In this article, we will consider the following questions:

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. Midpoint coordinates
6. Dot product of vectors
7. Angle between two vectors

I think you already guessed why the coordinate method is called that? It is true that he received such a name, since he operates not with geometric objects, but with their numerical characteristics (coordinates). And the transformation itself, which allows us to go 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 only consider 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 in solving problems on planimetry in part B of the USE). The next two sections on this topic are devoted to the 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 the 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 that way. For example, if, then, if, then, etc. What did you get in the end? And you received points with coordinates: and. Then you drew a "cross" (coordinate system), chose a scale on it (how many cells you will have as a unit segment) and marked on it the points you received, which you then connected with a straight line, the resulting line is the graph of the function.

There are several 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 nicely and compactly in the picture

2. It is assumed 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 recording the coordinates of a point, for example, on the left in brackets 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 set any point on the coordinate axis, you need to specify its coordinates (2 numbers)

6. For any point lying on the axis,

7. For any point on the axis,

8. The axis is called the abscissa axis

9. The axis is called the y-axis.

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

Remember, what else is a directed segment called? That's right, it's called a vector!

Thus, if we connect a point with a point, where the beginning will be point A, and the end will be point B, then we get a vector. You also did this formation in the 8th grade, remember?

It turns out that vectors, like points, can be denoted by two numbers: these numbers are called the coordinates of the vector. The question is: do you think it is enough for us to know the coordinates of the beginning and end of the vector to find its coordinates? It turns out that yes! And this is done very simply:

Thus, since the point in the vector is the beginning and a 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 the end: now the beginning of the vector will be at the point, and the end will be at the point. Then:

Look closely, how are vectors and? Their only difference is the signs in the coordinates. They are opposite. It is customary to write this fact like this:

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

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

Verification:

Now solve the problem a little harder:

Vektor with na-cha-lom at the point has co-or-di-na-ty. Nay-di-those abs-cis-su points.

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

I made up the system by definition of what the coordinates of a vector 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 cannot 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:

The vector expands or contracts or changes direction when multiplied or divided by a number:

However, here we will be interested in the question of what is happening with the coordinates.

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

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

For instance:

· Nay-di-te sum of co-or-di-nat vek-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.

Answer:

Now solve the following problem yourself:

Find the sum of the coordinates of a vector

We check:

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

What I've done? I, firstly, connected the points and, and also from the point I drew a line parallel to the axis, and from the point I drew a line parallel to the axis. Did they intersect at a point, thus forming a wonderful figure? What is it remarkable for? Yes, you and I know almost everything about a right-angled triangle. Well, the Pythagorean theorem - for sure. The sought segment is the hypotenuse of this triangle, and the segments are the legs. What are the coordinates of a point? Yes, they are easy to find from the picture: Since the segments are parallel to the axes and, accordingly, their lengths are easy to find: if you denote the lengths of the segments, respectively, by, 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 squares of the differences from the coordinates. Or, the distance between two points is the length of the line that connects them. It is easy to see that the distance between points is independent of direction. Then:

From this we draw three conclusions:

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

For example, if, then the distance between and is

Or let's go differently: find the coordinates of the vector

And find the length of the vector:

As you can see, the same thing!

Now do some practice yourself:

Task: find the distance between the specified points:

We check:

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

1. Nay-di-te square-rat of the length of the century-to-ra.

2. Nay-di-te square-rat of the length of the century-to-ra

I think you did with them easily? We check:

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

2. Find the coordinates of the vector

Then the square of its length is

Nothing complicated, right? Simple arithmetic, nothing more.

The following tasks cannot be categorized unambiguously, they are more likely to general erudition and the ability to draw simple pictures.

1. Nay-di-te sine of an angle on-to-on from-cut, co-single-nya-yu-shch-th point, with the abscissa axis.

and

What are we going to do here? You need to find the sine of the angle between and the axis. And where do we know how to look for a sine? Right in a right-angled triangle. So what do we need to do? Build this triangle!

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

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

Answer:

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

Objective 2. Per-pen-di-ku-lar is lowered from the point to the abs-ciss axis. Nay-di-te abs-cis-su os-no-va-nia per-pen-di-ku-la-ra.

Let's make a drawing:

The base of the perpendicular is the point at which it crosses the abscissa axis (axis), for me this is the point. The figure shows that it has coordinates:. We are interested in the abscissa - that is, the "X" component. It is equal.

Answer: .

Objective 3. Under the conditions of the previous problem, find the sum of the distances from a point to the coordinate axes.

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

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

Answer: .

Problem 4. In the conditions of problem 2, find the ordinate of the point symmetrical to the point about the abscissa axis.

I think you intuitively understand what symmetry is? Many objects have it: many buildings, tables, airplanes, many geometric shapes: a 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. What then is an axis? This is exactly the line along which a figure can, relatively speaking, be "cut" into equal halves (in this picture, the axis of symmetry is a straight line):

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

Did you do the same? Okay! At the found point, we are interested in the ordinate. She is equal

Answer:

Now tell me, after thinking about 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 coordinates:

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

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

Answer:

Now parallelogram problem:

Problem 5: The points are ver-shi-na-mi paral-le-lo-gram-ma. Nay-di-te or-di-na-tu points.

You can solve this problem in two ways: logic and the method of coordinates. I will first apply the coordinate method, and then I will tell you how you can decide otherwise.

It is quite clear that the abscissa of the point is. (it lies on the perpendicular drawn from a point to the abscissa). We need to find the ordinate. Let's take advantage of the fact that our figure is a parallelogram, which means that. 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. The intersection point will be marked with a letter.

The segment length is. (find the problem itself, where we discussed this point), then we find the length of the segment by the Pythagorean theorem:

The length of the segment is exactly the same as its ordinate.

Answer: .

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

Solution progress:

1. Conduct

2. Find point coordinates and length

3. Prove that.

One more segment length puzzle:

The points appear-la-are-Xia ver-shi-na-mi tre-coal-no-ka. Nay-di-te is the length of its middle line, paral-lel-noy.

Do you remember what the middle line of a triangle is? Then this task is elementary for you. If you don't remember, then I will 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 line segment. We had to search for its length earlier, it is equal. Then the length of the middle line is half and equal.

Answer: .

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

In the meantime - here are a few tasks for you, practice them, they are quite simple, but they help you to "get your hands on" using the method of coordinates!

1. The points are the ver-shi-na-mi tra-petsii. Nay-di-te is the length of its middle line.

2. Dots and are-la-sy-ver-shi-na-mi pa-ra-le-lo-gram-ma. Nay-di-te or-di-na-tu points.

3. Nay-di-te length from-cut, co-single-nya-yu-shch-go point and

4. Nay-di-te area of \u200b\u200bthe beautiful fi-gu-ry on the co-or-di-nat-noy plane.

5. The circle with the center at na-cha-le ko-or-di-nat passes through the point. Nay-di-te her ra-di-us.

6. Nay-di-te ra-di-us of the circle, described-san-noy around the rectangular-coal-nik, the vertices of the co-ro-go have co-op -di-na-you co-vet-but

Solutions:

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

Answer:

2. The easiest way to solve this problem is to notice that (parallelogram rule). Calculate the coordinates of vectors and is not difficult:. When adding vectors, the coordinates are added. Then has coordinates. The point has the same coordinates, since the origin of the vector is the point with coordinates. We are interested in the ordinate. It is equal.

Answer:

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

Answer:

4. Look at the picture and tell me, between which two shapes is the shaded area "sandwiched"? It is sandwiched between two squares. Then the area of \u200b\u200bthe required figure is equal to the area of \u200b\u200bthe large square minus the area of \u200b\u200bthe small one. The side of the small square is a line segment connecting the points and its length is

Then the area of \u200b\u200bthe 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 \u200b\u200bthe large square is

The area of \u200b\u200bthe required figure is found by the formula:

Answer:

5. If the circle has the origin of coordinates as its center and passes through a point, then its radius will be exactly equal to the length of the segment (draw a picture 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 of its diagonal. Find the length of any of the two diagonals (after all, they are equal in a rectangle!)

Answer:

Well, have you dealt with everything? It wasn't very hard to figure it out, was it? The rule here is one - to 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 midpoint, then it has coordinates:

I.e: midpoint coordinates \u003d arithmetic mean of the corresponding coordinates of the segment ends.

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

1. Nay-di-te or-di-na-tu-re-di-us from-cut, co-uni-nya-yu-shch-go point and

2. The points are-la-yut-sya ver-shi-na-mi-you-rekh-coal-no-ka. Nay-di-te or-di-na-tu points of pe-re-se-ch-niya his dia-go-na-lei.

3. Nay-di-those abs-cis-su center-tra of the circle, described-san-noy around the coal-no-ka, the vertices of the ko-to-ro-go have co-op-di-na-you co-vet-but.

Solutions:

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

Answer:

2. It is easy to see that the given quadrangle is a parallelogram (even a rhombus!). You yourself can prove this by calculating the lengths of the sides and comparing them with each other. What do I know about a parallelogram? Its diagonals are halved by the intersection point! Aha! So the point of intersection of the diagonals is what? 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.

Answer:

3.What does the center of the circumscribed circle about the rectangle coincide with? It coincides with the point of intersection of its diagonals. What do you know about the diagonals of a rectangle? They are equal and the intersection point is halved. The task was reduced to the previous one. Take the diagonal, for example. Then if is the center of the circumscribed circle, then is the middle. Looking for coordinates: Abscissa is equal.

Answer:

Now practice a little yourself, I will just give the answers to each problem so that you can test yourself.

1. Nay-di-te ra-di-us of the circle, described-san-noy around the triangle, the vertices of the co-to-ro-go have co-or-di -no misters

2. Nay-di-te or-di-na-tu center-tra of the circle, describe-san-noy around the triangle-nik, the vertices of ko-to-ro-go have coordinates

3. How-to-ra-di-u-sa should there be a circle with a center at the point so that it would be aligned with the abs-cissa axis?

4. Nay-di-te or-di-na-tu points of pe-re-seeding of the axis and from-cut, co-uni-nya-yu-shch-go point and

Answers:

Did you succeed? I really hope for it! Now - the last push. Be especially careful now. The material that I will now explain is directly related not only to simple problems on the method of coordinates from the B part, but also occurs everywhere in the problem C2.

Which of my promises have I not kept yet? Remember what operations on vectors I promised to introduce and what ones I eventually introduced? Have I forgotten anything? I forgot! 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 nature:

The vector product is pretty tricky. How to do it and what it is for, we will discuss with you in the next article. And in this one we will focus on the dot product.

There are two ways to calculate it:

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

Dot product in terms of coordinates

Find: - common dot product notation

The formula for the calculation is as follows:

That is, the dot product \u003d the sum of the products of the coordinates of the vectors!

Example:

Nai di te

Decision:

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

We calculate the dot product by the formula:

Answer:

See, absolutely nothing complicated!

Well, now try it yourself:

Nay-di-te scalar-noe pro-iz-ve-de-vek-to-moat and

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

The coordinates of the vectors are the same as in the previous task! Answer:.

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

Indicates the angle between vectors and.

That is, the dot product is equal to the product of the vector lengths by 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 we can derive from the first and second formulas 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 dot product formula, then I get:

But in other way:

So what did you and I get? We now have a formula 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 dot product in terms of 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. Nay-di-those angle between the century-to-ra-mi and. Give the answer in gra-du-sakh.

2. Under 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! I agree? Then let's get started!

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

Then we are looking for the cosine between the vectors:

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

Answer:

Now solve the second problem yourself, and then we'll compare! I will only give a very short solution:

2. has coordinates, has coordinates.

Let be the angle between vectors and, then

Answer:

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

COORDINATES AND VECTORS. MEDIUM ROVEN

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

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

Of course, the whole coordinate method does not fit into these 6 points. It lies at the heart of such a science as analytical geometry, which you will get to know at the university. I just want to build a foundation that will allow you to solve problems in a single state. exam. We figured out the tasks of Part B in Now it's time to move to a qualitatively new level! This article will be devoted to the method for solving those problems C2, in which it would be reasonable to switch to the method of coordinates. This rationality is determined by what is required to find 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 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 point to line
6. Find the distance from a straight line to a plane
7. Find the distance between two straight lines

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

Suitable shapes for the coordinate method are:

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

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

1. Finding the cross-sectional areas
2. Calculating body volumes

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

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

All of them are mutually perpendicular, intersect at one point, which we will call the origin. The abscissa axis, as before, will be denoted, the ordinate axis -, and the entered applicate axis -.

If earlier 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, ordinate, applicate. For instance:

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

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

is called the projection of a point onto a plane

is 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 look the same. For a little detail. I think you already guessed for which one. We will have to add one more term to all formulas, which is responsible for the applicate axis. Namely.

1. If two points are given:, then:

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

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

• Their dot product is:
• The cosine of the angle between vectors is:

However, space is not so simple. As you can imagine, the addition of one more coordinate introduces a significant variety in 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" is the plane. What do you know about a 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, it is a kind of endless "leaf" thrust into space. "Infinity" should be understood that the plane extends in all directions, that is, its area is equal to infinity. However, this explanation "on the fingers" does not give the slightest idea about the structure of the plane. And we will be interested in it.

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

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

Or its counterpart in space:

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

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

For example, a straight line passes through the points:

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

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

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

Once again, I will not be very interested in the equation of a line, but I really need you to remember what a direction vector is! Again: it is ANY nonzero vector lying on a line or parallel to it.

Withdraw equation of a plane at three given points is no longer so trivial, and usually this issue is not addressed in a high school course. But in vain! This technique is vital when we use 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 with the methodology that is usually studied in the course of analytical geometry. 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 the plane is not very different from the equation of a straight line (linear function). However, remember what you and I said? We said that if we have three points that do not lie on one straight line, then the equation of the plane is uniquely reconstructed from them. But how? I'll try to explain 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 get the correct identity:

Thus, it becomes necessary to solve three equations even with unknowns! Dilemma! However, you can always assume that (for this you need to divide by). Thus, we get three equations with three unknowns:

However, we will not solve such a system, but write out a 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 | \u003d 0 \\]

Stop! What is this? Some very unusual module! However, the object 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 come across these same determinants. What is a third-order determinant? Oddly enough, this is 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 line number, and by the index - the column number. For example, it means that this number is at the intersection of the second row and the third column. Let's pose the next question: how exactly are we going to calculate such a determinant? That is, what specific number will we match to it? For the determinant of the third order, there is a heuristic (visual) rule of the triangle, it looks as follows:

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 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 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 \u200b\u200bobtained at step and

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

Nevertheless, you do not need to memorize the calculation method in this form, it is enough to just keep the triangles and the very idea of \u200b\u200bwhat 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:

The terms that come with a "plus":

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

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

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

Add three numbers:

Terms that come with a "minus"

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

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

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

Add three numbers:

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

In this way,

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

We check:

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

Here's a couple more determinants, calculate their values \u200b\u200byourself and compare them with the answers:

Answers:

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

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

All you need is to calculate its value directly (using the triangles method) and set the result to zero. Naturally, since they 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 one straight line!

Let's illustrate this with a simple example:

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

We compose the determinant for these three points:

Let's simplify:

Now we calculate it directly by the rule of triangles:

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

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

We compose the determinant:

And we calculate its value:

Then the equation of the plane has the form:

Or, by reducing by, we get:

Now two tasks for self-control:

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

Answers:

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

However, with the help of determinants, we will construct not only the 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 dot 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 the area of \u200b\u200bthe parallelogram built on the vectors and. We will need this vector to calculate the distance from a point to a straight line. How do we calculate the cross product of vectors and, if their coordinates are given? The determinant of the third order comes to our aid again. However, before I move on to the vector product calculation algorithm, I have to make a small lyrical 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 product

Now I can start introducing the cross product:

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

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

Example 1: Find the cross product of vectors:

Solution: I compose a determinant:

And I calculate it:

Now, from writing in terms of basis vectors, I'll return to the usual notation of a vector:

In this way:

Now try it.

Are you ready? We check:

And traditionally two tasks for control:

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

Answers:

Mixed product of three vectors

The last construct I need is a 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 have three vectors:

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

1. - that is, the mixed product is the dot product of a vector by the cross product of two other vectors

For example, the mixed product of three vectors is:

Try to calculate it yourself through the cross product and make sure that the results match!

And again - two examples for an independent solution:

Answers:

Coordinate system selection

Well, now we have all the necessary knowledge base to solve complex stereometric problems in geometry. However, before proceeding directly to the examples and algorithms for their solution, I believe that it will be useful to dwell on another question: how exactly select a coordinate system for a particular figure. After all, it is the choice of the relative position of the coordinate system and the figure in space that will ultimately determine how cumbersome the calculations will be.

Let me remind you that in this section we are looking at the following shapes:

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

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

That is, I will place the figure "in the corner". The cube and the parallelepiped are very nice shapes. 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 do not need to remember this, but remembering how best to place a cube or rectangular parallelepiped is desirable.

Straight prism

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

Triangular prism:

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

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:

A situation similar to a cube: align the two sides of the base with the coordinate axes, align one of the vertices with the origin. The only small 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 in finding 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 getting down to solving 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 consider the problem of finding an angle. They, in turn, are divided into the following categories (as the difficulty increases):

Finding corners

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

Let's consider these tasks in sequence: start by finding the angle between two straight lines. Well, remember, did you and I solve similar examples before? Remember, we already had something similar ... We were looking for an angle between two vectors. I will remind you, if two vectors are given: and, then the angle between them is found from the ratio:

Now we have a goal - to find the angle between two straight lines. Let's turn to the "flat picture":

How many corners did we get when two straight lines intersect? As many things. True, only two of them are unequal, while others are vertical to them (and therefore coincide with them). So what 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 the two straight lines is equal. In order not to bother with finding the smallest of two angles every time, cunning mathematicians suggested using the module. Thus, the angle between two straight lines is determined by the formula:

As an attentive reader, you should have had a question: where, in fact, will we take these very numbers that we need to calculate the cosine of an angle? Answer: we will take them from the direction vectors of the straight 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 dot product
4. We are looking for the length of the first vector
5. We are looking for the length of the second vector
6. Multiplying the results from point 4 by the results from point 5
7. 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 calculate the angle exactly, look for it
9. Otherwise, we write through the inverse cosine

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

Tasks:

1. In the correct tet-ra-ed-re, nay-di-those angle between you-so-that tet-ra-ed-ra and the med-di-a-noy bo-kovy face.

2. In the right-handed six-coal-noy pi-ra-mi-de, the sides of the os-no-va-nia are equal, and the ribs are equal, find the angle between the straight lines and.

3. The lengths of all the edges of the correct four-you-rekh-coal pi-ra-mi-dy are equal to each other. Nay-di-those angle between the straight lines and if from-cut is you-co-that given pi-ra-mi-dy, the point is se-re-di-na her bo-ko- second rib

4. On the edge of the cube from-me-che-na point so that Nay-di-te is the angle between straight lines and

5. Point - se-re-di-on the edges of the cube Nay-di-te angle between straight lines and.

It is no coincidence that I have arranged the tasks in this order. While you have not yet had time to start navigating the method of coordinates, I myself will analyze the most "problematic" figures, 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 of its faces (including the base) are regular triangles. Since we are not given the length of the side, I can take it equal. I think you understand that the angle will not really 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 also need to find the coordinates of the points. Now we think: a point is the intersection point of the heights (or bisectors or medians) of the triangle. A point is a raised point. The point is the middle of the segment. Then finally we need to find: coordinates of points:.

Let's start with the simplest: point coordinates. Look at the picture: It is clear that the applicate of the point is equal to zero (the point lies on the plane). Its ordinate is equal to (since - 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 the 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 a point, that is. Let's find its abscissa. This is done quite trivially if you remember that the heights of an equilateral triangle are divided by the point of intersection in proportioncounting 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 equal:

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 searched for from the considerations that I have highlighted in bold:

Point is the midpoint of the line. Then we need to remember the formula for the coordinates of the midpoint of the segment:

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

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

In this way,

Answer:

You should not be intimidated by such "scary" answers: for C2 problems this is a common practice. I would rather be surprised at the "nice" answer in this part. Also, as you noticed, I practically did not resort to anything other than the Pythagorean theorem and the property of heights 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's draw a regular hexagonal pyramid together with a coordinate system, as well as its base:

We need to find the angle between the lines and. Thus, our task is reduced to finding the coordinates of points:. We will find the coordinates of the last three from the small picture, and we will find the coordinate of the vertex through the coordinate of the point. Work in bulk, but you need to start it!

a) Coordinate: it is clear that its applicate and ordinate are zero. Let's find the abscissa. To do this, consider a right-angled triangle. Alas, in it we know only the hypotenuse, which is equal to. We will try to find the leg (for it is clear that the doubled leg length will give us the abscissa of the point). How can we find her? 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 it has all sides and all angles. I ought to find one such corner. 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:

We 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.

Thus, it 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 to. Finding the ordinate is also not very difficult: if we connect the points and denote the intersection point of the straight line, say by. (DIY 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 find the coordinates of the point. Consider a 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 applicator. Since, then. Consider a right-angled triangle. By the statement of the problem, the side edge. This is the hypotenuse of my triangle. Then the height of the pyramid is the leg.

Then the point has coordinates:

All right, I have the coordinates of all the points of interest to me. Looking for the coordinates of the direction vectors of straight lines:

We are looking for the angle between these vectors:

Answer:

Again, when solving this problem, I did not use any sophisticated tricks, except for the formula for the sum of the angles of a regular n-gon, as well as determining the cosine and sine of a right triangle.

3. Since we are again not given the lengths of the ribs in the pyramid, I will consider them equal to one. Thus, since ALL edges, and not only 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's depict such a pyramid, as well as its base on a plane, marking all the data given in the text of the problem:

We are looking for the angle between and. I will be doing very brief calculations when looking for the coordinates of points. You will need to "decipher" them:

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

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

Coordinates:

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

e) Vector coordinates

f) Vector coordinates

g) Looking for an angle:

The cube is the simplest figure. I'm sure you can handle her 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 tasks 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. From three points we construct the equation of the plane
   ,
   using a third-order determinant.
2. Looking for the coordinates of the directing vector of the straight line by two points:
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 the angles between two lines. The structure of the right side is just the same, but on the left we are now looking for sine, not cosine, as before. Well, one nasty action was added - the search for the equation of the plane.

Let's not postpone solution of examples:

1. The main-but-va-no-em direct prize-we are-la-is-equal-but-poor-worn triangles You-so-that prize-we are equal. Nai di te angle between straight line and plane

2. In a rectangular pa-ra-le-le-pi-pe-de from the West Nay-di-te angle between straight line and plane

3. In the correct six-coal prism, all edges are equal. Nay-di-those angle between a straight line and a plane.

4. In the right-handed triangular pi-ra-mi-de with os-no-va-ni-e from the west of the ribs Nay-di-te angle, ob-ra-zo-van flat-to-bone os-no-va-nia and straight, pro-ho-dya-shi through se-re-di-us of the ribs and

5. The lengths of all the ribs of a regular four-corner pyramid with apex are equal to each other. Nay-di-te is the angle between a straight line and a plane, if the point is se-re-di-na bo-ko-th ribs pi-ra-mi-dy.

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

Solutions:

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

I apologize for some non-observance of the proportions, but for solving the problem, this, in fact, is not so important. The plane is just the "back wall" of my prism. It is easy enough to 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 compose the equation of the plane:

Exercise for you: calculate this determinant yourself. Did you do it? Then the equation of the plane has the form:

Or simply

In this way,

To solve the example, I need to find the coordinates of the direction vector of a straight line. Since the point coincides with the origin, 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 (it is the median and the bisector) from the vertex. Since, then the ordinate of the point is. In order to find the abscissa of this point, we need to calculate the length of the segment. By the Pythagorean theorem we have:

Then the point has coordinates:

A point is a “raised” point:

Then the coordinates of the vector:

Answer:

As you can see, there is nothing fundamentally difficult in solving such problems. In fact, the process further simplifies the "straightness" of a shape 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: Coordinates of three points lying in it:

(The first two coordinates were 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 direction vector: It is clear that its coordinates coincide with the coordinates of the point, isn't it? How do I find the coordinates? These are the coordinates of the point, raised along the axis of the application by one! ... Then we are looking for the required angle:

Answer:

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

Here even drawing a plane is problematic, not to mention solving this problem, but the coordinate method does not care! It is in its versatility that its main advantage lies!

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

1) . Draw the coordinates for the last two points yourself. The solution to the problem with a hexagonal pyramid is useful for you!

2) We build the equation of the plane:

We are looking for the coordinates of the vector:. (see the triangle 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. For the last two problems, I will only give answers:

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 in some formulas. It remains for us to consider one more class of problems for calculating angles, namely:

Calculating angles between two planes

The solution algorithm will be as follows:

1. We look for the equation of the first plane by three points:
2. For 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 searched for the angles between straight lines and between a straight line and a plane. So remembering this one will not be difficult for you. Let's go straight to the analysis of tasks:

1. One hundred-ro-na of the os-no-va-nia of the right-handed triangular prism is equal, and the dia-go-nal of the big face is equal. Nay-di-those are the angle between the plane of the prism and the plane of the prism.

2. In the correct four-you-rekh-coal-noy pi-ra-mi-de, all the edges of the swarm are equal, find the sine of the angle between the plane and the plane to-stu, pro-ho-dya-shchey through the point per-pen-di-ku-lar-but straight.

3. In the correct four-you-rekh-coal prism, the sides of the os-no-va-nia are equal, and the sides are equal. On the edge from-me-che-to point so that. Find the angle between the plane-to-sti-mi and

4. In the right four-corner prism, the sides of the os-no-va-nia are equal, and the side edges are equal. On the edge from-me-che-to point so that Nay-di-te is the angle between plane-to-st-mi and.

5. In the cube nai-di-te ko-si-nus of the angle between the plane-ko-sti-mi and

Problem solutions:

1. I draw a regular (at the base 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 by three points, but I will compose the equation at once:

Now we find the equation Point has coordinates Point - Since is the median and the height of the triangle, it is easy to find in a triangle using the Pythagorean theorem. Then the point has coordinates: Find the applicate of the point To do this, consider a right-angled triangle

Then we get these coordinates: We make the equation of the plane.

We calculate the angle between the planes:

Answer:

2. Make a drawing:

The most difficult thing is to understand what this mysterious plane is, passing through a point perpendicularly. Well, the main thing is what is this? The main thing is attentiveness! Indeed, 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 goes 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.

Find the coordinate of the point through the point. From the small figure it is easy to deduce that the coordinates of the point will be as follows: What is now left to find 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 special in calculating determinants. You can easily get:

Or else (if we multiply both parts by the root of two)

Now we find the equation of the plane:

(You have not forgotten how we get the equation of the plane, right? If you do not understand where this minus one came from, then go back to the definition of the equation of the plane! It's just that before that it turned out that the origin of coordinates belonged to my plane!)

We calculate the determinant:

(You can see that the equation of the plane coincides with the equation of the straight line passing through the points and! Think why!)

Now we calculate the angle:

We need to find the sine:

Answer:

3. Tricky question: what do you think is a rectangular prism? It's just a parallelepiped you know well! Make a drawing immediately! It is even possible not to depict the base separately, there is little benefit from it here:

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

Now we make a plane

We immediately compose 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 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 problems. Namely, we will consider the following cases:

1. Calculation of the distance between crossed lines.

I have ordered these tasks as their complexity increases. It turns out to be the 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 start considering 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 get all the necessary data, then we apply the formula:

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

Tasks:

1. Given a cube. The length of the edge of the cube is. Nay-di-te distance-i-ni from se-re-di-us from-cut to flat-to-sti

2. Given the right-vil-naya four-you-rekh-coal-naya pi-ra-mi-da Bo-ko-th edge of the side-ro-na os-no-va-nia is equal. Nay-di-those distance from point to plane-to-sti where - se-re-di-on ribs.

3. In the right-handed triangular pi-ra-mi-de with os-no-va-ni, the bo-kov edge is equal, and the side-ro-na is-no-va- is equal to. Nai-di-te distance-i-nye from the top to the plane.

4. In the correct six-angle prism, all edges are equal. Nay-di-te distance-i-nye from point to plane.

Solutions:

1. Draw a cube with unit edges, build a segment and a plane, denote the middle of the segment by the letter

.

Let's start with an easy one: find the coordinates of a point. Since then (remember the coordinates of the midpoint of the segment!)

Now we compose the equation of the plane by three points

\\ [\\ left | (\\ begin (array) (* (20) (c)) x & 0 & 1 \\\\ y & 1 & 0 \\\\ z & 1 & 1 \\ end (array)) \\ right | \u003d 0 \\]

Now I can start looking for distance:

2. Start again with the drawing, on which we mark all the data!

For the pyramid, it would be helpful to draw its base separately.

Even the fact that I am drawing like a chicken with a paw does 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 midpoint of the segment, then

We can also find the coordinates of two more points on the plane without any problems.We compose the equation of the plane and simplify it:

\\ [\\ left | (\\ left | (\\ begin (array) (* (20) (c)) x & 1 & (\\ frac (3) (2)) \\\\ y & 0 & (\\ frac (3) (2)) \\\\ z & 0 & (\\ frac ( (\\ sqrt 3)) (2)) \\ end (array)) \\ right |) \\ right | \u003d 0 \\]

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

Answer (very rare!):

Well, figured it out? It seems to me that everything here is as technical as in the examples that we considered with you 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 will just give the answers:

Calculating the distance from a straight line to a plane

In fact, there is nothing new here. How can a line and a plane be located relative to each other? They have all the possibilities: 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. An uninteresting case.

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

In this way:

And 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, we are looking for the equation of the plane, and we calculate the distance from a point to the plane. In fact, such tasks are extremely rare in the exam. I managed to find only one problem, and the data in it was 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. The coordinates of the point from which we are looking for the distance:

2. Coordinates of any point lying on a straight line

3. Coordinates of the direction vector of the straight line

What formula do we use?

What does the denominator of this fraction mean to you and so it should be clear: this is the length of the directing vector of a straight line. There is a very tricky numerator here! The expression means the modulus (length) of the vector product of vectors and How to calculate the cross product, we studied in the previous part of the work. Refresh your knowledge, they will be very useful to us 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 straight line to which we are looking for the distance:

3. Build a vector

4. Build the direction vector of the straight line

5. Calculate the cross product

6. We are looking for the length of the resulting vector:

7. Calculate the distance:

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

1. Dana is a right-vil-naya triangular pi-ra-mi-da with a top. One hundred-ro-na os-no-va-nia pi-ra-mi-dy is equal, you-so-that is equal. Nay-di-those distance from the se-re-di-ny of the bo-ko-th rib to the straight line, where the points and are the se-re-di-ny of the ribs and co- vet-but.

2. The lengths of the ribs and the right-angled pa-ral-le-le-pi-pe-da are equal co-vet-n-but and Nay-di-te distance from top to top to straight

3. In the correct six-coal prism, all the edges of a swarm are equal find-di-those 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 with you! First, I would like to describe in words what we will be looking 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. The length of the vector product

8. Distance from to

Well, we have a lot of work to do! We get down to it, rolling up our sleeves!

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 the ordinate is equal to the Abscissa, it is equal to the length of the segment. Since is the height of an equilateral triangle, it is divided in relation, counting from the top, hence. Finally, we got the coordinates:

Point coordinates

2. - the middle of the segment

3. - the middle of the segment

Mid-point

4.Coordinates

Vector coordinates

5. We calculate the cross product:

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

7. We consider the length of the vector product:

8. Finally, we find the distance:

Phew, that's it! Honestly, the solution to this problem by traditional methods (through constructions) would be much faster. But here I have reduced everything to a ready-made algorithm! I think that 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, and not resorting to the coordinate method. I have demonstrated this solution only to show you a universal method that allows you to "complete nothing".

Finally, consider the last class of problems:

Calculating the distance between crossed lines

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

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

How do we find the distance between straight lines?

The formula is as follows:

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

I will remind you that

then the formula for the distance can be rewritten as:

A sort of 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 rather complicated calculations. If I were you, I would only use it as a last resort!

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

1. In the correct triangular prism, all the edges of a swarm are equal, find-di-those distance between straight lines and.

2. Given the right-handed triangular prism, all the edges of the os-no-va-tion of the co-swarm are equal Se-c-c, passing through the god rib and se-re-di-well ribs yav-la-et-sya square-ra-tom. Nai di te distance between straight we and

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

1. Draw a prism and mark the straight lines and

Point C coordinates: then

Point coordinates

Vector coordinates

Point coordinates

Vector coordinates

Vector coordinates

\\ [\\ left ((B, \\ overrightarrow (A (A_1)) \\ overrightarrow (B (C_1))) \\ right) \u003d \\ left | (\\ begin (array) (* (20) (l)) (\\ begin (array) (* (20) (c)) 0 & 1 & 0 \\ end (array)) \\\\ (\\ begin (array) (* (20) (c)) 0 & 0 & 1 \\ end (array)) \\\\ (\\ begin (array) (* (20) (c)) (\\ frac ((\\ sqrt 3)) (2)) & (- \\ frac (1) (2)) & 1 \\ end (array)) \\ end (array)) \\ right | \u003d \\ frac ((\\ sqrt 3)) (2) \\]

We consider the cross product between vectors and

\\ [\\ overrightarrow (A (A_1)) \\ cdot \\ overrightarrow (B (C_1)) \u003d \\ left | \\ begin (array) (l) \\ begin (array) (* (20) (c)) (\\ overrightarrow i) & (\\ overrightarrow j) & (\\ overrightarrow k) \\ end (array) \\\\\\ begin (array ) (* (20) (c)) 0 & 0 & 1 \\ end (array) \\\\\\ begin (array) (* (20) (c)) (\\ frac ((\\ sqrt 3)) (2)) & (- \\ - \\ frac ((\\ sqrt 3)) (2) \\ overrightarrow k + \\ frac (1) (2) \\ overrightarrow i \\]

Now we 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

Vector is a directed line segment. - the beginning of the vector, - the end of the vector.
The vector is denoted by or.

Absolute valuevector - the length of the segment representing the vector. It is indicated 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 the product of their absolute values \u200b\u200bby the cosine of the angle between them:

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In addition to the previously considered operations of addition and subtraction of vectors, as well as multiplication of a vector by a scalar (see

In addition to the previously considered operations of addition and subtraction of vectors, as well as multiplication of a vector by a scalar (see §2), there are also operations of multiplication of vectors. Two vectors can be multiplied by each other in two ways: the first method results in some new vector, the second leads to a scalar value. Note that the operation of dividing a vector by a vector does not exist.

Now we will consider the sector product of vectors. We will introduce the scalar product of vectors later, when we need it.

A vector product of two vectors A and B is a vector C with the following properties:

1) the modulus of the vector C is equal to the product of the moduli of the vectors being multiplied by the sine of the angle α between them (Figure 35):

2) the vector C is perpendicular to the plane in which the vectors A and B lie, and its direction is connected with the directions A and B according to the rule of the right screw: if you look after the vector C, then the rotation made along the shortest path from the first factor to the second is clockwise arrow.

Symbolic cross product can be written in two ways:

| AB | or .

We will use the first of these methods, and sometimes we will put a comma between the factors to make it easier to read the formulas. You should not use both an oblique cross and square brackets: [AB], It is not allowed to write this type: [AB] \u003d ABsi nα. On the left there is a vector, on the right - the modulus of this vector, that is, a scalar. The following equality is valid:

Since the direction of the vector product is determined by the direction of rotation from the first factor to the second, the result of vector multiplication of two vectors depends on the order of the factors. Changing the order of the factors causes the direction of the resulting vector to change to the opposite (Fig. 35)

Thus, the cross product is not commutative.

One can prove that the vector product is distributive, that is, that

The vector product of two polar or two axial vectors is an axial vector. The vector product of an axial vector by a polar one (or vice versa) will, however, be a polar vector. Changing the condition that determines the direction of the axial vectors to the opposite will lead in this case to a change in sign before the vector product and at the same time to a change in sign in front of one of the factors.As a result, the value expressed by the vector product remains unchanged.

The vector product module can be given a simple geometric interpretation: the expression ABsi nα is numerically equal to the area of \u200b\u200bthe parallelogram built on vectors A and B (Fig. 36; vector C \u003d [AB] is directed in this case perpendicular to the plane of the drawing, per drawing).

Let the vectors A and B be mutually perpendicular (Fig. 37).

We form a double cross product of these vectors:

that is, we multiply the vector B by A, and then multiply the vector A by the vector resulting from the first multiplication. The vector [VA] has a modulus equal to , and forms angles equal to π / 2 with vectors A and B. Therefore, the modulus of the vector D is | A | * || \u003d A * BA \u003d A 2 B. The direction of the vector D, as is easy to see from Fig. 37, coincides with the direction of vector B. This gives us reason to write the following equality:

We will use formula (11.3) more than once in what follows. We emphasize that it is valid only if the vectors A and B are mutually perpendicular.

Equation (10.9) establishes a connection between the moduli of the vectors v and ω. With the help of a cross product, an expression can be written that gives a relation between the vectors themselves. Let the body rotate around the z axis with an angular velocity ω (Fig. 38). It is easy to see that the vector product ω and the radius vector of the point whose velocity v we want to find is a vector that coincides in direction with the vector v and has a modulus equal to ωr sinα \u003d ωR, i.e. v [see formula (10.9)]. Thus, the vector product [ωR] in both direction and modulus is equal to the vector v:

v \u003d [ωr]

Formula (11.4) can be given a different form. To do this, we represent the radius vector r as the sum of two components - a vector r z parallel to the z axis and a vector perpendicular to the z axis: r \u003d r z + R (see Fig. 38). Substituting this expression into formula (11.4) and using the distributivity of the vector product [see. (11.2)], we get:

The vectors ω and r z are collinear. Therefore their cross product is equal to zero (sinα \u003d 0). Hence, one can write that

In what follows, when considering rotational motion, we will always denote by R the component of the radius vector r, drawn from a point taken on the axis, perpendicular to the axis of rotation. The modulus of this vector gives the distance R of the point from the axis.

Vector length, angle between vectors - these concepts are naturally applicable and intuitive when defining a vector as a segment of a certain direction. Below we will learn how to determine the angle between vectors in three-dimensional space, its cosine and consider the theory with examples.

To consider the concept of an angle between vectors, let us turn to a graphic illustration: let us define two vectors a → and b → on a plane or in three-dimensional space, which are nonzero. We also define an arbitrary point O and set aside vectors O A → \u003d b → and O B → \u003d b →

Definition 1

Corner between the vectors a → and b → is the angle between the rays O A and O B.

The resulting angle will be denoted as follows: a →, b → ^

Obviously, the angle has the ability to take values \u200b\u200bfrom 0 to π or from 0 to 180 degrees.

a →, b → ^ \u003d 0 when the vectors are codirectional and a →, b → ^ \u003d π when the vectors are oppositely directed.

Definition 2

The vectors are called perpendicularif the angle between them is 90 degrees or π 2 radians.

If at least one of the vectors is zero, then the angle a →, b → ^ is undefined.

The cosine of the angle between two vectors, and hence the angle itself, can usually be determined either using the dot product of vectors, or by means of the cosine theorem for a triangle built on the basis of two given vectors.

By definition, the scalar product is a →, b → \u003d a → b → cos a →, b → ^.

If the given vectors a → and b → are nonzero, then we can divide the right and left sides of the equality by the product of the lengths of these vectors, thus obtaining a formula for finding the cosine of the angle between nonzero vectors:

cos a →, b → ^ \u003d a →, b → a → b →

This formula is used when the number of source data includes the lengths of vectors and their dot product.

Example 1

Initial data: vectors a → and b →. Their lengths are 3 and 6, respectively, and their dot product is - 9. It is necessary to calculate the cosine of the angle between the vectors and find the angle itself.

Decision

The initial data are enough to apply the formula obtained above, then cos a →, b → ^ \u003d - 9 3 6 \u003d - 1 2,

Now let's define the angle between vectors: a →, b → ^ \u003d a r c cos (- 1 2) \u003d 3 π 4

Answer: cos a →, b → ^ \u003d - 1 2, a →, b → ^ \u003d 3 π 4

Problems are more common where vectors are specified by coordinates in a rectangular coordinate system. For such cases, it is necessary to derive the same formula, but in coordinate form.

The length of a vector is defined as the square root of the sum of the squares of its coordinates, and the scalar product of vectors is equal to the sum of the products of the corresponding coordinates. Then the formula for finding the cosine of the angle between vectors on the plane a → \u003d (a x, a y), b → \u003d (b x, b y) looks like this:

cos a →, b → ^ \u003d a x b x + a y b y a x 2 + a y 2 b x 2 + b y 2

And the formula for finding the cosine of the angle between vectors in three-dimensional space a → \u003d (ax, ay, az), b → \u003d (bx, by, bz) will have the form: cos a →, b → ^ \u003d ax bx + ay by + az bzax 2 + ay 2 + az 2 bx 2 + by 2 + bz 2

Example 2

Initial data: vectors a → \u003d (2, 0, - 1), b → \u003d (1, 2, 3) in a rectangular coordinate system. It is necessary to determine the angle between them.

Decision

1. To solve the problem, we can immediately apply the formula:

cos a →, b → ^ \u003d 2 1 + 0 2 + (- 1) 3 2 2 + 0 2 + (- 1) 2 1 2 + 2 2 + 3 2 \u003d - 1 70 ⇒ a →, b → ^ \u003d arc cos (- 1 70) \u003d - arc cos 1 70

1. You can also determine the angle using the formula:

cos a →, b → ^ \u003d (a →, b →) a → b →,

but first calculate the lengths of the vectors and the dot product by coordinates: a → \u003d 2 2 + 0 2 + (- 1) 2 \u003d 5 b → \u003d 1 2 + 2 2 + 3 2 \u003d 14 a →, b → ^ \u003d 2 1 + 0 2 + (- 1) 3 \u003d - 1 cos a →, b → ^ \u003d a →, b → ^ a → b → \u003d - 1 5 14 \u003d - 1 70 ⇒ a →, b → ^ \u003d - arc cos 1 70

Answer: a →, b → ^ \u003d - a r c cos 1 70

Tasks are also common when the coordinates of three points in a rectangular coordinate system are given and it is necessary to determine some angle. And then, in order to determine the angle between the vectors with the given coordinates of the points, it is necessary to calculate the coordinates of the vectors in the form of the difference between the corresponding start and end points of the vector.

Example 3

Initial data: points A (2, - 1), B (3, 2), C (7, - 2) are specified on a plane in a rectangular coordinate system. It is necessary to determine the cosine of the angle between vectors A C → and B C →.

Decision

Let's find the coordinates of the vectors by the coordinates of the given points A C → \u003d (7 - 2, - 2 - (- 1)) \u003d (5, - 1) B C → \u003d (7 - 3, - 2 - 2) \u003d (4, - 4)

Now we use the formula to determine the cosine of the angle between vectors in the plane in coordinates: cos AC →, BC → ^ \u003d (AC →, BC →) AC → BC → \u003d 5.44 + (- 1) (- 4) 5 2 + (- 1) 2 4 2 + (- 4) 2 \u003d 24 26 32 \u003d 3 13

Answer: cos A C →, B C → ^ \u003d 3 13

The angle between the vectors can be determined by the cosine theorem. Let us set aside vectors O A → \u003d a → and O B → \u003d b → from the point O, then, according to the theorem of cosines in the triangle O AB, the equality will be true:

A B 2 \u003d O A 2 + O B 2 - 2 O A O B cos (∠ A O B),

which is equivalent to:

b → - a → 2 \u003d a → + b → - 2 a → b → cos (a →, b →) ^

and from here we derive the formula for the cosine of an angle:

cos (a →, b →) ^ \u003d 1 2 a → 2 + b → 2 - b → - a → 2 a → b →

To apply the resulting formula, we need the lengths of the vectors, which are easily determined by their coordinates.

Although this method takes place, the formula is still more often used:

cos (a →, b →) ^ \u003d a →, b → a → b →

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