In this lesson we will look at two more operations with vectors: vector product of vectors And mixed product of vectors (immediate link for those who need it). It’s okay, sometimes it happens that for complete happiness, in addition to scalar product of vectors, more and more are required. This is vector addiction. It may seem that we are getting into the jungle of analytical geometry. This is wrong. In this section of higher mathematics there is generally little wood, except perhaps enough for Pinocchio. In fact, the material is very common and simple - hardly more complicated than the same scalar product, even typical tasks there will be less. The main thing in analytical geometry, as many will be convinced or have already been convinced, is NOT TO MAKE MISTAKES IN CALCULATIONS. Repeat like a spell and you will be happy =)

If vectors sparkle somewhere far away, like lightning on the horizon, it doesn’t matter, start with the lesson Vectors for dummies to restore or reacquire basic knowledge about vectors. More prepared readers can get acquainted with the information selectively; I tried to collect the most complete collection of examples that are often found in practical work

What will make you happy right away? When I was little, I could juggle two or even three balls. It worked out well. Now you won't have to juggle at all, since we will consider only spatial vectors, and flat vectors with two coordinates will be left out. Why? This is how these actions were born - the vector and mixed product of vectors are defined and work in three-dimensional space. It's already easier!

This operation, just like the scalar product, involves two vectors. Let these be imperishable letters.

The action itself denoted by in the following way: . There are other options, but I’m used to denoting the vector product of vectors this way, in square brackets with a cross.

And right away question: if in scalar product of vectors two vectors are involved, and here two vectors are also multiplied, then what is the difference? The obvious difference is, first of all, in the RESULT:

The result of the scalar product of vectors is NUMBER:

The result of the cross product of vectors is VECTOR: , that is, we multiply the vectors and get a vector again. Closed club. Actually, this is where the name of the operation comes from. In different educational literature, designations may also vary; I will use the letter.

Definition of cross product

First there will be a definition with a picture, then comments.

Definition: Vector product non-collinear vectors, taken in this order, called VECTOR, length which is numerically equal to the area of ​​the parallelogram, built on these vectors; vector orthogonal to vectors, and is directed so that the basis has a right orientation:

Let’s break down the definition piece by piece, there’s a lot of interesting stuff here!

So, the following significant points can be highlighted:

1) The original vectors, indicated by red arrows, by definition not collinear. It will be appropriate to consider the case of collinear vectors a little later.

2) Vectors are taken in a strictly defined order: – "a" is multiplied by "be", not “be” with “a”. The result of vector multiplication is VECTOR, which is indicated in blue. If the vectors are multiplied by reverse order, then we get a vector equal in length and opposite in direction (raspberry color). That is, the equality is true .

3) Now let's get acquainted with the geometric meaning of the vector product. This is a very important point! The LENGTH of the blue vector (and, therefore, the crimson vector) is numerically equal to the AREA of the parallelogram built on the vectors. In the figure, this parallelogram is shaded black.

Note : the drawing is schematic, and, naturally, the nominal length of the vector product is not equal to the area of ​​the parallelogram.

Let us recall one of the geometric formulas: The area of ​​a parallelogram is equal to the product of adjacent sides and the sine of the angle between them. Therefore, based on the above, the formula for calculating the LENGTH of a vector product is valid:

I emphasize that the formula is about the LENGTH of the vector, and not about the vector itself. What is the practical meaning? And the meaning is that in problems of analytical geometry, the area of ​​a parallelogram is often found through the concept of a vector product:

Let us obtain the second important formula. The diagonal of a parallelogram (red dotted line) divides it into two equal triangles. Therefore, the area of ​​a triangle built on vectors (red shading) can be found using the formula:

4) Not less important fact is that the vector is orthogonal to the vectors, that is . Of course, the oppositely directed vector (raspberry arrow) is also orthogonal to the original vectors.

5) The vector is directed so that basis It has right orientation. In the lesson about transition to a new basis I spoke in sufficient detail about plane orientation, and now we will figure out what space orientation is. I will explain on your fingers right hand . Mentally combine forefinger with vector and middle finger with vector. Ring finger and little finger press it into your palm. As a result thumb – the vector product will look up. This is a right-oriented basis (it is this one in the figure). Now change the vectors ( index and middle fingers) in some places, as a result the thumb will turn around, and the vector product will already look down. This is also a right-oriented basis. You may have a question: which basis has left orientation? “Assign” to the same fingers left hand vectors, and get the left basis and left orientation of space (in this case, the thumb will be located in the direction of the lower vector). Figuratively speaking, these bases “twist” or orient space in different directions. And this concept should not be considered something far-fetched or abstract - for example, the orientation of space is changed by the most ordinary mirror, and if you “pull the reflected object out of the looking glass,” then in the general case it will not be possible to combine it with the “original.” By the way, hold three fingers up to the mirror and analyze the reflection ;-)

...how good it is that you now know about right- and left-oriented bases, because the statements of some lecturers about a change in orientation are scary =)

Cross product of collinear vectors

The definition has been discussed in detail, it remains to find out what happens when the vectors are collinear. If the vectors are collinear, then they can be placed on one straight line and our parallelogram also “folds” into one straight line. The area of ​​such, as mathematicians say, degenerate parallelogram is equal to zero. The same follows from the formula - the sine of zero or 180 degrees is equal to zero, which means the area is zero

Thus, if , then And . Please note that the vector product itself is equal to the zero vector, but in practice this is often neglected and they are written that it is also equal to zero.

A special case is the cross product of a vector with itself:

Using the vector product, you can check the collinearity of three-dimensional vectors, and we will also analyze this problem, among others.

To solve practical examples you may need trigonometric table to find the values ​​of sines from it.

Well, let's light the fire:

Example 1

a) Find the length of the vector product of vectors if

b) Find the area of ​​a parallelogram built on vectors if

Solution: No, this is not a typo, I deliberately made the initial data in the clauses the same. Because the design of the solutions will be different!

a) According to the condition, you need to find length vector (cross product). According to the corresponding formula:

Answer:

If you were asked about length, then in the answer we indicate the dimension - units.

b) According to the condition, you need to find square parallelogram built on vectors. The area of ​​this parallelogram is numerically equal to the length of the vector product:

Answer:

Please note that the answer does not talk about the vector product at all; we were asked about area of ​​the figure, accordingly, the dimension is square units.

We always look at WHAT we need to find according to the condition, and, based on this, we formulate clear answer. It may seem like literalism, but there are plenty of literalists among teachers, and the assignment has a good chance of being returned for revision. Although this is not a particularly far-fetched quibble - if the answer is incorrect, then one gets the impression that the person does not understand simple things and/or has not understood the essence of the task. This point must always be kept under control when solving any problem in higher mathematics, and in other subjects too.

Where did the big letter “en” go? In principle, it could have been additionally attached to the solution, but in order to shorten the entry, I did not do this. I hope everyone understands that and is a designation for the same thing.

A popular example for a DIY solution:

Example 2

Find the area of ​​a triangle built on vectors if

The formula for finding the area of ​​a triangle through the vector product is given in the comments to the definition. The solution and answer are at the end of the lesson.

In practice, the task is really very common; triangles can generally torment you.

To solve other problems we will need:

Properties of the vector product of vectors

We have already considered some properties of the vector product, however, I will include them in this list.

For arbitrary vectors and an arbitrary number, the following properties are true:

1) In other sources of information, this item is usually not highlighted in the properties, but it is very important in practical terms. So let it be.

2) – the property is also discussed above, sometimes it is called anticommutativity. In other words, the order of the vectors matters.

3) – associative or associative vector product laws. Constants can be easily moved outside the vector product. Really, what should they do there?

4) – distribution or distributive vector product laws. There are no problems with opening the brackets either.

To demonstrate, let's look at a short example:

Example 3

Find if

Solution: The condition again requires finding the length of the vector product. Let's paint our miniature:

(1) According to associative laws, we take the constants outside the scope of the vector product.

(2) We move the constant outside the module, and the module “eats” the minus sign. The length cannot be negative.

(3) The rest is clear.

Answer:

It's time to add more wood to the fire:

Example 4

Calculate the area of ​​a triangle built on vectors if

Solution: Find the area of ​​the triangle using the formula . The catch is that the vectors “tse” and “de” are themselves presented as sums of vectors. The algorithm here is standard and somewhat reminiscent of examples No. 3 and 4 of the lesson Dot product of vectors. For clarity, we will divide the solution into three stages:

1) At the first step, we express the vector product through the vector product, in fact, let's express a vector in terms of a vector. No word yet on lengths!

(1) Substitute the expressions of the vectors.

(2) Using distributive laws, we open the brackets according to the rule of multiplication of polynomials.

(3) Using associative laws, we move all constants beyond the vector products. With a little experience, steps 2 and 3 can be performed simultaneously.

(4) The first and last terms are equal to zero (zero vector) due to the nice property. In the second term we use the property of anticommutativity of a vector product:

(5) We present similar terms.

As a result, the vector turned out to be expressed through a vector, which is what was required to be achieved:

2) In the second step, we find the length of the vector product we need. This action is similar to Example 3:

3) Find the area of ​​the required triangle:

Stages 2-3 of the solution could have been written in one line.

Answer:

The problem considered is quite common in tests ah, here's an example to solve on your own:

Example 5

Find if

A short solution and answer at the end of the lesson. Let's see how attentive you were when studying the previous examples ;-)

Cross product of vectors in coordinates

, specified in an orthonormal basis, expressed by the formula:

The formula is really simple: in the top line of the determinant we write the coordinate vectors, in the second and third lines we “put” the coordinates of the vectors, and we put in strict order– first the coordinates of the “ve” vector, then the coordinates of the “double-ve” vector. If the vectors need to be multiplied in a different order, then the rows should be swapped:

Example 10

Check whether the following space vectors are collinear:
A)
b)

Solution: The check is based on one of the statements in this lesson: if the vectors are collinear, then their vector product is equal to zero (zero vector): .

a) Find the vector product:

Thus, the vectors are not collinear.

b) Find the vector product:

Answer: a) not collinear, b)

Here, perhaps, is all the basic information about the vector product of vectors.

This section will not be very large, since there are few problems where the mixed product of vectors is used. In fact, everything will depend on the definition, geometric meaning and a couple of working formulas.

A mixed product of vectors is the product of three vectors:

So they lined up like a train and can’t wait to be identified.

First, again, a definition and a picture:

Definition: Mixed work non-coplanar vectors, taken in this order, called parallelepiped volume, built on these vectors, equipped with a “+” sign if the basis is right, and a “–” sign if the basis is left.

Let's do the drawing. Lines invisible to us are drawn with dotted lines:

Let's dive into the definition:

2) Vectors are taken in a certain order, that is, the rearrangement of vectors in the product, as you might guess, does not occur without consequences.

3) Before commenting on the geometric meaning, I will note an obvious fact: the mixed product of vectors is a NUMBER: . In educational literature, the design may be slightly different; I am used to denoting a mixed product by , and the result of calculations by the letter “pe”.

A-priory the mixed product is the volume of the parallelepiped, built on vectors (the figure is drawn with red vectors and black lines). That is, the number is equal to the volume of a given parallelepiped.

Note : The drawing is schematic.

4) Let’s not worry again about the concept of orientation of the basis and space. The meaning of the final part is that a minus sign can be added to the volume. In simple words, the mixed product can be negative: .

Directly from the definition follows the formula for calculating the volume of a parallelepiped built on vectors.

In this article we will take a closer look at the concept of the cross product of two vectors. We will give the necessary definitions, write a formula for finding the coordinates of a vector product, list and justify its properties. After this, we will dwell on the geometric meaning of the vector product of two vectors and consider solutions to various typical examples.

Page navigation.

Definition of cross product.

Before defining a vector product, let's understand the orientation of an ordered triple of vectors in three-dimensional space.

Let's plot the vectors from one point. Depending on the direction of the vector, the three can be right or left. Let's look from the end of the vector at how the shortest turn from the vector to . If the shortest rotation occurs counterclockwise, then the triple of vectors is called right, otherwise - left.

Now let's take two non-collinear vectors and . Let us plot the vectors and from point A. Let's construct some vector perpendicular to both and and . Obviously, when constructing a vector, we can do two things, giving it either one direction or the opposite (see illustration).

Depending on the direction of the vector, the ordered triplet of vectors can be right-handed or left-handed.

This brings us close to the definition of a vector product. It is given for two vectors given in rectangular coordinate system three-dimensional space.

Definition.

The cross product of two vectors and , specified in a rectangular coordinate system of three-dimensional space, is called a vector such that

The cross product of vectors and is denoted as .

Coordinates of the vector product.

Now we will give the second definition of a vector product, which allows you to find its coordinates from the coordinates of given vectors and.

Definition.

In a rectangular coordinate system of three-dimensional space vector product of two vectors And is a vector , where are the coordinate vectors.

This definition gives us the cross product in coordinate form.

It is convenient to represent the cross product as a determinant square matrix third order, the first line of which is the unit vectors, the second line contains the coordinates of the vector, and the third line contains the coordinates of the vector in the given rectangular coordinate system:

If we expand this determinant into the elements of the first row, we obtain the equality from the definition of the vector product in coordinates (if necessary, refer to the article):

It should be noted that the coordinate form of the vector product is fully consistent with the definition given in the first paragraph of this article. Moreover, these two definitions of a cross product are equivalent. You can see the proof of this fact in the book listed at the end of the article.

Properties of a vector product.

Since the vector product in coordinates can be represented as a determinant of the matrix, the following can easily be justified on the basis properties of the cross product:

As an example, let us prove the anticommutative property of a vector product.

A-priory And . We know that the value of the determinant of a matrix is ​​reversed if two rows are swapped, therefore, , which proves the anticommutative property of a vector product.

Vector product - examples and solutions.

There are mainly three types of problems.

In problems of the first type, the lengths of two vectors and the angle between them are given, and you need to find the length of the vector product. In this case, the formula is used .

Example.

Find the length of the vector product of the vectors and , if known .

Solution.

We know from the definition that the length of the vector product of vectors and is equal to the product of the lengths of vectors and by the sine of the angle between them, therefore, .

Answer:

.

Problems of the second type are related to the coordinates of vectors, in which the vector product, its length or anything else is searched through the coordinates of given vectors And .

There are a lot of different options possible here. For example, not the coordinates of the vectors and can be specified, but their expansions into coordinate vectors of the form and , or vectors and can be specified by the coordinates of their start and end points.

Let's look at typical examples.

Example.

Two vectors are given in a rectangular coordinate system . Find their cross product.

Solution.

According to the second definition, the vector product of two vectors in coordinates is written as:

We would have arrived at the same result if the vector product had been written in terms of the determinant

Answer:

.

Example.

Find the length of the vector product of the vectors and , where are the unit vectors of the rectangular Cartesian coordinate system.

Solution.

First we find the coordinates of the vector product in a given rectangular coordinate system.

Since vectors and have coordinates and, respectively (if necessary, see the article vector coordinates in a rectangular coordinate system), then by the second definition of a vector product we have

That is, the vector product has coordinates in a given coordinate system.

We find the length of the vector product as the square root of the sum of the squares of its coordinates (we obtained this formula for the length of the vector in the section finding the length of a vector):

Answer:

.

Example.

In a rectangular Cartesian coordinate system, the coordinates of three points are given. Find some vector that is perpendicular and at the same time.

Solution.

Vectors and have coordinates and respectively (see article finding vector coordinates through point coordinates). If we find the vector product of the vectors and , then by definition it is a vector perpendicular to both to and to , that is, it is a solution to our problem. Let's find him

Answer:

- one of the perpendicular vectors.

In problems of the third type, the skill of using the properties of the vector product of vectors is tested. After applying the properties, the corresponding formulas are applied.

Example.

The vectors and are perpendicular and their lengths are 3 and 4, respectively. Find the length of the cross product .

Solution.

By the distributive property of a vector product, we can write

By virtue of associative properties Let’s take the numerical coefficients out of the sign of the vector products in the last expression:

The vector products and are equal to zero, since And , Then .

Since the vector product is anticommutative, then .

So, using the properties of the vector product, we arrived at the equality .

By condition, the vectors and are perpendicular, that is, the angle between them is equal to . That is, we have all the data to find the required length

Answer:

.

Geometric meaning of a vector product.

By definition, the length of the vector product of vectors is . And from the geometry course high school We know that the area of ​​a triangle is equal to half the product of the lengths of the two sides of the triangle and the sine of the angle between them. Consequently, the length of the vector product is equal to twice the area of ​​a triangle whose sides are the vectors and , if they are plotted from one point. In other words, the length of the vector product of vectors and is equal to the area of ​​a parallelogram with sides and and the angle between them equal to . This is the geometric meaning of the vector product.

Test No. 1

Vectors. Elements of higher algebra

1-20. The lengths of the vectors and and are known; – the angle between these vectors.

Calculate: 1) and, 2).3) Find the area of ​​the triangle built on the vectors and.

Make a drawing.

Solution. Using the definition of dot product of vectors:

And the properties of the scalar product: ,

1) find the scalar square of the vector:

that is, Then .

Arguing similarly, we get

that is, Then .

By definition of a vector product: ,

taking into account that

The area of ​​a triangle constructed from vectors and is equal to

21-40. Known coordinates of three vertices A, B, D parallelogram ABCD. Using vector algebra, you need:

A(3;0;-7), B(2;4;6), D(-7;-5;1)

Solution.

It is known that the diagonals of a parallelogram are divided in half at the point of intersection. Therefore, the coordinates of the point E- intersection of diagonals - find as coordinates of the middle of the segment BD. Denoting them by x E ,y E , z E we get that

We get.

Knowing the coordinates of the point E- midpoint of the diagonal BD and the coordinates of one of its ends A(3;0;-7), Using formulas we determine the required coordinates of the vertex WITH parallelogram:

So, the top.

2) To find the projection of a vector onto a vector, we find the coordinates of these vectors: ,

similarly . The projection of a vector onto a vector is found using the formula:

3) The angle between the diagonals of a parallelogram is found as the angle between the vectors

And by the property of the scalar product:

Then

4) Find the area of ​​the parallelogram as the modulus of the vector product:

5) The volume of the pyramid is found as one sixth of the modulus of the mixed product of vectors, where O(0;0;0), then

Then the required volume (cubic units)

41-60. Given matrices:

V C -1 +3A T

Designations:

First, we find the inverse matrix of matrix C.

To do this, we find its determinant:

The determinant is different from zero, therefore, the matrix is ​​non-singular and for it you can find the inverse matrix C -1

Let us find the algebraic complements using the formula , where is the minor of the element:

Then , .

61–80. Solve the system linear equations:

Cramer's method; 2. Matrix method.

Solution.

a) Cramer's method

Let's find the determinant of the system

Since , the system has a unique solution.

Let's find the determinants and by replacing the first, second, third columns in the coefficient matrix with a column of free terms, respectively.

According to Cramer's formulas:

b)matrix method (using an inverse matrix).

We write this system in matrix form and solve it using the inverse matrix.

Let A– matrix of coefficients for unknowns; X– matrix-column of unknowns x, y, z And N– matrix-column of free members:

The left side of system (1) can be written as a product of matrices , and the right side as a matrix N. Therefore we have the matrix equation

Since the determinant of the matrix A is different from zero (point “a”), then the matrix A has an inverse matrix. Let's multiply both sides of equality (2) on the left by the matrix, we get

Since where E – identity matrix, and , then

Let us have a non-singular matrix A:

Then we find the inverse matrix using the formula:

Where A ij- algebraic complement of an element a ij in the determinant of the matrix A, which is the product of (-1) i+j and the minor (determinant) n-1 order obtained by deleting i-th lines and jth column in the determinant of matrix A:

From here we get the inverse matrix:

Column X: X=A -1 H

81–100. Solve a system of linear equations using the Gauss method

Solution. Let's write the system in the form of an extended matrix:

We perform elementary transformations with strings.

From the 2nd line we subtract the first line multiplied by 2. From line 3 we subtract the first line multiplied by 4. From line 4 we subtract the first line, we get the matrix:

Next, we get zero in the first column of subsequent rows; to do this, subtract the third row from the second row. From the third row, subtract the second row, multiplied by 2. From the fourth row, subtract the second row, multiplied by 3. As a result, we obtain a matrix of the form:

From the fourth line we subtract the third.

Let's swap the penultimate and last lines:

The last matrix is ​​equivalent to the system of equations:

From the last equation of the system we find .

Substituting into the penultimate equation, we get .

From the second equation of the system it follows that

From the first equation we find x:

Answer:

Test No. 2

Analytic geometry

1-20. Given the coordinates of the vertices of the triangle ABC. Find:

1) side length AIN;

2) equations of the sides AB And Sun and their angular coefficients;

3) angle IN in radians accurate to two digits;

4) height equation CD and its length;

5) median equation AE

height CD;

TO parallel to the side AB,

7) make a drawing.

A(3;6), B(15;-3), C(13;11)

Solution.

Applying (1), we find the length of the side AB:

2) equations of the sides AB And Sun and their angular coefficients:

Equation of a line, passing through the points and , has the form

Substituting the coordinates of the points into (2) A And IN, we obtain the equation of the side AB:

(AB).

(B.C.).

3) angle IN in radians with an accuracy of two digits.

It is known that the tangent of the angle between two straight lines, the angular coefficients of which are respectively equal and is calculated by the formula

Required angle IN formed by straight lines AB And Sun, the angular coefficients of which are found: ; . Applying (3), we get

; , or

4) height equation CD and its length.

Distance from point C to straight line AB:

5) median equation AE and the coordinates of the point K of the intersection of this median with

height CD.

middle of the sun side:

Then the equation AE:

We solve the system of equations:

6) equation of a line passing through a point TO parallel to the side AB:

Since the desired line is parallel to the side AB, then her slope will be equal to the slope of the straight line AB. Substituting the coordinates of the found point into (4) TO and the slope, we get

; (KF).

The area of ​​the parallelogram is 12 square meters. units, its two vertices are points A(-1;3) And B(-2;4). Find the other two vertices of this parallelogram if it is known that the point of intersection of its diagonals lies on the x-axis. Make a drawing.

Solution. Let the point of intersection of the diagonals have coordinates .

Then it is obvious that

therefore, the coordinates of the vectors are .

We find the area of ​​a parallelogram using the formula

Then the coordinates of the other two vertices are .

In problems 51-60 the coordinates of the points are given A and B. Required:

Compose canonical equation hyperbola passing through these points A and B, if the foci of the hyperbola are located on the x-axis;

Find the semi-axes, foci, eccentricity and equations of asymptotes of this hyperbola;

Find all points of intersection of the hyperbola with a circle with center at the origin, if this circle passes through the foci of the hyperbola;

Construct a hyperbola, its asymptotes and circle.

A(6;-2), B(-8;12).

Solution. The equation of the desired hyperbola in canonical form is written

Where a- real semiaxis of the hyperbola, b- imaginary semi-axis. Substituting the coordinates of the points A And IN In this equation we find these semi-axes:

– hyperbola equation: .

Semi-axes a=4,

focal length Focuses (-8.0) and (8.0)

Eccentricity

Asyptotes:

If a circle passes through the origin, its equation is

Substituting one of the foci, we find the equation of the circle

Find the intersection points of the hyperbola and the circle:

We build a drawing:

In problems 61-80, construct a graph of a function in a polar coordinate system point by point, giving  values ​​through the interval  /8 (0   2). Find the equation of the line in a rectangular Cartesian coordinate system (the positive semi-axis of the abscissa coincides with the polar axis, and the pole with the origin).

Solution. Let's build a line by points, having first filled in the table of values ​​and φ.

Number	φ ,	φ, degrees			Number	φ , glad	degrees
								3∙(x 2 +2∙1x + 1) -3∙1 = 3(x+1) 2 - 3 we conclude that this equation defines an ellipse: Points are given A, IN , C, D . Need to find: 1. Plane equation (Q), passing through points A, B, C D in the plane (Q); 2. Line equation (I), passing through points IN and D; 3. Angle between plane (Q) and straight (I); 4. Plane equation (R), passing through a point A perpendicular to a straight line (I); 5. Angle between planes (R) And (Q) ; 6. Equation of a line (T), passing through a point A in the direction of its radius vector; 7. Angle between straight lines (I) And (T). A(9;-8;1), B(-9;4;5), C(9;-5;5),D(6;4;0) 1. Plane equation (Q), passing through points A, B, C and check if the point lies D in the plane is determined by the formula Find: 1) . 2) Square parallelogram, built on And. 3) Volume of the parallelepiped, built on vectors, And. Control Job on this topic " Elements theory of linear spaces... Methodological recommendations for completing tests for undergraduate part-time studies in qualification 080100. 62 in the direction Guidelines Parallelepiped and volume of the pyramid, built on vectors, And. Solution: 2-=2(1;1;1)-(2;1;4)= (2;2;2)-(2;1;4)=(0;1;-2)... . . . 4. TASKS FOR CONTROL WORKS Section I. Linear algebra. 1 – 10. Given...