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

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Definition of a vector product.

Before giving a definition of a cross product, let's deal with the orientation of an ordered triple of vectors in three-dimensional space.

Let's postpone vectors from one point. Depending on the direction of the vector, the triple 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 is counterclockwise, then the triple of vectors is called right, otherwise - left.

Now let's take two non-collinear vectors and . Set aside vectors and from point A. Let's construct some vector that is perpendicular to and and at the same time. 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 triple of vectors can be right or left.

So we came close to the definition of a vector product. It is given for two vectors given in a rectangular coordinate system of three-dimensional space.

Definition.

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

The cross product of vectors and is denoted as .

Vector product coordinates.

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

Definition.

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

This definition gives us the cross product in coordinate form.

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

If we expand this determinant by the elements of the first row, then we get 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 cross 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. The proof of this fact can be found in the book indicated at the end of the article.

Vector product properties.

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

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

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

Vector product - examples and solutions.

Basically there are three types of tasks.

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

Example.

Find the length of the cross product of vectors and if known .

Solution.

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

Answer:

.

Tasks of the second type are associated with the coordinates of vectors, in which the vector product, its length, or something else is searched through the coordinates of the given vectors And .

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

Let's consider typical examples.

Example.

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

Solution.

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

We would have come to the same result if we had written the vector product through the determinant

Answer:

.

Example.

Find the length of the cross product of vectors and , where are the orts of the rectangular Cartesian coordinate system.

Solution.

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

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

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

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

Answer:

.

Example.

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

Solution.

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

Answer:

is one of the perpendicular vectors.

In tasks of the third type, the skill of using the properties of the vector product of vectors is checked. After properties are applied, the corresponding formulas are applied.

Example.

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

Solution.

By the distributivity property of the vector product, we can write

By virtue of the associative property, we take out the numerical coefficients for the sign of vector products in the last expression:

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 have come to 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:

.

The geometric meaning of the vector product.

By definition, the length of the cross product of vectors is . And from the high school geometry course, 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. Therefore, the length of the cross product is equal to twice the area of ​​a triangle with sides of the vectors and , if they are postponed from one point. In other words, the length of the cross product of vectors and is equal to the area of ​​a parallelogram with sides and and an 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; is the angle between these vectors.

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

Make a drawing.

Solution. Using the definition of the 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 the fact that

The area of ​​a triangle built on vectors and is equal to

21-40. The coordinates of three vertices are known A, B, D parallelogram ABCD. By means of 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 at the point of intersection are divided in half. Therefore, the coordinates of the point E- intersections of the diagonals - find as the coordinates of the middle of the segment BD. Denoting them with x E ,y E , z E we get that

We get .

Knowing the coordinates of the point E- diagonal midpoints BD and the coordinates of one of its ends A(3;0;-7), by the formulas we determine the desired 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: ,

likewise . The projection of a vector onto a vector , we find by the formula:

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

And by the property of the scalar product:

Then

4) The area of ​​the parallelogram is found as the module 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 desired volume (cubic units)

41-60. Matrix data:

V C -1 +3A T

Designations:

First, we find the inverse of matrix C.

To do this, we find its determinant:

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

Let's find algebraic complements by the formula , where is the minor of the element :

Then , .

61–80. Solve the system of 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.

Find the determinants and , replacing the first, second, third columns in the matrix of coefficients, respectively, with a column of free members.

According to Cramer's formulas:

b)matrix method (using the inverse matrix).

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

Let A is the matrix of coefficients for unknowns; X is the column matrix of unknowns x, y, z And H is a column matrix of free members:

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

Since the matrix determinant A is different from zero (item "a"), then the matrix A has an inverse matrix. Multiplying both sides of equality (2) on the left by the matrix , we obtain

Since where E is the identity matrix, and , then

Let we have a non-singular matrix A:

Then the inverse matrix is ​​found by the formula:

Where A ij- algebraic complement of an element a ij in matrix determinant A, which is the product of (-1) i+j and the minor (determinant) n-1 order obtained by deletion i-th lines and j-th columns 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. We write the system in the form of an extended matrix:

We perform elementary transformations with strings.

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

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

Subtract the third from the fourth line.

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 obtain .

It follows from the second equation of the system that

From the first equation we find x:

Answer:

Examination No. 2

Analytic geometry

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

1) side length AIN;

2) side equations AB And sun and their slopes;

3) angle IN in radians to two decimal places;

4) height equation CD and its length

5) median equation AE

tall CD;

TO parallel to 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) side equations AB And sun and their slopes:

The equation of a straight line passing through the points and has the form

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

(AB).

(BC).

3) angle IN in radians to two decimal places.

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

Desired angle IN formed by direct AB And sun, whose angular coefficients are found: ; . Applying (3), we obtain

; , or

4) height equation CD and its length.

Distance from point C to line AB:

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

tall CD.

mid-side BC:

Then the equation AE:

We solve the system of equations:

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

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

; (KF).

The area of ​​a parallelogram is 12 square meters. units, two of its vertices are points A(-1;3) And B(-2;4). Find two other 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

hence the coordinates of the vectors .

The area of ​​a parallelogram is found by the formula

Then the coordinates of the other two vertices are .

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

Write the canonical equation of a hyperbola passing through given points A and B if the foci of the hyperbola are located on the x-axis;

Find semiaxes, foci, eccentricity and equations of asymptotes of this hyperbola;

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

Construct a hyperbola, its asymptotes and a circle.

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

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

Where a is the real semiaxis of the hyperbola, b- imaginary axle. Substituting point coordinates A And IN in this equation we find these semiaxes:

- the equation of the hyperbola: .

Semiaxes a=4,

focal length Foci (-8.0) and (8.0)

Eccentricity

Aciptotes:

If the circle passes through the origin, its equation

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

Find the intersection points of the hyperbola and the circle:

Building a drawing:

In problems 61-80 plot the function in the polar coordinate system by points, 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 coincides with the origin).

Solution. Let's build a line by points, having previously 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: Given points A, IN , C, D . Required to find: 1. Equation of the plane (Q), passing through points A, B, C D in plane (Q); 2. Equation of a straight line (I) passing through points IN and D; 3. Angle between plane (Q) and direct (I); 4. Equation of the plane (R), passing through a point A perpendicular to the line (I); 5. Angle between planes (R) And (Q) ; 6. Equation of a straight 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. Equation of the plane (Q), passing through points A, B, C and check if point lies D in the plane is determined by the formula Find : 1) . 2) Square parallelogram, built on And. 3) The volume of the parallelepiped, built on vectors, And. Control Job on this topic " Elements theory of linear spaces... Guidelines for the implementation of tests for undergraduate correspondence courses for qualification 080100. 62 in the direction Guidelines The parallelepiped and the 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. Dana...

In this lesson, we will look at two more operations with vectors: cross product of vectors And mixed product of vectors (immediate link for those who need it). It's okay, it sometimes happens that for complete happiness, in addition to dot product of vectors, more and more is needed. Such is vector addiction. One may get the impression that we are getting into the jungle of analytic geometry. This is wrong. In this section of higher mathematics, there is generally little firewood, except perhaps enough for Pinocchio. In fact, the material is very common and simple - hardly more difficult than the same scalar product, even there will be fewer typical tasks. The main thing in analytic geometry, as many will see or have already seen, is NOT TO MISTAKE CALCULATIONS. Repeat like a spell, and you will be happy =)

If the 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? When I was little, I could juggle two and even three balls. It worked out well. Now there is no need to juggle at all, since we will consider only space 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. Already easier!

In this operation, in the same way as in the scalar product, two vectors. Let it be imperishable letters.

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

And immediately question: if in dot product of vectors two vectors are involved, and here two vectors are also multiplied, then what is the difference? A clear difference, first of all, in the RESULT:

The result of the scalar product of vectors is a NUMBER:

The result of the cross product of vectors is a VECTOR: , that is, we multiply the vectors and get a vector again. Closed club. Actually, hence the name of the operation. In various educational literature, the 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: cross product non-collinear vectors , taken in this order, is 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:

We analyze the definition by bones, there is a lot of interesting things!

So, we can highlight the following significant points:

1) Source 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 taken in a strict order: – "a" is multiplied by "be", not "be" to "a". The result of vector multiplication is VECTOR , which is denoted in blue. If the vectors are multiplied in reverse order, then we get a vector equal in length and opposite in direction (crimson color). That is, the equality .

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 in black.

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

We 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 foregoing, the formula for calculating the LENGTH of a vector product is valid:

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

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

4) An equally important fact is that the vector is orthogonal to the vectors , that is . Of course, the oppositely directed vector (crimson arrow) is also orthogonal to the original vectors .

5) The vector is directed so that basis It has right orientation. In a lesson about transition to a new basis I have spoken in detail about plane orientation, and now we will figure out what the orientation of space 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 into your palm. As a result thumb- the vector product will look up. This is the right-oriented basis (it is in the figure). Now swap 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. Perhaps you have a question: what basis has a left orientation? "Assign" the same fingers left hand vectors , and get the left basis and left space orientation (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 most ordinary mirror changes the orientation of space, and if you “pull the reflected object out of the mirror”, then in general it will not be possible to combine it with the “original”. By the way, bring three fingers 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 the change of orientation are terrible =)

Vector product of collinear vectors

The definition has been worked out 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 zero. The same follows from the formula - the sine of zero or 180 degrees is equal to zero, which means that the area is zero

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

A special case is the vector product of a vector and itself:

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

To solve practical examples, it may be necessary trigonometric table to find the values ​​of the sines from it.

Well, let's start a 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 intentionally made the initial data in the condition items the same. Because the design of the solutions will be different!

a) According to the condition, it is required to find length vector (vector product). According to the corresponding formula:

Answer:

Since it was asked about the length, then in the answer we indicate the dimension - units.

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

Answer:

Please note that in the answer about the vector product there is no talk at all, we were asked about figure area, respectively, the dimension is square units.

We always look at WHAT is required to be found by the condition, and, based on this, we formulate clear answer. It may seem like literalism, but there are enough literalists among the teachers, and the task with good chances will be returned for revision. Although this is not a particularly strained nitpick - 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 moment should always be kept under control, solving any problem in higher mathematics, and in other subjects too.

Where did the big letter "en" go? In principle, it could be additionally stuck to the solution, but in order to shorten the record, I did not. I hope everyone understands that and is the designation of the same thing.

A popular example for a do-it-yourself 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. Solution and answer at the end of the lesson.

In practice, the task is really very common, triangles can generally be tortured.

To solve other problems, we need:

Properties of the cross 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 distinguished 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) - combination or associative vector product laws. The constants are easily taken out of the limits of the vector product. Really, what are they doing there?

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

As a demonstration, consider a short example:

Example 3

Find if

Solution: By condition, it is again required to find the length of the vector product. Let's paint our miniature:

(1) According to the associative laws, we take out the constants beyond the limits of the vector product.

(2) We take the constant out of the module, while the module “eats” the minus sign. The length cannot be negative.

(3) What follows is clear.

Answer:

It's time to throw wood on the fire:

Example 4

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

Solution: Find the area of ​​a triangle using the formula . The snag is that the vectors "ce" and "te" are themselves represented as sums of vectors. The algorithm here is standard and is somewhat reminiscent of examples No. 3 and 4 of the lesson. Dot product of vectors. Let's break it down into three steps for clarity:

1) At the first step, we express the vector product through the vector product, in fact, express the vector in terms of the vector. No word on length yet!

(1) We substitute expressions of vectors .

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

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

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

(5) We present similar terms.

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

2) At 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:

Steps 2-3 of the solution could be arranged in one line.

Answer:

The considered problem is quite common in tests, here is an example for an independent solution:

Example 5

Find if

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

, given in the orthonormal basis , is expressed by the formula:

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

Example 10

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

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

a) Find the vector product:

So 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 rest on the definition, geometric meaning and a couple of working formulas.

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

This is how they lined up like a train and wait, they can’t wait until they are calculated.

First again the definition and picture:

Definition: Mixed product non-coplanar vectors , taken in this order, is called volume of the parallelepiped, 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 by a dotted line:

Let's dive into the definition:

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

3) Before commenting on the geometric meaning, I will note the obvious fact: the mixed product of vectors is a NUMBER: . In educational literature, the design may be somewhat different, I used to designate a mixed product through, and the result of calculations with 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 the given parallelepiped.

Note : The drawing is schematic.

4) Let's not bother again with the concept of the 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 terms, the mixed product can be negative: .

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