MIXED PRODUCT OF THREE VECTORS AND ITS PROPERTIES

Mixed work three vectors is called a number equal to . Designated . Here the first two vectors are multiplied vectorially and then the resulting vector is multiplied scalarly by the third vector. Obviously, such a product is a certain number.

Let's consider the properties of a mixed product.

1. Geometric meaning mixed work. The mixed product of 3 vectors, up to a sign, is equal to the volume of the parallelepiped built on these vectors, as on edges, i.e. .

   Thus, and .

   

   Proof. Let's set aside the vectors from the common origin and construct a parallelepiped on them. Let us denote and note that . By definition of the scalar product

   Assuming that and denoting by h find the height of the parallelepiped.

   Thus, when

   If, then so. Hence, .

   Combining both of these cases, we get or .

   From the proof of this property, in particular, it follows that if the triple of vectors is right-handed, then the mixed product is , and if it is left-handed, then .

2. For any vectors , , the equality is true

   The proof of this property follows from Property 1. Indeed, it is easy to show that and . Moreover, the signs “+” and “–” are taken simultaneously, because the angles between the vectors and and and are both acute and obtuse.

3. When any two factors are rearranged, the mixed product changes sign.

   Indeed, if we consider a mixed product, then, for example, or

4. A mixed product if and only if one of the factors is equal to zero or the vectors are coplanar.

   Proof.

   Thus, a necessary and sufficient condition for the coplanarity of 3 vectors is that their mixed product is equal to zero. In addition, it follows that three vectors form a basis in space if .

   If the vectors are given in coordinate form, then it can be shown that their mixed product is found by the formula:

   .

   Thus, the mixed product is equal to the third-order determinant, which has the coordinates of the first vector in the first line, the coordinates of the second vector in the second line, and the coordinates of the third vector in the third line.

   Examples.

ANALYTICAL GEOMETRY IN SPACE

The equation F(x, y, z)= 0 defines in space Oxyz some surface, i.e. locus of points whose coordinates x, y, z satisfy this equation. This equation is called the surface equation, and x, y, z– current coordinates.

However, often the surface is not specified by an equation, but as a set of points in space that have one or another property. In this case, it is necessary to find the equation of the surface based on its geometric properties.

PLANE.

NORMAL PLANE VECTOR.

EQUATION OF A PLANE PASSING THROUGH A GIVEN POINT

Let us consider an arbitrary plane σ in space. Its position is determined by specifying a vector perpendicular to this plane and some fixed point M0(x 0, y 0, z 0), lying in the σ plane.

The vector perpendicular to the plane σ is called normal vector of this plane. Let the vector have coordinates .

Let us derive the equation of the plane σ passing through this point M0 and having a normal vector. To do this, take an arbitrary point on the plane σ M(x, y, z) and consider the vector .

For any point MО σ is a vector. Therefore, their scalar product is equal to zero. This equality is the condition that the point MО σ. It is valid for all points of this plane and is violated as soon as the point M will be outside the σ plane.

If we denote the points by the radius vector M, – radius vector of the point M0, then the equation can be written in the form

This equation is called vector plane equation. Let's write it in coordinate form. Since then

So, we have obtained the equation of the plane passing through this point. Thus, in order to create an equation of a plane, you need to know the coordinates of the normal vector and the coordinates of some point lying on the plane.

Note that the equation of the plane is an equation of the 1st degree with respect to the current coordinates x, y And z.

Examples.

GENERAL EQUATION OF THE PLANE

It can be shown that any first degree equation with respect to Cartesian coordinates x, y, z represents the equation of a certain plane. This equation is written as:

Ax+By+Cz+D=0

and is called general equation plane, and the coordinates A, B, C here are the coordinates of the normal vector of the plane.

Let us consider special cases of the general equation. Let's find out how the plane is located relative to the coordinate system if one or more coefficients of the equation become zero.

A is the length of the segment cut off by the plane on the axis Ox. Similarly, it can be shown that b And c– lengths of segments cut off by the plane under consideration on the axes Oy And Oz.

It is convenient to use the equation of a plane in segments to construct planes.

This online calculator calculates the mixed product of vectors. A detailed solution is given. To calculate a mixed product of vectors, select the method of representing vectors (by coordinates or by two points), enter data in the cells and click on the "Calculate" button.

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Data entry instructions. Numbers are entered as integers (examples: 487, 5, -7623, etc.), decimals (ex. 67., 102.54, etc.) or fractions. The fraction must be entered in the form a/b, where a and b (b>0) are integers or decimal numbers. Examples 45/5, 6.6/76.4, -7/6.7, etc.

Mixed product of vectors (theory)

Mixed work three vectors is the number that is obtained by scalar product of the result of the vector product of the first two vectors and the third vector. In other words, if three vectors are given a, b And c, then to obtain the mixed product of these vectors, first the first two vectors and the resulting vector [ ab] is scalarly multiplied by the vector c.

Mixed product of three vectors a, b And c denoted as follows: abc or so ( a,b,c). Then we can write:

abc=([ab],c)

Before formulating a theorem representing the geometric meaning of a mixed product, familiarize yourself with the concepts of right triple, left triple, right coordinate system, left coordinate system (definitions 2, 2" and 3 on the page vector product of vectors online).

For definiteness, in what follows we will consider only right-handed coordinate systems.

Theorem 1. Mixed product of vectors ([ab],c) is equal to the volume of a paralleliped constructed on vectors reduced to a common origin a, b, c, taken with a plus sign, if three a, b, c right, and with a minus sign if three a, b, c left If the vectors a, b, c are coplanar, then ([ ab],c) is equal to zero.

Corollary 1. The following equality holds:

Therefore, it is enough for us to prove that

([ab],c)=([bc],a)

(3)

From expression (3) it is clear that the left and right parts are equal to the volume of the paralleliped. But the signs of the right and left sides coincide, since the triples of vectors abc And bca have the same orientation.

The proven equality (1) allows us to write the mixed product of three vectors a, b, c just in the form abc, without specifying which two vectors are multiplied vectorially by the first two or the last two.

Corollary 2. A necessary and sufficient condition for the coplanarity of three vectors is that their mixed product is equal to zero.

The proof follows from Theorem 1. Indeed, if the vectors are coplanar, then the mixed product of these vectors is equal to zero. Conversely, if the mixed product is equal to zero, then the coplanarity of these vectors follows from Theorem 1 (since the volume of a paralleliped built on vectors reduced to a common origin is equal to zero).

Corollary 3. The mixed product of three vectors, two of which coincide, is equal to zero.

Really. If two of the three vectors coincide, then they are coplanar. Therefore, the mixed product of these vectors is equal to zero.

Mixed product of vectors in Cartesian coordinates

Theorem 2. Let three vectors a, b And c defined by their Cartesian rectangular coordinates

Proof. Mixed work abc equal to the scalar product of vectors [ ab] And c. Cross product of vectors [ ab] in Cartesian coordinates is calculated by the formula ():

The last expression can be written using second-order determinants:

it is necessary and sufficient for the determinant to be equal to zero, the rows of which are filled with the coordinates of these vectors, i.e.:

.

(7)

To prove the corollary, it is enough to consider formula (4) and Corollary 2.

Mixed product of vectors with examples

Example 1. Find a mixed product of vectors abс, Where

Mixed product of vectors a, b, c equal to the determinant of the matrix L. Let's calculate the determinant of the matrix L, expanding the determinant along line 1:

Vector end point a.

Mixed (or vector-scalar) product three vectors a, b, c (taken in the indicated order) is called the scalar product of vector a and the vector product b x c, i.e. the number a(b x c), or, what is the same, (b x c)a.
Designation: abc.

Purpose. The online calculator is designed to calculate the mixed product of vectors. The resulting solution is saved in a Word file. Additionally, a solution template is created in Excel.

Signs of coplanarity of vectors

Three vectors (or a larger number) are called coplanar if they, being reduced to a common origin, lie in the same plane.
If at least one of the three vectors is zero, then the three vectors are also considered coplanar.

Sign of coplanarity. If the system a, b, c is right-handed, then abc>0 ; if left, then abc Geometric meaning of mixed product. The mixed product abc of three non-coplanar vectors a, b, c is equal to the volume of the parallelepiped built on the vectors a, b, c, taken with a plus sign if the system a, b, c is right-handed, and with a minus sign if this system is left-handed.

Properties of a mixed product

1. When the factors are rearranged circularly, the mixed product does not change; when two factors are rearranged, the sign is reversed: abc=bca=cab=-(bac)=-(cba)=-(acb)
   It follows from the geometric meaning.
2. (a+b)cd=acd+bcd (distributive property). Extends to any number of terms.
   Follows from the definition of a mixed product.
3. (ma)bc=m(abc) (combinative property with respect to a scalar factor).
   Follows from the definition of a mixed product. These properties make it possible to apply transformations to mixed products that differ from ordinary algebraic ones only in that the order of the factors can be changed only taking into account the sign of the product.
4. A mixed product that has at least two equal factors is equal to zero: aab=0.

Example No. 1. Find a mixed product. ab(3a+2b-5c)=3aba+2abb-5abc=-5abc .

Example No. 2. (a+b)(b+c)(c+a)= (axb+axc+bxb+bxc)(c+a)= (axb+axc +bxc)(c+a)=abc+acc+aca+ aba+bcc+bca. All terms except the two extreme ones are equal to zero. Also, bca=abc . Therefore (a+b)(b+c)(c+a)=2abc .

Example No. 3. Calculate the mixed product of three vectors a=15i+20j+5k, b=2i-4j+14k, c=3i-6j+21k.
Solution. To calculate the mixed product of vectors, it is necessary to find the determinant of a system composed of vector coordinates. Let's write the system in the form.

Definition. The number [, ] is called the mixed product of an ordered triple of vectors, .

We denote: (,) = = [, ].

Since the vector and scalar products are involved in the definition of a mixed product, their common properties are the properties of a mixed product.

For example, () = ().

Theorem 1. The mixed product of three coplanar vectors is zero.

Proof. If a given triple of vectors is coplanar, then one of the following conditions is satisfied for the vectors.

• 1. In a given triple of vectors there is at least one zero vector. In this case, the proof of the theorem is obvious.
• 2. In a given triple of vectors there is at least one pair of collinear vectors. If ||, then [, ] = 0, since [, ]= . If

|| , then [, ] and [, ] = 0. Similarly, if || .

3. Let this triple of vectors be coplanar, but cases 1 and 2 do not hold. Then the vector [, ] will be perpendicular to the plane to which all three vectors are parallel.

Therefore, [, ] and (,) = 0.

Theorem 2. Let the vectors (), (), () be specified in the basis (). Then

Proof. According to the definition of a mixed product

(,) = [, ] = с 1 - с 2 + с 3 = .

Due to the properties of the determinant, we have:

The theorem is proven.

Theorem 3. (,) = [, ].

Proof. Because

and due to the properties of the determinant we have:

(,) = = = [, ] = [, ].

The theorem is proven.

Theorem 4. The modulus of the mixed product of a non-coplanar triple of vectors is numerically equal to the volume of a parallelepiped built on representatives of these vectors with a common origin.

Proof. Let's choose an arbitrary point O and set aside from it the representatives of these vectors, : , . In the plane OAB we will construct a parallelogram OADB and, adding the edge OS, we will construct a parallelepiped OADBCADB. The volume V of this parallelepiped is equal to the product of the area of ​​the base OADB and the length of the height of the parallelepiped OO.

The area of ​​the parallelogram OADB is |[, ]|. On the other side

|OO| = || |cos |, where is the angle between the vectors and [, ].

Consider the mixed product module:

|(,)| = | [, ]| = |[, ]||||cos | = |[, ]||OO| = V.

The theorem has been proven.

Note 1. If the mixed product of a triple of vectors is equal to zero, then this triple of vectors is linearly dependent.

Note 2. If the mixed product of a given triple of vectors is positive, then the triple of vectors is right, and if it is negative, then the triple of vectors is left. Indeed, the sign of the mixed product coincides with the sign of cos, and the magnitude of the angle determines the orientation of the triple, . If the angle is acute, then the three is right, and if it is an obtuse angle, then the three is left.

Example 1. Given the parallelepiped ABCDA 1 B 1 C 1 D 1 and the coordinates of the following vectors in the orthonormal basis: (4; 3; 0), (2; 1; 2), (-3; -2; 5).

Find: 1) volume of the parallelepiped;

• 2) areas of faces ABCD and CDD 1 C;
• 3) cosine of the dihedral angle between planes ABC and CDD 1.

Solution.

This parallelepiped is built on vectors

Thus, its volume is equal to the modulus of the mixed product of these vectors, i.e.

So, V steam = 12 cubic units.

Recall that the area of ​​a parallelogram is equal to the length of the vector product of the vectors on which it is constructed.

Let us introduce the notation: , then

Therefore, (6; - 8; - 2), whence

That. sq. units

Likewise,

Let it be then

whence (15; - 20; 1) and

This means sq. units.

Let us introduce the following notation: pl. (ABC)=, pl. (DCC 1)=.

According to the definition of a vector product, we have:

This means that the following equality is true:

From the second point of the solution we have:

Prove that if and are mutually perpendicular unit vectors, then for any vectors and the following equality holds:

Solution.

Let the coordinates of the vectors be given in an orthonormal basis: ; . Since, by the property of a mixed product we have:

Thus, equality (1) can be written in the following form: , and this is one of the proven properties of the vector product of vectors and. Thus, the validity of equality (1) is proven.

Solving the zero version of the test work

Task No. 1

The vector forms angles and with the basis vectors and, respectively. Determine the angle that the vector makes with the vector.

Solution.

Let's construct a parallelepiped on vectors and on a diagonal, such that the vectors and are equal.

Then in a right triangle with a right angle, the magnitude of the angle is equal to where.

Similarly, in a right triangle with a right angle, the magnitude is equal to, whence.

In a right triangle, using the Pythagorean theorem we find:

In a right triangle, the leg and the hypotenuse are right angles. So the angle is equal. But the angle is equal to the angle between the vectors and. Thus the problem is solved.

Task No. 2.

Three vectors are given in the basis. Prove that the quadrilateral is flat. Find its area.

Solution.

1. If the vectors and are coplanar, then it is a flat quadrilateral. Let's calculate the determinant made up of the coordinates of these vectors.

Since the determinant is equal to zero, the vectors and are coplanar, which means the quadrilateral is flat.

2. Note that, therefore and thus, the quadrilateral is a trapezoid with bases AB and CD.

By the vector product property we have:

Finding the vector product

Task No. 3. Find a vector collinear to the vector (2; 1; -2), whose length is 5.

Solution.

Let's denote the coordinates of the vector (x, y, z). As you know, collinear vectors have proportional coordinates, and therefore we have:

x = 2t, y = t, z = ? 2t.

According to the conditions of the problem || = 5, and in coordinate form:

Expressing variables through the parameter t, we get:

4t 2 +t 2 +4t 2 =25,

Thus,

x = , y = , z = .

We received two solutions.

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, there will even be fewer typical tasks. 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 in reverse order, we obtain 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) An equally 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, here is an example for solving it yourself:

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, a mixed product can be negative: .

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