The meaning of the vector product. Vector product of vectors

Definition. The vector product of vector a and vector b is a vector denoted by the symbol [α, b] (or l x b), such that 1) the length of the vector [a, b] is equal to (p, where y is the angle between vectors a and b ( Fig. 31); 2) vector [a, b) is perpendicular to vectors a and b, i.e. perpendicular to the plane of these vectors; 3) the vector [a, b] is directed in such a way that from the end of this vector the shortest turn from a to b is seen to occur counterclockwise (Fig. 32). Rice. 32 Fig.31 In other words, vectors a, b and [a, b) form a right-hand triplet of vectors, i.e. located like the thumb, index and middle fingers of the right hand. If the vectors a and b are collinear, we will assume that [a, b] = 0. By definition, the length of the vector product is numerically equal to the area Sa of a parallelogram (Fig. 33), constructed on the multiplied vectors a and b as sides: 6.1 . Properties of a vector product 1. A vector product is equal to a zero vector if and only if at least one of the multiplied vectors is zero or when these vectors are collinear (if vectors a and b are collinear, then the angle between them is either 0 or 7r). This can be easily obtained from the fact that If we consider the zero vector to be collinar with any vector, then the condition for the collinearity of vectors a and b can be expressed as follows: 2. The vector product is anticommutative, i.e., always. In fact, the vectors (a, b) have the same length and are collinear. The directions of these vectors are opposite, since from the end of the vector [a, b] the shortest turn from a to b will be seen occurring counterclockwise, and from the end of the vector [b, a] - clockwise (Fig. 34). 3. The vector product has a distributive property in relation to addition 4. The numerical factor A can be taken out of the sign of the vector product 6.2. Vector product of vectors specified by coordinates Let vectors a and b be specified by their coordinates in the basis. Using the distribution property of the vector product, we find the vector product of vectors given by coordinates. Mixed work. Let us write down the vector products of coordinate unit vectors (Fig. 35): Therefore, for the vector product of vectors a and b, we obtain from formula (3) the following expression Formula (4) can be written in a symbolic, easy-to-remember form if we use the 3rd order determinant: Expanding this determinant over the elements of the 1st row, we obtain (4). Examples. 1. Find the area of a parallelogram constructed on vectors. The required area. Therefore, we find = whence 2. Find the area of the triangle (Fig. 36). It is clear that the area b"d of the triangle OAO is equal to half the area S of the parallelogram O AC B. Calculating the vector product (a, b| of the vectors a = OA and b = ob, we obtain Hence Remark. The vector product is not associative, i.e. the equality ( (a, b),c) = [a, |b,c)) is not true in the general case. For example, for a = ss j we have § 7. Mixed product of vectors Let us multiply the vectors a, b, and c. and 1> vector. As a result, we obtain the vector [a, 1>]. Let us multiply it scalarly by the vector c: (k b), c).The number ([a, b], e) is called the mixed product of the vectors a, b. and is denoted by the symbol (a, 1), e). vectors a, b and c are called coplanar in this case), then the mixed product ([a, b], c) = 0. This follows from the fact that the vector [a, b| is perpendicular to the plane in which the vectors a and 1 lie. ", and therefore to the vector c. / If the points O, A, B, C do not lie in the same plane (vectors a, b and c are non-coplanar), we will construct a parallelepiped on the edges OA, OB and OS (Fig. 38 a). By the definition of a vector product, we have (a,b) = So c, where So is the area of the parallelogram OADB, and c is the unit vector perpendicular to the vectors a and b and such that the triple a, b, c is right-handed, i.e. vectors a, b and c are located respectively as the thumb, index and middle fingers of the right hand (Fig. 38 b). Multiplying both sides of the last equality on the right scalarly by the vector c, we obtain that the vector product of vectors given by coordinates. Mixed work. The number pc c is equal to the height h of the constructed parallelepiped, taken with the “+” sign if the angle between the vectors c and c is acute (triple a, b, c - right), and with the “-” sign if the angle is obtuse (triple a, b, c - left), so that Thus, the mixed product of vectors a, b and c is equal to the volume V of the parallelepiped built on these vectors as on edges, if the triple a, b, c is right, and -V, if the triple a , b, c - left. Based on the geometric meaning of the mixed product, we can conclude that by multiplying the same vectors a, b and c in any other order, we will always get either +7 or -K. Manufacturer's mark Fig. 38 reference will depend only on what kind of triple the multiplied vectors form - right or left. If the vectors a, b, c form a right-handed triple, then the triples b, c, a and c, a, b will also be right-handed. At the same time, all three triples b, a, c; a, c, b and c, b, a - left. Thus, (a,b, c) = (b,c, a) = (c,a,b) = -(b,a,c) = -(a,c,b) = -(c,b ,A). We emphasize again that the mixed product of vectors is equal to zero only if the multiplied vectors a, b, c are coplanar: (a, b, c are coplanar) 7.2. Mixed product in coordinates Let the vectors a, b, c be given by their coordinates in the basis i, j, k: a = (x\,y\,z]), b= (x2,y2>z2), c = (x3, uz, 23). Let us find an expression for their mixed product (a, b, c). We have a mixed product of vectors specified by their coordinates in the basis i, J, k, equal to the third-order determinant, the lines of which are composed respectively of the coordinates of the first, second and third of the multiplied vectors. The necessary and sufficient condition for the coplanarity of the vectors a y\, Z|), b = (хъ У2.22), с = (жз, з, 23) will be written in the following form У| z, ag2 y2 -2 =0. Uz Example. Check whether the vectors „ = (7,4,6), b = (2, 1,1), c = (19, II, 17) are coplanar. The vectors under consideration will be coplanar or non-coplanar, depending on whether the determinant is equal to zero or not. Expanding it into the elements of the first row, we obtain D = 7-6-4-15 + 6-3 = 0^ - vectors n, b, c are coplanar. 7.3. Double cross product The double cross product [a, [b, c]] is a vector perpendicular to the vectors a and [b, c]. Therefore, it lies in the plane of vectors b and c and can be expanded into these vectors. It can be shown that the formula [a, [!>, c]] = b(a, e) - c(a, b) is valid. Exercises 1. Three vectors AB = c, Ж? = o and CA = b serve as the sides of the triangle. Express in terms of a, b and c the vectors coinciding with the medians AM, DN, CP of the triangle. 2. What condition must the vectors p and q be connected so that the vector p + q divides the angle between them in half? It is assumed that all three vectors are related to a common origin. 3. Calculate the length of the diagonals of a parallelogram constructed on the vectors a = 5p + 2q and b = p - 3q, if it is known that |p| = 2v/2, |q| = 3 H-(p7ci) = f. 4. Denoting by a and b the sides of the rhombus that extend from the common vertex, prove that the diagonals of the rhombus are mutually perpendicular. 5. Calculate the scalar product of the vectors a = 4i + 7j + 3k and b = 31 - 5j + k. 6. Find the unit vector a0 parallel to the vector a = (6, 7, -6). 7. Find the projection of the vector a = l+ j- kHa vector b = 21 - j - 3k. 8. Find the cosine of the angle between the vectors IS “w, if A(-4,0,4), B(-1,6,7), C(1,10.9). 9. Find the unit vector p°, which is simultaneously perpendicular to the vector a = (3, 6, 8) and the Ox axis. 10. Calculate the sine of the angle between the diagonals of the parallelogram constructed on the vectors a = 2i+J-k, b=i-3j + k as on the sides. Calculate the height h of a parallelepiped built on vectors a = 31 + 2j - 5k, b = i- j + 4knc = i-3j + k, if a parallelogram built on vectors a and I is taken as the base. Answers

Vector artwork is a pseudovector perpendicular to a plane constructed from two factors, which is the result of the binary operation “vector multiplication” over vectors in three-dimensional Euclidean space. The vector product does not have the properties of commutativity and associativity (it is anticommutative) and, unlike the scalar product of vectors, is a vector. Widely used in many engineering and physics applications. For example, angular momentum and Lorentz force are written mathematically as a vector product. The cross product is useful for "measuring" the perpendicularity of vectors - the modulus of the cross product of two vectors is equal to the product of their moduli if they are perpendicular, and decreases to zero if the vectors are parallel or antiparallel.

The vector product can be defined in different ways, and theoretically, in a space of any dimension n, one can calculate the product of n-1 vectors, thereby obtaining a single vector perpendicular to them all. But if the product is limited to non-trivial binary products with vector results, then the traditional vector product is defined only in three-dimensional and seven-dimensional spaces. The result of a vector product, like a scalar product, depends on the metric of Euclidean space.

Unlike the formula for calculating the scalar product vectors from coordinates in a three-dimensional rectangular coordinate system, the formula for the cross product depends on the orientation of the rectangular coordinate system, or, in other words, its “chirality”.

Definition:
The vector product of vector a and vector b in space R3 is a vector c that satisfies the following requirements:
the length of vector c is equal to the product of the lengths of vectors a and b and the sine of the angle φ between them:
|c|=|a||b|sin φ;
vector c is orthogonal to each of vectors a and b;
vector c is directed so that the triple of vectors abc is right-handed;
in the case of the space R7, the associativity of the triple of vectors a, b, c is required.
Designation:
c===a × b

Rice. 1. The area of a parallelogram is equal to the modulus of the vector product

Geometric properties of a cross product:
A necessary and sufficient condition for the collinearity of two nonzero vectors is that their vector product is equal to zero.

Vector product module equals area S parallelogram constructed on vectors reduced to a common origin a And b(see Fig. 1).

If e- unit vector orthogonal to the vectors a And b and chosen so that three a,b,e- right, and S is the area of the parallelogram constructed on them (reduced to a common origin), then the formula for the vector product is valid:
=S e

Fig.2. Volume of a parallelepiped using the vector and scalar product of vectors; the dotted lines show the projections of vector c onto a × b and vector a onto b × c, the first step is to find the scalar products

If c- some vector, π - any plane containing this vector, e- unit vector lying in the plane π and orthogonal to c,g- unit vector orthogonal to the plane π and directed so that the triple of vectors ecg is right, then for any lying in the plane π vector a the formula is correct:
=Pr e a |c|g
where Pr e a is the projection of vector e onto a
|c|-modulus of vector c

When using vector and scalar products, you can calculate the volume of a parallelepiped built on vectors reduced to a common origin a, b And c. Such a product of three vectors is called mixed.
V=|a (b×c)|
The figure shows that this volume can be found in two ways: the geometric result is preserved even when the “scalar” and “vector” products are swapped:
V=a×b c=a b×c

The magnitude of the cross product depends on the sine of the angle between the original vectors, so the cross product can be perceived as the degree of “perpendicularity” of the vectors, just as the scalar product can be seen as the degree of “parallelism”. The vector product of two unit vectors is equal to 1 (unit vector) if the original vectors are perpendicular, and equal to 0 (zero vector) if the vectors are parallel or antiparallel.

Expression for the cross product in Cartesian coordinates
If two vectors a And b defined by their rectangular Cartesian coordinates, or more precisely, represented in an orthonormal basis
a=(a x ,a y ,a z)
b=(b x ,b y ,b z)
and the coordinate system is right-handed, then their vector product has the form
=(a y b z -a z b y ,a z b x -a x b z ,a x b y -a y b x)
To remember this formula:
i =∑ε ijk a j b k
Where ε ijk- symbol of Levi-Civita.

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This online calculator calculates the vector product of vectors. A detailed solution is given. To calculate the cross product of vectors, enter the coordinates of the vectors in the cells and click on the "Calculate" button.

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Vector product of vectors

Before moving on to the definition of the vector product of vectors, let's consider the concepts ordered vector triplet, left vector triplet, right vector triplet.

Definition 1. Three vectors are called ordered triple(or triple), if it is indicated which of these vectors is the first, which is the second and which is the third.

Record cba- means - the first is a vector c, the second is the vector b and the third is the vector a.

Definition 2. Triple of non-coplanar vectors abc is called right (left) if, when reduced to a common origin, these vectors are located in the same way as the large, unbent index and middle fingers of the right (left) hand are located, respectively.

Definition 2 can be formulated differently.

Definition 2". Triple of non-coplanar vectors abc is called right (left) if, when reduced to a common origin, the vector c is located on the other side of the plane defined by the vectors a And b, where is the shortest turn from a To b performed counterclockwise (clockwise).

Troika of vectors abc, shown in Fig. 1 is right, and three abc shown in Fig. 2 is the left one.

If two triplets of vectors are right or left, then they are said to be of the same orientation. Otherwise they are said to be of opposite orientation.

Definition 3. A Cartesian or affine coordinate system is called right (left) if three basis vectors form a right (left) triple.

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

Definition 4. Vector artwork vector a to vector b called a vector With, denoted by the symbol c=[ab] (or c=[a,b], or c=a×b) and satisfying the following three requirements:

vector length With equal to the product of vector lengths a And b by the sine of the angle φ between them:

|c|=|[ab]|=|a||b|sinφ;

(1)

vector With orthogonal to each of the vectors a And b;
vector c directed so that the three abc is right.

The cross product of vectors has the following properties:

[ab]=−[ba] (anti-permutability factors);
[(λa)b]=λ [ab] (combination relative to the numerical factor);
[(a+b)c]=[ac]+[bc] (distributiveness relative to the sum of vectors);
[aa]=0 for any vector a.

Geometric properties of the vector product of vectors

Theorem 1. For two vectors to be collinear, it is necessary and sufficient that their vector product be equal to zero.

Proof. Necessity. Let the vectors a And b collinear. Then the angle between them is 0 or 180° and sinφ=sin180=sin 0=0. Therefore, taking into account expression (1), the length of the vector c equal to zero. Then c zero vector.

Adequacy. Let the vector product of vectors a And b obviously zero: [ ab]=0. Let us prove that the vectors a And b collinear. If at least one of the vectors a And b zero, then these vectors are collinear (since the zero vector has an indefinite direction and can be considered collinear to any vector).

If both vectors a And b non-zero, then | a|>0, |b|>0. Then from [ ab]=0 and from (1) it follows that sinφ=0. Therefore the vectors a And b collinear.

The theorem has been proven.

Theorem 2. Length (modulus) of the vector product [ ab] equals area S parallelogram constructed on vectors reduced to a common origin a And b.

Proof. As you know, the area of a parallelogram is equal to the product of the adjacent sides of this parallelogram and the sine of the angle between them. Hence:

Then the vector product of these vectors has the form:

Expanding the determinant over the elements of the first row, we obtain the decomposition of the vector a×b by basis i, j, k, which is equivalent to formula (3).

Proof of Theorem 3. Let's create all possible pairs of basis vectors i, j, k and calculate their vector product. It should be taken into account that the basis vectors are mutually orthogonal, form a right-handed triple and have unit length (in other words, we can assume that i={1, 0, 0}, j={0, 1, 0}, k=(0, 0, 1)). Then we have:

From the last equality and relations (4), we obtain:

Let's create a 3x3 matrix, the first row of which is the basis vectors i, j, k, and the remaining lines are filled with vector elements a And b.

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.

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 build 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 .
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.
When any two factors are rearranged, the mixed product changes sign.
Indeed, if we consider a mixed product, then, for example, or
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

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 some 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.