Angle between vectors

In order for us to introduce the concept of a cross product of two vectors, we must first deal with such a concept as the angle between these vectors.

Let us be given two vectors $\overline(α)$ and $\overline(β)$. Let us take some point $O$ in space and set aside the vectors $\overline(α)=\overline(OA)$ and $\overline(β)=\overline(OB)$ from it, then the angle $AOB$ will be called angle between these vectors (Fig. 1).

Notation: $∠(\overline(α),\overline(β))$

The concept of the cross product of vectors and the formula for finding

Definition 1

The vector product of two vectors is a vector perpendicular to both given vectors, and its length will be equal to the product of the lengths of these vectors with the sine of the angle between these vectors, and this vector with two initial ones has the same orientation as the Cartesian coordinate system.

Notation: $\overline(α)х\overline(β)$.

Mathematically it looks like this:

1. $|\overline(α)x\overline(β)|=|\overline(α)||\overline(β)|sin⁡∠(\overline(α),\overline(β))$
2. $\overline(α)x\overline(β)⊥\overline(α)$, $\overline(α)x\overline(β)⊥\overline(β)$
3. $(\overline(α)x\overline(β),\overline(α),\overline(β))$ and $(\overline(i),\overline(j),\overline(k))$ are the same oriented (Fig. 2)

Obviously, the outer product of vectors will equal the zero vector in two cases:

1. If the length of one or both vectors is zero.
2. If the angle between these vectors is equal to $180^\circ$ or $0^\circ$ (because in this case the sine is equal to zero).

To clearly see how the cross product of vectors is found, consider the following solution examples.

Example 1

Find the length of the vector $\overline(δ)$, which will be the result of the cross product of vectors, with coordinates $\overline(α)=(0,4,0)$ and $\overline(β)=(3,0,0 )$.

Solution.

Let's depict these vectors in the Cartesian coordinate space (Fig. 3):

Figure 3. Vectors in Cartesian coordinate space. Author24 - online exchange of student papers

We see that these vectors lie on the $Ox$ and $Oy$ axes, respectively. Therefore, the angle between them will be equal to $90^\circ$. Let's find the lengths of these vectors:

$|\overline(α)|=\sqrt(0+16+0)=4$

$|\overline(β)|=\sqrt(9+0+0)=3$

Then, by Definition 1, we obtain the module $|\overline(δ)|$

$|\overline(δ)|=|\overline(α)||\overline(β)|sin90^\circ=4\cdot 3\cdot 1=12$

Answer: $12$.

Calculation of the cross product by the coordinates of the vectors

Definition 1 immediately implies a way to find the cross product for two vectors. Since a vector, in addition to a value, also has a direction, it is impossible to find it only using a scalar value. But besides it, there is another way to find the vectors given to us using the coordinates.

Let us be given vectors $\overline(α)$ and $\overline(β)$, which will have coordinates $(α_1,α_2,α_3)$ and $(β_1,β_2,β_3)$, respectively. Then the vector of the cross product (namely, its coordinates) can be found by the following formula:

$\overline(α)x\overline(β)=\begin(vmatrix)\overline(i)&\overline(j)&\overline(k)\\α_1&α_2&α_3\\β_1&β_2&β_3\end(vmatrix)$

Otherwise, expanding the determinant, we obtain the following coordinates

$\overline(α)х\overline(β)=(α_2 β_3-α_3 β_2,α_3 β_1-α_1 β_3,α_1 β_2-α_2 β_1)$

Example 2

Find the vector of the cross product of collinear vectors $\overline(α)$ and $\overline(β)$ with coordinates $(0,3,3)$ and $(-1,2,6)$.

Solution.

Let's use the formula above. Get

$\overline(α)x\overline(β)=\begin(vmatrix)\overline(i)&\overline(j)&\overline(k)\\0&3&3\\-1&2&6\end(vmatrix)=(18 -6)\overline(i)-(0+3)\overline(j)+(0+3)\overline(k)=12\overline(i)-3\overline(j)+3\overline(k )=(12,-3,3)$

Answer: $(12,-3,3)$.

Properties of the cross product of vectors

For arbitrary mixed three vectors $\overline(α)$, $\overline(β)$ and $\overline(γ)$, as well as $r∈R$, the following properties hold:

Example 3

Find the area of ​​a parallelogram whose vertices have coordinates $(3,0,0)$, $(0,0,0)$, $(0,8,0)$ and $(3,8,0)$.

Solution.

First, draw this parallelogram in coordinate space (Fig. 5):

Figure 5. Parallelogram in coordinate space. Author24 - online exchange of student papers

We see that the two sides of this parallelogram are constructed using collinear vectors with coordinates $\overline(α)=(3,0,0)$ and $\overline(β)=(0,8,0)$. Using the fourth property, we get:

$S=|\overline(α)x\overline(β)|$

Find the vector $\overline(α)х\overline(β)$:

$\overline(α)x\overline(β)=\begin(vmatrix)\overline(i)&\overline(j)&\overline(k)\\3&0&0\\0&8&0\end(vmatrix)=0\overline (i)-0\overline(j)+24\overline(k)=(0,0,24)$

Consequently

$S=|\overline(α)x\overline(β)|=\sqrt(0+0+24^2)=24$

Before giving the concept of a vector product, let us turn to the question of the orientation of the ordered triple of vectors a → , b → , c → in three-dimensional space.

To begin with, let's set aside the vectors a → , b → , c → from one point. The orientation of the triple a → , b → , c → is right or left, depending on the direction of the vector c → . From the direction in which the shortest turn is made from the vector a → to b → from the end of the vector c → , the form of the triple a → , b → , c → will be determined.

If the shortest rotation is counterclockwise, then the triple of vectors a → , b → , c → is called right if clockwise - left.

Next, take two non-collinear vectors a → and b → . Let us then postpone the vectors A B → = a → and A C → = b → from the point A. Let us construct a vector A D → = c → , which is simultaneously perpendicular to both A B → and A C → . Thus, when constructing the vector A D → = c →, we can do two things, giving it either one direction or the opposite (see illustration).

The ordered trio of vectors a → , b → , c → can be, as we found out, right or left depending on the direction of the vector.

From the above, we can introduce the definition of a vector product. This definition is given for two vectors defined in a rectangular coordinate system of three-dimensional space.

Definition 1

The vector product of two vectors a → and b → we will call such a vector given in a rectangular coordinate system of three-dimensional space such that:

• if the vectors a → and b → are collinear, it will be zero;
• it will be perpendicular to both vector a →​​ and vector b → i.e. ∠ a → c → = ∠ b → c → = π 2 ;
• its length is determined by the formula: c → = a → b → sin ∠ a → , b → ;
• the triplet of vectors a → , b → , c → has the same orientation as the given coordinate system.

The cross product of vectors a → and b → has the following notation: a → × b → .

Cross product coordinates

Since any vector has certain coordinates in the coordinate system, it is possible to introduce a second definition of the vector product, which will allow you to find its coordinates from the given coordinates of the vectors.

Definition 2

In a rectangular coordinate system of three-dimensional space vector product of two vectors a → = (a x ; a y ; a z) and b → = (b x ; b y ; b z) call the vector c → = a → × b → = (a y b z - a z b y) i → + (a z b x - a x b z) j → + (a x b y - a y b x) k → , where i → , j → , k → are coordinate vectors.

The vector product can be represented as a determinant square matrix of the third order, where the first line is the vectors i → , j → , k → , the second line contains the coordinates of the vector a → , and the third line contains the coordinates of the vector b → in a given rectangular coordinate system, this matrix determinant looks like this: c → = a → × b → = i → j → k → a x a y a z b x b y b z

Expanding this determinant over the elements of the first row, we obtain the equality: c → = a → × b → = i → j → k → a x a y a z b x b y b z = a y a z b y b z i → - a x a z b x b z j → + a x a y b x b y k → = = a → × b → = (a y b z - a z b y) i → + (a z b x - a x b z) j → + (a x b y - a y b x) k →

Cross product properties

It is known that the vector product in coordinates is represented as the determinant of the matrix c → = a → × b → = i → j → k → a x a y a z b x b y b z , then on the base matrix determinant properties the following vector product properties:

1. anticommutativity a → × b → = - b → × a → ;
2. distributivity a (1) → + a (2) → × b = a (1) → × b → + a (2) → × b → or a → × b (1) → + b (2) → = a → × b (1) → + a → × b (2) → ;
3. associativity λ a → × b → = λ a → × b → or a → × (λ b →) = λ a → × b → , where λ is an arbitrary real number.

These properties have not complicated proofs.

For example, we can prove the anticommutativity property of a vector product.

Proof of anticommutativity

By definition, a → × b → = i → j → k → a x a y a z b x b y b z and b → × a → = i → j → k → b x b y b z a x a y a z . And if two rows of the matrix are interchanged, then the value of the determinant of the matrix should change to the opposite, therefore, a → × b → = i → j → k → a x a y a z b x b y b z = - i → j → k → b x b y b z a x a y a z = - b → × a → , which and proves the anticommutativity of the vector product.

Vector Product - Examples and Solutions

In most cases, there are three types of tasks.

In problems of the first type, the lengths of two vectors and the angle between them are usually given, but you need to find the length of the cross product. In this case, use the following formula c → = a → b → sin ∠ a → , b → .

Example 1

Find the length of the cross product of vectors a → and b → if a → = 3 , b → = 5 , ∠ a → , b → = π 4 is known.

Solution

Using the definition of the length of the vector product of vectors a → and b →, we solve this problem: a → × b → = a → b → sin ∠ a → , b → = 3 5 sin π 4 = 15 2 2 .

Answer: 15 2 2 .

Tasks of the second type have a connection with the coordinates of vectors, they contain a vector product, its length, etc. are searched through the known coordinates of the given vectors a → = (a x ; a y ; a z) and b → = (b x ; b y ; b z) .

For this type of task, you can solve a lot of options for tasks. For example, not the coordinates of the vectors a → and b → , but their expansions in coordinate vectors of the form b → = b x i → + b y j → + b z k → and c → = a → × b → = (a y b z - a z b y) i → + (a z b x - a x b z) j → + (a x b y - a y b x) k → , or the vectors a → and b → can be given by the coordinates of their start and end points.

Consider the following examples.

Example 2

Two vectors are set in a rectangular coordinate system a → = (2 ; 1 ; - 3) , b → = (0 ; - 1 ; 1) . Find their vector product.

Solution

According to the second definition, we find the vector product of two vectors in given coordinates: a → × b → = (a y b z - a z b y) i → + (a z b x - a x b z) j → + (a x b y - a y b x) k → = = (1 1 - (- 3) (- 1)) i → + ((- 3) 0 - 2 1) j → + (2 (- 1) - 1 0) k → = = - 2 i → - 2 j → - 2 k → .

If we write the cross product in terms of the matrix determinant, then the solution this example looks like this: a → × b → = i → j → k → a x a y a z b x b y b z = i → j → k → 2 1 - 3 0 - 1 1 = - 2 i → - 2 j → - 2 k → .

Answer: a → × b → = - 2 i → - 2 j → - 2 k → .

Example 3

Find the length of the cross product of vectors i → - j → and i → + j → + k → , where i → , j → , k → - orts of a rectangular Cartesian coordinate system.

Solution

First, let's find the coordinates of the given vector product i → - j → × i → + j → + k → in the given rectangular coordinate system.

It is known that the vectors i → - j → and i → + j → + k → have coordinates (1 ; - 1 ; 0) and (1 ; 1 ; 1) respectively. Find the length of the vector product using the matrix determinant, then we have i → - j → × i → + j → + k → = i → j → k → 1 - 1 0 1 1 1 = - i → - j → + 2 k → .

Therefore, the vector product i → - j → × i → + j → + k → has coordinates (- 1 ; - 1 ; 2) in the given coordinate system.

We find the length of the vector product by the formula (see the section on finding the length of the vector): i → - j → × i → + j → + k → = - 1 2 + - 1 2 + 2 2 = 6 .

Answer: i → - j → × i → + j → + k → = 6 . .

Example 4

The coordinates of three points A (1 , 0 , 1) , B (0 , 2 , 3) ​​, C (1 , 4 , 2) are given in a rectangular Cartesian coordinate system. Find some vector perpendicular to A B → and A C → at the same time.

Solution

Vectors A B → and A C → have the following coordinates (- 1 ; 2 ; 2) and (0 ; 4 ; 1) respectively. Having found the vector product of the vectors A B → and A C → , it is obvious that it is a perpendicular vector by definition to both A B → and A C → , that is, it is the solution to our problem. Find it A B → × A C → = i → j → k → - 1 2 2 0 4 1 = - 6 i → + j → - 4 k → .

Answer: - 6 i → + j → - 4 k → . is one of the perpendicular vectors.

Problems of the third type are focused on using the properties of the vector product of vectors. After applying which, we will obtain a solution to the given problem.

Example 5

The vectors a → and b → are perpendicular and their lengths are 3 and 4 respectively. Find the length of the cross product 3 a → - b → × a → - 2 b → = 3 a → × a → - 2 b → + - b → × a → - 2 b → = = 3 a → × a → + 3 a → × - 2 b → + - b → × a → + - b → × - 2 b → .

Solution

By the distributivity property of the vector product, we can write 3 a → - b → × a → - 2 b → = 3 a → × a → - 2 b → + - b → × a → - 2 b → = = 3 a → × a → + 3 a → × - 2 b → + - b → × a → + - b → × - 2 b →

By the property of associativity, we take out the numerical coefficients beyond the sign of vector products in the last expression: 3 a → × a → + 3 a → × - 2 b → + - b → × a → + - b → × - 2 b → = = 3 a → × a → + 3 (- 2) a → × b → + (- 1) b → × a → + (- 1) (- 2) b → × b → = = 3 a → × a → - 6 a → × b → - b → × a → + 2 b → × b →

The vector products a → × a → and b → × b → are equal to 0, since a → × a → = a → a → sin 0 = 0 and b → × b → = b → b → sin 0 = 0 , then 3 a → × a → - 6 a → × b → - b → × a → + 2 b → × b → = - 6 a → × b → - b → × a → . .

From the anticommutativity of the vector product it follows - 6 a → × b → - b → × a → = - 6 a → × b → - (- 1) a → × b → = - 5 a → × b → . .

Using the properties of the vector product, we obtain the equality 3 · a → - b → × a → - 2 · b → = = - 5 · a → × b → .

By condition, the vectors a → and b → are perpendicular, that is, the angle between them is equal to π 2 . Now it remains only to substitute the found values ​​into the corresponding formulas: 3 a → - b → × a → - 2 b → = - 5 a → × b → = = 5 a → × b → = 5 a → b → sin (a →, b →) = 5 3 4 sin π 2 = 60.

Answer: 3 a → - b → × a → - 2 b → = 60 .

The length of the cross product of vectors by definition is a → × b → = a → · b → · sin ∠ a → , b → . Since it is already known (from school course) that the area of ​​a triangle is half the product of the lengths of its two sides multiplied by the sine of the angle between the given sides. Therefore, the length of the vector product is equal to the area of ​​a parallelogram - a doubled triangle, namely, the product of the sides in the form of vectors a → and b → , laid off from one point, by the sine of the angle between them sin ∠ a → , b → .

That's what it is geometric sense vector product.

The physical meaning of the vector product

In mechanics, one of the branches of physics, thanks to the vector product, you can determine the moment of force relative to a point in space.

Definition 3

Under the moment of force F → , applied to point B , relative to point A we will understand the following vector product A B → × F → .

If you notice a mistake in the text, please highlight it and press Ctrl+Enter

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 not true. 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 repurchase 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 practical sense? 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.

special case is the cross 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 reminiscent of Example 3:

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

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

Answer:

The considered problem is quite common in control work, here's an example for a do-it-yourself 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

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".

By definition 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 words, the mixed product can be negative: .

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

We will use the cross product table vectors i,j uk:

if the direction of the shortest path from the first vector to the second coincides with the direction of the arrow, then the product is equal to the third vector, if it does not match, the third vector is taken with a minus sign.

Let two vectors a=axi +ayj +azk and b =bxi +byj +bzk be given. Let's find the vector product of these vectors by multiplying them as polynomials (according to the properties of the vector product):
The resulting formula can be written even shorter: since the right side of equality (7.1) corresponds to the expansion of the third-order determinant in terms of the elements of the first row. Equality (7.2) is easy to remember.

7.4. Some applications of the cross product

Establishing collinearity of vectors.
Finding the area of ​​a parallelogram and a triangle

According to the definition of the cross product of vectors a and b | a xb | = |a| * |b |sing , i.e. S pairs = |a x b |. And, therefore, DS \u003d 1/2 | a x b |.

Determining the moment of force about a point

Let a force F = AB be applied at point A and let O be some point in space It is known from physics that the moment of force F relative to the point O is the vector M, which passes through the point O and:

1) perpendicular to the plane passing through the points O, A, B;

2) is numerically equal to the product of the force and the arm 3) forms a right triple with the vectors OA and A B.

So, M=OA x F. Finding the linear speed of rotation

Speed ​​v point M solid body, rotating with an angular velocity w around a fixed axis, is determined by the Euler formula v \u003d w x r, where r \u003d OM, where O is some fixed point of the axis (see Fig. 21).

Angle between vectors

It follows from the definition of the scalar product of two vectors that If the vectors and are given by the coordinates and , then formula (1.6.3.1) can be written as:

Area of ​​a parallelogram built on vectors

Tasks for measuring the lengths of segments, distances between points, surface areas and volumes of bodies belong to an important class of problems that are commonly called metric. In the previous section, we learned how to use vector algebra to calculate line lengths and distances between points. Now we are going to find ways to calculate areas and volumes. Vector algebra allows us to set and solve similar problems only for fairly simple cases. Methods of analysis are required to calculate areas of arbitrary surfaces and volumes of arbitrary bodies. But the methods of analysis, in turn, are essentially based on the results that vector algebra gives.

To solve the problem, we chose a rather long and difficult path, suggested by Gilbert Strang, associated with numerous geometric transformations and painstaking algebraic calculations. We chose this path despite the fact that there are other approaches that lead to the goal faster because it seemed to us direct and natural. The direct path in science is not always the easiest. Sophisticated people know about this and prefer detours, but if you do not try to go straight, then you can remain ignorant of some of the subtleties of the theory.

On the path we have chosen, such concepts as the orientation of space, the determinant, vector and mixed products naturally appear. Especially clearly, as under a microscope, the geometric meaning of the determinant and its properties is manifested. Traditionally, the concept of a determinant is introduced in the theory of systems of linear equations, but it is for solving such systems that the determinant is almost useless. The geometric meaning of the determinant is essential for vector and tensor algebra.

Now let's be patient and start with the simplest and most understandable cases.

1. The vectors are oriented along the coordinate axes of the Cartesian coordinate system.

Let the vector a be directed along the x-axis, and the vector b along the y-axis. On fig. 21 shows four different options for the arrangement of vectors with respect to the coordinate axes.

Vectors a and b in coordinate form: Where a and b denote the modulus of the corresponding vector and is the sign of the vector's coordinate.

Since the vectors are orthogonal, the parallelograms built on them are rectangles. Their areas are simply the product of their sides. Let us express these products in terms of the coordinates of the vectors for all four cases.

All four formulas for calculating the area are the same except for the sign. You could just close your eyes and write down, which is in all cases. However, another possibility turns out to be more productive: to give the sign some meaning. Let's look closely at Fig. 21. In cases where the rotation of the vector to the vector is carried out clockwise. In those cases when we are forced to use the minus sign in the formula, the rotation of the vector to the vector is carried out counterclockwise. This observation makes it possible to relate the sign in the expressions for the area to the orientation of the plane.

The area of ​​the rectangle built on the vectors a and b with a plus or minus sign will be considered the oriented area, while the sign will be associated with the orientation given by the vectors. For an oriented area, we can write a single formula for all four cases considered: . The sign of the "vector" line above the letter S is introduced in order to distinguish the usual area, which is always positive, from the oriented one.

In this case, it is obvious that the same vectors, taken in a different order, determine the opposite orientation, therefore, . It's just that the area will continue to be denoted by the letter S and, therefore, .

Now that, at what seems to be the price of expanding the concept of area, we have obtained a general expression, the attentive reader will say that we have not considered all the possibilities. Indeed, in addition to the four options for the location of the vectors shown in Fig. 21, there are four more (Fig. 22) Let's write again the vectors and in the coordinate form: Let's express the areas in terms of the coordinates of the vectors. four. . The signs in the new expressions have not changed, but, unfortunately, the orientation has changed in relation to the previous four cases. Therefore, for the oriented area, we are forced to write: . Although the hope for ingenious simplicity was not justified, nevertheless, we can still write down a general expression for all four cases.

That is, the oriented area of ​​a rectangle built on vectors, as on sides, is equal to the determinant, composed of the coordinates of the vectors, as from columns.

We believe that the reader is familiar with the theory of determinants, therefore, we do not dwell on this concept in detail. Nevertheless, we give the appropriate definitions in order to change the emphasis and show that this concept can be arrived at from purely geometric considerations. So, , , - different forms of designation for the same concept - the determinant, composed of the coordinates of the vectors, as from columns. Equality can be taken as its definition for the two-dimensional case.

2. The vector b is not parallel to the x axis; the vector a/ is an arbitrary vector.

In order to reduce this case to those already known, we consider some geometric transformations of a parallelogram built on vectors and (Fig. . mixed products of vectors and its properties