The area of ​​a parallelogram built on vectors is equal to the product of the lengths of these vectors and the angle of the angle that lies between them.

It’s good when the conditions give the lengths of these same vectors. However, it also happens that the formula for the area of ​​a parallelogram built on vectors can be applied only after calculations using coordinates.
If you are lucky and the conditions give the lengths of the vectors, then you just need to apply the formula, which we have already discussed in detail in the article. The area will be equal to the product of the modules and the sine of the angle between them:

Let's consider an example of calculating the area of ​​a parallelogram built on vectors.

Task: The parallelogram is built on the vectors and . Find the area if , and the angle between them is 30°.
Let's express the vectors through their values:

Perhaps you have a question - where do the zeros come from? It is worth remembering that we are working with vectors, and for them . also note that if the result is an expression, it will be converted to. Now we carry out the final calculations:

Let's return to the problem when the lengths of the vectors are not specified in the conditions. If your parallelogram lies in the Cartesian coordinate system, then you will need to do the following.

Calculation of the lengths of the sides of a figure given by coordinates

First, we find the coordinates of the vectors and subtract the corresponding coordinates of the beginning from the end coordinates. Let's say the coordinates of vector a are (x1;y1;z1), and vector b is (x3;y3;z3).
Now we find the length of each vector. To do this, each coordinate must be squared, then add the results obtained and from finite number extract the root. Based on our vectors there will be the following calculations:


Now we need to find the scalar product of our vectors. To do this, their corresponding coordinates are multiplied and added.

Having the lengths of the vectors and their scalar product, we can find the cosine of the angle lying between them .
Now we can find the sine of the same angle:
Now we have all the necessary quantities, and we can easily find the area of ​​a parallelogram built on vectors using the already known formula.

Let us first remember what a vector product is.

Note 1

Vector artwork for $\vec(a)$ and $\vec(b)$ is $\vec(c)$, which is some third vector $\vec(c)= ||$, and this vector has special properties:

• The scalar of the resulting vector is the product of $|\vec(a)|$ and $|\vec(b)|$ by the sine of the angle $\vec(c)= ||= |\vec(a)| \cdot |\vec(b)|\cdot \sin α \left(1\right)$;
• All $\vec(a), \vec(b)$ and $\vec(c)$ form a right triple;
• The resulting vector is orthogonal to $\vec(a)$ and $\vec(b)$.

If vectors have some coordinates ($\vec(a)=\(x_1; y_1; z_1\)$ and $\vec(b)= \(x_2; y_2; z_2\)$), then their vector product in Cartesian coordinate system can be determined by the formula:

$ = \(y_1 \cdot z_2 – y_2 \cdot z_1; z_1 \cdot x_2 – z_2 \cdot x_1; x_2 \cdot y_2 – x_2 \cdot y_1\)$

The easiest way to remember this formula is to write it in determinant form:

$ = \begin(array) (|ccc|) i & j & k \\ x_1 & y_1 & z_1 \\ x_2 & y_2 & z_2 \\ \end(array)$.

This formula is very convenient to use, but in order to understand how to use it, you should first become familiar with the topic of matrices and their determinants.

Area of ​​a parallelogram, whose sides are determined by two vectors $\vec(a)$ and $vec(b)$ is equal to scalar of the vector product of the given two vectors.

This relationship is not at all difficult to derive.

Let us recall the formula for finding the area of ​​an ordinary parallelogram, which can be characterized by the segments $a$ and $b$ that form it:

$S = a \cdot b \cdot \sin α$

In this case, the lengths of the sides are equal to the scalar values ​​of the vectors $\vec(a)$ and $\vec(b)$, which is quite suitable for us, that is, the scalar of the vector product of these vectors will be the area of ​​the figure under consideration.

Example 1

Given are vectors $\vec(c)$ with coordinates $\(5;3; 7\)$ and vector $\vec(g)$ with coordinates $\(3; 7;10\)$ in the Cartesian coordinate system. Find the area of ​​the parallelogram formed by $\vec(c)$ and $\vec(g)$.

Solution:

Let's find the vector product for these vectors:

$ = \begin(array) (|ccc|) i & j & k \\ 5 & 3 & 7 \\ 3 & 7 & 10 \\ \end(array)= i \cdot \begin(array) (|cc |) 3 & 7 \\ 7 & 10 \\ \end(array) - j \cdot \begin(array) (|cc|) 5 & 7 \\ 3 & 10 \\ \end(array) + k \cdot \begin(array) (|cc|) 5 & 3 \\ 3 & 7 \\ \end(array) = i \cdot (3 \cdot 10 – 49) – j \cdot (50 -21) + k \cdot (35-9) = -19i -29j + 26k=\(- 19; 29; 26\)$.

Now let’s find the modular value for the resulting directed segment, it is the value of the area of ​​the constructed parallelogram:

$S= \sqrt(|19|^2 + |29|^2 + |26|^2) = \sqrt(1878) ≈ 43.34$.

This line of reasoning is valid not only for finding area in 3-dimensional space, but also for two-dimensional space. Check out the following puzzle on this topic.

Example 2

Calculate the area of ​​a parallelogram if its generating segments are specified by the vectors $\vec(m)$ with coordinates $\(2; 3\)$ and $\vec(d)$ with coordinates $\(-5; 6\)$.

Solution:

This problem is a special example of problem 1, solved above, but both vectors lie in the same plane, which means that the third coordinate, $z$, can be taken as zero.

To summarize all of the above, the area of ​​the parallelogram will be:

$S = \begin(array) (||cc||) 2 & 3\\ -5 & 6 \\ \end(array) = \sqrt(12 + 15) =3 \sqrt3$.

Example 3

Given vectors $\vec(a) = 3i – j + k; \vec(b)= 5i$. Determine the area of ​​the parallelogram they form.

$[ \vec(a) \times \vec(b)] = (3i – j + k) \times 5i = 15 – 5 + $

Let us simplify according to the table below for unit vectors:

Figure 1. Decomposition of a vector by basis. Author24 - online exchange of student works

$[ \vec(a) \times \vec(b)] = 5 k + 5 j$.

Calculation time:

$S = \sqrt(|-5|^2 + |5|^2) = 5\sqrt(2)$.

The previous problems were about vectors whose coordinates are specified in the Cartesian coordinate system, but consider also the case if the angle between the basis vectors differs from $90°$:

Example 4

Vector $\vec(d) = 2a + 3b$, $\vec(f)= a – 4b$, lengths $\vec(a)$ and $\vec(b)$ are equal to each other and equal to one, and the angle between $\vec(a)$ and $\vec(b)$ is 45°.

Solution:

Let's calculate the vector product $\vec(d) \times \vec(f)$:

$[\vec(d) \times \vec(f) ]= (2a + 3b) \times (a – 4b) = 2 – 8 + 3 – 12 $.

For vector products, according to their properties, the following is true: $$ and $$ are equal to zero, $ = - $.

Let's use this to simplify:

$[\vec(d) \times \vec(f) ]= -8 + 3 = -8 - 3 =-11$.

Now let's use the formula $(1)$ :

$[\vec(d) \times \vec(f) ] = |-11 | = 11 \cdot |a| \cdot |b| \cdot \sin α = 11 \cdot 1 \cdot 1 \cdot \frac12=$5.5.

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 wilds analytical geometry. This is wrong. In this section of higher mathematics there is generally little wood, except perhaps enough for Pinocchio. In fact, the material is very common and simple - hardly more complicated than the same scalar product, even typical tasks there will be less. The main thing in analytical geometry, as many will be convinced or have already been convinced, is NOT TO MAKE MISTAKES IN CALCULATIONS. Repeat like a spell and you will be happy =)

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

What will make you happy right away? When I was little, I could juggle two and 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 various 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 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. Happening collinear vectors It will be appropriate to consider 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's remember one of 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 triangle. 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 sides. 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 it will general case cannot be combined 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.

Special case– vector 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 did not understand 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 an independent solution:

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.

Mixed work vectors is the product of three vectors:

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

First, again, a definition and a picture:

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

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

Let's dive into the definition:

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

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

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

Note : The drawing is schematic.

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

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