Математика. Материалы курса

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At this lecture, we introduce the concept of a scalar product of vectors. First, we define what is meant by the angle between the vectors. So, two vectors a and b are given. The angle between them is the value of the smaller angle by which one of the vectors can be rotated to the position of co-directivity with the other vector. If one of the vectors is equal to zero, the angle with such a vector can take any value, that is, this value cannot be determined identically.

So, two vectors are given. Let’s consider the angle between them, and then proceed to considering the cosine of the angle. After that, we will turn to some of the simplest consequences.

If the vectors are co-directed, that is, they are collinear and have the same direction, the angle between them will be zero, which means that the cosine of this angle is equal to one. If the value of the angle is less than 90 degrees, that is, the angle is sharp, the cosine of this angle will take a positive value. If the angle is straight, its cosine is zero. Further, by increasing the angle, if it is obtuse, its cosine is less than zero. Finally, for an expanded angle, that is, when the vectors are oppositely directed, the cosine of this angle is -1.

Let’s remember what operations we can perform on vectors. We will give the corresponding examples from physics.

Two vectors are given. These vectors can be added together. For example, let’s say that we have a certain body, and several forces act on it. In this case, instead of these forces, we can consider their sum, the so-called resultant force. At the same time, if we replace all the forces with their resultant, the movement of the body will not change in any way. So, for two vectors, we can construct a new vector – their sum.

We also know the operation that allows us to find a new vector by a vector and a number. For example, let’s remember Newton’s Second law. It relates the force vector to the acceleration vector. If the acceleration vector is multiplied by the mass of the body, we get the force vector. Or in other words: the acceleration is equal to the force multiplied by the inverse of the mass.

There are also problems when we get a certain number from two given vectors. Consider this example. Let a certain force f act on the body, while the body performs some movement s. So, we have two vector values – force and displacement. In this case, we can say that the force does some work. This work is determined by these vectors. In this case, the work is already a scalar value. It is measured in joules.

Let’s see how this value can be determined, or to be exact, what parameters it should depend on? So, as I’ve said, we have two vectors – force and displacement. First, the work should depend on the effort applied, i.e., the force modulus. It also depends on how far the body has moved, i.e., on the displacement module. Note that if our body has not moved anywhere, that is, it has made zero movement, we can say that zero work has been done (work is zero, there is no work). In this case, there must be a dependence on the angle between these vectors. In other words, the work should depend on the relative position of these vectors. Therefore, we take the following formula, the work is the product of the force modulus by the displacement modulus multiplied by the cosine of the angle between them. This expression is called the scalar product of these vectors.

It is clear that the more effort we make, the more work we do. At the same time, the greater the distance the body moves, the more work is done.

In general, the scalar product of these vectors a and b is understood as a similar expression, that is, the product of the lengths of these vectors by the cosine of the angle between them. Once again, please note that as a result of this operation for two given vectors, we got a number, that is, a scalar value.

Let’s consider the main properties of the entered operation.

First, let’s work with the sign of the scalar product. Let me remind you that each angle sets a certain cosine, and depending on the angle, the cosine has a certain sign, which determines the sign of the scalar product. So, let’s consider the option when the vectors are first co-directed, that is, consider the smallest value of the angle. For co-directional vectors, the angle is zero. In this case, if the cosine is 1, the scalar product is the product of the lengths of these vectors. If the angle is sharp, the scalar product will be greater than zero, that is, it will coincide with the cosine sign.

For orthogonal vectors, the cosine is zero, which means that the scalar product will turn to zero. For an obtuse angle, the scalar product is negative, and if the vectors are oppositely directed, their scalar product has the opposite value to the product of their lengths.

Now let’s consider more serious properties.

First, by changing the places of the multipliers, the scalar product will change. In other words, the scalar product operation is commutative. This property is an easy consequence of the definition.

Second, the number r can be taken out of the sign of the scalar product.

Next, we take two equal vectors, that is, we substitute a vector equal to a in the formula instead of b. The product of a vector by itself is called a scalar square. Since two equal vectors are co-directed, their scalar product, that is, the scalar square is equal to the product of their lengths. Thus, the scalar square of a vector is equal to the square of its length. In other words, the length of the vector is equal to the square root of the scalar square. We will frequently use this important property .

Finally, there is also a law that allows opening brackets. This property is called the distributivity of the multiplication operation with respect to the addition of vectors. That is, by multiplying (a + b)c, we get the sum of the products of ac+bc. From this property, we can formulate a consequence with a physical context. If several forces act on the body, the work of the resulting force is equal to the sum of the work of these forces. In particular, for two forces, we have this equality, similar to the fourth point of the above theorem.

Let’s consider the formula that allows us to calculate the scalar product with the given coordinates of these vectors. Let a rectangular coordinate system be defined on the plane, the vector a has the coordinates а₁, а₂ in it, and the vector b has the coordinates b₁, b₂. Let’s decompose the original vectors according to the orthonormal basis and expand the resulting expression using the last property, that is, using the distributive law. In this case, we get 4 terms, that is, we multiply each term of the first bracket by each term of the second bracket.

Now let’s remember that the basis is orthonormal, that is, the basis vectors are unit vectors, and at the same time, orthogonal. This means that the last two terms will be zero since the product of vector i and vector j will be zero, and the squares of these vectors will be equal to the squares of their lengths, which means they are equal to 1. We will get this simple form, which we will write as a theorem. So, if two vectors have the specified coordinates, their scalar product is equal to the sum of the products of the corresponding coordinates. We obtained this formula for a plane when the vector is given by a pair of coordinates. Similarly, it is possible to obtain the same fact for the three-dimensional space. The only difference is that instead of two coordinates, the vector will be set by a triple of coordinates.

Let’s consider the consequence. Two vectors in the coordinate system are given. We’ll use the same letters to indicate their coordinates. In this case, taking into account that the length of the vector is the square root of a scalar square, we get a formula for calculating the length of the vector in terms of its coordinates. Secondly, if we remember how the scalar product is defined and express the cosine of the angle between the vectors from this definition, we can easily get the formula shown on the slide. The numerator is the scalar product, written in terms of coordinates, and the denominator is the product of the lengths of these vectors. Thus, knowing the ordinates of the vectors, we can easily find the cosine of the angle between them, i.e., the angle itself. So, once again, both the theorem itself and its corollary can be transferred to three-dimensional space.

Let’s consider an example on calculating a scalar product. In this case, we will give an example for three-dimensional space. Let’s remeber that the concept of work is defined through the concept of a scalar product. We assume that a coordinate system is set in space, and the coordinates of the force vector are known in this system. In this case, the body moved to the specified vector s. Let’s calculate the work of this force. Since the vectors are set by coordinates, we just need to apply the specified formula. So, let’s remember. As I have already said, the work depends on the effort when moving the body, as well as on the angle. However, for the calculation, we will not use the definition, but the derived formula. We know the coordinates, multiply them and add them up. As a result, we get 8 joules.

Now let’s look at one interesting formula. Let’s assume that some coordinate system with an orthonormal basis i, j is again given on the plane. We take an arbitrary vector and consider two angles: one angle between the vector a and the Ox axis, and the second angle between the same vector and the Oy axis. The cosines of these angles are called direction cosines.

Let’s look at the figure. The first corner in the figure is marked with the letter α, the second corner – with the letter β. So, we have the following result: the cosine of the angle between the vector and the first axis is equal to the ratio of the first coordinate to the length of this vector, and the cosine of the angle between the vector a and the second axis is equal to the ratio of the second coordinate to the length of this vector. These formulas are the consequence of the previous properties. Note that the vector i and j in this basis have the coordinates (1, 0), (0, 1). Therefore, substituting the specified values in the formulas, we easily come to the following equalities. A similar formula holds for three-dimensional space.

Now let’s find the square and the sum of the equations. We get the following consequence: the denominators of equations are the same, so after reducing this sum to a common denominator, we get that the numerator and the denominator are equal numbers, which means that this expression is equal to 1. Thus, the sum of the squares of the direction cosines of an arbitrary vector is 1.The same statement holds for the space.

Finally, we consider the formula that allows us to express the square of both a triangle and a parallelogram in terms of a scalar product. Let two non-collinear vectors a and b be given. They define a certain triangle after we set these vectors aside from a single point. They also set a certain parallelogram, and the area of the parallelogram is twice the area of the triangle. Let’s remember that the area of a parallelogram is expressed in terms of the lengths of the two sides and the sine of the angle between them. This is a well-known school formula. In this case, the triangle has a similar formula, the only thing that appears is the coefficient ½. Let’s square this equality and, given that the area is a non-negative number, extract the square root. In the right part, we get the following expression. Now, if we write a square sine through a square cosine by well-known trigonometric identity and make some transformations, we obtain the second term square of the scalar product.

Note that if you put the square outside the common bracket, you get the product of the lengths of vectors a and b, which is multiplied by the cosine of the angle between them. In the first term, we have the product of squares of lengths, that is, the product of scalar squares of these vectors. So, the area of the parallelogram is equal to the square root of this difference: first we multiply the scalar squares of these vectors, and then we subtract the square of their scalar product. Thus, knowing the coordinates of the vectors, we can easily find the area of the parallelogram built on these vectors. For a triangle, we have a similar formula, only with a coefficient of ½ since the triangle is two times smaller than the parallelogram under consideration.