Lecture 2. Vector and mixed products of vectors
In this lecture, we will introduce another product of vectors, which is called a vector product, and introduce the concept of a mixed product.
In order to define this new product of two vectors, we need to distinguish the concept of the right basis from the left one.
Let's have a three-dimensional basis consisting of vectors a, b, and c. We call the basis right if three fingers of the right hand are in the direction of these vectors (see the video).
We can give a clearer definition: if we look from the end of vector a, the transition from vector b to vector c is counterclockwise.
If these 3 vectors correspond to the left hand, i.e. three fingers are in the direction of vectors a, b, c for the left hand (see the video), then the basis is called the left one.
Then, looking from the end of vector a, the transition from vector b to vector c is clockwise.
We have two orientations: right and left. One of these orientations is positive (this is usually done for the right basis). In future, we are going to use the term “positive orientation” for the right basis orientation.
Now we introduce the required concept. Let two non-collinear vectors a and b be given, their vector product is a vector that has the following properties.
First, this vector must be perpendicular to both vector a and vector b, i.e. it must be located along the line indicated on the slide (see the video).
Second, the length of this vector must be equal to the area of the parallelogram constructed on these vectors.
Since these two conditions set the required vector ambiguously, we add a third property.
We require that the orientation of the vectors a, b and their vector product be positive.
After that, the vector product becomes unambiguously defined.
These three properties define a new vector – the vector product of the original ones – unambiguously.
If the original vectors are collinear, then their vector product is considered equal to 0.
We have found the third vector to any two vectors a and b. That is why we call it a vector product.
Earlier, we studied the concept of a scalar product, when the result is a number.
This definition leads us to a number of consequences. First, point 2 causes the following two properties.
The area of a parallelogram is equal to the product of the length of its sides by the sin of the angle between them. Thus, the module of the vector product can be presented using the given formula (see the video).
Second, the area of a triangle set on vectors a and b is equal to half the area of the parallelogram, which means that the area of such a triangle is equal to half the module of the vector product.
Two vectors are collinear if and only if their vector product is 0 (see video).
If we change the order of the multipliers a and b, the vector product changes its sign. You can see how we do it here (see the video).
That is, by swapping a and b, we change the product orientation.
If we take an arbitrary number r, it can be extracted from the vector product according to the specified equations (see the video).
There is a distributive property that allows us to multiply the vector by the sum, or rather expand the brackets.
There is a property similar to the scalar product property, which allows us to expand the brackets.
That is, if vector b is multiplied by the sum of vectors a and b, then we get the sum of two products: the first vector by b and the second vector by b.
On the other hand, the same property holds for the case when we multiply the vector b.
Now we are going to get the formula that allows us to find the vector product in terms of coordinates.
We expand each of the vectors into basis vectors (see the video), i.e., we assume that the vectors i, j, and k form an orthonormal basis. Then the expansion coefficients are the coordinates of the original vectors, and we use the properties.
First, let's apply the last property that allows us to expand the brackets. To do this, we need to multiply each summand of the first bracket by each summand of the second, so we get 9 summands (see the video).
Now we simplify this large sum.
Since the product of two equal vectors is a zero vector, the number of summands will be re-zeroed.
Second, using another property, and changing the order of vectors, we get a vector product with the opposite sign.
After that, the equation can be converted to the given sum, that is, we give similar summands and get the sum of 3 summands.
At the same time, let's think about what the product of vectors i by j will be.
If we take two basis vectors in some order, their product is equal to the third vector.
That is, i by j is k, then k by i is equal to the vector j, and j by k is equal to the vector i.
We multiply in this order, and the minus sign does not appear anywhere. Substituting the specified equation in our expression, we get the desired decomposition of the vector through the basis vectors, while we have the k vector in the first summand and the j vector in the second vector, and the i vector in the third vector (see video).
Note that each difference can be represented as a second-order determinant, with a minus sign appearing in the second summand.
Check that if we write down the determinant, it gets the opposite value to the one in the second summand.
We have obtained a theorem that allows us to find the vector product in terms of coordinates, but the resulting formula is quite difficult to understand.
At the same time, it is easy to see that this long equation can be written in a very convenient form, namely as a third-order determinant, where the first line contains the basis vectors i, j, and k, and the second and third lines are the coordinates of the original vectors (see the video).
If we decompose this determinant by the first line, we get exactly the above sum.
In this case, the minus sign occurs due to the fact that when calculating the algebraic complement, the sum of its indices is odd.
It is useful to learn this formula, and if necessary you can decompose by the first line, and get the desired amount, the coefficients of which give the coordinates of the vector product (see the video).
Let's take an interesting example. Knowing the formula for the vector product allows us to get a very simple formula for the area of a parallelogram that is set on two vectors.
These vectors are set on a plane, so each vector has two coordinates (а1, а2) and (b1, b2).
The area of the parallelogram set on these vectors is equal to the modulus of the determinant made up of the specified coordinates.
What do we do?
Let us consider these vectors and the plane, and add a third vector to the basis vectors of the plane.
Thus, we get a three-dimensional coordinate system, and we add a vector that is orthogonal to the vectors i and j, that is, we get a rectangular coordinate system (see the video).
In this system, the vectors a and b are defined by a triple of numbers, and the third coordinate is equal to 0, then the area of the parallelogram is equal to the modulus of the vector product (by definition).
Let's apply the formula derived earlier by writing the zero coordinates in the third column (see the video).
After calculating this determinant, we get the following: the first two coefficients, that is, the first two coordinates are zeroed, and the third coordinate is equal to the required determinant (see the video). Thus, the length of this vector is equal to the modulus of the specified determinant (see the video).
Now let's take some examples, while trying to give examples with a physical plot.
Let's take this plot problem.
Let's assume that the ship's mast direction m and wind direction v are set in a certain rectangular coordinate system. You need to find the transverse direction of the sail in order to capture the wind in the best possible way.
In order to select the desired direction, note that it must be perpendicular to both the ship's mast and the wind direction, which means that this direction p must be parallel to their vector product.
Let's calculate the coordinates of this product. Due to the formula derived earlier, we need to find the determinant (see the video).
Let's decompose it by the first line.
Note that we don't have to remember the long formula, but we have to know the third-order determinant.
Next, we decompose and get a triple of coordinates that are equal to the numbers -2, -1 and 0 (see the video). This means that the desired direction has the specified coordinates.
Note that very often, among all vectors that have the same direction, we need to find one that has a unit length.
Thus, we can take this vector as the vector p (see the video).
Let's take another example. Let's assume we have a certain plane. A point is marked on this plane. Let's assume that this plane is illuminated by some source.
We have a vector – the direction of light – l; let it have a unit length.
We also consider a vector n that is perpendicular to this plane.
This vector is the normal vector.We are going to express in terms of vectors n and l the degree of illumination of this plane by the specified light source.
We fix these vectors n and l and define the illumination as the scalar product of the vectors n and l.
Let's consider a positive orientation basis on the plane, and let's make it a unit one (vectors e1, e2 have a unit length and are orthogonal).
In this case, the normal vector is equal to their vector product. Then the illumination coefficient s can be written as such an expression (see the video).
That is, the vector n is the vector product of the basis vectors, which is multiplied scalar by the vector l.
Note that if we have the direction of light parallel to this plane, then the degree of illumination is minimal, and the scalar product is 0.
To get maximum illumination, we must make the direction of light perpendicular to this plane, that is, parallel to the normal vector.
In this case, we have the highest illumination value.
While considering this example, we have obtained the expression, in which there are two types of product, both vector and scalar.
This example allows us to introduce another product, which is called a mixed product (because we use both a vector and a scalar product to calculate it).
The mixed product of three vectors a, b, and c is a number equal to the scalar product of two vectors: the first vector is the vector product of a by b, and the second vector is the vector c.
Note that the result is a number, because the second action is a scalar product.
Let's assume that there are three vectors a, b, and c in space. Let's consider what values their mixed product can have.
First, by multiplying the vectors a by b vector-wise, we get a vector perpendicular to each of them, i.e. a vector perpendicular to the plane, which is set by the vectors a and b.
If the vectors a, b, and c have a positive orientation, then the angle between c and the vector product of a by b, which is used to calculate the scalar product, is sharp.
This means that the mixed product is greater than 0.
If the specified angle is right, that is, the vectors are in the same plane, then the scalar product, and hence the mixed product of the original vectors, are 0.
If the vectors a, b and c have a negative orientation, that is, the specified angle is obtuse, the mixed product is less than 0.
These results can be put into this theorem (see the video).
Now let's consider the formula that connects the introduced concept of a mixed product and the volume of a parallelepiped. If we are given three noncoplanar vectors a, b, and c, then we can set a parallelepiped with the sides a, b, and c.
Its volume is equal to the modulus of the mixed product of these vectors (see the video).
Let's make an illustration.
Let's take vectors, set a parallelepiped on them, and consider one face as the base. In this case, the area of the parallelepiped is found by the formula: the area of the base is multiplied by the height.
The area of a parallelogram is the modulus of the vector product of the vectors a and b.
Let us consider this vector product, draw it in the figure and express the height h in terms of the angle between the product of a by b and the vector c. The figure shows (see the video) how this angle is obtained. In this case, the volume can be calculated using the specified formula (see the video). If we substitute the resulting expression for h, then the vector product of a by b is multiplied by the vector scalarly.
Note that the lengths of the specified vectors are multiplied, and the result is multiplied by the cosine of the corresponding angle. Thus, we obtain the required equation.
Second, if we consider a tetrahedron set on these vectors, it is easy to see that its volume is six times less than the volume of the corresponding parallelepiped. This means that the volume of the tetrahedron is equal to 1/6 of the modulus of the mixed product.
Now we are going to get the formula for calculating the mixed product using the coordinates of the original vectors. Let's assume that there is a rectangular coordinate system in which three vectors are defined by their coordinates. We get the formula written on the slide (see the video), which is easy to learn. To calculate the mixed product of vectors, you need to make a determinant from their coordinates and find it using any way method.
Let's see how this formula works. We set the mixed product by definition, then substitute the basis expansion instead of the first factor.
The vector c is also decomposed by the basis, given that it has the coordinates we know. Next, we multiply the 2 obtained brackets using the properties of the scalar product, paying attention that the brackets are multiplied scalarly (see the video). The squares of the basis vectors are 1, and the products of different basis vectors give 0 scalarly, because they are orthogonal. Thus, we can easily simplify this expression. Therefore, instead of nine summands, we have just three (see the video).
Those that correspond to the products of the corresponding vectors from the basis, but these vectors are reduced due to the above property. This sum is the determinant (see the video). Check it yourself: if we decompose it by the first line, we will get exactly this decomposition.
Here we also have a different order of lines than in the above formula. Note that if you swap the lines, the determinant changes the sign. Let's try to reduce the resulting determinant to the desired one by this transformation. We swap the first two lines, then the line a takes the first place, the line c becomes the second, and the sign changes. Now let me change two lines again: the second and third ones. Then c takes the last place, and b takes the second one, the sign changes again and becomes the original one. We have changed the sign twice, so it doesn’t change, and the lines are arranged in the correct order: a, b, and then c.
On the one hand, we have proved the formula, and on the other hand, we have obtained the properties according to which we understand when we change the sign of the mixed product. If in the triple a, b, c we swap just two vectors, for example, a and b or b and c, then we change the sign, and if we have a cyclic permutation, like here (see the video) instead of c, a, b, we take a, b, c, then the sign does not change.
Let's conclude by noting the main properties. The first property: if we fix the order of writing the vectors a, b, and c, then nothing changes if we take these vectors in any order along the specified cycle abc, or bca, or cab. If we swap two vectors, such as a and b, the sign changes. The same applies to b and c, and to a and c.
Further, as for all other products, the scalar r can be taken out of the sign of the mixed product from any multiplier.
The fourth point is that there is a distributive property, that is, when considering the mixed product, if the sum of vectors is in some multiplier, we can open this product and present it as the sum of the mixed products of the resulting vectors. In this case, the sum can be either in the first multiplier, or in the second, or in the third – all equations are correct. The lecture is over.