Practical lesson 2. vector and mixed product of vectors to solve them
At this lesson we’ll consider a few problems where it is possible to apply the concept of vector and mixed product of vectors to solve them.
Let's start with the following problem. Calculate the area of parallelogram formed on the two vectors: a + 3b and 3a + b, where a and b are unit vectors, with the angle 30 degrees.
Remember that the area parallelogram formed by vectors is equal to the module of the vector product of these vectors. Consider the vector product of vectors a + 3b and 3a + b, and try to simplify this expression by using the problem conditions, at the same time we will use the properties the vector product considered at the lectures.
We open the brackets, according to the distribution law we are entitled to open the brackets in a usual manner, and thus we get: (a + 3b) x (3a + b) = ax 3a + ax b + 3b x 3a + 3b xb (sign x denotes some vector product ).
Further, by the properties, the numerical coefficients can be taken off the sign of the vector product. For example, in the first summand (a x 3a), we get: 3 (a x a). As you know, similar vectors give a zero vector in the product, since they are collinear. In the last summand 3b x b there is a zero vector. By the relevant property, the third summand 3b x 3a = -9ba. Thus, this sum is equal to -8a x b.
So, we are to find the parallelogram area, we have a module of the vector product, which we convert to the form: -8 a x b. We get: S = | -8 a x b |. Numerical coefficient can be taken off with a plus sign: 8 | a x b |. We must find a module of the vector product | a x b |. By definition, this is the product of the given vectors lengths by the sine of the angle between them (namely between vectors a and b, by the condition the angle is 30 °): 8 ∙ | a | ∙ | b | ∙ sin30.
Thus, this expression is easy to calculate. Vectors are unit ones, we get 8 ∙ sin30 = 8 ∙ 1/2 = 4. Thus, the parallelogram area built on the given vector is equal to 4. The answer: S = 4. That's the way we used the concept of vector product, although initially it was not mentioned in the problem.
Consider the following problem. Suppose that in space there is a defined rectangular coordinate system. It is required to find unknown vector h, while it is known that it is orthogonal to the two given vectors a (2; 3; -2), b (2, 2, 0) and its length is equal to two: | h | = 2.
Again remember the concept of the vector product of vectors. If you find the vector product of vectors a and b, then we get a vector orthogonal to the given ones, which means that it is collinear to vectors h. We know that if the vectors are collinear, their coordinates are proportional. Thus, I propose firstly to calculate the vector product of a x b, using the coordinate formula, and then write the condition imposed on the coordinates of an unknown vector.
So, we remember that this expression can be written as a determinant: in the first line there are basis vectors (i, j, k), and then the coordinates of the given ones (in the order in which we find the product) (2, 3, -2) (2, 2, 0). Next, to calculate the required vector we decompose by the first line, thereby we get a linear combination of vectors i, j and k: | (3, -2) (2, 0) | i- | (2, -2) (2, 0 ) | j + | (2, 3) (2, 2) | k. Calculate the coordinates: 4i-4j-2k. Thus, the coordinates of product a x b is equal to (4, -4, -2). Let us denote the coordinates of the unknown vector h through (x; y; z) and find these numbers.
As we said earlier, vector h is collinear with the vector product a x b: h (x; y; z) || (a x b). Hence, its coordinates are proportional to the obtained coordinates, it is possible to introduce some coefficients, and write down the following system of equations: x = 4t, y = -4t, z = -2t, where t is some number, some parameter. Now we use the fact that we know the length of vector h, | h | = 2.
Hence, the squared length is equal to four: | h |2 = 4. Squared length of the vector is the sum of the squares of its coordinates: | h |2 = x2 + y2 + z2. Substitute the expression through t instead of x, y, z, t and simplify: | h |2 = (4t) 2 + (- 4t) 2 + (- 2t) 2 = 36t2. By conditions, the squared length is equal to four: 36t2 = 4, t2 = 1/9. It turns out that t = 1/3 or t = -1 / 3, then vector h is defined uniquely. So any of the vectors can be taken as vector h. Substituting t = 1/3, we obtain h1 (4/3; -4/3; -2/3), this is the first answer. And the second option is witht t = -1 / 3: h2 (-4/3; 4/3; 2/3), this is a vector opposite to vector h1.
We received two options. Note that if the condition stated, for example, that orientation of vectors a, b and h is positive, then we would have chosen only one answer. In this case, vector h would be the same direction as the vector product. If a, b and h orientation was positive, we would have chosen only the first option, which corresponds to the number t with a plus sign. If such conditions are not given, then the answer is both vectors. So, the problem is solved.
Now let’s consider the problem, which uses the concept of a mixed product. It is known that the mixed product of three vectors is equal to 3 (abc = 3), it is required to find the scalar triple product of vectors (a + b) (b + c) (c + a) and (a - b) (b - c) (c - a ).
Here it is necessary to use the properties of the mixed product. Remember what properties it has. Consider the (a + b) (b + c) (c + a). If we denote the two second sums asf p and q (p = (b + c), q = (a + a)), then we can open the brackets, that is, to introduce this product into the sum and we get:apq +bpq. Thus, we have the following expression: a [(b + c) (c + a)] + b [(b + c) (c + a)] (we work with vectors, but in the result there is a number). Now we apply a similar procedure within the brackets.
Remove the brackets (this is just an indication that this is one and the same vector), for example, (b + c) is denoted by p, we get the following mixed product: ap (c + a) + bp (c + a). Again open the brackets and get: a (b + c) c + a (b + c)a + b (b + c)c + b (b + c)a. We got a sufficiently long sum. Now, the final step we will open the last brackets. So, let's see what we get, after that. Decompose the first summand, and then understand how to simplify the following summands. So, you get: a (b + c) c = abc + acc.
Note that if there are two equal vector in the mixed product, it is equal to zero, as the number, because if we decompose the mixed product by he determinant, there will be two identical lines, then it turns zero. Similarly, if we reveal all the remaining brackets, the summands of the same type and will turn zero. For example, a (b + c) a= 0 and, it does not matter how we open, we will have two equal vectors a and a, then it is 0. We open the brackets b (b + c)c to get bbc + bcc, again similar vectors, again 0.
Finally, decompose the last product: b (b + c) a = bba + bca = 0a+bca. Thus, we see that from our large sum there are only two non-zero summands abc and bca. Another feature - perform a cyclic change: move vector a in the second summand to the first place, the order of the remaining vectors stays unchanged, we get abc. It is known that a cyclic change does not change the mixed product. There are two equal summands in this expression: 2abc. Please note, we know what it is equal to, by the condition, it is 3. So, we get 6. Thus, the desired mixed product is found, it is 6.
Let's move on to the next item. Of course, you can do all the same, for a long time, but you can complete the procedure. The only thing is that because of the minus sign somewhere we amy get negative coefficients. We will act in a different way, we note that in this mixed product the third vector (c - a) is the sum of the first two: (a - b) + (b - c) = (c - a). This suggests that these vectors are linearly dependent, since one of them is a linear combination of the other two.
The mixed product of linearly dependent vectors are known to be 0. Remember that the linearly dependent triple of vectors means that they are coplanar. Thus, the first answer is 6, the second answer can be written down as 0 without further changes.
Finally, another task, where it is appropriate to apply the learned concept. In a triangular pyramid with corners A (1; 1; 1), B (2, 0, 2), C (2, 2, 2), D (3; 4; 3), it is required to calculate the height lowered from corner D. Note that the there is no either a vector or a mixed product here, but let's try to apply the necessary formulas.
What do you need to calculate the height length? This, of course, can be done in different ways, one option is to use the standard formula for the volume where the height is present.
There is some pyramid, we need the height from corner D (denote by h), ABC is the base (see video.). As it is known, V is the volume of this pyramid equal to 1 / 3hS, where h is height, S is base area (the area of triangle ABC). From this formula we can be expressh: h = (3V) / S. So we need to find the volume and the base area of the pyramid.
Here again, there is the concept of vector product. The triangle area is 1/2 length of the vector product. Earlier we calculated the parallelogram area l, there is no coefficient 1/2, and it appears for the triangle. Volume is 1/6 module of the mixed product. Also, we will need vectors. Triangle ABC can be built on vectors AC and AB, we introduce them into consideration. Let AB be vector b, with coordinates (-1; -1; 1), and AC is denoted by vector c (1; 1; 1). Our pyramid or tetrahedron, is built on three vectors. There appears vector AD = d (2; 3; 4).
It remains only to apply coordinate formulas for finding the area and volume. The area is half the length of the vector product b x c, triangle ABC is built on them. Again we find this vector. At the same time I should mention that we could also use the scalar product for the area of a triangle, we talked about it at one of the previous sessions
So, to test you can apply a different formula to find the area d through the scalar product. So, we are not going to decompose in detail: b x c = | (i, j, k), (1, -1, 1), (1, 1, 1) | = -2i-0j + 2k. Hence, the area of triangle ABC is half the length of the vector product (the length is the square root of the sum of the squared coordinates): S = 1/2 (4 + 0 + 4)1/2 = 1/2 (8) 1/2 = 21/2 = √2.
It remains to calculate the volume, that is to find one sixth module of the mixed product: V = 1/6 | bcd | (order is not important, as there is a module). So on the one hand we can apply the formula for calculating the mixed product, on the other hand, by definition, bcd is (b x c) d (we already know b x c and may use it, d is multiplied by the bracket scalarly).
Thus, there are a few calculation options here again. Use the second fact, at the same time remember the formula for the scalar product (multiply their corresponding coordinates and add them up): V = 1/6 | bcd | = 1/6 | (-4-0-8) | = 12/6 = 2. Thus, the volume of the given pyramid is 2.
It remains to remember the derived formula and find the required height. The length of height h is equal to the tripled volume divided by the area: h = 6 / √2 = 3 ∙ √2. Thus, the length of the height built from corner D, is 3 ∙ √2. The problem is solved.