Математика. Материалы курса

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At this lesson, we’ll consider a number of problems using scalar product of vectors.

So, the first problem.

There are two vectors a and b, we know their length and the angle between these vectors, it is equal to 90º.

We are to find the scalar product of the two following vectors: 3a + 2b and 4b-a.

To do this we’ll use the scalar product properties and, possibly, will apply the basic definition.

So, to convert the given expression, you must use the following properties.

Multiply each summand in the brackets by each of the summands in the other bracket, we’ll get the following expression (see the video.): 3a is multiplied by 4b, wherein the numerical coefficients of this product can be taken out, you get 3 by 4, a by b, then 3a multiplied by -a, it gives minus 3a squared, that is, a is multiplied by a.

They say this is a scalar vector square.

Further 2b multiply by 4b, 2 by 4 is 8, 8b squared and, finally, 2b by –a, -2ba.

Let us understand what properties we are to use now.

First, ab and ba is one and the same thing, which obviously follows from the definition of the scalar product.

Secondly, the scalar vector square is equal to its length square.

Therefore, as we know the lengths of the vectors, we can easily find scalar squares.

So let's, first of all, group the summands containing product ab, and calculate 12ab - 2ab = 10ab and, given that the scalar vector square is equal to its length square, we write down -3*7², that is 49, plus 8 multiplied by the square vector b length, i.e. 5² by 25.

It remains to find the scalar product ab.

Remember the definition.

To find the scalar product, you are to multiply the these vectors length which we know, by the cosine of the angle between them.

Since the angle of 90º, the cosine of this angle is equal to 0.

By the way, it was possible to take advantage of the following property.

Since the vectors are orthogonal, it means that the scalar product is zero, it means that the whole first summand is zero.

Thus, it remains to calculate the numerical expression 3 by 49 is 147, and with a minus sign, minus 147, 8 by 25, 200, plus200.

The result is 53.

This is the answer, that is the scalar product of these vectors.

Now there is another question for this problem.

Suppose we want to find what the vector is equal to, or more precisely, what the length of vector 4b-a is equal to.

So, try to answer this question yourself.

Let's see what the error may hide here.

Of course, it is incorrectly to assume that the length of this vector is the difference of lengths 4b and a.

We cannot find length difference in such a simple way, as well as length sum.

To answer the question, you can again use the fact that the vector square is its length square, so instead of finding the vector length, let's find its square.

The length square is the same as the vector square.

That is, raise 4b-a to the second power, and we transform this expression in this way.

The difference square formula is true, because all the basic properties that are used to derive this formula is valid here.

So, we can write 16b² - 8ba + a².

So, we have already discussed, that the scalar product ab = 0, because the vectors are orthogonal.

So, the average summand is zero.

We found the scalar square of vector b, it is 25, and the scalar square of vector a = 49.

So, 25 * 16 + 49 = 449.

So, the squared difference of these vectors is equal to this number here, thus we found the vector length square.

So, to find the length itself, it is necessary to find the square root of the resulting number.

The length of vector 4b-a is √449. The problem is solved.

The next problem.

Note that in formulating the problem there is no concept of the scalar product, however, to solve this problem it is necessary to apply this concept.

We are given 4 points, it is necessary to find the angle between the straight lines which are set by the given points.

Thus, we have some points M and N, a line goes through them.

Further, there are points P and Q, through which a straight line also goes.

I remind that the angle between the lines is the least of angles that are formed at the intersection of the given lines.

I note that since we are given points in space, lines may intersect, so remind that the angle between the skew lines is understood as the angle between the intersecting lines, each of which is parallel to the given line.

That is, in the parallel transfer the angle does not change.

Therefore, we can always assume that lines intersect, when we define the angle between them.

So, in order to find the required angle we will need  so-called direction vectors, which are easily found through the coordinates of points.

Thus, to find coordinates of vector MN, we use the rule, we subtract the coordinates of point M from the coordinates of point N.

We get -1, then 0 and 1. This is a so-called direction vector  of line MN.

Further, line PQ is set by vector PQ.

We find its coordinates (1, 2, 3).

And now to find the angle between the lines, at the beginning find the angle between the vectors.

If we consider direction vectors for the given lines and put them from one point, then, we consider the fact that these vectors are parallel to the lines (that's in the picture angle φ is equal to angle α).

However, pay attention that the direction vector of the line line can be taken in the opposite direction for this picture.

And then, if angle φ is acute, its addition can be obtuse.

However, we know how to distinguish between an acute angle and an obtuse one, it is necessary to know cosine, the cosine sign coincides with the sign of the scalar product, so let's find the scalar product of the given vectors and understand what angle is between them, acute or obtuse.

Let us remember how to find the scalar product through the coordinates of the vectors.

To do this multiply the corresponding coordinates and summarize the obtained product.

We get -1 + 0-3, that is -4.

So, since the scalar product is negative, it means that we have chosen vectors in such a way that the angle between them turned out to be blunt.

That is, there is such a picture.

If we build the vectors from one point, the angle formed by these vectors is obtuse, so after we find the angle between vectors, for the angle between the lines, we should take this one, it is the addition of the angle to 180º.

This should be borne in mind.

So first, let us find angle φ.

We already know the scalar product, it is equal to -4.

Thus we have denoted , vector MN as a, vector PQ as b.

Given that the scalar product of signs is decomposed by definition as the product of the lengths by the cosine of the angle between them, we can easily express the cosine of angle φ.

To do this, the scalar product, in this case -4, is divided by the product of the lengths of vectors a and b.

To find their lengths, remember what coordinates these vectors are set by.

Thus, a (-1, 0, 1), and vector b (1, 2, 3).

Remembe the vector length formula.

By the way, this formula can give an explicit form of the formula for the distance between two points.

It was possible to apply this formula, but for those who have forgotten it, I show the general way to solve if you forgot the formula for the distance between points.

Thus, the length of segment MN is the length of vector MN, i.e. vector a, this is the square root of the sum of squares of this vector coordinates, i.e. √2.

The length of vector b is the square root of the sum of squares of its coordinates.

Calculate, √ (1 + 4 + 9) = √14.

Pay attention, we square, so minus disappears, it is √14.

Substitute into the formula, find cosine of the angle between the vectors, and ger -4 / (√2√2√7).

To slightly simplify the resulting number I multiply two identical roots and get 2, reduces by four, is -2√7.

Thus, if it is required to find the angle between the vectors, we would write down that φ = arccos (-2√7).

I remind that the angle is obtuse.

Since we need an acute angle, since we take the least of all the corners, we have to remove a minus sign in the cosine, in other words, the cosine of the angle between the lines is the cosine of the angle between the vectors.

Accordingly, we find for the problem that the cosine of angle α is 2√7.

The value is not the table one, so the answer is written down using the arccosine function.

Angle α = arccos 2√7, that is the answer to the question of the problem.

So, let's solve another problem.

Assume that we are given a triangle, we know the coordinates of radius vectors of its appexes.

In other words, we know the coordinates of points A, B and C.

Let's prove that the triangle is equilateral, and find its area.

I will make a schematic drawing, this is triangle ABC.

Again, we consider points in a three-dimensional space.

But as we know, any three points define a plane, so you always have a flat shape, triangle ABC.

Given that the coordinates of radius vectors of points A and B are set, we immediately obtain the coordinates of the given points A (1, 2, 3), B (3, 2, 1), G (1, 4, 1).

To check whether this is an equilateral triangle, it is necessary to ensure that all its sides are equal.

This requires to find the lengths of three vectors corresponding to the triangle sides, and make sure that these vectors are of equal length.

And again either we apply the formula of the distance between points, or if we have forgotten it, we find the coordinates of vectors AB, BC and AC, and calculate the lengths of these vectors.

Let's do this procedure.

Vector AB (2, 0, -2), this is the side lying against angle C (see. Video), it can be marked by a small letter c.

Further, let us denote side BC by a small letter a, and the side lying against B by letter b.

Vector BC.

Please note, in order to find the lengths, we do not care about the direction of the vectors.

I can take the CB. BC (-2, 2, 0).

Finally, vector AC (0, 2, -2).

So, here are three vectors.

The lengths of these vectors give us the lengths of the triangle sides.

Vectors of course are different.

However, if we calculate their lengths, then they will obviously be the same.

The fact is that under the root, we will have three squared numbers, two of the modules are the same, and everywhere there is one 0.

Thus, the length of side a will be equal to √8, i.e. 8 = 4 + 4 under the root.

Side b will have the same length

Check, 4 plus 4, it will be 8 under the root.

The same applies to side c.

If you find the length of vector AB, it will be equal to √8 too.

Thus, all three segments a, b and c has the same length.

This means that this is an equilateral triangle.

To answer the second question, we consider the formula that allows you to find the triangle area, which is built on two vectors.

Let's take a look at our picture, we have three vectors a, b and c, and vector b is AC.

So, vectors b and c are built from one point, and they define triangle ABC.

Accordingly, the triangle area can be found according to the following formula: 1/2 multiplied by the square root of the following equation (see the video.).

Multiply the scalar squares of vectors b, and c, I'll use small letters, but writing a line on the top to point out that we consider vectors.

So, multiply their squares and subtract the square of the scalar product of these vectors.

Note, that on two vectors built from one point, it is also possible to construct a parallelogram, its area will be twice larger than the triangle area, so if we are looking for the area of this parallelogram coefficient 1/2 disappears.

Since we have a triangle, we write down this factor.

So, to find squares of vectors b and c, we must know their length.

We know that all the sides are equal to the same number.

To find the scalar product bc, you can take two ways, either use the formula of the coordinates, or the definition.

I propose to use the coordinate formula, that is, consider the sum of the products of the corresponding coordinates.

So, bc = 4.

We get 4.

Since the vector length is √8, then the length square is equal to 8.

So, square b is the same as the square of vector c, and it is 8.

Substituting into the formula gives 2√3 (cm. Video).

Pay attention that for a scalene triangle there is a separate good formula, which also could be used, i.e. this formula is general, it is valid for any triangle, which is built on two non-collinear vectors.

However, perhaps someone remembers from school mathematics, that if we are given an equilateral triangle with the side, for example, equal to b, then its area can be found as follows: b²√3 / 4.

So, this formula is exclusively for the equilateral triangle.

Let's check.

We obtain a true equality (see. Video).

Finally, one more problem.

It is necessary to find the coordinates of vector a, it is unknown, if it is collinear to the known vector b (3, 2, 1), and wherein the scalar product of these vectors is equal to 3.

So, let's write the problem statement, the scalar product ab = three, vector b (3, 2, 1), and the vector a || b.

It is required to find the coordinates of vector a.

Let us remember the characteristic property of the collinear vectors.

In collinear vectors coordinates are proportional, then, if we denote unknown coordinates of vector a by x, y, z, we can argue that the three numbers x, y, z is proportional to the triple 3, 2, 1.

This means that each coordinate of vector a is equal to the corresponding coordinate of vector b, multiplied by some number k, x = 3k, y = 2k and finally, z = k.

Since vector b is non-zero, then vector a and is also non-zero.

Use the condition that the scalar product is equal to 3, wherein remember the coordinate formula.

Multiply and add up the corresponding coordinates.

We get, 3x + 2y + z = 3.

We have an equation with three unknowns, however, remember that these three unknowns are expressed by one parameter, we substitute and get the equation for the unknown number k.

From this equation it is easy to find k.

So (see. Video) 14k = 3.

Hence, the proportionality factor k = 3/14.

So, knowing k, we can easily determine the coordinates of the unknown vector, 3 by 3/14 is 9/14, 2 by 3/14 is 6/14, or we can reduce by 2 and get 3/7 and finally, the third coordinate 3/14 .

Thus, the desired vector is found, it has the given coordinates.

That's all.