Математика. Материалы курса

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Today in class we’ll consider the solution of some problems in coordinates.

Let's start with a simple task.

Assume that in a coordinate system some points are set.

You want to find the coordinates of the sum of two vectors: AB + CD.

To solve this problem, we must remember that when adding vectors, their coordinates are added.

In this case, in order to find the coordinates of vector AB, if the coordinates of its endpoints are known, it is necessary to subtract the coordinates of its beginning from the coordinates of the end of the vector.

Thus, vector AB has the following coordinates.

We will denote the vector, writing over the segment either the arrow, or just a line sign.

Thus, the coordinates of vector AB (-1-0, -2-2, 1-1) AB = (-1, -4, 0).

Similarly, we find vector CD: CD (1-3, 0 - (- 4), 3-5) = CD (0, 4, -2).

Now the problem becomes very simple, we summarize termwise corresponding coordinates and obtain the coordinates of the sum AB + CD.

Thus, (-3, 0, -2).

Thus, the problem is solved, and we did not have to represent the vector itself.

The answer is (-3, 0, -2).

The next problem.

In this task, we are given a rectangular coordinate system as follows.

We consider a cuboid.

Consider one of its corners, point O, and direct the basis vectors i, j along the edges, and k as indicated in the figure.

We obtain a rectangular coordinate system in which  each vector point gets its coordinates.

Perform some constructions, note point K - the middle of edge С1D1 and point M – the middle of segment AK and we find the coordinates of the desired vectors and the coordinates of point M.

So, note that we are given 4 vectors, they are radius vectors of points K, A, D and D1.

Let us first consider vector OA.

Its coordinates are easy to find, if we note that the vector is the same direction as the basis vector i, thus it can be expressed as 5i.

At the same time, to find the coordinates, we need coefficients for each basis vector.

We write down that we have a zero coefficient for the vector j, the same for vector k.

Thus, vector OA has coordinates (5, 0, 0), and point A will have the same coordinates.

I remind you that point coordinates coincide with coordinates of its radius vector.

So now we look at some other corners, for example, corner D.

To find vector OD, more precisely, to find its coordinates, again decompose this vector into basis ones.

Given that it lies in the plane of OADC, it will be expressed in terms of vectors i, k.

Vector j is with zero coefficient.

Thus, according to the parallelogram rule vector OD (I can represent it here), is the sum of vectors OA plus OS.

we already know OA=5i, OS is 3k.

We get, 5i, the zero coefficient before j, and 3 before k, 3k.

Thus, vector OD, like point D has coordinates (5, 0, 3).

So, now we have not found vector OK yet.

Let's get back to vector OK.

In order to find its coordinates, in fact, you may first find the coordinates of point K.

It is the middle of segment С1D1.

If we know the coordinates С1D1, it is easy to find the coordinate of point K.

Therefore, let's find the coordinates C1.

Similarly, vector OC1 is the sum of OB + OC, respectively, ОС1 looks like 4j + 3k.

So, the vector has the coordinates (0, 4, 3).

Hence point C1 has the coordinates (0, 4, 3).

Thus, we find the coordinates of point D1 and coordinates of vector OD1.

You can proceed as follows.

We can remember that the coordinates of point D1 coincide with the projections of vector OD1 on the coordinate axes. 

And since we are given a cuboid, we can say that points A, B and C are projections of vector OD1 on our axis.

Accordingly, the coordinates are (5, 4, 3), because the vector OA length equals 5, OB = 4, OC = 3, these vectors are co-directional with the basis ones.

So, point  D1 has coordinates (5, 4, 3).

To find the coordinates of point D1, and therefore, the coordinates of OD1, vector OD1 can be expressed in terms of basis vectors i, j, k.

Thus, we know the coordinates of the points, let us write down that vector OD1 has coordinates (5, 4, 3).

Now it is easy to find the coordinates of point K.

According to the formula of the midpoint we add up the corresponding coordinates and divide by 2, 0 plus 5 gives two and a half, dividing 5 in, 4 + 4 half = 4, 3 + 3 half 3.

Accordingly, vector OK has the same coordinates, the radius vector of point K.

Since M is the midpoint of AK, then to find its coordinates, you can apply the same rule.

Thus, we calculate the abscissa of point M, this is half the sum of numbers 2,5 and 5, it is equal to 3.75.

Further, the ordinate of point M is equal to 2 and z applicate of this same point equals number 1.5.

Thus, the coordinates of the point are found.

We may write M (3.5, 2, 1.15).

So, the problem is solved, we found all the required coordinates of the vectors and the point.

Let us solve the following problem.

Again in some coordinate system vectors are set up.

It is required to deduce whether are coplanar or not.

Vectors are set in space, because they are determined by three numbers.

Let me remind you that vectors coplanarity means that they are parallel to the same plane.

Let us remember that the three vectors a, b and c are coplanar if and only if they are linearly dependent.

This, in turn, is equivalent to a matrix composed of their coordinates is degenerate.

The matrix can be formed by arranging coordinates either in columns or in lines.

In order to check whether the matrix is degenerate, it is enough to calculate its determinant.

Let's do it.

Arrange the coordinates of the vectors in a line.

We calculate the determinant.

Thereby we give the answer.

Remember alongside, how to calculate determinants.

One variant of calculation is based on elementary transformations.

Let's subtract 3 first lines from the second, and subtract 1 from 3.

I remind you that this transformation does not change the determinant sign, therefore, the determinant does not change.

The first line is 1 8 3, the second 0 -22 -14, and finally convert the third line 0 -11 -7.

Perhaps someone has already noticed that the last two lines are proportional.

If you did not notice that at once, you can continue reducing the matrix to echelon form and, for example, to divide the second line by 2 or subtract the first two lines from the second.

After this conversion, if you subtract two third lines from the second, we get a zero line.

This means that the last two lines are proportional.

Thus, 0 0 0.

Once a null line appeared in the determinant, then it is equal to zero.

I repeat, if someone noticed immediately that lines are proportional, you can stop, indicating that the given lines are proportional, then the matrix is degenerate, hence, the vectors are linearly dependent, i.e. coplanar.

The answer is  vectors are coplanar.

The next problem.

Consider triangle OAB and mark point M on side AB.

And let us assume that we know the ratio of the lengths of segments AM to MB, let it be some number λ.

We express vector OM in terms of vectors OA and OB.

Vector OA is denoted by letter a, vector OM by letter b.

So, I propose to take the next step.

I propose to consider vector OM as a sum of vectors OA and AM according to the addition rule (the rule of triangles) used for vector addition.

Vector OM is the sum of vectors OA + AM.

Thus, vector OA is set, vectors a and b can be viewed as basis ones and we are faced with such a problem - we need to find the coordinates of vector OM in basis a, b.

Thus, vector OA is given, and it is vector a.

It is necessary to deal with vector AM.

Remember that we know the ratio of segments AM to MB.

We can write down the following ratio: the length of segment AM is equal to λMV.

These vectors AM and MB are co-directional, and therefore vector AM can be represented as number λMV.

Now we replace summand AM through this expression, and present MB as the difference OB and OM.

So, , let's write MB is the difference of vectors OB and OM.

If someone forgot this rule of subtracting vectors, I remind that it can be derived from the addition formula.

Vector OB is equal to OM plus MB.

We transfer OM to another part, and get the required equality.

So, here is vector MB.

Let's make a substitution here.

We obtain a + λ (b-OM).

Remember, from what we started?

We moved from vector OM, thus having this equality, let's open the brackets and rewrite the equation in another form.

OM = a + λb.

So now our task is to express the vector OM from this equation.

To do this, I first open the brackets on the right side, a+ λb-λOM.

We transfer the expression containing vector OM to the right side and take it out the bracket.

Now, in order to express vector OM, we move the expression containing this vector from the right side to the left side, and take it out the bracket.

We get (1 + λ) OM, there are two summands in the right side: a + λb.

It is easy to express vector OM from this equation.

So expressing vector OM, we get the following formula: (a + λb) / (1 + λ).

This formula can be used, it looks fairly nice.

Thus, vector OM is expressed by this equation (see. Video).

This equation can be written a little differently, highlighting clearly the coefficients before vectors a and b.

Let's write down.

Thus, the coefficient before a is 1 / (1 + λ), the coefficient of vector b is equal to λ / (1 + λ).

Thereby, it can be argued that in base a, b vector OM has coordinates (see. Video).

Note, in particular, if point M is the midpoint of AB, parameter λ is equal to 1, because AM and MB are equal, which means that from this formula, you can get a particular case, the formula of the segment midpoint which we already know.

Vector OM, where M is the midpoint, is equal to 1/2 + 1/2 b.

The problem is solved.