Difference of vectors coming from one point. Addition and subtraction of vectors

Standard definition: “A vector is a directed segment.” This is usually the extent of a graduate’s knowledge about vectors. Who needs any “directional segments”?

But really, what are vectors and what are they for?
Weather forecast. “Wind northwest, speed 18 meters per second.” Agree, both the direction of the wind (where it blows from) and the module (that is, the absolute value) of its speed matter.

Quantities that have no direction are called scalar. Mass, work, electric charge not directed anywhere. They are characterized only numerical value- “how many kilograms” or “how many joules”.

Physical quantities that have not only an absolute value, but also a direction are called vector quantities.

Speed, force, acceleration - vectors. For them, “how much” is important and “where” is important. For example, acceleration due to gravity directed towards the surface of the Earth, and its value is 9.8 m/s 2. Impulse, tension electric field, induction magnetic field- also vector quantities.

Do you remember that physical quantities denoted by letters, Latin or Greek. The arrow above the letter indicates that the quantity is vector:

Here's another example.
A car moves from A to B. The end result is its movement from point A to point B, that is, movement by a vector.

Now it’s clear why a vector is a directed segment. Please note that the end of the vector is where the arrow is. Vector length is called the length of this segment. Indicated by: or

Until now, we have worked with scalar quantities, according to the rules of arithmetic and elementary algebra. Vectors are a new concept. This is another class of mathematical objects. They have their own rules.

Once upon a time we didn’t even know anything about numbers. My acquaintance with them began in elementary school. It turned out that numbers can be compared with each other, added, subtracted, multiplied and divided. We learned that there is a number one and a number zero.
Now we are introduced to vectors.

The concepts of “more” and “less” for vectors do not exist - after all, their directions can be different. Only vector lengths can be compared.

But there is a concept of equality for vectors.
Equal vectors that have the same length and the same direction are called. This means that the vector can be transferred parallel to itself to any point in the plane.
Single is a vector whose length is 1. Zero is a vector whose length is zero, that is, its beginning coincides with the end.

It is most convenient to work with vectors in a rectangular coordinate system - the same one in which we draw graphs of functions. Each point in the coordinate system corresponds to two numbers - its x and y coordinates, abscissa and ordinate.
The vector is also specified by two coordinates:

Here the coordinates of the vector are written in parentheses - in x and y.
They are found simply: the coordinate of the end of the vector minus the coordinate of its beginning.

If the vector coordinates are given, its length is found by the formula

Vector addition

There are two ways to add vectors.

1 . Parallelogram rule. To add the vectors and , we place the origins of both at the same point. We build up to a parallelogram and from the same point we draw a diagonal of the parallelogram. This will be the sum of the vectors and .

Remember the fable about the swan, crayfish and pike? They tried very hard, but they never moved the cart. After all, the vector sum of the forces they applied to the cart was equal to zero.

2. The second way to add vectors is the triangle rule. Let's take the same vectors and . We will add the beginning of the second to the end of the first vector. Now let's connect the beginning of the first and the end of the second. This is the sum of the vectors and .

Using the same rule, you can add several vectors. We arrange them one after another, and then connect the beginning of the first with the end of the last.

Imagine that you are going from point A to point B, from B to C, from C to D, then to E and to F. The end result of these actions is movement from A to F.

When adding vectors and we get:

Vector subtraction

The vector is directed opposite to the vector. The lengths of the vectors and are equal.

Now it’s clear what vector subtraction is. The vector difference and is the sum of the vector and the vector .

Multiplying a vector by a number

When a vector is multiplied by the number k, a vector is obtained whose length is k times different from the length . It is codirectional with the vector if k is greater than zero, and opposite if k is less than zero.

Dot product of vectors

Vectors can be multiplied not only by numbers, but also by each other.

The scalar product of vectors is the product of the lengths of the vectors and the cosine of the angle between them.

Please note that we multiplied two vectors, and the result was a scalar, that is, a number. For example, in physics mechanical work equal to the scalar product of two vectors - force and displacement:

If the vectors are perpendicular, their scalar product is zero.
And this is how the scalar product is expressed through the coordinates of the vectors and:

From the formula for the scalar product you can find the angle between the vectors:

This formula is especially convenient in stereometry. For example, in problem 14 Profile Unified State Examination in mathematics you need to find the angle between intersecting lines or between a line and a plane. Problem 14 is often solved several times faster using the vector method than using the classical method.

IN school curriculum in mathematics they study only the scalar product of vectors.
It turns out that, in addition to the scalar product, there is also a vector product, when the result of multiplying two vectors is a vector. Anyone who takes the Unified State Exam in physics knows what the Lorentz force and the Ampere force are. The formulas for finding these forces include vector products.

Vectors are a very useful mathematical tool. You will see this in your first year.

How vector addition occurs is not always clear to students. Children have no idea what is hidden behind them. You just have to remember the rules, and not think about the essence. Therefore, it is precisely the principles of addition and subtraction of vector quantities that require a lot of knowledge.

The addition of two or more vectors always results in one more. Moreover, it will always be the same, regardless of how it is found.

Most often in school course geometry considers the addition of two vectors. It can be performed according to the triangle or parallelogram rule. These drawings look different, but the result of the action is the same.

How does addition occur using the triangle rule?

It is used when the vectors are non-collinear. That is, they do not lie on the same straight line or on parallel ones.

In this case, the first vector must be plotted from some arbitrary point. From its end it is required to draw parallel and equal to the second. The result will be a vector starting from the beginning of the first and ending at the end of the second. The pattern resembles a triangle. Hence the name of the rule.

If the vectors are collinear, then this rule can also be applied. Only the drawing will be located along one line.

How is addition performed using the parallelogram rule?

Yet again? applies only to non-collinear vectors. The construction is carried out according to a different principle. Although the beginning is the same. We need to set aside the first vector. And from its beginning - the second. Based on them, complete the parallelogram and draw a diagonal from the beginning of both vectors. This will be the result. This is how vector addition is performed according to the parallelogram rule.

So far there have been two. But what if there are 3 or 10 of them? Use the following technique.

How and when does the polygon rule apply?

If you need to perform addition of vectors, the number of which is more than two, do not be afraid. It is enough to put them all aside sequentially and connect the beginning of the chain with its end. This vector will be the required sum.

What properties are valid for operations with vectors?

About the zero vector. Which states that when added to it, the original is obtained.

About the opposite vector. That is, about one that has the opposite direction and equal magnitude. Their sum will be zero.

On the commutativity of addition. What has been known since primary school. Changing the positions of the terms does not change the result. In other words, it doesn't matter which vector to put off first. The answer will still be correct and unique.

On the associativity of addition. This law allows you to add any vectors from a triple in pairs and add a third one to them. If you write this using symbols, you get the following:

first + (second + third) = second + (first + third) = third + (first + second).

What is known about vector difference?

There is no separate subtraction operation. This is due to the fact that it is essentially addition. Only the second of them is given the opposite direction. And then everything is done as if adding vectors were considered. Therefore, there is practically no talk about their difference.

In order to simplify the work with their subtraction, the triangle rule is modified. Now (when subtracting) the second vector must be set aside from the beginning of the first. The answer will be the one that connects the end point of the minuend with the same one as the subtrahend. Although you can postpone it as described earlier, simply by changing the direction of the second.

How to find the sum and difference of vectors in coordinates?

The problem gives the coordinates of the vectors and requires finding out their values for the final result. In this case, there is no need to perform constructions. That is, you can use simple formulas that describe the rule for adding vectors. They look like this:

a (x, y, z) + b (k, l, m) = c (x + k, y + l, z + m);

a (x, y, z) -b (k, l, m) = c (x-k, y-l, z-m).

It is easy to see that the coordinates simply need to be added or subtracted depending on the specific task.

First example with solution

Condition. Given a rectangle ABCD. Its sides are equal to 6 and 8 cm. The intersection point of the diagonals is designated by the letter O. It is required to calculate the difference between the vectors AO and VO.

Solution. First you need to draw these vectors. They are directed from the vertices of the rectangle to the point of intersection of the diagonals.

If you look closely at the drawing, you can see that the vectors are already combined so that the second of them is in contact with the end of the first. It's just that his direction is wrong. It should start from this point. This is if the vectors are adding, but the problem involves subtraction. Stop. This action means that you need to add the oppositely directed vector. This means that VO needs to be replaced with OV. And it turns out that the two vectors have already formed a pair of sides from the triangle rule. Therefore, the result of their addition, that is, the desired difference, is the vector AB.

And it coincides with the side of the rectangle. To write down your numerical answer, you will need the following. Draw a rectangle lengthwise so that the larger side is horizontal. Start numbering the vertices from the bottom left and go counterclockwise. Then the length of vector AB will be 8 cm.

Answer. The difference between AO and VO is 8 cm.

Second example and its detailed solution

Condition. The diagonals of the rhombus ABCD are 12 and 16 cm. The point of their intersection is indicated by the letter O. Calculate the length of the vector formed by the difference between the vectors AO and VO.

Solution. Let the designation of the vertices of the rhombus be the same as in the previous problem. Similar to the solution to the first example, it turns out that the required difference is equal to the vector AB. And its length is unknown. Solving the problem came down to calculating one of the sides of the rhombus.

For this purpose, you will need to consider the triangle ABO. It is rectangular because the diagonals of a rhombus intersect at an angle of 90 degrees. And its legs are equal to half the diagonals. That is, 6 and 8 cm. The side sought in the problem coincides with the hypotenuse in this triangle.

To find it you will need the Pythagorean theorem. The square of the hypotenuse will be equal to the sum numbers 6 2 and 8 2. After squaring, the values obtained are: 36 and 64. Their sum is 100. It follows that the hypotenuse is equal to 10 cm.

Answer. The difference between the vectors AO and VO is 10 cm.

Third example with detailed solution

Condition. Calculate the difference and sum of two vectors. Their coordinates are known: the first one has 1 and 2, the second one has 4 and 8.

Solution. To find the sum you will need to add the first and second coordinates in pairs. The result will be the numbers 5 and 10. The answer will be a vector with coordinates (5; 10).

For the difference, you need to subtract the coordinates. After performing this action, the numbers -3 and -6 will be obtained. They will be the coordinates of the desired vector.

Answer. The sum of the vectors is (5; 10), their difference is (-3; -6).

Fourth example

Condition. The length of the vector AB is 6 cm, BC is 8 cm. The second is laid off from the end of the first at an angle of 90 degrees. Calculate: a) the difference between the modules of the vectors VA and BC and the module of the difference between VA and BC; b) the sum of the same modules and the module of the sum.

Solution: a) The lengths of the vectors are already given in the problem. Therefore, calculating their difference is not difficult. 6 - 8 = -2. The situation with the difference module is somewhat more complicated. First you need to find out which vector will be the result of the subtraction. For this purpose, the vector BA, which is directed in the opposite direction AB, should be set aside. Then draw the vector BC from its end, directing it in the direction opposite to the original one. The result of subtraction is the vector CA. Its modulus can be calculated using the Pythagorean theorem. Simple calculations lead to a value of 10 cm.

b) The sum of the moduli of the vectors is equal to 14 cm. To find the second answer, some transformation will be required. Vector BA is oppositely directed to that given - AB. Both vectors are directed from the same point. In this situation, you can use the parallelogram rule. The result of the addition will be a diagonal, and not just a parallelogram, but a rectangle. Its diagonals are equal, which means that the modulus of the sum is the same as in the previous paragraph.

Answer: a) -2 and 10 cm; b) 14 and 10 cm.

Let $\overrightarrow(a)$ and $\overrightarrow(b)$ be two vectors (Fig. 1, a).

Let's take an arbitrary point O and construct a vector $\overrightarrow(OA) = \overrightarrow(a)$ . Then from point A we plot the vector $\overrightarrow(AB) = \overrightarrow(b)$. The vector $\overrightarrow(OB)$ connecting the beginning of the first term of the vector with the end of the second (Fig. 1, b) is called the sum of these vectors and is denoted $\overrightarrow(a) + \overrightarrow(b)$$ ( triangle rule).

The same sum of vectors can be obtained in another way. Let us plot the vectors $\overrightarrow(OA) = \overrightarrow(a) \,and\, \overrightarrow(OS) = \overrightarrow(b) $ from point O (Fig. 1, c). Let's construct a parallelogram OABC on these vectors as on the sides. The vector $\overrightarrow(OB)$, which serves as the diagonal of this parallelogram drawn from the vertex O, is obviously the sum of the vectors $\overrightarrow(a) + \overrightarrow(b)$ ( parallelogram rule). From Figure 1, in It immediately follows that the sum of two vectors has the commutative property: $\overrightarrow(a) + \overrightarrow(b) = \overrightarrow(b) + \overrightarrow(a)$

Indeed, each of the vectors $\overrightarrow(a) + \overrightarrow(b) \,and\, = \overrightarrow(b) + \overrightarrow(a)$ is equal to the same vector $\overrightarrow(OB)$ .

Example 1. In triangle ABC AB = 3, BC = 4, ∠ B = 90°. Find: $a)\,\ \overrightarrow(|AB|) + \overrightarrow(|BC|);\,\,\ b)\,\ |\overrightarrow(AB) + \overrightarrow(BC)|$ .

Solution

b) Since $\overrightarrow(AB) + \overrightarrow(ВС) = \overrightarrow(АС) \,\,\,\, then\,\, |\overrightarrow(АВ) + \overrightarrow(ВС)| = |\overrightarrow(AC)| = AC$ .

Now, applying the Pythagorean theorem, we find $$ AC = \sqrt(AB^2 + BC^2) = \sqrt(9 + 16) = 5 \\ i.e.\, |\overrightarrow(AB) + \overrightarrow( Sun)| = 5. $$

The concept of a sum of vectors can be generalized to the case of any finite number of summand vectors.

Let, for example, be given three vectors $\overrightarrow(a), \overrightarrow(b) \,and\, \overrightarrow(c)$ (Fig. 2).

By first constructing the sum of vectors $\overrightarrow(a) + \overrightarrow(b)$ , and then adding the vector $\overrightarrow(c)$ to this sum, we obtain the vector $(\overrightarrow(a) + \overrightarrow(b)) + \overrightarrow(c)$ . In Figure 2 $$ \overrightarrow(OA) = \overrightarrow(a)\,; \overrightarrow(AB) = b\,; \overrightarrow(OB) = \overrightarrow(a) + \overrightarrow(b)\,; \overrightarrow(BC) = \overrightarrow(c) \\ and \\ \overrightarrow(OS) = \overrightarrow(OB) + \overrightarrow(BC) = (\overrightarrow(a) + \overrightarrow(b)) + \overrightarrow (c) $$ From Figure 2 it is clear that we will obtain the same vector $\overrightarrow(OS)$ if we add the vector $\overrightarrow(AB) = \to the vector $\overrightarrow(АВ) = \overrightarrow(a)$ overrightarrow(b) + \overrightarrow(c)$ . Thus, $(\overrightarrow(a) + \overrightarrow(b)) + \overrightarrow(c) = \overrightarrow(a) + (\overrightarrow(b) + \overrightarrow(c))$ , i.e. the sum vectors have a combining property. Therefore, the sum of three vectors $\overrightarrow(a)\,\,\overrightarrow(b)\,\,\overrightarrow(c)$ is written simply $\overrightarrow(a) + \overrightarrow(b) + \overrightarrow(c)$ .

By difference two vectors $\overrightarrow(a) \,and\, \overrightarrow(b)$ is called the third vector $\overrightarrow(c) = \overrightarrow(a) - \overrightarrow(b)$ , the sum of which with the subtrahend vector $\overrightarrow (b)$ gives the vector $\overrightarrow(a)$. Thus, if $\overrightarrow(c) = \overrightarrow(a) - \overrightarrow(b)\,\ then\, \overrightarrow(c) + \overrightarrow(b) = \overrightarrow(a)$ .

From the definition of the sum of two vectors, the rule for constructing a difference vector follows (Fig. 3).

We plot the vectors $\overrightarrow(OA) = \overrightarrow(a) \,and\, \overrightarrow(OB) = \overrightarrow(b)$ from the common point O. Vector $\overrightarrow(BA)$ connecting the ends of the reduced vector $ \overrightarrow(a)$ and the subtrahend vector $\overrightarrow(b)$ and directed from the subtrahend to the minuend is the difference $\overrightarrow(c) = \overrightarrow(a) - \overrightarrow(b)$ . Indeed, according to the rule of vector addition $\overrightarrow(OB) + \overrightarrow(BA) = \overrightarrow(OA) \text( , or ) \overrightarrow(b) + \overrightarrow(c) = \overrightarrow(a)$ .

Example 2. The side of an equilateral triangle ABC is equal to a. Find: $a) |\overrightarrow(BA) - \overrightarrow(BC)|\,;\,\ b)\,\,\ |\overrightarrow(AB) - \overrightarrow(AC)|$ .

Solution a) Since $\overrightarrow(BA) - \overrightarrow(BC) = \overrightarrow(CA)\text( , a )|\overrightarrow(CA)| = a\text( , then )|\overrightarrow(BA) - \overrightarrow(BC)| = a$ .

b) Since $\overrightarrow(AB) - \overrightarrow(AC) = \overrightarrow(CB)\text( , a )|\overrightarrow(CB)| = a\text( , then )|\overrightarrow(AB) - \overrightarrow(AC)| = a$ .

The product of the vector $\overrightarrow(a)$ (denoted $=\lambda\overrightarrow(a)$ or $\overrightarrow(a)\lambda$) by the real number $\lambda$ is the vector $\overrightarrow(b)$, collinear vector $\overrightarrow(a)$ having length equal to $|\lambda||\overrightarrow(a)|$ and the same direction as vector $\overrightarrow(a)$ if $\lambda > 0$ , and direction, opposite direction vector $\overrightarrow(a)$ if $\lambda< 0$ . Так, например, $2\overrightarrow{a}$ есть вектор, имеющий то же направление, что и вектор $\overrightarrow{a}$ , а длину, вдвое большую, чем вектор $\overrightarrow{a}$ (рис.4).

In the case where $\lambda = 0$ or $\overrightarrow(a) = 0$ , the product $\lambda\overrightarrow(a)$ represents the null vector. The opposite vector $-\overrightarrow(a)$ can be considered as the result of multiplying the vector $\overrightarrow(a)$ by $\lambda = -1$ (see Fig. 4): $$ -\overrightarrow(a) = \ ( -1)\overrightarrow(a) $$ Obviously, $\overrightarrow(a) + (-\overrightarrow(a)) = \overrightarrow(0)$ .

Example 3. Prove that if O, A, B and C are arbitrary points, then $\overrightarrow(OA) + \overrightarrow(AB) + \overrightarrow(BC) + \overrightarrow(СО) = 0$.

Solution. The sum of vectors $\overrightarrow(OA) + \overrightarrow(AB) + \overrightarrow(CB) = \overrightarrow(OS)$, vector $\overrightarrow(CO)$ is the opposite of vector $\overrightarrow(OS)$. Therefore $\overrightarrow(OS) + \overrightarrow(СО) = \overrightarrow(0)$ .

Let the vector $\overrightarrow(a)$ be given. Consider a unit vector $\overrightarrow(a_0)$, collinear to the vector $\overrightarrow(a)$ and directed in the same direction. From the definition of multiplying a vector by a number it follows that $$ \overrightarrow(a) = |\overrightarrow(a)|\,\ \overrightarrow(a_0) $$ , i.e. each vector equal to the product its modulus by a unit vector of the same direction. Further, from the same definition it follows that if $\overrightarrow(b) = \lambda\overrightarrow(a)$ , where $\overrightarrow(a)$ is a nonzero vector, then the vectors $\overrightarrow(a) \, and\, \overrightarrow(b)$ are collinear. Obviously, conversely, from the collinearity of the vectors $\overrightarrow(a) \,and\, \overrightarrow(b)$ it follows that $\overrightarrow(b) = \lambda\overrightarrow(a)$.

Example 4. The length of the vector AB is 3, the length of the vector AC is 5. The cosine of the angle between these vectors is 1/15. Find the length of the vector AB + AC.

Video solution.

The vector $\overrightarrow(AB)$ can be considered as the movement of a point from position $A$ (beginning of movement) to position $B$ (end of movement). That is, the trajectory of movement in this case is not important, only the beginning and the end are important!

$\blacktriangleright$ Two vectors are collinear if they lie on the same line or on two parallel lines.
Otherwise, the vectors are called non-collinear.

$\blacktriangleright$ Two collinear vectors are called codirectional if their directions coincide.
If their directions are opposite, then they are called oppositely directed.

Rules for adding collinear vectors:

co-directed end first. Then their sum is a vector, the beginning of which coincides with the beginning of the first vector, and the end with the end of the second (Fig. 1).

$\blacktriangleright$ To add two oppositely directed vector, we can postpone the second vector from started first. Then their sum is a vector, the beginning of which coincides with the beginning of both vectors, the length is equal to the difference in the lengths of the vectors, the direction coincides with the direction of the longer vector (Fig. 2).

Rules for adding non-collinear vectors $\overrightarrow (a)$ and $\overrightarrow(b)$ :

$\blacktriangleright$ Triangle rule (Fig. 3).

It is necessary to set aside the vector $\overrightarrow (b)$ from the end of the vector $\overrightarrow (a)$. Then the sum is a vector, the beginning of which coincides with the beginning of the vector $\overrightarrow (a)$ , and the end with the end of the vector $\overrightarrow (b)$ .

$\blacktriangleright$ Parallelogram rule (Fig. 4).

It is necessary to set aside the vector $\overrightarrow (b)$ from the beginning of the vector $\overrightarrow (a)$. Then the amount $\overrightarrow (a)+\overrightarrow (b)$– a vector coinciding with the diagonal of a parallelogram constructed on the vectors $\overrightarrow (a)$ and $\overrightarrow (b)$ (the beginning of which coincides with the beginning of both vectors).

$\blacktriangleright$ In order to find the difference of two vectors $\overrightarrow (a)-\overrightarrow(b)$, you need to find the sum of the vectors $\overrightarrow (a)$ and $-\overrightarrow(b)$ : $\overrightarrow(a)-\overrightarrow(b)=\overrightarrow(a)+(-\overrightarrow(b))$(Fig. 5).

Task 1 #2638

Task level: More difficult than the Unified State Exam

Dan right triangle$ABC$ with right angle $A$ , point $O$ is the center of the circumscribed given triangle circles. Vector coordinates $\overrightarrow(AB)=\(1;1$\), $\overrightarrow(AC)=\(-1;1$\). Find the sum of the coordinates of the vector $\overrightarrow(OC)$ .

Because triangle $ABC$ is rectangular, then the center of the circumscribed circle lies on the middle of the hypotenuse, i.e. $O$ is the middle of $BC$ .

notice, that $\overrightarrow(BC)=\overrightarrow(AC)-\overrightarrow(AB)$, hence, $\overrightarrow(BC)=\(-1-1;1-1$=$-2;0$\).

Because $\overrightarrow(OC)=\dfrac12 \overrightarrow(BC)$, That $\overrightarrow(OC)=\(-1;0$\).

This means that the sum of the coordinates of the vector $\overrightarrow(OC)$ is equal to $-1+0=-1$ .

Answer: -1

Task 2 #674

Task level: More difficult than the Unified State Exam

$ABCD$ – a quadrilateral on the sides of which the vectors $\overrightarrow(AB)$ , $\overrightarrow(BC)$ , $\overrightarrow(CD)$ , $\overrightarrow(DA) $ . Find the length of the vector $\overrightarrow(AB) + \overrightarrow(BC) + \overrightarrow(CD) + \overrightarrow(DA)$.

$\overrightarrow(AB) + \overrightarrow(BC) = \overrightarrow(AC)$, $\overrightarrow(AC) + \overrightarrow(CD) = \overrightarrow(AD)$, Then
$\overrightarrow(AB) + \overrightarrow(BC) + \overrightarrow(CD) + \overrightarrow(DA) = \overrightarrow(AC) + \overrightarrow(CD) + \overrightarrow(DA)= \overrightarrow(AD) + \overrightarrow(DA) = \overrightarrow(AD) - \overrightarrow(AD) = \vec(0)$.
The null vector has length equal to $0$ .

A vector can be perceived as displacement, then $\overrightarrow(AB) + \overrightarrow(BC)$– moving from $A$ to $B$ and then from $B$ to $C$ – ultimately this is moving from $A$ to $C$ .

With this interpretation, it becomes obvious that $\overrightarrow(AB) + \overrightarrow(BC) + \overrightarrow(CD) + \overrightarrow(DA) = \vec(0)$, because in the end here we moved from point $A$ to point $A$, that is, the length of such a movement is $0$, which means that the vector of such movement itself is $\vec(0)$ .

Answer: 0

Task 3 #1805

Task level: More difficult than the Unified State Exam

Given a parallelogram $ABCD$ . Diagonals $AC$ and $BD$ intersect at point $O$ . Let , , then $\overrightarrow(OA) = x\cdot\vec(a) + y\cdot\vec(b)$

\[\overrightarrow(OA) = \frac(1)(2)\overrightarrow(CA) = \frac(1)(2)(\overrightarrow(CB) + \overrightarrow(BA)) = \frac(1)( 2)(\overrightarrow(DA) + \overrightarrow(BA)) = \frac(1)(2)(-\vec(b) - \vec(a)) = - \frac(1)(2)\vec (a) - \frac(1)(2)\vec(b)\]$\Rightarrow$ $x = - \frac(1)(2)$ , $y = - \frac(1)(2)$ $\Rightarrow$ $x + y = - 1$ .

Answer: -1

Task 4 #1806

Task level: More difficult than the Unified State Exam

Given a parallelogram $ABCD$ . The points $K$ and $L$ lie on the sides $BC$ and $CD$, respectively, and $BK:KC = 3:1$ and $L$ is the middle of \ (CD\) . Let $\overrightarrow(AB) = \vec(a)$, $\overrightarrow(AD) = \vec(b)$, Then $\overrightarrow(KL) = x\cdot\vec(a) + y\cdot\vec(b)$, where $x$ and $y$ are some numbers. Find the number equal to $x + y$ .

\[\overrightarrow(KL) = \overrightarrow(KC) + \overrightarrow(CL) = \frac(1)(4)\overrightarrow(BC) + \frac(1)(2)\overrightarrow(CD) = \frac (1)(4)\overrightarrow(AD) + \frac(1)(2)\overrightarrow(BA) = \frac(1)(4)\vec(b) - \frac(1)(2)\vec (a)\]$\Rightarrow$ $x = -\frac(1)(2)$ , $y = \frac(1)(4)$ $\Rightarrow$ $x + y = -0 ,25$ .

Answer: -0.25

Task 5 #1807

Task level: More difficult than the Unified State Exam

Given a parallelogram $ABCD$ . The points $M$ and $N$ lie on the sides $AD$ and $BC$, respectively, with $AM:MD = 2:3$ and $BN:NC = 3: 1$ . Let $\overrightarrow(AB) = \vec(a)$, $\overrightarrow(AD) = \vec(b)$, Then $\overrightarrow(MN) = x\cdot\vec(a) + y\cdot\vec(b)$

\[\overrightarrow(MN) = \overrightarrow(MA) + \overrightarrow(AB) + \overrightarrow(BN) = \frac(2)(5)\overrightarrow(DA) + \overrightarrow(AB) + \frac(3 )(4)\overrightarrow(BC) = - \frac(2)(5)\overrightarrow(AD) + \overrightarrow(AB) + \frac(3)(4)\overrightarrow(BC) = -\frac(2 )(5)\vec(b) + \vec(a) + \frac(3)(4)\vec(b) = \vec(a) + \frac(7)(20)\vec(b)\ ]$\Rightarrow$ $x = 1$ , $y = \frac(7)(20)$ $\Rightarrow$ $x\cdot y = 0.35$ .

Answer: 0.35

Task 6 #1808

Task level: More difficult than the Unified State Exam

Given a parallelogram $ABCD$ . The point $P$ lies on the diagonal $BD$, the point $Q$ lies on the side $CD$, and $BP:PD = 4:1$, and $CQ:QD = 1:9$ . Let $\overrightarrow(AB) = \vec(a)$, $\overrightarrow(AD) = \vec(b)$, Then $\overrightarrow(PQ) = x\cdot\vec(a) + y\cdot\vec(b)$, where $x$ and $y$ are some numbers. Find the number equal to $x\cdot y$ .

\[\begin(gathered) \overrightarrow(PQ) = \overrightarrow(PD) + \overrightarrow(DQ) = \frac(1)(5)\overrightarrow(BD) + \frac(9)(10)\overrightarrow( DC) = \frac(1)(5)(\overrightarrow(BC) + \overrightarrow(CD)) + \frac(9)(10)\overrightarrow(AB) =\\ = \frac(1)(5) (\overrightarrow(AD) + \overrightarrow(BA)) + \frac(9)(10)\overrightarrow(AB) = \frac(1)(5)(\overrightarrow(AD) - \overrightarrow(AB)) + \frac(9)(10)\overrightarrow(AB) = \frac(1)(5)\overrightarrow(AD) + \frac(7)(10)\overrightarrow(AB) = \frac(1)(5) \vec(b) + \frac(7)(10)\vec(a)\end(gathered)\]

$\Rightarrow$ $x = \frac(7)(10)$ , $y = \frac(1)(5)$ $\Rightarrow$ $x\cdot y = 0, 14$ . and $ABCO$ – parallelogram; $AF \parallel BE$ and $ABOF$ – parallelogram $\Rightarrow$ \[\overrightarrow(BC) = \overrightarrow(AO) = \overrightarrow(AB) + \overrightarrow(BO) = \overrightarrow(AB) + \overrightarrow(AF) = \vec(a) + \vec(b)\ ]$\Rightarrow$ $x = 1$ , $y = 1$ $\Rightarrow$ $x + y = 2$ .

Answer: 2

High school students preparing for passing the Unified State Exam in mathematics and at the same time expect to receive decent scores, they must definitely repeat the topic “Rules for adding and subtracting several vectors.” As can be seen from many years of practice, such tasks are included in the certification test every year. If a graduate has difficulties with problems from the section “Plane Geometry”, for example, in which it is necessary to apply the rules of addition and subtraction of vectors, he should definitely repeat or re-understand the material in order to successfully pass the Unified State Exam.

The Shkolkovo educational project offers a new approach to preparing for the certification test. Our resource is built in such a way that students can identify the most difficult sections for themselves and fill gaps in knowledge. Shkolkovo specialists prepared and systematized all required material to prepare for passing the certification test.

To ensure that USE problems in which you need to apply the rules for adding and subtracting two vectors do not cause difficulties, we recommend first of all brushing up on the basic concepts. Students will be able to find this material in the “Theoretical Information” section.

If you already remember the rule for subtracting vectors and the basic definitions on this topic, we suggest you consolidate your knowledge by completing the appropriate exercises, which were selected by experts educational portal"Shkolkovo". For each problem, the site presents a solution algorithm and gives the correct answer. The topic “Rules for Vector Addition” presents various exercises; Having completed two or three relatively easy tasks, students can successively move on to more complex ones.

Schoolchildren have the opportunity to hone their own skills on such tasks, for example, online, while in Moscow or any other city in Russia. If necessary, the task can be saved in the “Favorites” section. Thanks to this, you can quickly find examples of interest and discuss algorithms for finding the correct answer with your teacher.

No one will argue that it is impossible to get to your destination without knowing the direction of travel. In physics this concept is called vector. Up to this point, we have been operating with some numbers and values, which are called quantities. A vector differs from a quantity in that it has a direction.

When working with a vector, they operate on it direction And size. Physical parameter without regard to direction is called scalar.

Visually, the vector is displayed as an arrow. The length of the arrow is the magnitude of the vector.

In physics, vectors are represented by a capital letter with an arrow at the top.

Vectors can be compared. Two vectors will be equal if they have the same magnitude and direction.

Vectors can be added. The resulting vector is the sum of both vectors and determines the distance and direction. For example, you live in Kyiv and decided to visit old friends in Moscow, and from there make a visit to your beloved mother-in-law in Lviv. How far will you be from your home while visiting your wife’s mother?

To answer this question, you need to draw a vector from the starting point of the trip (Kyiv) to the final point (Lviv). The new vector determines the result of the entire journey from beginning to end.

Vector A - Kyiv-Moscow
Vector B - Moscow-Lviv
Vector C - Kyiv-Lviv

C = A+B, where C - vector sum or the resulting vector

Vectors can not only be added, but also subtracted! To do this, you need to combine the bases of the subtrahend and subtracting vectors and connect their ends with arrows:

Vector A = C-B
Vector B = C-A

Let's put it on our vectors coordinate grid. For vector A we can say that it is directed 5 cells up ( positive value Y axis) and 3 cells to the left ( negative meaning X axis): X=-3; Y=5.

For vector B: direction 4 cells to the left and 7 cells down: X=-4; Y=-7.

Thus, to add vectors along the X and Y axes, you need to add their coordinates. To get the coordinates of the resulting vector along the X and Y axes:

Let's consider the problem: the ball moves at a speed of 10 m/s along an inclined plane with a base length X = 1 m, located at 30° to the horizontal. It is required to determine the time during which the ball moves from the beginning to the end of the plane.

In this problem, speed is a vector V with magnitude 10m/s and direction α=30° to the horizontal. To determine the speed of the ball's movement along the base of the inclined plane, we need to determine the X-component of the ball's movement, which is a scalar (only has a value, not a direction) and is denoted Vx. Similarly, the Y-component of velocity is also a scalar and is denoted V y. Velocity vector through components: V = (V x ;V y)

Let's determine the components (V x ;V y). Let's remember trigonometry:

V x = V cosα
V y = V sinα

X-component of the ball speed:

V x = V cosα = V cos30° = 10.0 0.866 = 8.66 m/s

The horizontal speed of the ball is 8.66 m/s.

Because the length of the base of the inclined plane is 1 m, then the ball will cover this distance in:

1.00(m)/8.66(m/s) = 0.12 s

Thus, the ball will need 0.12 s to move along the inclined plane. Answer: 0.12s

For the sake of interest, let’s define the Y-component of speed:

V y = V sinα = 10 1/2 = 5.0 m/s

Since the ball’s “travel” time is the same for both components, we can determine the height Y from which the ball rolled:

5.0(m/s)·0.12(s) = 0.6 m

Distance traveled by the ball:

Inverse problem

Let's consider the inverse problem of the previous one:

The ball moved along the inclined plane to a height of 0.6 m, while in the horizontal plane its movement was 1.0 m. It is necessary to find the distance traveled by the ball and the angle.

We calculate the distance using the Pythagorean theorem:

L = √1.00 2 + 0.60 2 = √1.36 = 1.16m

For trigonometry:

X = L cosα; Y = L sinα

X/L = cosα; Y/L = sinα

Now you can find the angle:

α = arccos(X/L); α = arcsin(Y/L)

Let's substitute the numbers:

α = arccos(1/1.16) = 30°

The intermediate calculation of L can be eliminated:

Y = X tanα