How to calculate the dot product of modules of vectors. Dot product of vectors. Vector length. Dot product in coordinates

Lecture: Vector coordinates; dot product of vectors; angle between vectors

Vector coordinates

So, as mentioned earlier, vectors are a directional segment, which has its own beginning and end. If the beginning and end are represented by some points, then they have their own coordinates on a plane or in space.

If each point has its own coordinates, then we can get the coordinates of the whole vector.

Suppose we have some vector, whose beginning and end of the vector have the following designations and coordinates: A (A x; Ay) and B (B x; By)

To get the coordinates of this vector, it is necessary to subtract the corresponding coordinates of the beginning from the coordinates of the end of the vector:

To determine the coordinates of a vector in space, use the following formula:

Dot product of vectors

There are two ways to define the dot product:

• Geometric way. According to him, the dot product is equal to the product of the values \u200b\u200bof these modules by the cosine of the angle between them.

• Algebraic meaning. From the point of view of algebra, the dot product of two vectors is a certain quantity that is obtained as a result of the sum of the products of the corresponding vectors.

If vectors are given in space, then you should use a similar formula:

Properties:

• If you multiply two identical vectors scalarly, then their dot product will not be negative:

• If the scalar product of two identical vectors turned out to be equal to zero, then these vectors are considered to be zero:

• If a vector is multiplied by itself, then the scalar product will be equal to the square of its modulus:

• The scalar product has a communicative property, that is, the scalar product will not change from the permutation of the vectors:

• The scalar product of nonzero vectors can be zero only if the vectors are perpendicular to each other:

• For the scalar product of vectors, the displacement law is valid in the case of multiplying one of the vectors by a number:

• With the dot product, you can also use the distributive property of multiplication:

Angle between vectors

In the case of the plane problem, the scalar product of vectors a \u003d (a x; a y) and b \u003d (b x; b y) can be found using the following formula:

a b \u003d a x b x + a y b y

Vector dot product formula for spatial problems

In the case of the spatial problem, the scalar product of vectors a \u003d (a x; a y; a z) and b \u003d (b x; b y; b z) can be found using the following formula:

a b \u003d a x b x + a y b y + a z b z

Dot product formula of n -dimensional vectors

In the case of n-dimensional space, the scalar product of vectors a \u003d (a 1; a 2; ...; a n) and b \u003d (b 1; b 2; ...; b n) can be found using the following formula:

a b \u003d a 1 b 1 + a 2 b 2 + ... + a n b n

Dot product properties of vectors

1. The scalar product of a vector by itself is always greater than or equal to zero:

2. The scalar product of a vector by itself is equal to zero if and only if the vector is equal to the zero vector:

a a \u003d 0<=> a \u003d 0

3. The scalar product of a vector by itself is equal to the square of its modulus:

4. Operation of scalar multiplication is communicative:

5. If the scalar product of two nonzero vectors is equal to zero, then these vectors are orthogonal:

a ≠ 0, b ≠ 0, a b \u003d 0<=> a ┴ b

6. (αa) b \u003d α (a b)

7. The operation of scalar multiplication is distributive:

(a + b) c \u003d a c + b c

Examples of problems for calculating the dot product of vectors

Examples of calculating the dot product of vectors for plane problems

Find the dot product of vectors a \u003d (1; 2) and b \u003d (4; 8).

Decision: a b \u003d 1 4 + 2 8 \u003d 4 + 16 \u003d 20.

Find the scalar product of vectors a and b if their lengths | a | \u003d 3, | b | \u003d 6, and the angle between the vectors is 60˚.

Decision: a b \u003d | a | · | B | cos α \u003d 3 6 cos 60˚ \u003d 9.

Find the scalar product of vectors p \u003d a + 3b and q \u003d 5a - 3 b if their lengths | a | \u003d 3, | b | \u003d 2, and the angle between vectors a and b is 60˚.

Decision:

p q \u003d (a + 3b) (5a - 3b) \u003d 5 a a - 3 a b + 15 b a - 9 b b \u003d

5 | a | 2 + 12 a b - 9 | b | 2 \u003d 5 3 2 + 12 3 2 cos 60˚ - 9 2 2 \u003d 45 +36 -36 \u003d 45.

An example of calculating the dot product of vectors for spatial problems

Find the dot product of vectors a \u003d (1; 2; -5) and b \u003d (4; 8; 1).

Decision: a b \u003d 1 4 + 2 8 + (-5) 1 \u003d 4 + 16 - 5 \u003d 15.

An example of calculating the dot product for n -dimensional vectors

Find the dot product of vectors a \u003d (1; 2; -5; 2) and b \u003d (4; 8; 1; -2).

Decision: a b \u003d 1 4 + 2 8 + (-5) 1 + 2 (-2) \u003d 4 + 16 - 5 -4 \u003d 11.

13. The vector product of vectors and a vector is called third vector defined as follows:

2) perpendicular, perpendicular. (1"")

3) vectors are oriented in the same way as the basis of the entire space (positively or negatively).

Designate:.

The physical meaning of a vector product

- moment of force relative to point O; - radius is the vector of the point of application of the force, then

moreover, if transferred to point O, then the triplet must be oriented as a basis vector.

Definition 1

The scalar product of vectors is a number equal to the product of dyn of these vectors and the cosine of the angle between them.

The notation for the product of vectors a → and b → has the form a →, b →. Let's convert to the formula:

a →, b → \u003d a → b → cos a →, b → ^. a → and b → denote the lengths of vectors, a →, b → ^ denote the angle between the given vectors. If at least one vector is zero, that is, has a value of 0, then the result will be zero, a →, b → \u003d 0

When multiplying the vector by itself, we get the square of its length:

a →, b → \u003d a → b → cos a →, a → ^ \u003d a → 2 cos 0 \u003d a → 2

Definition 2

Scalar multiplication of a vector by itself is called a scalar square.

Calculated by the formula:

a →, b → \u003d a → b → cos a →, b → ^.

The notation a →, b → \u003d a → b → cos a →, b → ^ \u003d a → npa → b → \u003d b → npb → a → shows that npb → a → is the numerical projection of a → on b →, npa → a → is the projection of b → onto a →, respectively.

Let us formulate the definition of a product for two vectors:

The scalar product of two vectors a → by b → is called the product of the length of the vector a → by the projection b → by the direction a → or the product of the length b → by the projection a → respectively.

Dot product in coordinates

The dot product can be calculated through the coordinates of vectors in a given plane or in space.

The scalar product of two vectors on a plane, in three-dimensional space, is called the sum of the coordinates of the given vectors a → and b →.

When calculating the scalar product of the given vectors a → \u003d (a x, a y), b → \u003d (b x, b y) in the Cartesian system, use:

a →, b → \u003d a x b x + a y b y,

for three-dimensional space, the following expression applies:

a →, b → \u003d a x b x + a y b y + a z b z.

In fact, this is the third definition of the dot product.

Let's prove it.

Proof 1

For the proof, use a →, b → \u003d a → b → cos a →, b → ^ \u003d ax bx + ay by for vectors a → \u003d (ax, ay), b → \u003d (bx, by) on Cartesian system.

Vectors should be postponed

O A → \u003d a → \u003d a x, a y and O B → \u003d b → \u003d b x, b y.

Then the length of the vector A B → will be equal to A B → \u003d O B → - O A → \u003d b → - a → \u003d (b x - a x, b y - a y).

Consider a triangle O A B.

A B 2 \u003d O A 2 + O B 2 - 2 O A O B cos (∠ A O B) is true based on the cosine theorem.

By the condition, it can be seen that O A \u003d a →, O B \u003d b →, A B \u003d b → - a →, ∠ A O B \u003d a →, b → ^, hence the formula for finding the angle between vectors is written differently

b → - a → 2 \u003d a → 2 + b → 2 - 2 a → b → cos (a →, b → ^).

Then it follows from the first definition that b → - a → 2 \u003d a → 2 + b → 2 - 2 (a →, b →), hence (a →, b →) \u003d 1 2 (a → 2 + b → 2 - b → - a → 2).

Applying the formula for calculating the length of vectors, we get:
a →, b → \u003d 1 2 ((a 2 x + ay 2) 2 + (b 2 x + by 2) 2 - ((bx - ax) 2 + (by - ay) 2) 2) \u003d \u003d 1 2 (a 2 x + a 2 y + b 2 x + b 2 y - (bx - ax) 2 - (by - ay) 2) \u003d \u003d ax bx + ay by

Let us prove the equalities:

(a →, b →) \u003d a → b → cos (a →, b → ^) \u003d \u003d a x b x + a y b y + a z b z

- respectively for vectors of three-dimensional space.

The scalar product of vectors with coordinates says that the scalar square of a vector is equal to the sum of the squares of its coordinates in space and on a plane, respectively. a → \u003d (a x, a y, a z), b → \u003d (b x, b y, b z) and (a →, a →) \u003d a x 2 + a y 2.

Dot product and its properties

There are dot product properties that are applicable for a →, b → and c →:

1. commutativity (a →, b →) \u003d (b →, a →);
2. distributivity (a → + b →, c →) \u003d (a →, c →) + (b →, c →), (a → + b →, c →) \u003d (a →, b →) + (a → , c →);
3. the combination property (λ a →, b →) \u003d λ (a →, b →), (a →, λ b →) \u003d λ (a →, b →), λ is any number;
4. the scalar square is always greater than zero (a →, a →) ≥ 0, where (a →, a →) \u003d 0 in the case when a → is zero.

Example 1

The properties are explicable due to the definition of the dot product on the plane and the properties of the addition and multiplication of real numbers.

Prove the commutativity property (a →, b →) \u003d (b →, a →). From the definition, we have that (a →, b →) \u003d a y b y + a y b y and (b →, a →) \u003d b x a x + b y a y.

By the commutativity property, the equalities a x b x \u003d b x a x and a y b y \u003d b y a y are true, so a x b x + a y b y \u003d b x a x + b y a y.

It follows that (a →, b →) \u003d (b →, a →). Q.E.D.

Distributivity is valid for any numbers:

(a (1) → + a (2) → +.. + a (n) →, b →) \u003d (a (1) →, b →) + (a (2) →, b →) +. ... ... + (a (n) →, b →)

and (a →, b (1) → + b (2) → +.. + b (n) →) \u003d (a →, b (1) →) + (a →, b (2) →) + ... ... ... + (a →, b → (n)),

hence we have

(a (1) → + a (2) → +.. + a (n) →, b (1) → + b (2) → +... + b (m) →) \u003d (a ( 1) →, b (1) →) + (a (1) →, b (2) →) +. ... ... + (a (1) →, b (m) →) + + (a (2) →, b (1) →) + (a (2) →, b (2) →) +. ... ... + (a (2) →, b (m) →) +. ... ... + + (a (n) →, b (1) →) + (a (n) →, b (2) →) +. ... ... + (a (n) →, b (m) →)

Dot product with examples and solutions

Any problem of such a plan is solved using properties and formulas related to the dot product:

1. (a →, b →) \u003d a → b → cos (a →, b → ^);
2. (a →, b →) \u003d a → n p a → b → \u003d b → n p b → a →;
3. (a →, b →) \u003d a x b x + a y b y or (a →, b →) \u003d a x b x + a y b y + a z b z;
4. (a →, a →) \u003d a → 2.

Let's consider some solution examples.

Example 2

Length a → is 3, length b → is 7. Find the dot product if the angle is 60 degrees.

Decision

By condition, we have all the data, so we calculate by the formula:

(a →, b →) \u003d a → b → cos (a →, b → ^) \u003d 3 7 cos 60 ° \u003d 3 7 1 2 \u003d 21 2

Answer: (a →, b →) \u003d 21 2.

Example 3

Given vectors a → \u003d (1, - 1, 2 - 3), b → \u003d (0, 2, 2 + 3). What is the dot product.

Decision

In this example, the formula for calculating by coordinates is considered, since they are specified in the problem statement:

(a →, b →) \u003d ax bx + ay by + az bz \u003d \u003d 1 0 + (- 1) 2 + (2 + 3) (2 + 3) \u003d \u003d 0 - 2 + ( 2 - 9) \u003d - 9

Answer: (a →, b →) \u003d - 9

Example 4

Find the dot product A B → and A C →. Points A (1, - 3), B (5, 4), C (1, 1) are given on the coordinate plane.

Decision

To begin with, the coordinates of the vectors are calculated, since the coordinates of the points are given by the condition:

A B → \u003d (5 - 1, 4 - (- 3)) \u003d (4, 7) A C → \u003d (1 - 1, 1 - (- 3)) \u003d (0, 4)

Substituting into the formula using coordinates, we get:

(A B →, A C →) \u003d 4 0 + 7 4 \u003d 0 + 28 \u003d 28.

Answer: (A B →, A C →) \u003d 28.

Example 5

Given vectors a → \u003d 7 m → + 3 n → and b → \u003d 5 m → + 8 n →, find their product. m → is equal to 3 and n → is equal to 2 units, they are perpendicular.

Decision

(a →, b →) \u003d (7 m → + 3 n →, 5 m → + 8 n →). Applying the distributive property, we get:

(7 m → + 3 n →, 5 m → + 8 n →) \u003d \u003d (7 m →, 5 m →) + (7 m →, 8 n →) + (3 n →, 5 m →) + (3 n →, 8 n →)

We take out the coefficient for the sign of the product and get:

(7 m →, 5 m →) + (7 m →, 8 n →) + (3 n →, 5 m →) + (3 n →, 8 n →) \u003d \u003d 7 5 (m →, m →) + 7 8 (m →, n →) + 3 5 (n →, m →) + 3 8 (n →, n →) \u003d \u003d 35 (m →, m →) + 56 (m →, n →) + 15 (n →, m →) + 24 (n →, n →)

By the commutativity property, we transform:

35 (m →, m →) + 56 (m →, n →) + 15 (n →, m →) + 24 (n →, n →) \u003d 35 (m →, m →) + 56 (m →, n →) + 15 (m →, n →) + 24 (n →, n →) \u003d 35 (m →, m →) + 71 (m →, n → ) + 24 (n →, n →)

As a result, we get:

(a →, b →) \u003d 35 (m →, m →) + 71 (m →, n →) + 24 (n →, n →).

Now let's apply the formula for the dot product with the angle specified by the condition:

(a →, b →) \u003d 35 (m →, m →) + 71 (m →, n →) + 24 (n →, n →) \u003d \u003d 35 m → 2 + 71 m → n → cos (m →, n → ^) + 24 n → 2 \u003d \u003d 35 3 2 + 71 3 2 cos π 2 + 24 2 2 \u003d 411.

Answer: (a →, b →) \u003d 411

If there is a numerical projection.

Example 6

Find the dot product a → and b →. Vector a → has coordinates a → \u003d (9, 3, - 3), projection b → with coordinates (- 3, - 1, 1).

Decision

By hypothesis, the vectors a → and the projection b → are oppositely directed, because a → \u003d - 1 3 · n p a → b → →, so the projection b → corresponds to the length n p a → b → →, and with the sign "-":

n p a → b → → \u003d - n p a → b → → \u003d - (- 3) 2 + (- 1) 2 + 1 2 \u003d - 11,

Substituting into the formula, we get the expression:

(a →, b →) \u003d a → n p a → b → → \u003d 9 2 + 3 2 + (- 3) 2 (- 11) \u003d - 33.

Answer: (a →, b →) \u003d - 33.

Problems with a known dot product, where it is necessary to find the length of a vector or a numerical projection.

Example 7

What value should λ take for a given scalar product a → \u003d (1, 0, λ + 1) and b → \u003d (λ, 1, λ) will be equal to -1.

Decision

The formula shows that it is necessary to find the sum of the products of coordinates:

(a →, b →) \u003d 1 λ + 0 1 + (λ + 1) λ \u003d λ 2 + 2 λ.

Given we have (a →, b →) \u003d - 1.

To find λ, we calculate the equation:

λ 2 + 2 λ \u003d - 1, hence λ \u003d - 1.

Answer: λ \u003d - 1.

The physical meaning of the dot product

Mechanics considers the application of the dot product.

When working A with a constant force F → the body moved from point M to N, you can find the product of the lengths of the vectors F → and M N → with the cosine of the angle between them, which means that the work is equal to the product of the vectors of force and displacement:

A \u003d (F →, M N →).

Example 8

The movement of a material point by 3 meters under the influence of a force equal to 5 ntons is directed at an angle of 45 degrees relative to the axis. Find A.

Decision

Since work is the product of the force vector and displacement, it means, based on the condition F → \u003d 5, S → \u003d 3, (F →, S → ^) \u003d 45 °, we get A \u003d (F →, S →) \u003d F → S → cos (F →, S → ^) \u003d 5 3 cos (45 °) \u003d 15 2 2.

Answer: A \u003d 15 2 2.

Example 9

The material point, moving from M (2, - 1, - 3) to N (5, 3 λ - 2, 4) under the force F → \u003d (3, 1, 2), performed work equal to 13 J. Calculate the length of the movement.

Decision

For the given coordinates of the vector M N → we have M N → \u003d (5 - 2, 3 λ - 2 - (- 1), 4 - (- 3)) \u003d (3, 3 λ - 1, 7).

By the formula for finding work with vectors F → \u003d (3, 1, 2) and MN → \u003d (3, 3 λ - 1, 7), we obtain A \u003d (F ⇒, MN →) \u003d 3 3 + 1 (3 λ - 1) + 2 7 \u003d 22 + 3 λ.

By hypothesis, it is given that A \u003d 13 J, which means 22 + 3 λ \u003d 13. Hence λ \u003d - 3, hence M N → \u003d (3, 3 λ - 1, 7) \u003d (3, - 10, 7).

To find the length of displacement M N →, apply the formula and substitute the values:

M N → \u003d 3 2 + (- 10) 2 + 7 2 \u003d 158.

Answer: 158.

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Angle between vectors

Consider two given vectors $ \\ overrightarrow (a) $ and $ \\ overrightarrow (b) $. Let us set aside vectors $ \\ overrightarrow (a) \u003d \\ overrightarrow (OA) $ and $ \\ overrightarrow (b) \u003d \\ overrightarrow (OB) $ from an arbitrarily chosen point $ O $, then the angle $ AOB $ is called the angle between vectors $ \\ overrightarrow ( a) $ and $ \\ overrightarrow (b) $ (fig. 1).

Picture 1.

Note here that if the vectors $ \\ overrightarrow (a) $ and $ \\ overrightarrow (b) $ are codirectional or one of them is a zero vector, then the angle between the vectors is $ 0 ^ 0 $.

Designation: $ \\ widehat (\\ overrightarrow (a), \\ overrightarrow (b)) $

Dot product of vectors

Mathematically, this definition can be written as follows:

The dot product can be zero in two cases:

If one of the vectors is a zero vector (Since then its length is zero).

If the vectors are mutually perpendicular (i.e. $ cos (90) ^ 0 \u003d 0 $).

Note also that the dot product is greater than zero if the angle between these vectors is acute (since $ (cos \\ left (\\ widehat (\\ overrightarrow (a), \\ overrightarrow (b)) \\ \u200b\u200bright) \\)\u003e 0 $), and less than zero if the angle between these vectors is obtuse (since $ (cos \\ left (\\ widehat (\\ overrightarrow (a), \\ overrightarrow (b)) \\ \u200b\u200bright) \\)

The concept of a scalar square is associated with the concept of a scalar product.

Definition 2

The scalar square of the vector $ \\ overrightarrow (a) $ is the scalar product of this vector by itself.

We get that the scalar square is

\\ [\\ overrightarrow (a) \\ overrightarrow (a) \u003d \\ left | \\ overrightarrow (a) \\ right | \\ left | \\ overrightarrow (a) \\ right | (cos 0 ^ 0 \\) \u003d \\ left | \\ overrightarrow (a ) \\ right | \\ left | \\ overrightarrow (a) \\ right | \u003d (\\ left | \\ overrightarrow (a) \\ right |) ^ 2 \\]

Calculating the dot product from the coordinates of the vectors

In addition to the standard way of finding the dot product value, which follows from the definition, there is another way.

Let's consider it.

Let the vectors $ \\ overrightarrow (a) $ and $ \\ overrightarrow (b) $ have coordinates $ \\ left (a_1, b_1 \\ right) $ and $ \\ left (a_2, b_2 \\ right) $, respectively.

Theorem 1

The scalar product of the vectors $ \\ overrightarrow (a) $ and $ \\ overrightarrow (b) $ is equal to the sum of the products of the corresponding coordinates.

Mathematically, this can be written as follows

\\ [\\ overrightarrow (a) \\ overrightarrow (b) \u003d a_1a_2 + b_1b_2 \\]

Evidence.

The theorem is proved.

This theorem has several consequences:

Corollary 1: The vectors $ \\ overrightarrow (a) $ and $ \\ overrightarrow (b) $ are perpendicular if and only if $ a_1a_2 + b_1b_2 \u003d 0 $

Corollary 2: The cosine of the angle between vectors is $ cos \\ alpha \u003d \\ frac (a_1a_2 + b_1b_2) (\\ sqrt (a ^ 2_1 + b ^ 2_1) \\ cdot \\ sqrt (a ^ 2_2 + b ^ 2_2)) $

Dot product properties of vectors

For any three vectors and a real number $ k $ it is true:

$ (\\ overrightarrow (a)) ^ 2 \\ ge 0 $

This property follows from the definition of a scalar square (Definition 2).

Traveling law: $ \\ overrightarrow (a) \\ overrightarrow (b) \u003d \\ overrightarrow (b) \\ overrightarrow (a) $.

This property follows from the definition of the dot product (Definition 1).

Distribution law:

$ \\ left (\\ overrightarrow (a) + \\ overrightarrow (b) \\ right) \\ overrightarrow (c) \u003d \\ overrightarrow (a) \\ overrightarrow (c) + \\ overrightarrow (b) \\ overrightarrow (c) $. \\ end (enumerate)

By Theorem 1, we have:

\\ [\\ left (\\ overrightarrow (a) + \\ overrightarrow (b) \\ right) \\ overrightarrow (c) \u003d \\ left (a_1 + a_2 \\ right) a_3 + \\ left (b_1 + b_2 \\ right) b_3 \u003d a_1a_3 + a_2a_3 + b_1b_3 + b_2b_3 \u003d\u003d \\ overrightarrow (a) \\ overrightarrow (c) + \\ overrightarrow (b) \\ overrightarrow (c) \\]

Combination law: $ \\ left (k \\ overrightarrow (a) \\ right) \\ overrightarrow (b) \u003d k (\\ overrightarrow (a) \\ overrightarrow (b)) $. \\ end (enumerate)

By Theorem 1, we have:

\\ [\\ left (k \\ overrightarrow (a) \\ right) \\ overrightarrow (b) \u003d ka_1a_2 + kb_1b_2 \u003d k \\ left (a_1a_2 + b_1b_2 \\ right) \u003d k (\\ overrightarrow (a) \\ overrightarrow (b)) \\]

An example of a problem for calculating the dot product of vectors

Example 1

Find the dot product of vectors $ \\ overrightarrow (a) $ and $ \\ overrightarrow (b) $ if $ \\ left | \\ overrightarrow (a) \\ right | \u003d 3 $ and $ \\ left | \\ overrightarrow (b) \\ right | \u003d 2 $, and the angle between them is $ ((30) ^ 0, \\ 45) ^ 0, \\ (90) ^ 0, \\ (135) ^ 0 $.

Decision.

Using Definition 1, we obtain

For $ (30) ^ 0: $

\\ [\\ overrightarrow (a) \\ overrightarrow (b) \u003d 6 (cos \\ left ((30) ^ 0 \\ right) \\) \u003d 6 \\ cdot \\ frac (\\ sqrt (3)) (2) \u003d 3 \\ sqrt ( 3) \\]

For $ (45) ^ 0: $

\\ [\\ overrightarrow (a) \\ overrightarrow (b) \u003d 6 (cos \\ left ((45) ^ 0 \\ right) \\) \u003d 6 \\ cdot \\ frac (\\ sqrt (2)) (2) \u003d 3 \\ sqrt ( 2) \\]

For $ (90) ^ 0: $

\\ [\\ overrightarrow (a) \\ overrightarrow (b) \u003d 6 (cos \\ left ((90) ^ 0 \\ right) \\) \u003d 6 \\ cdot 0 \u003d 0 \\]

For $ (135) ^ 0: $

\\ [\\ overrightarrow (a) \\ overrightarrow (b) \u003d 6 (cos \\ left ((135) ^ 0 \\ right) \\) \u003d 6 \\ cdot \\ left (- \\ frac (\\ sqrt (2)) (2) \\ Similar articles