Records labeled "dot product of vectors". Dot product of vectors Test 6 dot product of vectors

24. Let's simplify this function. By Vieta's theorem, we find the roots of the square trinomial x 2 -x-6(x 1 = -2 , x 2 = 3 ), we expand the denominator of the fraction into linear factors (x-3) (x + 2) and cancel the fraction by (x-3)... Find the antiderivative H (x) the resulting function 1 / (x + 2).

25. So 126 players will play 63 games, of which 63 participants will qualify as winners in the second round. In total, 63 + 1 = 64 participants will fight in the second round. They will play 32 games, hence 32 more winners who will play 16 games. 16 winners will play 8 games, 8 winners will play 4 games. The four winners will play 2 games, and finally, the two winners will need to play last game... We count matches: 63+32+16+8+4+2+1=126.

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Option 1 A template for creating tests in PowerPoint MCOU "Pogorelskaya Secondary School" Koshcheev MM was used.

Option 1 b) blunt a) acute c) straight

Option 1 c) is equal to zero a) greater than zero b) less than zero

Option 1 b) -½ ∙ a² c) ½ ∙ a²

Option 1 4. D ABC - tetrahedron, AB = BC = AC = A D = BD = CD. Then it is not true that….

Option 1 5. Which statement is correct?

Option 1 b) a ₁ b ₁ + a ₂ b ₂ + a ₃ b ₃ c) a ₁ b ₂ b ₃ + b ₁ a ₂ b ₃ + b ₁ b ₂ a ₃ a) a ₁а₂а₃ + b ₁ b ₂ b ₃

Option 1 b) - a ² a) 0 c) a²

Option 1 a) a b) o

Option 1

Option 1 a) 7 c) -7 b) -9

Option 1 b) -4 a) 4 c) 2

Option 1 b) 120 ° a) 90 ° c) 60 °

Option 1 c) 0.7 a) -0.7 b) 1 13. The coordinates of the points are given: A (1; -1; -4), B (-3; -1; 0), C (-1; 2 ; 5), D (2; -3; 1). Then the cosine of the angle between lines AB and CD is equal to ……

Option 1 c) 4

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Variant 2 A template for creating tests in PowerPoint MCOU "Pogorelskaya Secondary School" Koshcheev MM was used.

Test result Correct: 14 Errors: 0 Mark: 5 Time: 1 min. 40 sec. still fix

Option 2 a) acute b) blunt c) straight

Option 2 a) is greater than zero c) is equal to zero b) is less than zero

Option 2 b) -½ ∙ a² a) ½ ∙ a²

Option 2 4. АВСА "В₁С" - prism,

Option 2 5. Which statement is correct?

Option 2 a) m ₁ n ₁ + m ₂ n ₂ + m ₃ n ₃ c) m ₁ m ₂ m ₃ + n ₁ n ₂ n ₃ b) (n ₁- m ₁) ² + (n ₂- m ₂ ) ² + (n ₃- m ₃) ²

Option 2 c) - a ² a) 0 b) a²

Option 2 a) o c) a²

Option 2

Option 2 b) 3 c) -3 a) 19

Option 2 a) - 0.5 b) -1 c) 0.5

Option 2 b) 6 0 ° a) 90 ° c) 12 0 °

Option 2 a) 0.7 c) -0.7 b) 1 13. The coordinates of the points are given: C (3; - 2; 1), D (- 1; 2; 1), M (2; -3; 3 ), N (-1; 1; -2). Then the cosine of the angle between straight lines CD and MN is equal to ……

Option 2 c) 4

Keys to the test: Vector dot product. Option 1 1 2 3 4 5 6 7 8 9 10 11 12 13 14 Repl. b c b c a b b a c a b b c b References G.I. Kovaleva, N.I. Mazurova Geometry 10-11 grades. Tests for current and generalized control. Publishing house "Teacher", 2009 Option 2 1 2 3 4 5 6 7 8 9 10 11 12 13 14 Repl. a a b b b a c a c b a b a b

Dot product a  b two nonzero vectors a and b is a number equal to the product of the lengths of these vectors by the cosine of the angle between them. If at least one of these vectors is equal to zero, the scalar product is equal to zero. Thus, by definition, we have

where  is the angle between vectors a and b .

Dot product of vectors a , b also denoted by symbols ab .

The sign of the dot product is determined by the value :

if 0    then a  b  0,

if    , then a  b  0.

The dot product is defined for only two vectors.

Operations on vectors in coordinate form

Let in the coordinate system Ooh given vectors a = (x 1 ; y 1) = x 1 i + y 1 j and b = (x 2 ; y 2) = x 2 i + y 2 j .

1. Each coordinate of the sum of two (or more) vectors is equal to the sum of the corresponding coordinates of the vector-summands, i.e. a + b = = (x 1 + x 2 ; y 1 + y 2).

2. Each coordinate of the difference of two vectors is equal to the difference of the corresponding coordinates of these vectors, i.e. a – b = (x 1 – x 2 ; y 1 – y 2).

3. Each coordinate of the product of a vector and a number  is equal to the product of the corresponding coordinate of this vector by , that is,  a = ( NS 1 ;  at 1).

4. The scalar product of two vectors is equal to the sum of the products of the corresponding coordinates of these vectors, ie. a  b = x 1  x 2 + + y 1  y 2 .

Consequence. Vector length a = (x; y) is equal to the square root of the sum of the squares of its coordinates, i.e.

=
(5)

Example 4. Given vectors
b = 3i – j .

Required:

1. Find

2. Find the dot product of vectors with , d .

3. Find the length of the vector with .

Solution

1. By property 3, we find the coordinates of vectors 2 a , –a , 3b , 2b : 2a = = 2(–2; 3) = (–4; 6), –a = –(–2; 3) = (2; –3), 3b = 3(3; –1) = (9; –3), 2b = = 2(3; –1) = = (6; –2).

By properties 2, 1, we find the coordinates of the vectors with , d : with = 2a – 3b = = (–4; 6) – (9; –3) = (–13; 9), d = –a + 2b = (2; –3) + (6; –2) = (8; –5).

2. By property 4 cd = –13  8 + 9  (–5) = –104 – 45 = –149.

3. By the corollary to property 4 | with | =
=
.

Test 3 . Determine vector coordinates a + b , if a = (–3; 4), b = = (5; –2):

Test 4. Determine vector coordinates a – b , if a = (2; –1), b = = (3; –4):

Test 5 . Find coordinates of vector 3 a , if a = (2; –1):

Test 6 . Find dot product a , b vectors a = (1; –4), b = (–2; 3):

Test 7 . Find the length of a vector a = (–12; 5):

3)
;

Answers to test tasks

1.3. Elements of analytical geometry in space

A rectangular coordinate system in space consists of three mutually perpendicular coordinate axes intersecting at the same point (origin 0) and having a direction, as well as a scale unit along each axis (Figure 17).

Figure 17

Point position M on the plane is uniquely determined by three numbers - its coordinates M(NS T ; at T ; z T), where NS T- abscissa, at T- ordinate, z T- applicate.

Each of them gives a distance from a point M to one of the coordinate planes with a sign that takes into account which side of this plane the point is located: whether it is taken in the direction of the positive or negative direction of the third axis.

Three coordinate planes divide the space into 8 parts (octants).

Distance between two points A(NS A ; at A ; z A) and B(NS V ; at V ; z V) is calculated by the formula

Given points A(NS 1 ; at 1 ; z 1) and B(NS 2 ; at 2 ; z 2). Then the coordinates of the point WITH(NS; at; z) dividing the segment
in relation to, are expressed by the following formulas:

Example 1 . Find distance AB, if A(3; 2; –10) and V(–1; 4; –5).

Solution

Distance AB calculated by the formula

The set of all points whose coordinates satisfy the equation with three variables makes up a certain surface.

The set of points, the coordinates of which satisfy two equations, constitutes a certain line - the line of intersection of the corresponding two surfaces.

Any equation of the first degree represents a plane, and, conversely, any plane can be represented by equations of the first degree.

Options A, B, C are the coordinates of the normal vector perpendicular to the plane, i.e. n = (A; B; C).

Equation of the plane in the segments cut off on the axes: a- along the axis ОX, b- along the axis OY, with- along the axis ОZ:

Let two planes be given A 1 x + B 1 y + C 1 z + D 1 = 0, A 2 x + B 2 y + C 2 z + + D 2 = 0.

Condition of parallelism of planes:
.

Condition of perpendicularity of planes:

The angle between the planes is determined by the following formula:

.

Let the plane pass through the points M 1 (x 1 ; y 1 ; z 1), M 2 (x 2 ; y 2 ; z 2), M 3 (x 3 ; y 3 ; z 3).

Then its equation has the form:

Distance from point M 0 (x 0 ; y 0 ; z 0) to the plane Ax + By + Cz + D= 0 is found by the formula

.

Test 1. Plane
goes through the point:

1) A(–1; 6; 3);

2) B(3; –2; –5);

3) C(0; 4; –1);

4) D(2; 0; 5).

Test 2 . Plane equation ОXY following:

1) z = 0;

2) x = 0;

3) y = 0.

Example 2 . Write the equation of the plane parallel to the plane ОXY and passing through the point (2; –5; 3).

Solution

Since the plane is parallel to the plane ОXY, its equation has the form Cz + D= 0 (vector  = (0; 0; WITH)  OHY).

Since the plane passes through the point (2; –5; 3), then C  3 + D= 0 or as D = –3C.

Thus, CZ – 3C= 0. Since WITH≠ 0, then z – 3 = 0.

Answer: z – 3 = 0.

Test 3 . The equation of the plane passing through the origin and perpendicular to the vector (3; –1; –4) has the form:

1)

2)

3)

4)

Test 4 . The value of the line cut along the axis OY plane
is equal to:

Example 3 . Write the equation of the plane:

1. Parallel plane
and passing through the point A(2; 0; –1).

2. Perpendicular plane
and passing through the point B(0; 2; 0).

Solution

The plane equations will be sought in the form A 1 x + B 1 y + C 1 z + D 1 = 0.

1. Since the planes are parallel, then
From here A= 3t,B= –t,C= 2t, where tR... Let be t= 1. Then A = 3, B = –1, C= 2. Therefore, the equation takes the form
Point coordinates A belonging to the plane turn the equation into true equality. Therefore, 32 - 10 + 2 (–1) + D= 0. Where from D= 4.

Answer:

2. Since the planes are perpendicular, then 3  A – 1  B + 2  C = 0.

Since there are three variables, and the equation is one, the two variables take on arbitrary values ​​simultaneously not equal to zero. Let be A = 1, B= 3. Then C= 0. The equation takes the form
D= –6.

Answer:

Test 5 . Pick plane parallel to plane x – 2y + 7z – 2 = 0:

1)

4)

Test 6 . Pick plane perpendicular to plane x– 2y+ + 6z– 2 = 0:

1)

4)

Test 7 . Cosine of the angle between planes 3 x + y – z- 1 = 0 and x – 4y – – 5z+ 3 = 0 is determined by the formula:

1)

2)

3)

Test 8 . Distance from point (3; 1; –1) to plane 3 x – y + 5z+ 1 = 0 is determined by the formula:

1)

2)

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Option 1 Option 2 We used a template for creating tests in PowerPoint MCOU "Pogorelskaya Secondary School" MM Koscheev

Test result Correct: 14 Errors: 0 Mark: 5 Time: 3 min. 29 sec. still fix

Option 1 b) 360 ° a) 180 ° c) 246 ° d) 274 ° e) 454 °

Option 1 c) 22 a) -22 b) 0 d) 8 e) 1

Option 1 e) 5 d) 0 a) 7

Option 1 b) stupid e) do not exist, since their origins do not coincide c) 0 ° d) acute a) straight

Option 1 b) 10.5 e) for no a) -10.5

Option 1 a) -10.5 b) 10.5 e) under no circumstances

Option 1 e) 0 b) it is impossible to determine a) -6 d) 4 c) 6

Option 1 b) 28 e) impossible to determine a) 70 d) -45.5 c) 91

Option 1 9. The two sides of the triangle are 16 and 5, and the angle between them is 120 °. Which of the specified intervals belongs to the third side length? d) e) (19; 31] a) (0; 7] b) (7; 11] c) a) (0; 7] b) (7; 11] d)

Option 1 13. The radius of the circle circumscribed about the triangle ABC is 0.5. Find the ratio of the sine of angle B to the length of the AC side. e) 1 c) 1, 3 a) 0.5 d) 2

Option 1 14. In a triangle ABC, the lengths of the sides BC and AB are 5 and 7, respectively, and

Option 2 c) 360 ° a) 180 ° b) 246 ° d) 274 ° e) 454 °

Option 2 e) 22 a) -22 b) 0 d) 8 c) 4

Option 2 a) 10 d) 17 e) 15

Option 2 c) is equal to 0 ° e) do not exist, since their origins do not coincide c) blunt d) acute a) straight

Option 2 b) 10.5 e) for no a) -10.5

Option 2 a) - 10.5 e) for no c) 10.5

Option 2 d) 0 b) it is impossible to determine a) -6 e) 4 c) 6

Option 2 a) 70 e) impossible to determine b) 28 d) -45.5 c) 91

Option 2 9. The two sides of the triangle are 12 and 7, and the angle between them is 60 °. Which of the specified intervals belongs to the third side length? e) (7; 11) d) (19; 31] a) (0; 7] b) c) e) (19; 31] c)

Option 2 13. The radius of the circle circumscribed about the triangle ABC is equal to 2. Find the ratio of the sine of angle B to the length of the AC side. a) 0.25 c) 1, 3 e) 1 d) 2

Option 2 14. In a triangle ABC, the lengths of the sides AC and AB are 9 and 7, respectively, and

Keys to the test: “Dot product of vectors. Triangle theorems ”. Option 1 1 2 3 4 5 6 7 8 9 10 11 12 13 14 Repl. b c e b c a e b d a c c e d 2 option 1 2 3 4 5 6 7 8 9 10 11 12 13 14 Otv. c d a c d b d a d d c a a d Literature L.I. Zvavich, E, V. Potoskuev Geometry tests Grade 9 to the textbook L.S. Atanasyan et al. M.: "Exam" publishing house 2013 - 128 p.