Dot product of vectors. Dot product of vectors Operations on vectors in coordinate form

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Option 1 Option 2 Used a template for creating tests in PowerPoint MKOU "Pogorelskaya Secondary School" Koshcheev M.M.

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

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

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

Option 1 e) 5 d) 0 a) 7

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

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

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

Option 1 e) 0 b) 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. Two sides of a triangle are equal to 16 and 5, and the angle between them is 120°. To which of the indicated intervals does the length of the third side belong? 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 triangle ABC is 0.5. Find the ratio of the sine of angle B to the length of side AC. e) 1 c) 1.3 a) 0.5 d) 2

Option 1 14. In triangle ABC, the lengths of sides BC and AB are equal to 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) equal to 0 ° e) do not exist, since their origins do not coincide c) obtuse d) acute a) straight

Option 2 b) 10.5 d) under no circumstances a) -10.5

Option 2 a) - 10.5 d) under no circumstances c) 10.5

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

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

Option 2 9. Two sides of the triangle are equal to 12 and 7, and the angle between them is 60°. To which of the indicated intervals does the length of the third side belong? 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 triangle ABC is equal to 2. Find the ratio of the sine of angle B to the length of side AC. a) 0.25 c) 1.3 d) 1 d) 2

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

Keys to the test: “Scalar product of vectors. Triangle theorems". Option 1 1 2 3 4 5 6 7 8 9 10 11 12 13 14 Answer. b c d b c a d b d a c c d d Option 2 1 2 3 4 5 6 7 8 9 10 11 12 13 14 Answer. c d a c d b d a d d c a a g Literature L.I. Zvavich, E, V. Potoskuev Tests in geometry 9th grade for the textbook L.S. Atanasyan and others. M.: “Exam” publishing house, 2013 - 128 p.

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Option 1 A template for creating tests in PowerPoint was used by MKOU "Pogorelskaya Secondary School" Koshcheev M.M.

Option 1 b) blunt a) sharp c) straight

Option 1 c) 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 true?

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 ₁a₂a₃+ b ₁ b ₂ b ₃

Option 1 b) - a² 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. Given the coordinates of the points: 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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Option 2 A template for creating tests in PowerPoint was used by MKOU "Pogorelskaya Secondary School" Koshcheev M.M.

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

Option 2 a) sharp b) obtuse c) straight

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

Option 2 b) -½∙a² a) ½∙a²

Option 2 4. ABCA ₁В₁С₁ – prism,

Option 2 5. Which statement is true?

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) - a² 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. Given the coordinates of the points: C(3 ; - 2 ; 1) , D(- 1 ; 2 ; 1) , M(2 ; -3 ; 3 ) , N(-1 ; 1 ; -2) . Then the cosine of the angle between the lines CD and MN is equal to......

Option 2 c) 4

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

Option 1.

Option 2.

e) Is this angle acute, right or obtuse (justify your answer)?

Option 1.

1. Given points A(1; 3), B(4; 7), C(-1; -1), D(7; 5), Q(x; 3)

a) Find the coordinates of the vectors AB and CD.

b) Find the lengths of the vectors AB and CD.

c) Find the scalar product of the vectors AB and CD.

d) Find the cosine of the angle between vectors AB and CD.

e) Is this angle acute, right or obtuse (justify your answer)?

f) At what value of x are the vectors CB and DQ perpendicular?

2. In an isosceles triangle ABC, angle B is a right angle, AC = 2√2, ВD is the median of the triangle. Calculate the scalar products of the vectors BD AC, BD BC, BD BD.

Option 2.

1. Given points M(2; 3), P(-2; 0), O(0; 0), K(-5; -12), R(4; y).

a) Find the coordinates of the vectors MR and OK.

b) Find the lengths of the vectors MR and OK.

c) Find the scalar product of the vectors MR and OK.

d) Find the cosine of the angle between the vectors MR and OK.

e) Is this angle acute, right or obtuse (justify your answer)?

f) At what value of y are the vectors PK and MR perpendicular?

2. In the equilateral triangle MNR NK is the bisector, MN = 2. Calculate the scalar products of the vectors NK MR, NK NR, RM RM

Option 1.

1. Given points A(1; 3), B(4; 7), C(-1; -1), D(7; 5), Q(x; 3)

a) Find the coordinates of the vectors AB and CD.

b) Find the lengths of the vectors AB and CD.

c) Find the scalar product of the vectors AB and CD.

d) Find the cosine of the angle between vectors AB and CD.

e) Is this angle acute, right or obtuse (justify your answer)?

f) At what value of x are the vectors CB and DQ perpendicular?

2. In an isosceles triangle ABC, angle B is a right angle, AC = 2√2, ВD is the median of the triangle. Calculate the scalar products of the vectors BD AC, BD BC, BD BD.

Option 2.

1. Given points M(2; 3), P(-2; 0), O(0; 0), K(-5; -12), R(4; y).

a) Find the coordinates of the vectors MR and OK.

b) Find the lengths of the vectors MR and OK.

c) Find the scalar product of the vectors MR and OK.

d) Find the cosine of the angle between the vectors MR and OK.

e) Is this angle acute, right or obtuse (justify your answer)?

f) At what value of y are the vectors PK and MR perpendicular?

2. In the equilateral triangle MNR NK is the bisector, MN = 2. Calculate the scalar products of the vectors NK MR, NK NR, RM RM

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Dot product a  b two non-zero vectors a And b is a number equal to the product of the lengths of these vectors and 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 the vectors a And b .

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

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

if 0    That 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 Ohoo vectors are given 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 component vectors, 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 by a number  is equal to the product of the corresponding coordinate of this vector by , i.e.  A = ( X 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, i.e. 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. Vectors are given
b = 3i – j .

Required:

1. Find

2. Find the scalar product of vectors With , d .

3. Find the length of the vector With .

Solution

1. Using 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).

Using 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 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 the coordinates of vector 3 A , If A = (2; –1):

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

Test 7 . Find the length of the 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 determined uniquely by three numbers - its coordinates M(X T ; at T ; z T), Where X T– abscissa, at T– ordinate, z T– applicate.

Each of them gives the distance from the 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 space into 8 parts (octants).

Distance between two points A(X A ; at A ; z A) And B(X IN ; at IN ; z IN) is calculated by the formula

Let points be given A(X 1 ; at 1 ; z 1) and B(X 2 ; at 2 ; z 2). Then the coordinates of the point WITH(X; at; z), dividing the segment
in relation, are expressed by the following formulas:

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

Solution

Distance AB calculated by the formula

The set of all points whose coordinates satisfy an equation with three variables constitutes a certain surface.

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

Every equation of the first degree represents a plane, and, conversely, every 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 segments cut off on the axes: a– along the axis OX, b– along the axis OY, With– along the axis OZ:

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 for parallel planes:
.

Condition for planes to be perpendicular:

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 looks like:

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

Test 1. Plane
passes 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 OXY following:

1) z = 0;

2) x = 0;

3) y = 0.

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

Solution

Since the plane is parallel to the plane OXY, 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 whatever 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:

Test 4 . The size of the segment cut off 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 to the plane
and passing through the point B(0; 2; 0).

Solution

We will look for plane equations 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 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 a true equality. Therefore, 32 – 10 + 2(–1) + D= 0. From D= 4.

Answer:

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

Since there are three variables, but one equation, two variables take arbitrary values that are not equal to zero at the same time. Let A = 1, B= 3. Then C= 0. The equation becomes
D= –6.