Records labeled "dot product of vectors". Dot product of vectors Test 6 dot product of vectors
This test can be used in the classroom of intermediate, generalizing or final control of students' knowledge. For the test to work correctly, you must set a low security level (service-macro-security)
Download:
Preview:
To use the preview of presentations, create yourself a Google account (account) and log into it: https://accounts.google.com
Slide captions:
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) blunt 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.
2. Simplify the equation by multiplying both sides by 7. We get 7y 2 -9y + 2 \u003d 0. By Vieta's theorem, the sum of the roots of the quadratic equation ax 2 + bx + c \u003d 0 is –b / a. Means:
3. A total of 880 passengers. Of these, 35% are men, which means women and children 100% -35% \u003d 65%. Find 65% of 880. To find the percentage of the number, you need to convert the percentage to a decimal and multiply by the given number.
65% \u003d 0.65; multiply 880 by 0.65, we get 572. So many women and children, and 75% of them are women, the remaining 25% of 572 are children. Find the percentage of the number again. 25% of 572. We convert 25% to a decimal fraction (will be 0.25) and multiply by 572. We consider: 572 · 0.25 \u003d 143. These are kids. Women: 572-143 \u003d 429 .
Is it shorter?
25% is a quarter of 100%, therefore, we reason like this: divide 572 by 4, we get 143 (dividing by 4 is easier than multiplying by 0.25) - these are children, and 75% of women are three quarters, therefore, 143 is multiplied by 3 and we get 429.
4. By condition, we compose the inequality:
11x + 3<5x-6; слагаемые с переменной х соберем в левой части неравенства, а свободные члены — в правой:
11x-5x<-6-3; приводим подобные слагаемые:
6x<-9; делим обе части неравенства на 6:
x<-1,5. Ответ: E).
5. We write 990 ° as 2 · 360 ° + 270 °. Then cos 990 °\u003d cos (2 360 ° + 270 °) \u003d cos 270 ° \u003d 0.
6. Let us apply the formula for solving the simplest equation tg t \u003d a.
t \u003d arctan a + πn, nєZ. We have t \u003d 4x.
7. We have: the first term of the arithmetic progression a 1 \u003d 25... Difference of arithmetic progression d\u003d a 2 -a 1 \u003d 30-25 =5. Let's apply the formula to find the sum of the first n members of the arithmetic progression and substitute our values \u200b\u200binto it a 1 \u003d 25, d \u003d 5 and n \u003d 22, since it is required to find the amount 22 members of the progression.
8. The graph of this quadratic function y \u003d x 2 -x-6 serves as a parabola, the branches of which are directed upwards, and the apex of the parabola is at the point O '(m; n)... This is the lowest point of the graph, therefore, its lowest value n the function will have at x \u003d m \u003d -b / (2a) \u003d 1/2. Answer: D).
9. In an isosceles triangle, the sides are equal to each other. We denote the base by x... Then each side will be equal to (x + 3)... Knowing that the perimeter of a triangle is 15.6 cm, compose the equation:
x + (x + 3) + (x + 3) \u003d 15.6;
3x \u003d 9.6 → x \u003d 3.2 Is the base of the triangle, and each side will be 3.2 + 3 \u003d 6,2 ... Answer: the sides of the triangle are equal 6.2 cm; 6.2 cm vs 3.2 cm.
10. Everything is clear with the first inequality of the system. We solve the second inequality by the method of intervals. To do this, we find the roots of the square trinomial 4x 2 + 5x-6 and expand it into linear factors.
11. To the right by the main logarithmic identity, we obtain 7 ... Omitting the bases of the degrees (7) on the left and right sides of the equality. Remains: x 2 \u003d 1, from here x \u003d ± 1. Answer: C).
12. Let's square both sides of the equality. Applying the formulas for the logarithm of the degree and the logarithm of the product, we obtain a quadratic equation with respect to the logarithm of the number 5 by reason x... Let's introduce the variable at, we solve the quadratic equation with respect to at and back to the variable x... Find the values x and analyze the answers.
13. Task: solve the system. We will not decide - we will make a check. Let's substitute the proposed answers into the second equation of the system, since it is simpler: x + y \u003d 35... Of all the proposed pairs of solutions to the system, only the answer is suitable D).
8+27=35 and 27+8=35 ... It is not worth substituting these pairs in the first equation of the system, but if one more of the answers came to the second equation, then one would have to substitute in the first equality of the system.
14. Function scope is the set of argument values x, for which the right-hand side of equality makes sense. Since the arithmetic square root can only be extracted from a non-negative number, the following condition must be met: 6 + 2x≥0, it follows that 2x≥-6 or x≥-3. Since the denominator of the fraction must be different from zero, then we write: x ≠ 5... It turns out that you can take all numbers greater than or equal -3 but not equal 5 . Answer: [-3; 5) U (5; + ∞).
15. To find the largest and smallest values \u200b\u200bof a function on a given segment, you need to find the values \u200b\u200bof this function at the ends of the segment and at those critical points that belong to this segment, and then select the largest and smallest from all the obtained values \u200b\u200bof the function.
16 ... Consider a circle inscribed in a regular hexagon and recall how the radius of an inscribed circle is expressed r across the side of a regular hexagon and... Find the radius, then the side and perimeter of the hexagon.
17 ... Since all the side edges of the pyramid are inclined to the base at the same angle, the top of the pyramid is projected to a point ABOUT - the intersection of the diagonals of the rectangle lying at the base of the pyramid, because the point ABOUT must be equidistant from all tops of the base of the pyramid.
Find the diagonal AC of the rectangle ABCD. AC 2 \u003d AD 2 + CD 2;
AC 2 \u003d 32 2 +24 2 \u003d 1024 + 576 \u003d 1600 → AC \u003d 40cm. Then OS \u003d 20cm. Since Δ MOS is rectangular and isosceles (/ OSM \u003d 45 °), then MO \u003d OS \u003d 20cm. Let's apply the formula for the volume of the pyramid, substituting the required values.
18. Any section of a sphere by a plane is a circle.
Let a circle centered at point O 1 and radius OA be perpendicular to the radius of the ball OB and pass through its midpoint O 1. Then in a right-angled triangle AO 1 O hypotenuse OA \u003d 10 cm (ball radius), leg OO 1 \u003d 5 cm. By the Pythagorean theorem О 1 А 2 \u003d ОА 2 -ОО 1 2. Hence O 1 A 2 \u003d 10 2 -5 2 \u003d 100-25 \u003d 75. The cross-sectional area is the area of \u200b\u200bour circle, we find by the formula S \u003d πr 2 \u003d π ∙ O 1 A 2 \u003d 75π cm 2.
19. Let be a 1and a 2 - the required coordinates of the vector. Since the vectors are mutually perpendicular, their dot product is zero. Let's write down: 2a 1 + 7a 2 \u003d 0. Let us express a 1 through a 2. Then a 1 \u003d -3.5a 2. Since the lengths of the vectors are equal, we have the equality: a 1 2 + a 2 2 \u003d 2 2 +7 2... Substitute in this equality the value a 1. We get: (3.5a 2) 2 + a 2 2 \u003d 4 + 49; simplify: 12.25a 2 2 + a 2 2 \u003d 53;
13.25a 2 2 \u003d 53, hence a 2 2 \u003d 53: 13.25 \u003d 4. It turns out two values a 2 \u003d ± 2. If a 2 \u003d -2, then a 1 \u003d -3.5 ∙ (-2) \u003d 7. If a 2 \u003d 2, then a 1 \u003d -7. Searched coordinates (7; -2) or (-7; 2) ... Answer: IN).
20. Let's simplify the denominator of the fraction. To do this, we open the brackets and bring the fractions under the root sign to a common denominator.
21. Let us bring the expression in brackets to a common denominator. Division is replaced by multiplication by the inverse of the divisor. We apply the formulas for the square of the difference between two expressions and the difference between the squares of two expressions. Let's reduce the fraction.
22. To solve this system of inequalities, you need to solve each inequality separately and find a general solution to the two inequalities. We solve 1st inequality. Move all the terms to the left, take the common factor outside the bracket.
x 2 ∙ 4 x -4 x +1\u003e 0;
x 2 ∙ 4 x -4 x ∙ 4\u003e 0;
4 x (x 2 -4)\u003e 0. Since the exponential function for any indicator takes only positive values, then 4 x\u003e 0, therefore, x 2 -4\u003e 0.
(x-2) (x + 2)\u003e 0.
We solve 2nd inequality.
Represent the left and right sides as degrees with base 2.
2 - x ≥2 3. Since the exponential function with a base greater than one, increases by R, we omit the bases, keeping the inequality sign.
X≥3 → x≤-3.
We find a general solution.
Answer: (-∞; -3].
23. According to the casting formula, cosine is converted to sine 3x... After reducing similar terms and dividing both sides of the inequality by 2 , we obtain the simplest inequality of the form: sin t\u003e a... We find the solution to this inequality by the formula:
arcsin a + 2πn
24. Let's simplify this function. By Vieta's theorem, we find the roots of the square trinomial x 2 -x-6 (x 1 \u003d -2 , x 2 \u003d 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 \u003d 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.
Option 1.
Option 2.
e) Is this angle acute, straight or obtuse (justify the 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 vectors AB and CD.
b) Find the lengths of the vectors AB and CD.
c) Find the dot product of vectors AB and CD.
d) Find the cosine of the angle between vectors AB and CD.
e) Is this angle acute, straight or obtuse (justify the answer)?
f) At what value of x are the vectors CB and DQ perpendicular?
2. In an isosceles triangle ABC, the angle B is straight, AC \u003d 2√2, BD is the median of the triangle. Calculate the dot products of vectors BD AC, BD BC, BD BD.
Option 2.
1. Points M (2; 3), P (-2; 0), O (0; 0), K (-5; -12), R (4; y) are given.
a) Find the coordinates of the vectors МР and OK.
b) Find the lengths of the vectors МР and ОК.
c) Find the dot product of the vectors МР and OK.
d) Find the cosine of the angle between the vectors МР and OK.
e) Is this angle acute, straight or obtuse (justify the answer)?
f) At what value of y are the vectors RK and MR perpendicular?
2. In an equilateral triangle МНР НК - bisector, МН \u003d 2. Calculate the scalar products of vectors НК МР, НК НР, РМ РМ
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 vectors AB and CD.
b) Find the lengths of the vectors AB and CD.
c) Find the dot product of vectors AB and CD.
d) Find the cosine of the angle between vectors AB and CD.
e) Is this angle acute, straight or obtuse (justify the answer)?
f) At what value of x are the vectors CB and DQ perpendicular?
2. In an isosceles triangle ABC, the angle B is straight, AC \u003d 2√2, BD is the median of the triangle. Calculate the dot products of vectors BD AC, BD BC, BD BD.
Option 2.
1. Points M (2; 3), P (-2; 0), O (0; 0), K (-5; -12), R (4; y) are given.
a) Find the coordinates of the vectors МР and OK.
b) Find the lengths of the vectors МР and ОК.
c) Find the dot product of the vectors МР and OK.
d) Find the cosine of the angle between the vectors МР and OK.
e) Is this angle acute, straight or obtuse (justify the answer)?
f) At what value of y are the vectors RK and MR perpendicular?
2. In an equilateral triangle МНР НК - bisector, МН \u003d 2. Calculate the scalar products of vectors НК МР, НК НР, РМ РМ
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 indicated 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 Oohgiven 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, and = ( 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, ie. a b = x 1 x 2 + + y 1 y 2 .
Consequence. Vector length and = (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 from , d .
3. Find the length of the vector from .
Decision
1. By property 3, we find the coordinates of vectors 2 and , –and , 3b , 2b : 2and = = 2(–2; 3) = (–4; 6), –and = –(–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 from , d : from = 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 |
from
|
=
=
.
Test 3 . Determine vector coordinates and + b , if a and = (–3; 4), b = = (5; –2):
Test 4. Determine vector coordinates and – b , if a and = (2; –1), b = = (3; –4):
Test 5 . Find coordinates of vector 3 and , if a and = (2; –1):
Test 6 . Find dot product a , b vectors and = (1; –4), b = (–2; 3):
Test 7 . Find the length of a vector and = (–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(x t ; at t ; z t), where x 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(x AND ; at AND ; z AND) and B(x IN ; at IN ; z IN) is calculated by the formula
Given points A(x 1 ;
at 1 ;
z 1) and B(x 2 ;
at 2 ;
z 2). Then the coordinates of the point FROM(x;
at;
z) dividing the segment 
in relation to, are expressed by the following formulas:
Example 1 . Find distance AB, if a AND(3; 2; –10) and IN(–1; 4; –5).
Decision
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,
from - 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 \u003d 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).
Decision
Since the plane is parallel to the plane ОXY, its equation has the form Cz + D \u003d 0 (vector = (0; 0; FROM) OHY).
Since the plane passes through the point (2; –5; 3), then C 3 + D \u003d 0 or as D = –3C.
Thus, CZ – 3C \u003d 0. Since FROM ≠ 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).
Decision
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= 2twhere tR... Let be t\u003d 1. Then A
= 3, B =
–1, C \u003d 2. Therefore, the equation takes the form 
Point coordinates ANDbelonging to the plane turn the equation into true equality. Therefore, 32 - 10 + 2 (–1) + D\u003d 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 \u200b\u200bsimultaneously not equal to zero. Let be A
= 1, B \u003d 3. Then C\u003d 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 \u003d 0 and x – 4y – – 5z + 3 \u003d 0 is determined by the formula:
1)
2)
3)
Test 8 . Distance from point (3; 1; –1) to plane 3 x – y + 5z + 1 \u003d 0 is determined by the formula:
1)
2)
