2.2 · 4/7. Зарби вектор ба адад - Геометрия — синфи 9
Зарби вектор ба адад
Масалан, 2 ⋅ ( 3 ; − 1 ) = ( 6 ; − 2 ) , 2\cdot(3;\,-1) = (6;\,-2), 2 ⋅ ( 3 ; − 1 ) = ( 6 ; − 2 ) , − 3 ⋅ ( 1 ; 4 ) = ( − 3 ; − 12 ) , -3\cdot(1;\,4) = (-3;\,-12), − 3 ⋅ ( 1 ; 4
навбатӣ 2.2 · 5/7. Векторҳои коллинеарӣ. Векторҳои воҳидӣ 1 соат
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( − 3 ; − 12 ) ,
0 ⋅ a ⃗ = 0 ⃗ . 0\cdot\vec a = \vec 0. 0 ⋅ a = 0 . a ⃗ + a ⃗ = ( 2 x ; 2 y ) = 2 a ⃗ . \vec a + \vec a = (2x;\,2y) = 2\vec a. a + a = ( 2 x ; 2 y ) = 2 a . Мисоли 1. Вектори a ⃗ \vec a a 4 a ⃗ 4\vec a 4 a − 3 a ⃗ -3\vec a − 3 a
Дарозӣ ва самти λ a ⃗ \lambda\vec a λ a Исбот. Дарозӣ. ∣ λ a ⃗ ∣ = ( λ x ) 2 + ( λ y ) 2 = λ 2 ( x 2 + y 2 ) = ∣ λ ∣ x 2 + y 2 = ∣ λ ∣ ⋅ ∣ a ⃗ ∣ . |\lambda\vec a| = \sqrt{(\lambda x)^2 + (\lambda y)^2} = \sqrt{\lambda^2(x^2 + y^2)} = |\lambda|\sqrt{x^2 + y^2} = |\lambda|\cdot|\vec a|. ∣ λ a ∣ = ( λ x ) 2 + ( λ y ) 2 = λ 2 ( x 2 + y 2 ) = ∣ λ ∣ x 2 + y 2 = ∣ λ ∣ ⋅ ∣ a ∣.
Самт. Дар системаи координатаҳо аз O ( 0 ; 0 ) O(0;\,0) O ( 0 ; 0 ) O A → = a ⃗ \overrightarrow{OA} = \vec a O A = a O B → = λ a ⃗ \overrightarrow{OB} = \lambda\vec a OB = λ a A ( x ; y ) , A(x;\,y), A ( x ; y ) , B ( λ x ; λ y ) . B(\lambda x;\,\lambda y). B ( λ x ; λ y ) . r = ∣ O A ∣ > 0. r = |OA| > 0. r = ∣ O A ∣ > 0. O B = ∣ λ ∣ r . OB = |\lambda|\,r. OB = ∣ λ ∣ r . A B AB A B A B = ( ( λ − 1 ) x ) 2 + ( ( λ − 1 ) y ) 2 = ∣ λ − 1 ∣ r . AB = \sqrt{((\lambda - 1)x)^2 + ((\lambda - 1)y)^2} = |\lambda - 1|\,r. A B = (( λ − 1 ) x ) 2 + (( λ − 1 ) y )
Ҳолати λ = 0 \lambda = 0 λ = 0 a ⃗ = 0 ⃗ \vec a = \vec 0 a = 0
Хосиятҳои зарб Исбот. Бигузор a ⃗ = ( x 1 ; y 1 ) \vec a = (x_1;\,y_1) a = ( x 1 ; y 1 ) b ⃗ = ( x 2 ; y 2 ) . \vec b = (x_2;\,y_2). b = ( x 2 ; y 2
( λ + μ ) a ⃗ = ( ( λ + μ ) x 1 ; ( λ + μ ) y 1 ) = ( λ x 1 + μ x 1 ; λ y 1 + μ y 1 ) = λ a ⃗ + μ a ⃗ . (\lambda + \mu)\vec a = ((\lambda + \mu)x_1;\; (\lambda + \mu)y_1) = (\lambda x_1 + \mu x_1;\; \lambda y_1 + \mu y_1) = \lambda\vec a + \mu\vec a. ( λ + μ ) a = (( λ + μ ) x 1 ; ( λ + μ ) y 1 ) = ( λ x 1 + μ x 1 ; λ y 1 + μ y 1 ) = λ a + μ a . Исботи хосияти 2 λ ( a ⃗ + b ⃗ ) = ( λ ( x 1 + x 2 ) ; λ ( y 1 + y 2 ) ) = ( λ x 1 + λ x 2 ; λ y 1 + λ y 2 ) = λ a ⃗ + λ b ⃗ . \lambda(\vec a + \vec b) = (\lambda(x_1 + x_2);\; \lambda(y_1 + y_2)) = (\lambda x_1 + \lambda x_2;\; \lambda y_1 + \lambda y_2) = \lambda\vec a + \lambda\vec b. λ ( a + b ) = ( λ ( x 1 + x 2 ) ; λ ( y 1 + y 2 )) = ( λ x 1 + λ x 2 ; λ y 1 + λ y 2 ) = λ a + λ b .
Аз хосиятҳо: 1 ⋅ a ⃗ = a ⃗ , 1\cdot\vec a = \vec a, 1 ⋅ a = a , ( − 1 ) a ⃗ = − a ⃗ , (-1)\vec a = -\vec a, ( − 1 ) a = − a , λ ( μ a ⃗ ) = ( λ μ ) a ⃗ . \lambda(\mu\vec a) = (\lambda\mu)\vec a. λ ( μ a ) = ( λ μ ) a 2 ( a ⃗ + 3 b ⃗ ) − 5 b ⃗ = 2 a ⃗ + b ⃗ . 2(\vec a + 3\vec b) - 5\vec b = 2\vec a + \vec b. 2 ( a + 3 b
Мисоли 2. a ⃗ = ( 2 ; 1 ) \vec a = (2;\,1) a = ( 2 ; 1 ) b ⃗ = ( − 1 ; 1 ) \vec b = (-1;\,1) b = ( − 1 ; 1 ) 2 a ⃗ + 3 b ⃗ 2\vec a + 3\vec b 2 a + 3 b
Ҳал. 2 a ⃗ = ( 4 ; 2 ) , 2\vec a = (4;\,2), 2 a = ( 4 ; 2 ) , 3 b ⃗ = ( − 3 ; 3 ) . 3\vec b = (-3;\,3). 3 b = ( − 3 ; 3 ) . O O O 2 a ⃗ 2\vec a 2 a 3 b ⃗ 3\vec b 3 b 2 a ⃗ + 3 b ⃗ = ( 4 − 3 ; 2 + 3 ) = ( 1 ; 5 ) . 2\vec a + 3\vec b = (4 - 3;\; 2 + 3) = (1;\,5). 2 a + 3 b
Худатон санҷед Вектори a ⃗ \vec a a λ \lambda λ 1 2 \tfrac12 2 1 λ a ⃗ \lambda\vec a λ a λ \lambda λ b ⃗ \vec b b
Вектор ба адад санҷида бин! Аз нав
Кор дар синф
Машқҳо 1. Агар λ \lambda λ a ⃗ \vec a a ∣ λ a ⃗ ∣ |\lambda\vec a| ∣ λ a ∣ λ = 3 , \lambda = 3, λ = 3 , a ⃗ = ( 3 ; 4 ) ; \vec a = (3;\,4); a = ( 3 ; 4 ) ; λ = − 2 , \lambda = -2, λ = − 2 , a ⃗ = ( − 5 ; 12 ) ; \vec a = (-5;\,12); a = ( − 5 ; 12 ) ; λ = − 5 , \lambda = -5, λ = − 5 , a ⃗ = ( − 6 ; 8 ) ; \vec a = (-6;\,8); a = ( − 6 ; 8 ) ; λ = 1 2 , \lambda = \dfrac12, λ = 2 1 , a ⃗ = ( 3 4 ; 1 ) . \vec a = \left(\dfrac34;\,1\right). a = ( 4 3 ; 1 ) .
2. Векторҳои a ⃗ \vec a a b ⃗ \vec b b 2 a ⃗ + 3 b ⃗ 2\vec a + 3\vec b 2 a + 3 b
3. Агар a ⃗ = ( 3 ; 2 ) \vec a = (3;\,2) a = ( 3 ; 2 ) b ⃗ = ( 0 ; − 1 ) \vec b = (0;\,-1) b = ( 0 ; − 1 ) − 2 a ⃗ + 4 b ⃗ -2\vec a + 4\vec b − 2 a + 4 b ∣ − 2 a ⃗ + 4 b ⃗ ∣ |-2\vec a + 4\vec b| ∣ − 2 a + 4 b
4. Агар a ⃗ = ( 3 ; 4 ) \vec a = (3;\,4) a = ( 3 ; 4 ) b ⃗ = ( 5 ; 12 ) \vec b = (5;\,12) b = ( 5 ; 12 ) 5 a ⃗ + 2 b ⃗ 5\vec a + 2\vec b 5 a + 2 b ∣ 5 a ⃗ + 2 b ⃗ ∣ |5\vec a + 2\vec b| ∣5 a + 2 b ∣
5. Адади λ \lambda λ ∣ λ a ⃗ ∣ = 5 |\lambda\vec a| = 5 ∣ λ a ∣ = 5 a ⃗ = ( − 6 ; 8 ) ; \vec a = (-6;\,8); a = ( − 6 ; 8 ) ; a ⃗ = ( 3 ; − 4 ) . \vec a = (3;\,-4). a = ( 3 ; − 4 ) .
6. Нуқтаи M M M A B AB A B O O O O M → = 1 2 ( O A → + O B → ) . \overrightarrow{OM} = \dfrac12\left(\overrightarrow{OA} + \overrightarrow{OB}\right). OM = 2 1 ( O A + OB
Нишондод A M → = M B → , \overrightarrow{AM} = \overrightarrow{MB}, A M = MB , O M → − O A → = O B → − O M → . \overrightarrow{OM} - \overrightarrow{OA} = \overrightarrow{OB} - \overrightarrow{OM}. OM − O A = OB
Вазифаи хонагӣ Машқҳои 4 ва 5; ҳавасмандон — машқи 6.
Дар ин дарс
Ман метавонам векторро ба адад зарб кунам (бо координатаҳо ва бо сохтан).Ман медонам ва исбот карда метавонам: ∣ λ a ⃗ ∣ = ∣ λ ∣ ⋅ ∣ a ⃗ ∣ |\lambda\vec a| = |\lambda|\cdot|\vec a| ∣ λ a ∣ = ∣ λ ∣ ⋅ ∣ a ∣ λ a ⃗ . \lambda\vec a. λ a . Ман метавонам хосиятҳои зарби вектор ба ададро барои ҳисоб истифода барам. Такрор (2 дақиқа)
a ⃗ = ( 3 ; − 2 ) . \vec a = (3;\,-2). a = ( 3 ; − 2 ) . a ⃗ + a ⃗ \vec a + \vec a a + a Дарозии вектори ( 6 ; 8 ) (6;\,8) ( 6 ; 8 )
Кай ду вектор ҳамсамт аст? Кай муқобилсамт?
Ҷавобҳо
( 6 ; − 4 ) ; (6;\,-4); ( 6 ; − 4 ) ; 36 + 64 = 10 ; \sqrt{36 + 64} = 10; 36 + 64 = 10 ; Вазъият барои андешаронӣ Агар a ⃗ \vec a a a ⃗ + a ⃗ + a ⃗ \vec a + \vec a + \vec a a + a + a
Таъриф Ҳосили зарби вектори a ⃗ = ( x ; y ) \vec a = (x;\,y) a = ( x ; y ) λ \lambda λ λ a ⃗ \lambda\vec a λ a
λ a ⃗ = ( λ x ; λ y ) . \lambda\vec a = (\lambda x;\; \lambda y). λ a = ( λ x ; λ y ) . Низоми сохтан
Вектори a ⃗ \vec a a O O O
Агар λ > 0 \lambda > 0 λ > 0 O A OA
Расми 1. Векторҳои 4a (аз O то D) ва −3a (аз C то O). Теорема ∣ λ a ⃗ ∣ = ∣ λ ∣ ⋅ ∣ a ⃗ ∣ . |\lambda\vec a| = |\lambda|\cdot|\vec a|. ∣ λ a ∣ = ∣ λ ∣ ⋅ ∣ a
Агар λ ≥ 1 \lambda \ge 1 λ ≥ 1 O A + A B = r + ( λ − 1 ) r = λ r = O B . OA + AB = r + (\lambda - 1)r = \lambda r = OB. O A + A B = r + ( λ − 1 ) r = λ
Хосият Барои ҳар ду адади λ , \lambda, λ , μ \mu μ a ⃗ , \vec a, a , b ⃗ \vec b
Расми 2. Сохтани 2a + 3b: аввал 2a, баъд аз интиҳои он 3b. Расми 3. Векторҳои a ва b. 0 ⋅ a ⃗ = 0 ⃗ 0\cdot\vec a = \vec 0 0 ⋅ a = 0
Вектори − 2 a ⃗ -2\vec a − 2 a (a ⃗ ≠ 0 ⃗ \vec a \ne \vec 0 a нисбат ба
Як ҷавобро интихоб кунед A B C D
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Барои 4 a ⃗ 4\vec a 4 a O A → = a ⃗ , \overrightarrow{OA} = \vec a, O A = a , A B → = a ⃗ , \overrightarrow{AB} = \vec a, A B = a , B C → = a ⃗ , \overrightarrow{BC} = \vec a, BC = a , C D → = a ⃗ \overrightarrow{CD} = \vec a C D = a O D → = O A → + A B → + B C → + C D → = 4 a ⃗ . \overrightarrow{OD} = \overrightarrow{OA} + \overrightarrow{AB} + \overrightarrow{BC} + \overrightarrow{CD} = 4\vec a. O D = O A Барои − 3 a ⃗ -3\vec a − 3 a A O → = − a ⃗ , \overrightarrow{AO} = -\vec a, A O = − a , B A → = − a ⃗ , \overrightarrow{BA} = -\vec a, B A = − a , C B → = − a ⃗ \overrightarrow{CB} = -\vec a CB = − a C O → = C B → + B A → + A O → = − 3 a ⃗ . \overrightarrow{CO} = \overrightarrow{CB} + \overrightarrow{BA} + \overrightarrow{AO} = -3\vec a. CO = CB ∣.
a ⃗ ≠ 0 ⃗ \vec a \ne \vec 0 a = 0 r
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OB .
A B + O A = O B AB + OA = OB A B + O A = OB Агар 0 < λ < 1 0 < \lambda < 1 0 < λ < 1 O B + B A = λ r + ( 1 − λ ) r = r = O A . OB + BA = \lambda r + (1 - \lambda)r = r = OA. OB + B A = λ r + ( 1 − λ ) r = r = O A . B B B O A OA O A B B B O A OA O A Агар λ < 0 \lambda < 0 λ < 0 A B = ( 1 − λ ) r = r + ∣ λ ∣ r = O A + O B . AB = (1 - \lambda)r = r + |\lambda|r = OA + OB. A B = ( 1 − λ ) r = r + ∣ λ ∣ r = O A + OB . O O O A B AB A B A A A B B B O O O O A OA O A O B OB OB
( λ + μ ) a ⃗ = λ a ⃗ + μ a ⃗ ; (\lambda + \mu)\vec a = \lambda\vec a + \mu\vec a; ( λ + μ ) a = λ a + μ a ; λ ( a ⃗ + b ⃗ ) = λ a ⃗ + λ b ⃗ . \lambda(\vec a + \vec b) = \lambda\vec a + \lambda\vec b. λ ( a + b ∣ λ a ⃗ ∣ |\lambda\vec a|
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Муқобилсамт, дарозиаш 2 ∣ a ⃗ ∣ 2|\vec a| 2∣ a ∣
Ҳамсамт, дарозиаш ∣ a ⃗ ∣ 2 \dfrac{|\vec a|}{2} 2 ∣ a ∣
Ҳамсамт, дарозиаш 2 ∣ a ⃗ ∣ 2|\vec a| 2∣ a ∣
Муқобилсамт, дарозиаш ∣ a ⃗ ∣ 2 \dfrac{|\vec a|}{2} 2 ∣ a ∣
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