2.2 · 5/7. Векторҳои коллинеарӣ. Векторҳои воҳидӣ - Геометрия — синфи 9
Шарти коллинеарӣ
Исбот. (⇒) Бигузор a ⃗ \vec a a b ⃗ \vec b b
навбатӣ 2.2 · 6/7. Зарби скалярии векторҳо 1 соат
λ = ∣ b ⃗ ∣ ∣ a ⃗ ∣ > 0 \lambda = \dfrac{|\vec b|}{|\vec a|} > 0 ∣ λ ∣ ⋅ ∣ a ⃗ ∣ = ∣ b ⃗ ∣ |\lambda|\cdot|\vec a| = |\vec b| ∣ λ ∣ ⋅ ∣ a ∣ = ∣ b ∣ теоремаи 4/7.
b ⃗ = λ a ⃗ . \vec b = \lambda\vec a. b = λ a . λ = − ∣ b ⃗ ∣ ∣ a ⃗ ∣ \lambda = -\dfrac{|\vec b|}{|\vec a|} λ = − ∣ a ∣ ∣ b ∣ (⇐) Бигузор b ⃗ = λ a ⃗ . \vec b = \lambda\vec a. b = λ a . b ⃗ ≠ 0 ⃗ , \vec b \ne \vec 0, b = 0 , λ ≠ 0. \lambda \ne 0. λ = 0. b ⃗ \vec b b a ⃗ \vec a a (λ > 0 \lambda > 0 λ > 0 ё муқобилсамт (λ < 0 \lambda < 0 λ < 0 аст; яъне коллинеаранд. ∎
Шарти координатавӣ. 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 )
x 1 y 2 − x 2 y 1 = 0. x_1y_2 - x_2y_1 = 0. x 1 y 2 − x 2 y 1 = 0. Исбот. (⇒) b ⃗ = λ a ⃗ \vec b = \lambda\vec a b = λ a x 2 = λ x 1 , x_2 = \lambda x_1, x 2 = λ x 1 , y 2 = λ y 1 y_2 = \lambda y_1 y 2 = λ y 1 x 1 y 2 − x 2 y 1 = x 1 ⋅ λ y 1 − λ x 1 ⋅ y 1 = 0. x_1y_2 - x_2y_1 = x_1\cdot\lambda y_1 - \lambda x_1\cdot y_1 = 0. x 1 y 2 − x 2 y
(⇐) Бигузор x 1 y 2 = x 2 y 1 . x_1y_2 = x_2y_1. x 1 y 2 = x 2 y 1 . x 1 ≠ 0 x_1 \ne 0 x 1 = 0 λ = x 2 x 1 \lambda = \dfrac{x_2}{x_1} λ = x 1 x 2 x 2 = λ x 1 x_2 = \lambda x_1 x 2 = λ x 1 y 2 = x 2 y 1 x 1 = λ y 1 , y_2 = \dfrac{x_2y_1}{x_1} = \lambda y_1, y 2 = x 1 x 2 b ⃗ = λ a ⃗ . \vec b = \lambda\vec a. b = λ a . x 1 = 0 , x_1 = 0, x 1 = 0 , y 1 ≠ 0 y_1 \ne 0 y 1 = 0 a ⃗ ≠ 0 ⃗ \vec a \ne \vec 0 a = 0 0 = x 2 y 1 0 = x_2y_1 0 = x 2 y 1 x 2 = 0 ; x_2 = 0; x 2 = 0 ; λ = y 2 y 1 \lambda = \dfrac{y_2}{y_1} λ = y 1 y 2 y 2 = λ y 1 , y_2 = \lambda y_1, y 2 = λ y 1 , x 2 = 0 = λ ⋅ 0. x_2 = 0 = \lambda\cdot 0. x 2 = 0 = λ ⋅ 0. b ⃗ = λ a ⃗ , \vec b = \lambda\vec a, b = λ a ,
Нисбати x 2 x 1 = y 2 y 1 \dfrac{x_2}{x_1} = \dfrac{y_2}{y_1} x 1 x 2 = y 1 y 2 ( 0 ; 3 ) (0;\,3) ( 0 ; 3 ) ( 0 ; − 6 ) (0;\,-6) ( 0 ; − 6 ) 0 ⋅ ( − 6 ) − 0 ⋅ 3 = 0 0\cdot(-6) - 0\cdot 3 = 0 0 ⋅ ( − 6 ) − 0 ⋅ 3 = 0 x 1 y 2 − x 2 y 1 = 0 x_1y_2 - x_2y_1 = 0 x 1 y 2 − x 2 y 1 =
Мисоли 1. Исбот кунед, ки a ⃗ = ( 5 ; 6 ) \vec a = (5;\,6) a = ( 5 ; 6 ) b ⃗ = ( 10 ; 12 ) \vec b = (10;\,12) b = ( 10 ; 12 )
Ҳал. b ⃗ = ( 10 ; 12 ) = ( 2 ⋅ 5 ; 2 ⋅ 6 ) = 2 a ⃗ . \vec b = (10;\,12) = (2\cdot 5;\; 2\cdot 6) = 2\vec a. b = ( 10 ; 12 ) = ( 2 ⋅ 5 ; 2 ⋅ 6 ) = 2 a . b ⃗ = 2 a ⃗ , \vec b = 2\vec a, b = 2 a , 5 ⋅ 12 − 10 ⋅ 6 = 0. 5\cdot 12 - 10\cdot 6 = 0. 5 ⋅ 12 − 10 ⋅ 6 = 0.
Худатон санҷед Вектори a ⃗ \vec a a λ \lambda λ b ⃗ = λ a ⃗ \vec b = \lambda\vec a b = λ a
Коллинеарӣ: b = λa санҷида бин! Аз нав
Векторҳои воҳидӣ Вектори дилхоҳи a ⃗ = ( x ; y ) \vec a = (x;\,y) a = ( x ; y )
a ⃗ = x e ⃗ 1 + y e ⃗ 2 , зеро x ( 1 ; 0 ) + y ( 0 ; 1 ) = ( x ; 0 ) + ( 0 ; y ) = ( x ; y ) . \vec a = x\vec e_1 + y\vec e_2, \qquad \text{зеро}\quad x(1;\,0) + y(0;\,1) = (x;\,0) + (0;\,y) = (x;\,y). a = x e 1 + y e 2 , зеро x ( 1 ; 0 ) + y ( 0 ; 1 ) = ( x ; 0 ) + ( 0 ; y ) = ( x ; y ) . Яъне координатаҳои вектор — коэффитсиентҳои ҷудокунӣ ба e ⃗ 1 \vec e_1 e 1 e ⃗ 2 . \vec e_2. e 2 . ( 3 ; − 5 ) = 3 e ⃗ 1 − 5 e ⃗ 2 . (3;\,-5) = 3\vec e_1 - 5\vec e_2. ( 3 ; − 5 ) = 3 e 1 − 5
Вектори воҳидии ҳамсамт. Агар a ⃗ ≠ 0 ⃗ \vec a \ne \vec 0 a = 0 e ⃗ = 1 ∣ a ⃗ ∣ a ⃗ \vec e = \dfrac{1}{|\vec a|}\vec a e = ∣ a a ⃗ \vec a a ∣ e ⃗ ∣ = 1 ∣ a ⃗ ∣ ⋅ ∣ a ⃗ ∣ = 1 |\vec e| = \dfrac{1}{|\vec a|}\cdot|\vec a| = 1 ∣ e ∣ = ∣ a λ = 1 ∣ a ⃗ ∣ > 0. \lambda = \dfrac{1}{|\vec a|} > 0. λ = ∣ a ∣ 1 > 0. a ⃗ = ( 3 ; 4 ) \vec a = (3;\,4) a = ( 3 ; 4 ) ∣ a ⃗ ∣ = 5 , |\vec a| = 5, ∣ a ∣ = 5 , e ⃗ = ( 3 5 ; 4 5 ) . \vec e = \left(\dfrac35;\,\dfrac45\right). e = ( 5 3 ;
Кор дар синф
Машқҳо 1. Исбот кунед, ки векторҳои a ⃗ = ( 8 ; 6 ) \vec a = (8;\,6) a = ( 8 ; 6 ) b ⃗ = ( 4 ; 3 ) \vec b = (4;\,3) b = ( 4 ; 3 )
2. λ \lambda λ b ⃗ = λ a ⃗ , \vec b = \lambda\vec a, b = λ a , a ⃗ = ( 24 ; 32 ) , \vec a = (24;\,32), a = ( 24 ; 32 ) , b ⃗ = ( 3 ; 4 ) \vec b = (3;\,4) b = ( 3 ; 4 )
3. Векторҳои a ⃗ = ( 2 ; − 4 ) , \vec a = (2;\,-4), a = ( 2 ; − 4 ) , b ⃗ = ( 1 ; 1 ) , \vec b = (1;\,1), b = ( 1 ; 1 ) , c ⃗ = ( 1 ; − 2 ) , \vec c = (1;\,-2), c = ( 1 ; − 2 ) , d ⃗ = ( − 2 ; 4 ) \vec d = (-2;\,4) d = ( − 2 ; 4 )
4. Векторҳои a ⃗ = ( 1 ; 4 ) \vec a = (1;\,4) a = ( 1 ; 4 ) b ⃗ = ( − 2 ; m ) \vec b = (-2;\,m) b = ( − 2 ; m ) m m m
5. Кадоме аз векторҳои зерин воҳидӣ аст: a ⃗ = ( − 3 5 ; 4 5 ) , \vec a = \left(-\dfrac35;\,\dfrac45\right), a = ( − 5 3 ; 5 4 ) , b ⃗ = ( 2 3 ; 2 3 ) , \vec b = \left(\dfrac23;\,\dfrac23\right), b = ( 3 2 ; c ⃗ = ( 0 ; 1 ) , \vec c = (0;\,1), c = ( 0 ; 1 ) , d ⃗ = ( 3 5 ; 4 5 ) \vec d = \left(\dfrac35;\,\dfrac45\right) d = ( 5 3 ;
6. Вектори воҳидии e ⃗ \vec e e a ⃗ = ( 6 ; 8 ) \vec a = (6;\,8) a = ( 6 ; 8 )
7. Нуқтаҳои M M M E E E A B AB A B D C DC D C A B C D ABCD A BC D M E → = 1 2 ( A D → + B C → ) . \overrightarrow{ME} = \dfrac12\left(\overrightarrow{AD} + \overrightarrow{BC}\right). ME = 2 1 ( A D
Нишондод Вектори M E → \overrightarrow{ME} ME M E → = M A → + A D → + D E → \overrightarrow{ME} = \overrightarrow{MA} + \overrightarrow{AD} + \overrightarrow{DE} ME = M A + A D + D E M E → = M B → + B C → + C E → . \overrightarrow{ME} = \overrightarrow{MB} + \overrightarrow{BC} + \overrightarrow{CE}. ME = MB M A → + M B → = 0 ⃗ \overrightarrow{MA} + \overrightarrow{MB} = \vec 0 M A + MB = D E → + C E → = 0 ⃗ . \overrightarrow{DE} + \overrightarrow{CE} = \vec 0. D E + CE =
Вазифаи хонагӣ
Машқҳои 5 ва 6; ҳавасмандон — машқи 7.
Вазифаи хонагии электронӣ «Геометрия 9 · Вазифаи хонагӣ · 2.2 · 2/3» (онро аз омӯзгор гиред): 6 савол. 2 кӯшиш.
Дар ин дарс
Ман медонам шарти коллинеарии ду векторро (b ⃗ = λ a ⃗ \vec b = \lambda\vec a b = λ a x 1 y 2 − x 2 y 1 = 0 x_1y_2 - x_2y_1 = 0 x 1 y 2 − x 2 y 1 = Ман метавонам санҷам, ки ду вектор бо координатаҳояшон коллинеаранд ё не.Ман метавонам векторро ба векторҳои воҳидии e ⃗ 1 , \vec e_1, e 1 , e ⃗ 2 \vec e_2 e Такрор (2 дақиқа)
∣ λ a ⃗ ∣ |\lambda\vec a| ∣ λ a ∣ λ a ⃗ \lambda\vec a λ a a ⃗ \vec a a a ⃗ = ( 4 ; − 6 ) . \vec a = (4;\,-6). a = ( 4 ; − 6 ) . − 1 2 a ⃗ -\tfrac12\vec a − 2 Кадом векторҳо коллинеарӣ мебошанд?
Ҷавобҳо
∣ λ ∣ ⋅ ∣ a ⃗ ∣ ; |\lambda|\cdot|\vec a|; ∣ λ ∣ ⋅ ∣ a ∣ ; λ > 0 ; \lambda > 0; λ > 0 ; ( − 2 ; Теорема Ду вектори ғайрисифри a ⃗ \vec a a b ⃗ \vec b b λ \lambda λ b ⃗ = λ a ⃗ . \vec b = \lambda\vec a. b = λ a .
Расми 1. b = λa: λ > 0 — ҳамсамт, λ < 0 — муқобилсамт. Таъриф Вектори дарозиаш 1 1 1 вектори воҳидӣ меноманд. Векторҳои воҳидии тирҳои координатӣ: e ⃗ 1 = ( 1 ; 0 ) \vec e_1 = (1;\,0) e 1 = ( 1
Расми 2. Векторҳои воҳидӣ e₁ ва e₂; a = 3e₁ + 2e₂. Расми 3. Трапетсияи ABCD; M ва E — миёнаҷойҳои асосҳои AB ва DC. Агар b ⃗ = − 2 a ⃗ \vec b = -2\vec a b = − 2 a a
Кадом вектор бо a ⃗ = ( 2 ; − 3 ) \vec a = (2;\,-3) a = ( 2 ; − 3 )
Як ҷавобро интихоб кунед A ( − 2 ; − 3 ) (-2;\,-3) ( − 2 ; − 3 )
B
C D Кадом векторҳо воҳидианд? (Якчанд ҷавоб дуруст аст.)
Ҳамаи мувофиқро интихоб кунед A ( 5 13 ; 12 13 ) \left(\dfrac{5}{13};\,\dfrac{12}{13}\right) ( 13 5 ; 13 12 )
B ( 0 ; − 1 ) (0;\,-1)
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