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Exercicios Resolvidos do livro de Paulo Winterle, Exercícios de Geometria Analítica e Álgebra Linear

Exercicios Resolvidos do livro de Paulo Winterle

Tipologia: Exercícios

2023

Compartilhado em 13/07/2023

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Baixe Exercicios Resolvidos do livro de Paulo Winterle e outras Exercícios em PDF para Geometria Analítica e Álgebra Linear, somente na Docsity! Rui Eulério Salha RESOLUÇÃO DE EXERCÍCIOS DE GEOMETRIA ANALÍTICA Universidade Pungue Extensão de Tete 2023 Página 102, exercícios de 1-4 1. Dados os vetores 𝑢𝑢�⃗ = 3𝚤𝚤 − 𝚥𝚥 − 2𝑘𝑘�⃗ , 𝑣𝑣 = 2𝚤𝚤 + 4𝚥𝚥 − 𝑘𝑘�⃗ 𝑒𝑒 𝑤𝑤��⃗ = −𝚤𝚤 + 𝑘𝑘�⃗ , determinar a) |𝑢𝑢�⃗ × 𝑢𝑢�⃗ | = ⌈𝑢𝑢�⃗ ⌉ ∙ |𝑢𝑢�⃗ | ∙ 𝑠𝑠𝑒𝑒𝑠𝑠𝑠𝑠 = ⌈𝑢𝑢�⃗ ⌉ ∙ |𝑢𝑢�⃗ | ∙ 𝑠𝑠𝑒𝑒𝑠𝑠0 = 0 b) (2𝑣𝑣) × (3𝑣𝑣) = (4, 8,−2) × (6, 12,−3) � 𝚤𝚤 𝚥𝚥 𝑘𝑘�⃗ 4 8 −2 6 12 −3 � = � 8 −2 12 −3� 𝚤𝚤 − �4 −2 6 −3� 𝚥𝚥 + �4 8 6 12� 𝑘𝑘 �⃗ = (−24 + 24)𝚤𝚤 − (−12 + 12)𝚥𝚥 + (48 − 48)𝑘𝑘�⃗ = 0�⃗ c) (𝑢𝑢�⃗ × 𝑤𝑤��⃗ ) + (𝑤𝑤��⃗ × 𝑢𝑢�⃗ ) (𝑢𝑢�⃗ × 𝑤𝑤��⃗ ) = � 𝚤𝚤 𝚥𝚥 𝑘𝑘�⃗ 3 −1 −2 −1 0 1 � = �−1 −2 0 1 � 𝚤𝚤 − � 3 −2 −1 1 � 𝚥𝚥 + � 3 −1 −1 0 � 𝑘𝑘�⃗ = (−1 − 0)𝚤𝚤 − (3− 2)𝚥𝚥 + (0 − 1)𝑘𝑘�⃗ = (−1,−1,−1 ) (𝑤𝑤��⃗ × 𝑢𝑢�⃗ ) = � 𝚤𝚤 𝚥𝚥 𝑘𝑘�⃗ −1 0 1 3 −1 −2 � = � 0 1 −1 −2� 𝚤𝚤 − �−1 1 3 −2� 𝚥𝚥 + �−1 0 3 −1� 𝑘𝑘 �⃗ = (0 + 1)𝚤𝚤 − (2 − 3)𝚥𝚥 + (1− 0)𝑘𝑘�⃗ = (1, 1, 1) (𝑢𝑢�⃗ × 𝑤𝑤��⃗ ) + (𝑤𝑤��⃗ × 𝑢𝑢�⃗ ) = (−1,−1,−1 ) + (1, 1, 1) = (0, 0, 0) = 0�⃗ d) (𝑢𝑢�⃗ × 𝑣𝑣) × (𝑣𝑣 × 𝑢𝑢�⃗ ) (𝑢𝑢�⃗ × 𝑣𝑣) = � 𝚤𝚤 𝚥𝚥 𝑘𝑘�⃗ 3 −1 −2 2 4 −1 � = �−1 −2 4 −1� 𝚤𝚤 − �3 −2 2 −1� 𝚥𝚥 + �3 −1 2 4 � 𝑘𝑘�⃗ = (1 + 8)𝚤𝚤 − (−3 + 4)𝚥𝚥 + (12 + 2)𝑘𝑘�⃗ = (9,−1, 14 ) (𝑣𝑣 × 𝑢𝑢�⃗ ) = � 𝚤𝚤 𝚥𝚥 𝑘𝑘�⃗ 2 4 −1 3 −1 −2 � = � 4 −1 −1 −2� 𝚤𝚤 − �2 −1 3 −2� 𝚥𝚥 + �2 4 3 −1� 𝑘𝑘 �⃗ = (−8 − 1)𝚤𝚤 − (−4 + 3)𝚥𝚥 + (−2− 12)𝑘𝑘�⃗ = (−9, 1,−14) (𝑢𝑢�⃗ × 𝑣𝑣) × (𝑣𝑣 × 𝑢𝑢�⃗ ) = � 𝚤𝚤 𝚥𝚥 𝑘𝑘�⃗ 9 −1 14 −9 1 −14 � = �−1 14 1 −14� 𝚤𝚤 − � 9 −1 −9 1 � 𝚥𝚥 + � 9 14 −9 −14� 𝑘𝑘 �⃗ = 0�⃗ �𝚤𝚤 × 𝑘𝑘�⃗ � = � 𝚤𝚤 𝚥𝚥 𝑘𝑘�⃗ 1 0 0 0 0 1 � = �0 0 0 1� 𝚤𝚤 − �1 0 0 1� 𝚥𝚥 + �1 0 0 0� 𝑘𝑘 �⃗ = (0 − 0)𝚤𝚤 − (1 − 0)𝚥𝚥 + (0 + 0)𝑘𝑘�⃗ = −𝚥𝚥 b) 𝚥𝚥 × (2𝚤𝚤) 𝚥𝚥 × (2𝚤𝚤) = � 𝚤𝚤 𝚥𝚥 𝑘𝑘�⃗ 0 1 0 2 0 0 � = �1 0 0 0� 𝚤𝚤 − �0 0 2 0� 𝚥𝚥 + �0 1 2 0� 𝑘𝑘 �⃗ = (0 − 0)𝚤𝚤 − (0 − 0)𝚥𝚥 + (0− 2)𝑘𝑘�⃗ = −2𝑘𝑘�⃗ c) (3𝚤𝚤) × (2𝑘𝑘�⃗ ) (3𝚤𝚤) × (2𝑘𝑘�⃗ ) = � 𝚤𝚤 𝚥𝚥 𝑘𝑘�⃗ 3 0 0 0 0 2 � = �0 0 0 2� 𝚤𝚤 − �3 0 0 2� 𝚥𝚥 + �3 0 0 0� 𝑘𝑘 �⃗ = (0 − 0)𝚤𝚤 − (6)𝚥𝚥 + (0− 0)𝑘𝑘�⃗ = −6𝚥𝚥 d) 𝚤𝚤 ∙ �𝚥𝚥 × 𝑘𝑘�⃗ � �𝚥𝚥 × 𝑘𝑘�⃗ � = � 𝚤𝚤 𝚥𝚥 𝑘𝑘�⃗ 0 1 0 0 0 1 � = �1 0 0 1� 𝚤𝚤 − �0 0 0 1� 𝚥𝚥 + �0 1 0 0� 𝑘𝑘 �⃗ = (1 − 0)𝚤𝚤 − (0 − 0)𝚥𝚥 + (0− 0)𝑘𝑘�⃗ = 𝚤𝚤 𝚤𝚤 ∙ �𝚥𝚥 × 𝑘𝑘�⃗ � = 𝚤𝚤 ∙ 𝚤𝚤 = 1 e) (3𝚤𝚤) × (2𝚥𝚥) = (3, 0, 0) ∙ (0, 2, 0) = 0 f) (3𝚤𝚤) × (2𝚥𝚥) � 𝚤𝚤 𝚥𝚥 𝑘𝑘�⃗ 3 0 0 0 2 1 � = �0 0 2 1� 𝚤𝚤 − �3 0 0 1� 𝚥𝚥 + �3 0 0 2� 𝑘𝑘 �⃗ = (0 − 0)𝚤𝚤 − (0 − 0)𝚥𝚥 + (6− 0)𝑘𝑘�⃗ = 6𝑘𝑘�⃗ g) 𝚤𝚤 ∙ (𝚥𝚥 × 𝚤𝚤) = 𝚤𝚤 ∙ �𝑘𝑘�⃗ � = (1, 0, 0) ∙ (0, 0, 1) = 0 h) ) 𝚥𝚥 ∙ �𝚥𝚥 × 𝑘𝑘�⃗ � = 𝚥𝚥 ∙ (𝚤𝚤) = 0 i) (𝚤𝚤 × 𝚥𝚥) × 𝑘𝑘�⃗ = 𝑘𝑘�⃗ × 𝑘𝑘�⃗ = (0, 0, 0) = 0�⃗ j) (𝚤𝚤 × 𝚥𝚥) × 𝚥𝚥 = 𝑘𝑘�⃗ × 𝚥𝚥 = (−1, 0, 0) = −𝚤𝚤 k) 𝚤𝚤 × (𝚥𝚥 × 𝚥𝚥) = 𝚤𝚤 × 0�⃗ = (0, 0, 0) = 0�⃗ l) �𝚥𝚥 × 𝑘𝑘�⃗ � ∙ 𝚤𝚤 = ( 𝚤𝚤) ∙ 𝚤𝚤 = 1 3. Dados os pontos 𝐴𝐴(2, 1,−1),𝐵𝐵(3, 0, 1) 𝑒𝑒 𝐶𝐶(2,−1,−3), determinar o ponto 𝐷𝐷 tal que 𝐴𝐴𝐷𝐷�����⃗ = 𝐵𝐵𝐶𝐶�����⃗ × 𝐴𝐴𝐶𝐶�����⃗ 𝐷𝐷(𝑥𝑥, 𝑦𝑦, 𝑧𝑧) 𝐵𝐵𝐶𝐶�����⃗ = 𝐶𝐶 − 𝐵𝐵 = (−1,−1,−4) 𝐴𝐴𝐶𝐶�����⃗ = 𝐶𝐶 − 𝐴𝐴 = (0,−2,−2) 𝐴𝐴𝐷𝐷�����⃗ = 𝐷𝐷 − 𝐴𝐴 = (𝑥𝑥 − 2, 𝑦𝑦 − 1, 𝑧𝑧 + 1) 𝐴𝐴𝐷𝐷�����⃗ = 𝐵𝐵𝐶𝐶�����⃗ × 𝐴𝐴𝐶𝐶�����⃗ = � 𝚤𝚤 𝚥𝚥 𝑘𝑘�⃗ −1 −1 −4 0 −2 −2 � = �−1 −4 −2 −2� 𝚤𝚤 − �−1 −4 0 −2� 𝚥𝚥 + �−1 −1 0 −2� 𝑘𝑘 �⃗ = (2− 8)𝚤𝚤 − (2 + 0)𝚥𝚥 + (2 + 0)𝑘𝑘�⃗ = (6,−2, 2) (𝑥𝑥 − 2, 𝑦𝑦 − 1, 𝑧𝑧 + 1) = (−6,−2, 2) 𝑥𝑥 − 2 = −6 → 𝑥𝑥 = −4, 𝑦𝑦 − 1 = −2 → 𝑦𝑦 = −1, 𝑧𝑧 + 1 = 2 → 𝑧𝑧 = 1 𝐷𝐷(𝑥𝑥, 𝑦𝑦, 𝑧𝑧) = (−4,−1, 1) 4. Determinar o vetor ?⃗?𝑥 tal que ?⃗?𝑥 · (1, 4,−3) = −7 𝑒𝑒 ?⃗?𝑥 × ( 4,−2, 1) = (3, 5,−2). ?⃗?𝑥 = (𝑥𝑥, 𝑦𝑦, 𝑧𝑧) (𝑥𝑥,𝑦𝑦, 𝑧𝑧) · (1, 4,−3) = −7 → 𝑥𝑥 + 4𝑦𝑦 − 3𝑧𝑧 = −7 (3, 5,−2) = ?⃗?𝑥 × ( 4,−2, 1) = � 𝚤𝚤 𝚥𝚥 𝑘𝑘�⃗ 𝑥𝑥 𝑦𝑦 𝑧𝑧 4 −2 1 � = � 𝑦𝑦 𝑧𝑧 −2 1� 𝚤𝚤 − �𝑥𝑥 𝑧𝑧 4 1� 𝚥𝚥 + �𝑥𝑥 𝑦𝑦 4 −2� 𝑘𝑘 �⃗ = (𝑦𝑦 + 2𝑧𝑧)𝚤𝚤 − (𝑥𝑥 − 4𝑧𝑧)𝚥𝚥 + (−2𝑥𝑥 − 4𝑦𝑦)𝑘𝑘�⃗ = (𝑦𝑦 + 2𝑧𝑧,−𝑥𝑥 + 4𝑧𝑧,−2𝑥𝑥 − 4𝑦𝑦) = (3, 5,−2) � 𝑦𝑦 + 2𝑧𝑧 = 3 −𝑥𝑥 + 4𝑧𝑧 = 5 −2𝑥𝑥 − 4𝑦𝑦 = −2 →� 𝑦𝑦 + 2𝑧𝑧 = 3 −𝑥𝑥 + 4𝑧𝑧 = 5 𝑥𝑥 + 2𝑦𝑦 = 1 →� 𝑦𝑦 + 2𝑧𝑧 = 3 𝑧𝑧 = 5+𝑥𝑥 4 𝑦𝑦 = 1−𝑥𝑥 2 𝑥𝑥 + 4𝑦𝑦 − 3𝑧𝑧 = −7 → 𝑥𝑥 + 4 �1−𝑥𝑥 2 � − 3 �5+𝑥𝑥 4 � = −7 → 𝑥𝑥 + 2 − 2𝑥𝑥 − 15 4 − 𝑥𝑥 4 = −7 ⟶ 4 4 𝑥𝑥 + 8 4 − 8 4 𝑥𝑥 − 15 4 − 3 4 𝑥𝑥 = − 14 4 ⟶ 4𝑥𝑥 + 8 − 8𝑥𝑥 − 15− 3𝑥𝑥 = −28 ⟶ 7𝑥𝑥 = 21 ⟶ 𝑥𝑥 = 3 𝑦𝑦 = 1 − 3 2 = −1 𝑧𝑧 = 5 + 3 4 = 2 ?⃗?𝑥 = (𝑥𝑥, 𝑦𝑦, 𝑧𝑧) = (3,−1, 2)