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Enunciado

Álgebra Linear e Geometria Analítica - Vetores e uma iniciacao a geometria analitica mello e watanabe- Ed: 2º - Capítulo Exercícios Adicionais.3 - Ex. 87

B →   = ( a → , b → , c → ) é uma base com | a → | = | b → | = | c → |   = 2 . Os ângulos são todos iguais a 60 graus. Sendo x →   =   ( 2,1 , 3 ) B e y → =   ( - 1,0 , 2 ) B , calcule o ângulo entre x →   e y →   .

Passo 1

Fala, galerinha! Vocês já estão ligados que precisamos da fórmula

c o s   α = x → ⋅ y → | x → | | y → |

Assim, precisamos calcular os termos

x → ⋅ y → = ( 2 a → + b → + 3 c → ) ⋅ ( - a → + 2 c → )

x → ⋅ y → = 2 a → ⋅ ( - a → + 2 c → ) + b → ⋅ ( - a → + 2 c → ) + 3 c → ⋅ ( - a → + 2 c → )

x → ⋅ y → = - 2 a → ⋅ a → + 4 a → ⋅ c → - a → ⋅ b → + 2 b → ⋅ c → - 3 a → ⋅ c → + 6 c → ⋅ c →

x → ⋅ y → = - 2 a → ⋅ a → - a → ⋅ b → + a → ⋅ c → + 2 b → ⋅ c → + 6 c → ⋅ c →

x → ⋅ y → = - 2 | a → | 2 - a → ⋅ b → + a → ⋅ c → + 2 b → ⋅ c → + 6 | c → | 2

| x → | 2 = x → ⋅ x → = ( 2 a → + b → + 3 c → ) ⋅ ( 2 a → + b → + 3 c → )

| x → | 2 = 2 a → ⋅ ( 2 a → + b → + 3 c → ) + b → ⋅ ( 2 a → + b → + 3 c → )

| x → | 2 = 2 a → ⋅ ( 2 a → + b → + 3 c → ) + b → ⋅ ( 2 a → + b → + 3 c → ) + 3 c → ⋅ ( 2 a → + b → + 3 c → )

| x → | 2 = 4 a → ⋅ a → + 2 a → ⋅ b → + 6 a → ⋅ c → + 2 a → ⋅ b → + b → ⋅ b → + 3 b → ⋅ c → + 6 a → ⋅ c → + 3 b → ⋅ c → + 9 c → ⋅ c →

| x → | 2 = 4 | a → | 2 + 4 a → ⋅ b → + 12 a → ⋅ c → + | b → | 2 + 6 b → ⋅ c → + 9 | c → | 2

| y → | 2 = y → ⋅ y → = ( - a → + 2 c → ) ⋅ ( - a → + 2 c → ) = a → ⋅ a → - 4 a → ⋅ c → + 4 c → ⋅ c → = | a → | 2 - 4 a → ⋅ c → + 4 | c → | 2

Passo 2

Como | a → | = | b → | = | c → |   = 2 e os ângulos são todos iguais a 60 graus, então

c o s   60 ° = a   → ⋅   c → | a → | | c → | ⇒   1 2 = a   → ⋅   c → | a → | 2 ⇒ a → ⋅ c → = | a → | 2 2 = 2

c o s   60 ° = b   → ⋅   c → | b → | | c → | ⇒   1 2 = b   → ⋅   c → | b → | 2 ⇒ c → ⋅ b → = | b → | 2 2 = 2

c o s   60 ° = b   → ⋅   a → | b → | | a → | ⇒   1 2 = b   → ⋅   a → | b → | 2 ⇒ a → ⋅ b → = | b → | 2 2 = 2

Assim, ficamos com

x → ⋅ y → = - 2 | a → | 2 - a → ⋅ b → + a → ⋅ c → + 2 b → ⋅ c → + 6 | c → | 2 = - 2 ⋅ 4 - 2 + 2 + 2 ⋅ 2 + 6 ⋅ 4

x → ⋅ y → = - 8 + 4 + 24 = 20

| x → | 2 = 4 | a → | 2 + 4 a → ⋅ b → + 12 a → ⋅ c → + | b → | 2 + 6 b → ⋅ c → + 9 | c → | 2

| x → | 2 = 4 ⋅ 4 + 4 ⋅ 2 + 12 ⋅ 2 + 4 + 6 ⋅ 2 + 9 ⋅ 4

| x → | 2 = 16 + 8 + 24 + 4 + 12 + 36 = 100 ⇒ | x → | = 100 = 10

| y → | 2 = | a → | 2 - 4 a → ⋅ c → + 4 | c → | 2 = 4 - 4 ⋅ 2 + 4 ⋅ 4 = 4 - 8 + 16 = 12 ⇒ | y → | = 12

Voltando a fórmula do cosseno, ficamos com

c o s   α = x → ⋅ y → | x → | | y → | = 20 10 12 = 2 12 12   =   12 6

Resposta

c o s   α = 12 6