Considere os vetores a, b e c dados por:
a=-3,0 imath+3,0 jmath+2,0 k
b=-2,0 imath-4,0 jmath+2,0 k
c=2,0 imath+3,0 jmath+1,0 k

Calcule:

c) (a times b) times(b+c).

Passo 1

Beleza, vamos resolver o exercício que envolve vetores e produtos vetoriais. A gente tem três vetores, a, b e c, com componentes dadas. O que precisamos calcular é (a times b) times (b + c). Para isso, vamos precisar entender como calcular produtos vetoriais e como somar vetores.

Passo 2

Primeiro, vamos calcular o produto vetorial a times b. O produto vetorial entre dois vetores u = ux imath + uy jmath + uz k e v = vx imath + vy jmath + vz k é dado por:
u times v = imath & jmath & k ux & uy & uz vx & vy & vz
Aplicando isso aos vetores a e b:
a times b = imath & jmath & k -3,0 & 3,0 & 2,0 -2,0 & -4,0 & 2,0
Calculando o determinante:
a times b = [(3,0 times 2,0) - (2,0 times -4,0)] imath - [(-3,0 times 2,0) - (2,0 times -2,0)] jmath + [(-3,0 times -4,0) - (3,0 times -2,0)] k
Simplificando:
a times b = (6,0 - (-8,0)) imath - (-6,0 - (-4,0)) jmath + (12,0 - (-6,0)) k = 14,0 imath - (-2,0) jmath + 18,0 k
Então:
a times b = 14,0 imath + 2,0 jmath + 18,0 k

Passo 3

Agora, vamos somar os vetores b e c:
b + c = (-2,0 imath -4,0 jmath + 2,0 k) + (2,0 imath + 3,0 jmath + 1,0 k)
Somando componente a componente:
b + c = (-2,0 + 2,0)imath + (-4,0 + 3,0)jmath + (2,0 + 1,0)k = 0,0 imath -1,0 jmath + 3,0 k
Então:
b + c = 0,0 imath -1,0 jmath + 3,0 k

Passo 4

Agora, vamos calcular o produto vetorial (a times b) times (b + c). Já temos a times b = 14,0 imath + 2,0 jmath + 18,0 k e b + c = 0,0 imath -1,0 jmath + 3,0 k. Aplicando a mesma fórmula do produto vetorial:
(a times b) times (b + c) = imath & jmath & k 14,0 & 2,0 & 18,0 0,0 & -1,0 & 3,0
Calculando o determinante:
(a times b) times (b + c) = [(2,0 times 3,0) - (18,0 times -1,0)] imath - [(14,0 times 3,0) - (18,0 times 0,0)] jmath + [(14,0 times -1,0) - (2,0 times 0,0)] k
Simplificando:
(a times b) times (b + c) = (6,0 + 18,0) imath - (42,0 - 0,0) jmath + (-14,0 - 0,0) k = 24,0 imath -42,0 jmath -14,0 k

Resposta

Portanto, o resultado de (a times b) times (b + c) é:
24,0 imath -42,0 jmath -14,0 k

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