Dados \mathrm{A}=(-2,-3,-4) \quad \mathrm{B}=(-5,8,-2) e \mathrm{C}=(1,-3,-5), calcular os seguintes vetores
-2 \mathrm{~A}-\frac{4}{3} \mathrm{~B}+\frac{1}{2} \mathrm{C}

Passo 1

O enunciado dá os vetores \mathrm{A}=(-2,-3,-4), \mathrm{B}=(-5,8,-2) e \mathrm{C}=(1,-3,-5) e pede o cálculo de
-2\,\mathrm{A}-\frac{4}{3}\,\mathrm{B}+\frac{1}{2}\,\mathrm{C}.
O conteúdo é operação com vetores em \mathbb{R}^3: multiplicação escalar e adição vetorial.

Passo 2

Calculamos cada termo separadamente:
-2\,\mathrm{A} = -2\cdot(-2,-3,-4) = (4,6,8),
-\frac{4}{3}\,\mathrm{B} = -\frac{4}{3}\cdot(-5,8,-2) =\Bigl(\tfrac{20}{3},-\tfrac{32}{3},\tfrac{8}{3}\Bigr),
\frac{1}{2}\,\mathrm{C} =\frac{1}{2}\cdot(1,-3,-5) =\Bigl(\tfrac12,-\tfrac32,-\tfrac52\Bigr).

Passo 3

Somamos essas parcelas coordenada a coordenada:
(4,6,8) +\Bigl(\tfrac{20}{3},-\tfrac{32}{3},\tfrac{8}{3}\Bigr) +\Bigl(\tfrac12,-\tfrac32,-\tfrac52\Bigr).

Passo 4

Para cada coordenada, usamos denominador comum 6:
coordenada x:
4 + \tfrac{20}{3} + \tfrac12 =\frac{24+40+3}{6} =\frac{67}{6},
coordenada y:
6 - \tfrac{32}{3} - \tfrac32 =\frac{36-64-9}{6} =-\frac{37}{6},
coordenada z:
8 + \tfrac{8}{3} - \tfrac{5}{2} =\frac{48+16-15}{6} =\frac{49}{6}.

Resposta

\bigl(-2\,\mathrm{A}-\tfrac{4}{3}\,\mathrm{B}+\tfrac12\,\mathrm{C}\bigr) =\Bigl(\tfrac{67}{6},-\tfrac{37}{6},\tfrac{49}{6}\Bigr).

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