1.5.10 Três vetores, \mathbf{A}, \mathbf{B} e \mathbf{C}, são dados por
\begin{aligned} & \mathbf{A}=3 \hat{\mathbf{x}}-2 \hat{\mathbf{y}}+2 \hat{\mathbf{z}} \\\ & \mathbf{B}=6 \hat{\mathbf{x}}+4 \hat{\mathbf{y}}-2 \hat{\mathbf{z}} \\\ & \mathbf{C}=-3 \hat{\mathbf{x}}-2 \hat{\mathbf{y}}-4 \hat{\mathbf{z}} \end{aligned}
Calcule os valores de \mathbf{A} \cdot \mathbf{B} \times \mathbf{C} e \mathbf{A} \times(\mathbf{B} \times \mathbf{C}), \mathbf{C} \times(\mathbf{A} \times \mathbf{B}) e \mathbf{B} \times(\mathbf{C} \times \mathbf{A}).

Passo 1

Temos três vetores em base cartesiana:
\mathbf{A} = 3\hat{\mathbf{x}} - 2\hat{\mathbf{y}} + 2\hat{\mathbf{z}},\quad \mathbf{B} = 6\hat{\mathbf{x}} + 4\hat{\mathbf{y}} - 2\hat{\mathbf{z}},\quad \mathbf{C} = -3\hat{\mathbf{x}} - 2\hat{\mathbf{y}} - 4\hat{\mathbf{z}}.
Precisamos calcular o produto escalar misto \mathbf{A}\cdot(\mathbf{B}\times\mathbf{C}) e os produtos vetoriais aninhados \mathbf{A}\times(\mathbf{B}\times\mathbf{C}), \mathbf{C}\times(\mathbf{A}\times\mathbf{B}) e \mathbf{B}\times(\mathbf{C}\times\mathbf{A}). Vamos usar determinantes para produto vetorial e a identidade de duplo produto vetorial
\mathbf{X}\times(\mathbf{Y}\times\mathbf{Z}) = \mathbf{Y}\,(\mathbf{X}\cdot\mathbf{Z})-\mathbf{Z}\,(\mathbf{X}\cdot\mathbf{Y}).

Passo 2

Calculamos \mathbf{B}\times\mathbf{C} via determinante e depois o produto escalar com \mathbf{A}.
\mathbf{B}\times\mathbf{C} =\begin{vmatrix} \hat{\mathbf{x}} & \hat{\mathbf{y}} & \hat{\mathbf{z}}\\\\ 6 & 4 & -2\\\\ -3 & -2 & -4 \end{vmatrix} =-20\,\hat{\mathbf{x}}+30\,\hat{\mathbf{y}}+0\,\hat{\mathbf{z}}.
Logo,
\mathbf{A}\cdot(\mathbf{B}\times\mathbf{C}) =3\cdot(-20)+(-2)\cdot30+2\cdot0=-120.

Passo 3

Usamos a identidade de duplo produto vetorial para \mathbf{A}\times(\mathbf{B}\times\mathbf{C}). Primeiro calculamos
\mathbf{A}\cdot\mathbf{C} =3\cdot(-3)+(-2)\cdot(-2)+2\cdot(-4)=-13, \quad \mathbf{A}\cdot\mathbf{B} =3\cdot6+(-2)\cdot4+2\cdot(-2)=6.
Então
\mathbf{A}\times(\mathbf{B}\times\mathbf{C}) =\mathbf{B}\,(\mathbf{A}\cdot\mathbf{C})-\mathbf{C}\,(\mathbf{A}\cdot\mathbf{B}) =(-13)\mathbf{B}-6\,\mathbf{C} =(-60)\hat{\mathbf{x}}-40\hat{\mathbf{y}}+50\hat{\mathbf{z}}.

Passo 4

Calculamos diretamente \mathbf{A}\times\mathbf{B} via determinante:
\mathbf{A}\times\mathbf{B} =\begin{vmatrix} \hat{\mathbf{x}} & \hat{\mathbf{y}} & \hat{\mathbf{z}}\\\\ 3 & -2 & 2\\\\ 6 & 4 & -2 \end{vmatrix} =-4\,\hat{\mathbf{x}}+18\,\hat{\mathbf{y}}+24\,\hat{\mathbf{z}}.

Passo 5

Para \mathbf{C}\times(\mathbf{A}\times\mathbf{B}) usamos a identidade. Calculamos
\mathbf{C}\cdot\mathbf{B} = -3\cdot6+(-2)\cdot4+(-4)\cdot(-2)=-18, \quad \mathbf{C}\cdot\mathbf{A} = -13.
Assim
\mathbf{C}\times(\mathbf{A}\times\mathbf{B}) =\mathbf{A}\,(\mathbf{C}\cdot\mathbf{B})-\mathbf{B}\,(\mathbf{C}\cdot\mathbf{A}) =(-18)\mathbf{A}-(-13)\mathbf{B} =24\hat{\mathbf{x}}+88\hat{\mathbf{y}}-62\hat{\mathbf{z}}.

Passo 6

Calculamos \mathbf{C}\times\mathbf{A} via determinante:
\mathbf{C}\times\mathbf{A} =\begin{vmatrix} \hat{\mathbf{x}} & \hat{\mathbf{y}} & \hat{\mathbf{z}}\\\\ -3 & -2 & -4\\\\ 3 & -2 & 2 \end{vmatrix} =-12\,\hat{\mathbf{x}}-6\,\hat{\mathbf{y}}+12\,\hat{\mathbf{z}}.

Passo 7

Para \mathbf{B}\times(\mathbf{C}\times\mathbf{A}) aplicamos a identidade. Calculamos
\mathbf{B}\cdot\mathbf{A}=6, \quad \mathbf{B}\cdot\mathbf{C}=-18.
Logo
\mathbf{B}\times(\mathbf{C}\times\mathbf{A}) =\mathbf{C}\,(\mathbf{B}\cdot\mathbf{A})-\mathbf{A}\,(\mathbf{B}\cdot\mathbf{C}) =6\,\mathbf{C}-(-18)\mathbf{A} =36\hat{\mathbf{x}}-48\hat{\mathbf{y}}+12\hat{\mathbf{z}}.

Resposta

\mathbf{A}\cdot(\mathbf{B}\times\mathbf{C})=-120,
\mathbf{A}\times(\mathbf{B}\times\mathbf{C}) =-60\hat{\mathbf{x}}-40\hat{\mathbf{y}}+50\hat{\mathbf{z}},
\mathbf{C}\times(\mathbf{A}\times\mathbf{B}) =24\hat{\mathbf{x}}+88\hat{\mathbf{y}}-62\hat{\mathbf{z}},
\mathbf{B}\times(\mathbf{C}\times\mathbf{A}) =36\hat{\mathbf{x}}-48\hat{\mathbf{y}}+12\hat{\mathbf{z}}.

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