1. (1,0 ponto) Considerando a matriz A e o vetor \mathbf{v} abaixo, calcule o ângulo formado entre o vetor ve o vetor \mathbf{u}=A \mathbf{v}.
A=\left[\begin{array}{ccc} \sqrt{3} / 2 & 0 & 1 / 2 \\\\ 0 & 1 & 0 \\\\ -1 / 2 & 0 & \sqrt{3} / 2 \end{array}\right] \text { e } \mathbf{v}=\left[\begin{array}{l} 0 \\\\ 0 \\\\ 1 \end{array}\right]

Passo 1

O enunciado dá a matriz
A = \begin{pmatrix} \frac{\sqrt{3}}{2} & 0 & \frac12\\\\ 0 & 1 & 0\\\\ -\frac12 & 0 & \frac{\sqrt{3}}{2} \end{pmatrix}
e o vetor
\mathbf{v} = \begin{pmatrix}0\\\ 0\\\ 1\end{pmatrix},
pede o ângulo \theta entre \mathbf{v} e \mathbf{u}=A\mathbf{v}.
Para isso vamos:
  1. Calcular \mathbf{u}=A\mathbf{v}.
  2. Usar a fórmula do produto escalar para achar \cos\theta:
\cos\theta = \frac{\mathbf{v}\cdot\mathbf{u}}{\|\mathbf{v}\|\;\|\mathbf{u}\|}.

Passo 2

Calcular \mathbf{u}=A\mathbf{v}. Como \mathbf{v} é a terceira coluna da matriz padrão, basta extrair a terceira coluna de A:
\mathbf{u} = A\mathbf{v} = \begin{pmatrix} \frac{\sqrt{3}}{2} & 0 & \frac12\\\\ 0 & 1 & 0\\\\ -\frac12 & 0 & \frac{\sqrt{3}}{2} \end{pmatrix} \begin{pmatrix}0\\\ 0\\\ 1\end{pmatrix} = \begin{pmatrix}\tfrac12\\\ 0\\tfrac{\sqrt{3}}{2}\end{pmatrix}.

Passo 3

Calcular o produto escalar e as normas:
\mathbf{v}\cdot\mathbf{u} = (0)\cdot\tfrac12 + (0)\cdot 0 + (1)\cdot\tfrac{\sqrt{3}}{2} = \frac{\sqrt{3}}{2},
\|\mathbf{v}\| = \sqrt{0^2 + 0^2 + 1^2} = 1, \quad \|\mathbf{u}\| = \sqrt{\bigl(\tfrac12\bigr)^2 + 0^2 + \bigl(\tfrac{\sqrt{3}}{2}\bigr)^2} = \sqrt{\frac14 + \frac34} = 1.
Logo,
\cos\theta = \frac{\mathbf{v}\cdot\mathbf{u}}{\|\mathbf{v}\|\;\|\mathbf{u}\|} = \frac{\tfrac{\sqrt{3}}{2}}{1\cdot1} = \frac{\sqrt{3}}{2}.

Passo 4

Determinar \theta a partir de \cos\theta = \tfrac{\sqrt{3}}{2}. Sabemos que
\theta = \operatorname{arccos}\!\bigl(\tfrac{\sqrt{3}}{2}\bigr) = \frac{\pi}{6}.

Resposta

\displaystyle \theta = \frac{\pi}{6}.

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