Question

Sabe-se que a matriz da transformação linear correspondente à rotação de um
vetor no espaço, em torno do eixo dos z é determinado pela matriz canônica:

Answer 1

Resposta:

$\boxed{v_{\pi/4}=\bigg(-\dfrac{1}{\sqrt{2}},\dfrac{7}{\sqrt{2}},0\bigg)}$

Explicação passo-a-passo:

Rotação em torno do eixo Z:

$\boxed{T_{\theta}=\left[\begin{array}{ccc}\cos{\theta}&-\sin{\theta}&0\\\sin{\theta}&\cos{\theta}&0\\0&0&1\end{array}\right] }$

$v=(3,4,0)=\left[\begin{array}{c}3&4&0\end{array}\right],\ \theta=\dfrac{\pi}{4}$

$\boxed{v_{\theta}=T_{\theta}v}\ \therefore\ v_{\theta}=\left[\begin{array}{ccc}\cos{\theta}&-\sin{\theta}&0\\\sin{\theta}&\cos{\theta}&0\\0&0&1\end{array}\right] \left[\begin{array}{c}3&4&0\end{array}\right]\ \therefore$

$v_{\theta}=\left[\begin{array}{c}3\cos{\theta}-4\sin{\theta}&3\sin{\theta}+4\cos{\theta}&0\end{array}\right] \ \therefore\ v_{\pi/4}=\left[\begin{array}{c}3\cos{\pi/4}-4\sin{\pi/4}&3\sin{\pi/4}+4\cos{\pi/4}&0\end{array}\right]\ \therefore$

$v_{\pi/4}=\left[\begin{array}{c}\tfrac{3}{\sqrt{2}}-\tfrac{4}{\sqrt{2}}&\tfrac{3}{\sqrt{2}}+\tfrac{4}{\sqrt{2}}&0\end{array}\right] \ \therefore\ \boxed{v_{\pi/4}=\left[\begin{array}{c}-\tfrac{1}{\sqrt{2}}&\tfrac{7}{\sqrt{2}}&0\end{array}\right]}$