Question

Determinar um vetor unitário simultaneamente ortogonal aos vetores u=(2,-6,3) e v=(4,3,1).

Answer 1

Para encontrar um vetor simultaneamente ortogonal a $\overset{\rightarrow}{\mathbf{u}}=\left \langle 2,\,-6,\,3 \right \rangle$ e $\overset{\rightarrow}{\mathbf{v}}=\left \langle 4,\,3,\,1 \right \rangle$, basta tomarmos qualquer vetor paralelo ao vetor resultado do produto vetorial entre $\overset{\rightarrow}{\mathbf{u}}$ e $\overset{\rightarrow}{\mathbf{v}}$:

$\overset{\rightarrow}{\mathbf{u}} \times \overset{\rightarrow}{\mathbf{v}}=\det \left[ \begin{array}{ccc} \overset{\rightarrow}{\mathbf{i}}&\overset{\rightarrow}{\mathbf{j}}&\overset{\rightarrow}{\mathbf{k}}\\ u_{1}&u_{2}&u_{3}\\ v_{1}&v_{2}&v_{3} \end{array} \right ]\\ \\ \\ \overset{\rightarrow}{\mathbf{u}} \times \overset{\rightarrow}{\mathbf{v}}=\det \left[ \begin{array}{ccc} \overset{\rightarrow}{\mathbf{i}}&\overset{\rightarrow}{\mathbf{j}}&\overset{\rightarrow}{\mathbf{k}}\\ 2&-6&3\\ 4&3&1 \end{array} \right ]\\ \\ \\ \overset{\rightarrow}{\mathbf{u}} \times \overset{\rightarrow}{\mathbf{v}}=-6\overset{\rightarrow}{\mathbf{i}}+12\overset{\rightarrow}{\mathbf{j}}+6\overset{\rightarrow}{\mathbf{k}}-\left(-24\overset{\rightarrow}{\mathbf{k}}+9\overset{\rightarrow}{\mathbf{i}}+2\overset{\rightarrow}{\mathbf{j}} \right )$


$\overset{\rightarrow}{\mathbf{u}} \times \overset{\rightarrow}{\mathbf{v}}=-6\overset{\rightarrow}{\mathbf{i}}+12\overset{\rightarrow}{\mathbf{j}}+6\overset{\rightarrow}{\mathbf{k}}+24\overset{\rightarrow}{\mathbf{k}}-9\overset{\rightarrow}{\mathbf{i}}-2\overset{\rightarrow}{\mathbf{j}}\\ \\ \overset{\rightarrow}{\mathbf{u}} \times \overset{\rightarrow}{\mathbf{v}}=-15\overset{\rightarrow}{\mathbf{i}}+10\overset{\rightarrow}{\mathbf{j}}+30\overset{\rightarrow}{\mathbf{k}}\\ \\ \overset{\rightarrow}{\mathbf{u}} \times \overset{\rightarrow}{\mathbf{v}}=5\cdot\left(-3\overset{\rightarrow}{\mathbf{i}}+2\overset{\rightarrow}{\mathbf{j}}+6\overset{\rightarrow}{\mathbf{k}} \right )\\ \\ \boxed{\overset{\rightarrow}{\mathbf{u}} \times \overset{\rightarrow}{\mathbf{v}}=5\cdot\left \langle -3,\,2,\,6 \right \rangle}$


Então o vetor $\left \langle -3,\,2,\,6 \right \rangle$ é simultaneamente ortogonal a $\overset{\rightarrow}{\mathbf{u}}$ e $\overset{\rightarrow}{\mathbf{v}}$.


O módulo deste vetor é

$\left\|\left \langle -3,\,2,\,6 \right \rangle\right\|\\ \\ =\sqrt{\left(-3 \right )^{2}+2^{2}+6^{2}}\\ \\ =\sqrt{9+4+36}\\ \\ =\sqrt{49}\\ \\ =7$


Então um vetor unitário simultaneamente ortogonal a $\overset{\rightarrow}{\mathbf{u}}$ e $\overset{\rightarrow}{\mathbf{v}}$ é

$\dfrac{\left \langle -3,\,2,\,6 \right \rangle}{\left\|\left \langle -3,\,2,\,6 \right \rangle\right\|}\\ \\ =\dfrac{1}{7}\cdot \left \langle -3,\,2,\,6 \right \rangle\\ \\ =\boxed{\left \langle -\dfrac{3}{7},\,\dfrac{2}{7},\,\dfrac{6}{7} \right \rangle}$