Question

38)Dados a =(2,1,–3) e b=(1,–2,1), determinar o vetor v perpendicular a, v perpendicular b e v=5.

Answer 1

$\vec{\mathbf{a}}=(2,\,1,\,-3),\;\;\;\vec{\mathbf{b}}=(1,\,-2,\,1).$


$\bullet\;\;$ Para que o vetor $\vec{\mathbf{v}}$ seja perpendicular ao vetor $\vec{\mathbf{a}}$ e ao vetor $\vec{\mathbf{b}},$ basta que $\vec{\mathbf{v}}$ tenha a mesma direção que o vetor $\vec{\mathbf{a}}\times \vec{\mathbf{b}}:$

$\vec{\mathbf{v}}=\lambda\cdot (\vec{\mathbf{a}}\times \vec{\mathbf{b}})$

para algum $\lambda$ real.


$\bullet\;\;$ Calculando o produto vetorial entre os vetores $\vec{\mathbf{a}}$ e $\vec{\mathbf{b}},$ temos:

$\vec{\mathbf{a}}\times \vec{\mathbf{b}}=\det\left[ \begin{array}{ccc} \vec{\mathbf{i}}&\vec{\mathbf{j}}&\vec{\mathbf{k}}\\ a_{1}&a_{2}&a_{3}\\ b_{1}&b_{2}&b_{3} \end{array} \right ]\\ \\ \\ \vec{\mathbf{a}}\times \vec{\mathbf{b}}=\det\left[ \begin{array}{ccc} \vec{\mathbf{i}}&\vec{\mathbf{j}}&\vec{\mathbf{k}}\\ 2&1&-3\\ 1&-2&1 \end{array} \right ]\\ \\ \\ \vec{\mathbf{a}}\times \vec{\mathbf{b}}=[1\cdot 1-(-2)\cdot (-3)]\,\vec{\mathbf{i}}-[2\cdot 1-1\cdot (-3)]\,\vec{\mathbf{j}}+[2\cdot (-2)-1\cdot 1]\,\vec{\mathbf{k}}\\ \\ \vec{\mathbf{a}}\times \vec{\mathbf{b}}=[1-6]\,\vec{\mathbf{i}}-[2+3]\,\vec{\mathbf{j}}+[-4-1]\,\vec{\mathbf{k}}\\ \\ \vec{\mathbf{a}}\times \vec{\mathbf{b}}=-5\vec{\mathbf{i}}-5\vec{\mathbf{j}}-5\vec{\mathbf{k}}\\ \\ \vec{\mathbf{a}}\times \vec{\mathbf{b}}=(-5,\,-5,\,-5)$


Logo, devemos ter

$\vec{\mathbf{v}}=\lambda \cdot (\vec{\mathbf{a}}\times \vec{\mathbf{b}})\\ \\ \vec{\mathbf{v}}=\lambda\cdot (-5,\,-5,\,-5)\\ \\ \vec{\mathbf{v}}=\lambda\cdot (-5)\cdot (1,\,1,\,1)\\ \\ \vec{\mathbf{v}}=-5\lambda\cdot (1,\,1,\,1)$


$\bullet\;\;$ Queremos que o módulo de $\vec{\mathbf{v}}$ seja igual a $5:$

$\|\vec{\mathbf{v}}\|=5\\ \\ \|-5\lambda\cdot (1,\,1,\,1)\|=5\\ \\ |-5\lambda|\cdot \|(1,\,1,\,1)\|=5\\ \\ 5|\lambda|\cdot \sqrt{1^{2}+1^{2}+1^{2}}=5\\ \\ 5|\lambda|\cdot \sqrt{3}=5\\ \\ |\lambda|=\frac{5}{5\sqrt{3}}\\ \\ |\lambda|=\frac{1}{\sqrt{3}}\\ \\ \lambda=\frac{1}{\sqrt{3}}\;\;\text{ ou }\;\;\lambda=-\frac{1}{\sqrt{3}}$


Sendo assim, temos duas possibilidades para o vetor $\vec{\mathbf{v}}:$

$\bullet\;\;$ Para $\lambda=\frac{1}{\sqrt{3}}:$

$\vec{\mathbf{v}}=-5\lambda\cdot (1,\,1,\,1)\\ \\ \vec{\mathbf{v}}=-5\cdot \frac{1}{\sqrt{3}}\cdot (1,\,1,\,1)\\ \\ \vec{\mathbf{v}}=-\frac{5}{\sqrt{3}}\cdot (1,\,1,\,1)\\ \\ \boxed{\begin{array}{c}\vec{\mathbf{v}}=\left(-\frac{5}{\sqrt{3}},\,-\frac{5}{\sqrt{3}},\,-\frac{5}{\sqrt{3}} \right ) \end{array}}$


$\bullet\;\;$ Para $\lambda=-\frac{1}{\sqrt{3}}:$

$\vec{\mathbf{v}}=-5\lambda\cdot (1,\,1,\,1)\\ \\ \vec{\mathbf{v}}=-5\cdot (-\frac{1}{\sqrt{3}})\cdot (1,\,1,\,1)\\ \\ \vec{\mathbf{v}}=\frac{5}{\sqrt{3}}\cdot (1,\,1,\,1)\\ \\ \boxed{\begin{array}{c}\vec{\mathbf{v}}=\left(\frac{5}{\sqrt{3}},\,\frac{5}{\sqrt{3}},\,\frac{5}{\sqrt{3}} \right ) \end{array}}$