Question

Obtenha um vetor ortogonal aos vetores u = (4, 2, 5) e v = (0, 3, 9) que meça 5 unidades?

Answer 1

São dados dois vetores $\overset{\to}{\mathbf{u}}=(4,\,2,\,5)$ e $\overset{\to}{\mathbf{v}}=(0,\,3,\,9).$ Queremos encontrar os vetores $\overset{\to}{\mathbf{n}}=(a,\,b,\,c),$ tais que

     •  $\overset{\to}{\mathbf{n}}$ é ortogonal a $\overset{\to}{\mathbf{u}};$

     •  $\overset{\to}{\mathbf{n}}$ é ortogonal a $\overset{\to}{\mathbf{v}};$

     •  $\|\overset{\to}{\mathbf{n}}\|=5.$


Dois vetores são orgononais somente se o produto escalar entre eles é igual a zero. Logo, devemos ter

     $\qquad~~~\overset{\to}{\mathbf{n}}\perp \overset{\to}{\mathbf{u}}\\\\ \Longleftrightarrow\quad \overset{\to}{\mathbf{n}}\cdot \overset{\to}{\mathbf{u}}=0\\\\ \Longleftrightarrow\quad (a,\,b,\,c)\cdot (4,\,2,\,5)=0\\\\ \Longleftrightarrow\quad 4a+2b+5c=0\qquad\quad\mathsf{(i)}$


     $\qquad~~~\overset{\to}{\mathbf{n}}\perp \overset{\to}{\mathbf{v}}\\\\ \Longleftrightarrow\quad \overset{\to}{\mathbf{n}}\cdot \overset{\to}{\mathbf{v}}=0\\\\ \Longleftrightarrow\quad (a,\,b,\,c)\cdot (0,\,3,\,9)=0\\\\ \Longleftrightarrow\quad 0a+3b+9c=0\\\\ \Longleftrightarrow\quad 3b+9c=0\qquad\quad\mathsf{(ii)}$


     $\qquad~~~\|\overset{\to}{\mathbf{n}}\|=5\\\\ \Longleftrightarrow\quad \|(a,\,b,\,c)\|=5\\\\ \Longleftrightarrow\quad a^2+b^2+c^2=5^2\\\\ \Longleftrightarrow\quad a^2+b^2+c^2=25\qquad\quad\mathsf{(iii)}$



Resolva o sistema formado pelas equações (i), (ii) e (iii):

     $\left\{ \!\begin{array}{rcrc} 4a+2b+5c&\!\!=\!\!&0&\qquad\mathsf{(i)}\\\\ 3b+9c&\!\!=\!\!&0&\qquad\mathsf{(ii)}\\\\ a^2+b^2+c^2&\!\!=\!\!&25&\qquad\mathsf{(iii)} \end{array} \right.$



Isole b na equação (ii) e substitua nas outras equações (i) e (iii):

     $3b+9c=0\\\\ 3b=-9c\\\\ b=\dfrac{-9c}{3}\\\\ b=-3c\\\\\\\\ \left\{ \!\begin{array}{ccrc} 4a+2\cdot (-3c)+5c&\!\!=\!\!&0\\\\ a^2+(-3c)^2+c^2&\!\!=\!\!&25 \end{array}\right.\\\\\\\\ \left\{ \!\begin{array}{ccrc} 4a-6c+5c&\!\!=\!\!&0\\\\ a^2+9c^2+c^2&\!\!=\!\!&25 \end{array}\right.\\\\\\\\ \left\{ \!\begin{array}{ccrc} 4a-c&\!\!=\!\!&0&\qquad\mathsf{(iv)}\\\\ a^2+10c^2&\!\!=\!\!&25&\qquad\mathsf{(v)} \end{array}\right.$



Isole c na equação (iv) e substitua na equação (v) acima:

     $4a-c=0\\\\ c=4a\\\\\\\\ a^2+10\cdot (4a)^2=25\\\\ a^2+10\cdot 16a^2=25\\\\ a^2+160a^2=25\\\\ 161a^2=25\\\\ a^2=\dfrac{25}{161}\\\\\\ a=\pm\,\sqrt{\dfrac{25}{161}}$

     $a=\pm\,\dfrac{5}{\sqrt{161}}$        ✔



     •  Para $a=-\,\dfrac{5}{\sqrt{161}}\,,$ encontramos

        $c=4a\\\\ c=4\cdot \Big(\!-\dfrac{5}{\sqrt{161}}\Big)$

        $c=-\,\dfrac{20}{\sqrt{161}}$        ✔


        $b=-3c\\\\ b=-3\cdot \Big(\!-\dfrac{20}{\sqrt{161}}\Big)$

        $b=\dfrac{60}{\sqrt{161}}$        ✔



     •  Para $a=\dfrac{5}{\sqrt{161}}\,,$ encontramos valores semelhantes, apenas com os sinais trocados:

        $c=4a\\\\ c=4\cdot \dfrac{5}{\sqrt{161}}$

        $c=\dfrac{20}{\sqrt{161}}$        ✔


        $b=-3c\\\\ b=-3\cdot \dfrac{20}{\sqrt{161}}$

        $b=-\,\dfrac{60}{\sqrt{161}}$        ✔


Portanto, há duas possibilidades para o vetor procurado:

     $\overset{\to}{\mathbf{n}}_1=\Big(\!-\dfrac{5}{\sqrt{161}},\,\dfrac{60}{\sqrt{161}},\,-\,\dfrac{20}{\sqrt{161}}\Big)$    ou    $\overset{\to}{\mathbf{n}}_2=\Big(\dfrac{5}{\sqrt{161}},\,-\,\dfrac{60}{\sqrt{161}},\,\dfrac{20}{\sqrt{161}}\Big)$


ou caso prefira, pode colocar o escalar $a=\dfrac{5}{\sqrt{161}}$ em evidência, e a resposta fica

     $\overset{\to}{\mathbf{n}}_1=\dfrac{5}{\sqrt{161}}\,(-1,\,12,\,-4)$    ou    $\overset{\to}{\mathbf{n}}_2=\dfrac{5}{\sqrt{161}}\,(1,\,-12,\,4).$


Dúvidas? Comente.


Bons estudos! :-)