Question

sejam os vetores u= (2,a,-1), v= (3,1,-2) e w= (2a-1,-2,4). Determine a de modo que u.v= (u+v).(v+w)

Answer 1

São dados os vetores

     $\overset{\to}{\mathbf{u}}=(2,\,a,\,-1),~~\overset{\to}{\mathbf{v}}=(3,\,1,\,-2),~~\overset{\to}{\mathbf{w}}=(2a-1,\,-2,\,4).$


Queremos resolver a equação abaixo para a variável $a:$

     $\overset{\to}{\mathbf{u}}\cdot \overset{\to}{\mathbf{v}}=(\overset{\to}{\mathbf{u}}+\overset{\to}{\mathbf{v}})\cdot(\overset{\to}{\mathbf{v}}+\overset{\to}{\mathbf{w}})$


Podemos tentar simplificar a expressão, aplicando propriedades operatórias do produto escalar:

     $\overset{\to}{\mathbf{u}}\cdot \overset{\to}{\mathbf{v}}=(\overset{\to}{\mathbf{u}}+\overset{\to}{\mathbf{v}})\cdot \overset{\to}{\mathbf{v}}+(\overset{\to}{\mathbf{u}}+\overset{\to}{\mathbf{v}})\cdot \overset{\to}{\mathbf{w}}\\\\ \overset{\to}{\mathbf{u}}\cdot \overset{\to}{\mathbf{v}}=\overset{\to}{\mathbf{u}}\cdot \overset{\to}{\mathbf{v}}+\overset{\to}{\mathbf{v}}\cdot \overset{\to}{\mathbf{v}}+\overset{\to}{\mathbf{u}}\cdot \overset{\to}{\mathbf{w}}+\overset{\to}{\mathbf{v}}\cdot \overset{\to}{\mathbf{w}}\\\\ \overset{\to}{\mathbf{u}}\cdot \overset{\to}{\mathbf{v}}-\overset{\to}{\mathbf{u}}\cdot \overset{\to}{\mathbf{v}}=\overset{\to}{\mathbf{v}}\cdot \overset{\to}{\mathbf{v}}+\overset{\to}{\mathbf{u}}\cdot \overset{\to}{\mathbf{w}}+\overset{\to}{\mathbf{v}}\cdot \overset{\to}{\mathbf{w}}\\\\ \overset{\to}{\mathbf{v}}\cdot \overset{\to}{\mathbf{v}}+\overset{\to}{\mathbf{u}}\cdot \overset{\to}{\mathbf{w}}+\overset{\to}{\mathbf{v}}\cdot \overset{\to}{\mathbf{w}}=0$

     $\|\overset{\to}{\mathbf{v}}\|^2+\overset{\to}{\mathbf{u}}\cdot \overset{\to}{\mathbf{w}}+\overset{\to}{\mathbf{v}}\cdot \overset{\to}{\mathbf{w}}=0$

     $\|(3,\,1,\,-2)\|^2+(2,\,a,\,-1)\cdot(2a-1,\,-2,\,4)+(3,\,1,\,-2)\cdot(2a-1,\,-2,\,4)=0$

     $(3^2+1^2+(-2)^2)+(2\cdot (2a-1)+a\cdot (-2)+(-1)\cdot 4)+(3\cdot (2a-1)+1\cdot (-2)+(-2)\cdot 4)=0$

     $(9+1+4)+(2(2a-1)-2a-4)+(3(2a-1)-2-8)=0\\\\ (14)+(4a-2-2a-4)+(6a-3-10)=0\\\\ (14)+(2a-6)+(6a-13)=0\\\\ 2a+6a+14-6-13=0\\\\ 8a-5=0\\\\ 8a=5$

     $a=\dfrac{5}{8}\quad\longleftarrow\quad \mathsf{resposta.}$


Dúvidas? Comente.


Bons estudos! :-)