Question

Dados os vetores u(-1,-3) e v(2,-1) calcule
||u+v||² + ||u-v||²

Answer 1

$\mathbf{u}\left(u_{1},\,u_{2} \right)=\left(-1,\,-3 \right )\;\;\Rightarrow\;\;\left\{ \begin{array}{l} u_{1}=-1\\ u_{2}=-3 \end{array} \right.\\ \\ \\ \mathbf{v}\left(v_{1},\,v_{2} \right )=\left(2,\,-1 \right )\;\;\Rightarrow\;\; \left\{ \begin{array}{l} v_{1}=2\\ v_{2}=-1 \end{array} \right.$


$\bullet\;\;\mathbf{u}+\mathbf{v}\\ \\ =\left(u_{1},\,u_{2} \right )+\left(v_{1},\,v_{2} \right )\\ \\ =\left(u_{1}+v_{1},\,u_{2}+v_{2} \right )\\ \\ =\left(-1+2,\,-3-1 \right )\\ \\ =\left(1,\,-4 \right )\\ \\ \\ \bullet\;\;\mathbf{u}-\mathbf{v}\\ \\ =\left(u_{1},\,u_{2} \right )-\left(v_{1},\,v_{2} \right )\\ \\ =\left(u_{1}-v_{1},\,u_{2}-v_{2} \right )\\ \\ =\left(-1-2,\,-3-\left(-1 \right ) \right )\\ \\ =\left(-1-2,\,-3+1 \right )\\ \\ =\left(-3,\,-2 \right )$


A norma (ou módulo) de um vetor é a raiz quadrada da soma dos quadrados de suas coordenadas: Então,

$\bullet\;\;\left\|\mathbf{u}+\mathbf{v}\right\|\\ \\ =\sqrt{\left(u_{1}+v_{1} \right )^{2}+\left(u_{2}+v_{2} \right )^{2}}\\ \\ =\sqrt{\left(1\right )^{2}+\left(-4 \right )^{2}}\\ \\ =\sqrt{1+16}\\ \\ =\sqrt{17}\\ \\ \\ \bullet\;\;\left\|\mathbf{u}-\mathbf{v}\right\|\\ \\ =\sqrt{\left(u_{1}-v_{1} \right )^{2}+\left(u_{2}-v_{2} \right )^{2}}\\ \\ =\sqrt{\left(-3\right )^{2}+\left(-2 \right )^{2}}\\ \\ =\sqrt{9+4}\\ \\ =\sqrt{13}$


Finalmente, temos que

$\left\|\mathbf{u}+\mathbf{v}\right\|^{2}+\left\|\mathbf{u}-\mathbf{v}\right\|^{2}\\ \\ =\left(\sqrt{17} \right )^{2}+\left(\sqrt{13} \right )^{2}\\ \\ =17+13\\ \\ =30\\ \\ \\ \boxed{\left\|\mathbf{u}+\mathbf{v}\right\|^{2}+\left\|\mathbf{u}-\mathbf{v}\right\|^{2}=30}$