Question

demonstrar que |u+v+w|^2=|u|^2+|v|^2+|w|^2

Answer 1

Seja, os vetores

$\\ u = ( u_{1} , u_{2} , u_{3} ) \\ \\ v = ( v_{1} , v_{2}, v_{3} ) \\ \\ w = ( w_{1} , w_{2} , w_{3} )$

Por definição de vetores, temos que:

$\\ u+v+w = ( u_{1} , u_{2} , u_{3})+ ( v_{1} , v_{2} , v_{3})+ ( w_{1} , w_{2} , w_{3}) \\ \\ u+v+w = ( u_{1} + v_{1} + w_{1} ,u_{2} + v_{2} + w_{2} , u_{3} + v_{3} + w_{3} )$


Aplicando o módulo teremos o seguinte:

$|u+v+w| = \sqrt{ ( u_{1} + v_{1} + w_{1})^2+( u_{2} + v_{2} + w_{2} )^2+( u_{3}+ v_{3} + w_{3} )^2}$



Então, 

$\\ |u+v+w|^2 = [ \sqrt{ ( u_{1} + v_{1} + w_{1})^2+( u_{2} + v_{2} + w_{2} )^2+( u_{3}+ v_{3} + w_{3} )^2} ]^2 \\ \\ |u+v+w|^2 = ( u_{1} + v_{1} + w_{1})^2+( u_{2} + v_{2} + w_{2} )^2+( u_{3}+ v_{3} + w_{3} )^2$

Substituindo essa informação em nossa igualdade:


$\\ |u+v+w|^2= |u|^2+|v|^2+|w|^2 \\ \\ ( u_{1} + v_{1} + w_{1})^2+( u_{2} + v_{2} + w_{2} )^2+( u_{3}+ v_{3} + w_{3} )^2=|u|^2+|v|^2+|w|^2$

E como nó já sabemos,

$\\ |u| = \sqrt{ (u_{1})^2+ (u_{2} )^2+ (u_{3})^2 } \\ \\ |v| =\sqrt{ (v_{1})^2+ (v_{2} )^2+ (v_{3})^2 } \\ \\ |w| = \sqrt{ (w_{1})^2+ (w_{2} )^2+ (w_{3})^2 }$

Logo,

$\\ |u|^2 = (u_{1})^2+ (u_{2} )^2+ (u_{3})^2 \\ \\ |v|^2 =(v_{1})^2+ (v_{2} )^2+ (v_{3})^2 \\ \\ |w|^2 = (w_{1})^2+ (w_{2} )^2+ (w_{3})^2$

Comparando a igualdade...

$\\ |u+v+w|^2 =|u|^2+|v|^2+|w|^2 \\ \\ ( u_{1} + v_{1} + w_{1} )^2+( u_{2} + v_{2} + w_{2} )^2+( u_{3} + v{3} + w_3)^2 \neq (u_{1})^2+( v_{1} )^2+ \\ ( w_{1} )^2+ (u_{2} )^2+(v_{2})^2+(w_{2})^2+(u_{3})^2+(v_{3})^2+(W_{3})^2 \\ \\ \left \{ {{ ( u_{1} + v_{1} + w_{1} )^2 \neq(u_{1})^2+( v_{1} )^2+ ( w_{1} )^2 } \atop {( u_{2} + v_{2} + w_{2} )^2 \neq (u_{2} )^2+(v_{2})^2+(w_{2})^2} \\ } \right. \\ e,( u_{3}+v{3}+w_3)^2\neq(u_{3})^2+(v_{3})^2+(w_{2})^2$

Falso!