Question

sendo U=(0,3,3) e V=(-2,2,1) dê em Radianos e medida do angulo entre U e V.

Answer 1

Boa noite!


Solução!


Vamos usar essa formula para determinarmos o ângulo.



$Cos~\theta= \dfrac{|\vec{u}\times \vec{v}|}{|\vec{u}|\times |\vec{v}|} \\\\\\\\\\\\\vec{u}=(0,3,3)~~~~~~~\vec{v}=(-2,2,1}\\\\\\\\ Cos~\theta= \dfrac{(0,3,3)\times(-2,2,1)}{ \sqrt{0^{2} +3^{2}+3^{2} } \times \sqrt{-2 ^{2}+2^{2} +1^{2} }}\\\\\\\\\\ Cos~\theta= \dfrac{(0,3,3)\times(-2,2,1)}{ \sqrt{18} \times \sqrt{9 }}\\\\\\\\ Cos~\theta= \dfrac{(0+6+3)}{ \sqrt{18} \times \sqrt{9 }}\\\\\\\\ Cos~\theta= \dfrac{9}{ \sqrt{18} \times \sqrt{9 }}$



$Cos~\theta= \dfrac{9}{ \sqrt{18} \times 3}\\\\\\ Cos~\theta= \dfrac{3}{ \sqrt{18}}\\\\\\ Cos~\theta= \dfrac{3}{ 3\sqrt{2}}\\\\\\ Cos~\theta= \dfrac{1}{ \sqrt{2}}\\\\\\ Racionalizando~~o denominador!$



$Cos~\theta= \dfrac{1}{ \sqrt{2}}\times \dfrac{ \sqrt{2} }{ \sqrt{2} } \\\\\\ Cos~\theta= \dfrac{ \sqrt{2} }{ \sqrt{4}} \\\\\\ Cos~\theta= \dfrac{ \sqrt{2} }{ 2} \\\\\\ Cos~\theta= 45\º\\\\\\\\ Em ~~radianos!\\\\\\ Cos~\theta= \dfrac{45 \pi }{180}\\\\\\ Simplificando~~por~~45!\\\\\\ \theta= arc~~cos \dfrac{ \pi }{4}$


$\boxed{Resposta:\theta=arc~~cos \dfrac{ \pi }{4} }$



Boa noite!
Bons estudos!