Calcular o ângulo entre os vetores u=(1,1,4) e v= ( -1,2,2)
Soluções para a tarefa
Respondido por
5
segue em anexo o desenvolvimento da questão
utilizar na calculadora cos⁻¹
utilizar na calculadora cos⁻¹
Anexos:
Respondido por
9
Boa noite Luiz!
Solução!
![\vec{u}=(1,1,4)\\\\\
\vec{v}=(-1,2,2)\\\\\\\\
Cos~ \alpha= \dfrac{\vec{u} \times \vec{v}}{|\vec{u}|\times|\vec{v}|}\\\\\\\
Cos~ \alpha= \dfrac{(1,1,4) \times (-1,2,2)}{ \sqrt{1^{2}+1^{2}+4^{2} } \times \sqrt{(-1)^{2}+2^{2} +2^{2}} }\\\\\\\\
Cos~ \alpha= \dfrac{(-1+2+8) }{ \sqrt{18 } \times \sqrt{9} }\\\\\\\\\\
Cos~ \alpha= \dfrac{9 }{ \sqrt{18 } \times 3 }\\\\\\\\\\
Cos~ \alpha= \dfrac{3 }{ \sqrt{18} }\\\\\\\\\\
Cos~ \alpha= \dfrac{3 }{ 3\sqrt{2} }](https://tex.z-dn.net/?f=%5Cvec%7Bu%7D%3D%281%2C1%2C4%29%5C%5C%5C%5C%5C%0A%5Cvec%7Bv%7D%3D%28-1%2C2%2C2%29%5C%5C%5C%5C%5C%5C%5C%5C%0ACos%7E+%5Calpha%3D+%5Cdfrac%7B%5Cvec%7Bu%7D+%5Ctimes+%5Cvec%7Bv%7D%7D%7B%7C%5Cvec%7Bu%7D%7C%5Ctimes%7C%5Cvec%7Bv%7D%7C%7D%5C%5C%5C%5C%5C%5C%5C%0ACos%7E+%5Calpha%3D+%5Cdfrac%7B%281%2C1%2C4%29+%5Ctimes+%28-1%2C2%2C2%29%7D%7B+%5Csqrt%7B1%5E%7B2%7D%2B1%5E%7B2%7D%2B4%5E%7B2%7D+++%7D+%5Ctimes+%5Csqrt%7B%28-1%29%5E%7B2%7D%2B2%5E%7B2%7D+%2B2%5E%7B2%7D%7D+%7D%5C%5C%5C%5C%5C%5C%5C%5C%0ACos%7E+%5Calpha%3D+%5Cdfrac%7B%28-1%2B2%2B8%29+%7D%7B+%5Csqrt%7B18+++%7D+%5Ctimes+%5Csqrt%7B9%7D+%7D%5C%5C%5C%5C%5C%5C%5C%5C%5C%5C%0A+Cos%7E+%5Calpha%3D+%5Cdfrac%7B9+%7D%7B+%5Csqrt%7B18+++%7D+%5Ctimes+3+%7D%5C%5C%5C%5C%5C%5C%5C%5C%5C%5C%0ACos%7E+%5Calpha%3D+%5Cdfrac%7B3+%7D%7B+%5Csqrt%7B18%7D++%7D%5C%5C%5C%5C%5C%5C%5C%5C%5C%5C%0ACos%7E+%5Calpha%3D+%5Cdfrac%7B3+%7D%7B+3%5Csqrt%7B2%7D++%7D%0A%0A+ "\vec{u}=(1,1,4)\\\\\
\vec{v}=(-1,2,2)\\\\\\\\
Cos~ \alpha= \dfrac{\vec{u} \times \vec{v}}{|\vec{u}|\times|\vec{v}|}\\\\\\\
Cos~ \alpha= \dfrac{(1,1,4) \times (-1,2,2)}{ \sqrt{1^{2}+1^{2}+4^{2} } \times \sqrt{(-1)^{2}+2^{2} +2^{2}} }\\\\\\\\
Cos~ \alpha= \dfrac{(-1+2+8) }{ \sqrt{18 } \times \sqrt{9} }\\\\\\\\\\
Cos~ \alpha= \dfrac{9 }{ \sqrt{18 } \times 3 }\\\\\\\\\\
Cos~ \alpha= \dfrac{3 }{ \sqrt{18} }\\\\\\\\\\
Cos~ \alpha= \dfrac{3 }{ 3\sqrt{2} }")
![Cos~ \alpha= \dfrac{1 }{\sqrt{2}}\\\\\\
Racionalizando~~o ~~denominador~~resulta!\\\\\\\
Cos~ \alpha= \dfrac{1 }{\sqrt{2}} \times \dfrac{ \sqrt{2} }{ \sqrt{2} } \\\\\\\\
Cos~ \alpha= \dfrac{ \sqrt{2} }{\sqrt{4}} \\\\\\\\
Cos~ \alpha= \dfrac{ \sqrt{2} }{2} \\\\\\\\
Cos~ \alpha= 45\º \\\\\\
\alpha=arc~cos~45\º\\\\\\\\
\boxed{Resposta:\alpha=arc~cos~45\º}](https://tex.z-dn.net/?f=Cos%7E+%5Calpha%3D+%5Cdfrac%7B1+%7D%7B%5Csqrt%7B2%7D%7D%5C%5C%5C%5C%5C%5C%0ARacionalizando%7E%7Eo+%7E%7Edenominador%7E%7Eresulta%21%5C%5C%5C%5C%5C%5C%5C%0ACos%7E+%5Calpha%3D+%5Cdfrac%7B1+%7D%7B%5Csqrt%7B2%7D%7D+%5Ctimes+%5Cdfrac%7B+%5Csqrt%7B2%7D+%7D%7B+%5Csqrt%7B2%7D+%7D+%5C%5C%5C%5C%5C%5C%5C%5C%0ACos%7E+%5Calpha%3D+%5Cdfrac%7B+%5Csqrt%7B2%7D++%7D%7B%5Csqrt%7B4%7D%7D+%5C%5C%5C%5C%5C%5C%5C%5C%0ACos%7E+%5Calpha%3D+%5Cdfrac%7B+%5Csqrt%7B2%7D++%7D%7B2%7D+%5C%5C%5C%5C%5C%5C%5C%5C%0ACos%7E+%5Calpha%3D+45%5C%C2%BA+%5C%5C%5C%5C%5C%5C%0A%5Calpha%3Darc%7Ecos%7E45%5C%C2%BA%5C%5C%5C%5C%5C%5C%5C%5C%0A%5Cboxed%7BResposta%3A%5Calpha%3Darc%7Ecos%7E45%5C%C2%BA%7D%0A%0A%0A "Cos~ \alpha= \dfrac{1 }{\sqrt{2}}\\\\\\
Racionalizando~~o ~~denominador~~resulta!\\\\\\\
Cos~ \alpha= \dfrac{1 }{\sqrt{2}} \times \dfrac{ \sqrt{2} }{ \sqrt{2} } \\\\\\\\
Cos~ \alpha= \dfrac{ \sqrt{2} }{\sqrt{4}} \\\\\\\\
Cos~ \alpha= \dfrac{ \sqrt{2} }{2} \\\\\\\\
Cos~ \alpha= 45\º \\\\\\
\alpha=arc~cos~45\º\\\\\\\\
\boxed{Resposta:\alpha=arc~cos~45\º}")
Boa noite!
Bons estudos!
Solução!
Boa noite!
Bons estudos!
Perguntas interessantes