Question

O vetor u=(-1,-1,-2) tem angulação de 60º com o vetor MN definido pelos pontos M(0,3,4) e N( x,-1,2). Com base no exposto, o valor de x está representado na alternativa x= –34 ou x=-2 x= –34 ou x=2 x= –30 ou x=2 x= –34 ou x=3 x= 34 ou x=2

Answer 1

Boa noite!

Solução!

Dados!

$\vec{u}=(-1,-1,-2)\\\\\\ M(0,3,4)\\\\\\ N(x,-1,2)\\\\\\ \theta=60\º$

Vamos transformar o ponto M e N em vetores fazendo a sua diferença.


$\vec{v}=(N-M)\\\\\ \vec{v}=(x,-1,2)-(0,3,4)\\\\ \vec{v}=(x,-1,2)+(0,-3,-4)\\\\ \boxed{\vec{v}=(x,-4,-2)}\\\\\\\ \boxed{\vec{u}=(-1,-1,-2)}\\\\\\ Cos\theta= \dfrac{1}{2}$


$Cos\theta= \dfrac{\vec{u}.\vec{v}}{|\vec{u}|.|\vec{v}|}\\\\\\\ \dfrac{1}{2}= \dfrac{(-x,-1,-2).(x,-4,-2)}{ \sqrt{(-1)^{2}+(-1)^{2}+(-2)^{2}. }\sqrt{(x)^{2}+(-4)^{2} +(-2)^{2}}}\\\\\\\\ \dfrac{1}{2}= \dfrac{(-x+4+4)}{ \sqrt{(1+1+4. }\sqrt{x^{2}+16 +4}}\\\\\\\\ \dfrac{1}{2}= \dfrac{(-x+8)}{ \sqrt{6. }\sqrt{x^{2}+20}}\\\\\\\\ Vamos~~elevar~~ambos~~os~~membros~~ ao ~~quadrado.$

$\left (\dfrac{1}{2}\right )^{2}= \dfrac{(-x+8)^{2} }{ (\sqrt{6})^{2} .(\sqrt{x^{2}+20} )^{2} }\\\\\\\\ \dfrac{1}{4}= \dfrac{ x^{2} -8x-8x+64}{6( x^{2} +20}$

$\dfrac{1}{4}= \dfrac{ x^{2} -16x+64}{6x^{2} +120}$


$6 x^{2} +120=4 x^{2} -64x+256\\\\\\\ 6 x^{2} -4 x^{2} +64x+120-256=0\\\\\\ 2 x^{2} +64x-136=0\\\\\\ Aplicando~~Bhaskara!\\\\\\ x= \dfrac{-64\pm \sqrt{(-64)^{2}-4.2.-136} }{4}\\\\\\\ x= \dfrac{-64\pm \sqrt{4096+1088} }{4}$

$x= \dfrac{-64\pm \sqrt{5184} }{4}\\\\\\\ x= \dfrac{-64\pm 72}{4}\\\\\\\ Raizes!\\\\\\ x_{1}= \dfrac{-64+72}{4}= \dfrac{8}{4}=2\\\\\\ x_{2}= \dfrac{-64-72}{4}= \dfrac{-136}{4}=-34\\\\\\ \boxed{Resposta:~~x=2~~~~x=-34}\\\\\\\\\ Boa noite!\\\\\\ Bons estudos!$