Question

O ângulo entre
r1: 
{x = 7 - t
{y = 1 + t         e           r2 : (x - 5)/2 = y/1 = (z-3)/1 é igual a :
{z = -2t

a) 45°
b) 70°
c) 60°
d) 65°
e) 90°

Answer 1

Boa tarde Roger!

Solução!

Sendo as retas!


$r1:\begin{cases} x=7-t\\ y=1+t\\ z=0-2t \end{cases}\\\\\\ (x,y,z)=(7,1,0)+t(-1,1,-2)\\\\ \vec{V}1=(-1,1,-2)$



$r2: \dfrac{(x+5)}{2}= \dfrac{y}{1} = \dfrac{z-3}{1} \\\\\\\ \vec{ V}2=(2,1,1)~~Normal$


$Cos~~ \alpha = \dfrac{\vec{V1}.\vec{V2}}{|\vec{V1}|.|\vec{V2}|}\\\\\\\\ Cos~~ \alpha = \dfrac{(-1,1,-2).(2,1,1)}{ \sqrt{((-1)^{2}+1^{2} +(-2)^{2}}) .\sqrt{(2 ^{2} +1^{2}+1^{2})}}\\\\\\\\ Cos~~ \alpha = \dfrac{(-2+1,-2)}{ \sqrt{(1+1+4}) .( \sqrt{(4+1+1)}}\\\\\\\\ Cos~~ \alpha = \dfrac{(-4+1)}{ \sqrt{6} . \sqrt{6}}\\\\\\\\ Cos~~ \alpha = \dfrac{(-3)}{ \sqrt{36}}\\\\\\\\ Cos~~ \alpha = \dfrac{-3}{6}\\\\\\\\ Cos~~ \alpha = -\dfrac{1}{2}$



Veja que na resposta não temos 120º,mas no circulo trigonométrico temos que o cosseno de 60 =1/2 e esta localizado no primeiro quadrante.


$Cos~~120\º= -\dfrac{1}{2}\\\\\\ Cos~~60\º= \dfrac{1}{2}\\\\\\\\ \boxed {Resposta:~~60\º~~Alternativa~~C}$






Boa tarde!
Bons estudos!