Question

Dados os vetores v=2.i + 2j -3.k e u = -1.i + 3.j + 1.k, calcule o cosseno do ângulo entre os respectivos vetores.

Answer 1

$\boxed{\boxed{cos(\theta)= \frac{\vec U . \vec V}{|\vec U|*|\vec V|} }}}$

temos
$V=(2,2,-3)\\|V|= \sqrt{(2)^2+(2)^2+(-3)^2} = \sqrt{17} \\\\U=(-1,3,1)\\|U|= \sqrt{(-1)^2+(3)^2+(1)^2} = \sqrt{11}$

encontrando o cosseno do angulo
$cos(\theta)= \frac{(2,2,-3).(-1,3,1)}{ \sqrt{17}* \sqrt{10} } \\\\cos(\theta)= \frac{(2*-1)+(2*3)+(-3*1)}{ \sqrt{17*10} } \\\\cos(\theta)= \frac{-2+6-3}{ \sqrt{170} } = \frac{1}{ \sqrt{170} }$