Question

Considere os vetores u=(2,-1,1) e v=(1,1,2). Encontre u.v e determine o ângulo 'teta" entre eles.

Answer 1

$\overrightarrow{\mathbf{u}}=(2,\,-1,\,1)~~\text{ e }~~\overrightarrow{\mathbf{v}}=(1,\,1,\,2)$


$\bullet\;\;$ Produto escalar:

$\overrightarrow{\mathbf{u}}\cdot \overrightarrow{\mathbf{v}}\\\\ =(2,\,-1,\,1)\cdot (1,\,1,\,2)\\\\ =2\cdot 1+(-1)\cdot 1+1\cdot 2\\\\ =2-1+2\\\\ =3$


$\bullet\;\;$ Encontrando as normas (módulos) dos vetores:

$\|\overrightarrow{\mathbf{u}}\|=\|(2,\,-1,\,1)\|\\\\ =\sqrt{2^2+(-1)^2+1^2}\\\\ =\sqrt{4+1+1}\\\\ =\sqrt{6}\\\\\\ \|\overrightarrow{\mathbf{v}}\|=\|(1,\,1,\,2)\|\\\\ =\sqrt{1^2+1^2+2^2}\\\\ =\sqrt{1+1+4}\\\\ =\sqrt{6}$


$\bullet\;\;$ Sendo $\theta$ o ângulo entre os vetores, devemos ter

$\overrightarrow{\mathbf{u}}\cdot \overrightarrow{\mathbf{v}}=\|\overrightarrow{\mathbf{u}}\|\cdot \|\overrightarrow{\mathbf{v}}\|\cdot \cos \theta\\\\ \cos \theta=\dfrac{\overrightarrow{\mathbf{u}}\cdot \overrightarrow{\mathbf{v}}}{\|\overrightarrow{\mathbf{u}}\|\cdot \|\overrightarrow{\mathbf{v}}\|}~~~~~~(0^\circ \le \theta \le 180^\circ)$


Substituindo os valores conhecidos, temos

$\cos \theta=\dfrac{3}{\sqrt{6}\cdot \sqrt{6}}\\\\\\ \cos \theta=\dfrac{3}{6}\\\\\\ \cos \theta=\dfrac{1}{2}\\\\\\ \theta=\arccos\!\left(\dfrac{1}{2}\right)\\\\\\ \boxed{\begin{array}{c}\theta=60^\circ \end{array}}$


Bons estudos! :-)