Respuesta :
[tex]\bf ~~~~~~~~~~~~\textit{angle between two vectors }
\\\\
cos(\theta)=\cfrac{\stackrel{\textit{dot product}}{u \cdot v}}{\stackrel{\textit{magnitude product}}{||u||~||v||}} \implies
\measuredangle \theta = cos^{-1}\left(\cfrac{u \cdot v}{||u||~||v||}\right)\\\\
-------------------------------[/tex]
[tex]\bf \begin{cases} v1=\ \textless \ -6,4\ \textgreater \ \\ v2=\ \textless \ -3,6\ \textgreater \ \\ ------------\\ v1\cdot v2=(-6\cdot -3)+(4\cdot 6)\\ \qquad \qquad 42\\ ||v1||=\sqrt{(-6)^2+4^2}\\ \qquad \sqrt{52}\\ ||v2||=\sqrt{(-3)^2+6^2}\\ \qquad \sqrt{45} \end{cases}\implies \measuredangle \theta =cos^{-1}\left( \cfrac{42}{\sqrt{52}\cdot \sqrt{45}} \right) \\\\\\ \measuredangle \theta =cos^{-1}\left( \cfrac{42}{\sqrt{2340}} \right)\implies \measuredangle \theta \approx 29.74488129694^o[/tex]
[tex]\bf \begin{cases} v1=\ \textless \ -6,4\ \textgreater \ \\ v2=\ \textless \ -3,6\ \textgreater \ \\ ------------\\ v1\cdot v2=(-6\cdot -3)+(4\cdot 6)\\ \qquad \qquad 42\\ ||v1||=\sqrt{(-6)^2+4^2}\\ \qquad \sqrt{52}\\ ||v2||=\sqrt{(-3)^2+6^2}\\ \qquad \sqrt{45} \end{cases}\implies \measuredangle \theta =cos^{-1}\left( \cfrac{42}{\sqrt{52}\cdot \sqrt{45}} \right) \\\\\\ \measuredangle \theta =cos^{-1}\left( \cfrac{42}{\sqrt{2340}} \right)\implies \measuredangle \theta \approx 29.74488129694^o[/tex]
