Respuesta :

[tex]~~~~~~~~~~~~\textit{angle between two vectors } \\\\ cos(\theta)=\cfrac{\stackrel{\textit{dot product}}{u \cdot v}}{\underset{\textit{magnitude product}}{||u||~||v||}} \implies \measuredangle \theta = cos^{-1}\left(\cfrac{u \cdot v}{||u||~||v||}\right) \\\\[-0.35em] ~\dotfill\\\\ \begin{cases} 6i-4j\\ 2i-6j \end{cases}\implies \stackrel{\qquad a~\hfill b\qquad }{\begin{array}{llll} < 6~~,~~-4 > \\ < 2~~,~~-6 > \end{array}}[/tex]

[tex]\stackrel{\textit{\Large dot product}}{ < 6~~,~~-4 > \cdot < 2~~,~~-6 > \implies (6\cdot 2)+(-4\cdot -6)}\implies 12+24\implies 36 \\\\\\ \stackrel{\textit{\Large magnitudes}}{|| < 6~~,~~-4 > ||\implies \sqrt{6^2+(-4)^2}\implies \sqrt{52}} \\\\\\ || < 2~~,~~-6 > ||\implies \sqrt{2^2+(-6)^2}\implies \sqrt{40} \\\\[-0.35em] ~\dotfill[/tex]

[tex]\measuredangle \theta =cos^{-1}\left( \cfrac{36}{\sqrt{52}\cdot \sqrt{40}} \right)\implies \measuredangle \theta =cos^{-1}\left( \cfrac{36}{\sqrt{52}\cdot \sqrt{40}} \right) \\\\\\ \measuredangle \theta =cos^{-1}\left( \cfrac{36}{\sqrt{2080}} \right)\implies \measuredangle \theta \approx 37.9^o[/tex]