Question

Find the angle between u = i+sqr of 7j and v = -i+9j. Round to the nearest tenth of a degree.

Answer 1

$\bf ~~~~~~~~~~~~\textit{angle between two vectors } \\\\ cos(\theta)=\cfrac{\stackrel{\textit{dot product}}{u \cdot v}}{\stackrel{\textit{magnitude product}}{||u||~||v||}} \implies \measuredangle \theta = cos^{-1}\left(\cfrac{u \cdot v}{||u||~||v||}\right) \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ \begin{cases} u=i+\sqrt{7}j\implies &\\\\ v=-i+9j\implies & \end{cases} \\\\[-0.35em] ~\dotfill$

$\bf u\cdot v\implies (1)(-1)~+~(\sqrt{7})(9)\implies -1+9\sqrt{7}\implies 9\sqrt{7}-1~\hfill dot~product \\\\[-0.35em] ~\dotfill\\\\ ||u||\implies \sqrt{1^2+(\sqrt{7})^2}\implies \sqrt{1+7}\implies \sqrt{8}~\hfill magnitudes \\\\\\ ||v||\implies \sqrt{(-1)^2+9^2}\implies \sqrt{1+81}\implies \sqrt{82} \\\\[-0.35em] ~\dotfill$

$\bf \theta =cos^{-1}\left( \cfrac{9\sqrt{7}-1}{\sqrt{8}\cdot \sqrt{82}} \right)\implies \theta =cos^{-1}\left( \cfrac{9\sqrt{7}-1}{\sqrt{656}} \right) \\\\\\ \theta \approx cos^{-1}(0.8906496638868531)\implies \theta \approx 27.05$

make sure your calculator is in Degree mode.