![\bf \textit{angle between two vectors }\\ \quad \\ cos(\theta)=\cfrac{u \cdot v}{||u||\ ||v||} \implies \cfrac{\text{dot product}}{\text{product of magnitudes}}\\ \quad \\\\ \theta = cos^{-1}\left(\cfrac{u \cdot v}{||u||\ ||v||}\right)\\\\ -----------------------------\\\\ \theta=cos^{-1}\left[ \cfrac{\ \textless \ 8,4\ \textgreater \ \quad \cdot \quad \ \textless \ 9,-9\ \textgreater \ }{(\sqrt{8^2+4^2})(\sqrt{9^2+(-9)^2})} \right]](/tpl/images/0450/7575/84eea.png)
![\bf \theta=cos^{-1}\left[ \cfrac{(8\cdot 9)+(4\cdot -9)}{(\sqrt{64+16})(\sqrt{81+81})} \right]\implies \theta=cos^{-1}\left[ \cfrac{36}{(\sqrt{80})(\sqrt{162})} \right] \\\\\\ \theta=cos^{-1}\left[ \cfrac{36}{(\sqrt{80})(\sqrt{162})} \right]\implies \theta=cos^{-1}\left[ \cfrac{36}{\sqrt{12960}}\right] \\\\\\ \theta=cos^{-1}\left[ \cfrac{36}{36\sqrt{10}}\right]\implies \theta\approx 71.565^o](/tpl/images/0450/7575/18c9a.png)
Find the angle between the given vectors to the nearest tenth of a degree.
u = < 8,...
Mathematics, 11.01.2020 07:31 cookies1164
Find the angle between the given vectors to the nearest tenth of a degree.
u = < 8, 4> , v = < 9, -9>
Answers: 2
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