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}}} \implies \measuredangle \theta = cos^(-1)\left(\cfrac{u \cdot v}\right) \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ \begin{cases} u=i+√(7)j\implies &<1,√(7)>\\\\ v=-i+9j\implies &<-1,9> \end{cases} \\\\[-0.35em] ~\dotfill`

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

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

make sure your calculator is in Degree mode.