Question

Find the angle between the pair of vectors to the nearest tenth of a degree
6i-4j, 2i-6j

Answer 1

`~~~~~~~~~~~~\textit{angle between two vectors } \\\\ cos(\theta)=\cfrac{\stackrel{\textit{dot product}}{u \cdot v}}{\underset{\textit{magnitude product}}u} \implies \measuredangle \theta = cos^(-1)\left(\cfrac{u \cdot v}u\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}}`

`\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}} \\\\\\ || < 2~~,~~-6 > ||\implies √(2^2+(-6)^2)\implies √(40) \\\\[-0.35em] ~\dotfill`

`\measuredangle \theta =cos^(-1)\left( \cfrac{36}{√(52)\cdot √(40)} \right)\implies \measuredangle \theta =cos^(-1)\left( \cfrac{36}{√(52)\cdot √(40)} \right) \\\\\\ \measuredangle \theta =cos^(-1)\left( \cfrac{36}{√(2080)} \right)\implies \measuredangle \theta \approx 37.9^o`