Question

Let v1 =(-6,4) and v2=(-3,6) compute the following what i sthe angle between v1 and v2

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)\\\\ -------------------------------`


`\bf \begin{cases} v1=\ \textless \ -6,4\ \textgreater \ \\ v2=\ \textless \ -3,6\ \textgreater \ \\ ------------\\ v1\cdot v2=(-6\cdot -3)+(4\cdot 6)\\ \qquad \qquad 42\\ ||v1||=√((-6)^2+4^2)\\ \qquad √(52)\\ ||v2||=√((-3)^2+6^2)\\ \qquad √(45) \end{cases}\implies \measuredangle \theta =cos^(-1)\left( \cfrac{42}{√(52)\cdot √(45)} \right) \\\\\\ \measuredangle \theta =cos^(-1)\left( \cfrac{42}{√(2340)} \right)\implies \measuredangle \theta \approx 29.74488129694^o`