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Verdich [7]
3 years ago
7

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

Mathematics
1 answer:
Agata [3.3K]3 years ago
8 0
\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)\\\\
-------------------------------

\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||=\sqrt{(-6)^2+4^2}\\
\qquad \sqrt{52}\\
||v2||=\sqrt{(-3)^2+6^2}\\
\qquad \sqrt{45}
\end{cases}\implies \measuredangle \theta =cos^{-1}\left( \cfrac{42}{\sqrt{52}\cdot \sqrt{45}} \right)
\\\\\\
\measuredangle \theta =cos^{-1}\left(  \cfrac{42}{\sqrt{2340}} \right)\implies \measuredangle \theta \approx 29.74488129694^o
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