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3 years ago

Find the angle between u = i+sqr of 7j and v = -i+9j. Round to the nearest tenth of a degree.

1 answer:

6 0

$\bf ~~~~~~~~~~~~\textit{angle between two vectors } \\\\ cos(\theta)=\cfrac{\stackrel{\textit{dot product}}{u \cdot v}}{\stackrel{\textit{magnitude product}}{||u||~||v||}} \implies \measuredangle \theta = cos^{-1}\left(\cfrac{u \cdot v}{||u||~||v||}\right) \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ \begin{cases} u=i+\sqrt{7}j\implies &\\\\ v=-i+9j\implies & \end{cases} \\\\[-0.35em] ~\dotfill$

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

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

make sure your calculator is in Degree mode.

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$\rm\displaystyle 0,\pm\pi$

Step-by-step explanation:

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===========================

we want to find all possible values of α+β+γ when tanα+tanβ+tanγ = tanαtanβtanγ to do so we can use algebra and trigonometric skills first

cancel tanγ from both sides which yields:

$\rm\displaystyle \tan( \alpha ) + \tan( \beta ) = \tan( \alpha ) \tan( \beta ) \tan( \gamma ) - \tan( \gamma )$

factor out tanγ:

$\rm\displaystyle \tan( \alpha ) + \tan( \beta ) = \tan( \gamma ) (\tan( \alpha ) \tan( \beta ) - 1)$

divide both sides by tanαtanβ-1 and that yields:

$\rm\displaystyle \tan( \gamma ) = \frac{ \tan( \alpha ) + \tan( \beta ) }{ \tan( \alpha ) \tan( \beta ) - 1}$

multiply both numerator and denominator by-1 which yields:

$\rm\displaystyle \tan( \gamma ) = - \bigg(\frac{ \tan( \alpha ) + \tan( \beta ) }{ 1 - \tan( \alpha ) \tan( \beta ) } \bigg)$

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$\rm\displaystyle \tan( \gamma ) = - \tan( \alpha + \beta )$

let α+β be t and transform:

$\rm\displaystyle \tan( \gamma ) = - \tan( t)$

remember that tan(t)=tan(t±kπ) so

$\rm\displaystyle \tan( \gamma ) = -\tan( \alpha +\beta\pm k\pi )$

therefore when k is 1 we obtain:

$\rm\displaystyle \tan( \gamma ) = -\tan( \alpha +\beta\pm \pi )$

remember Opposite Angle identity of tan function i.e -tan(x)=tan(-x) thus

$\rm\displaystyle \tan( \gamma ) = \tan( -\alpha -\beta\pm \pi )$

recall that if we have common trigonometric function in both sides then the angle must equal which yields:

$\rm\displaystyle \gamma = - \alpha - \beta \pm \pi$

isolate -α-β to left hand side and change its sign:

$\rm\displaystyle \alpha + \beta + \gamma = \boxed{ \pm \pi }$

when is 0:

$\rm\displaystyle \tan( \gamma ) = -\tan( \alpha +\beta \pm 0 )$

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$\rm\displaystyle \tan( \gamma ) = \tan( -\alpha -\beta\pm 0 )$

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isolate -α-β to left hand side and change its sign:

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and we're done!

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