The angle between two vectors given by $6\hat i + 6\hat j - 3\hat k$ and $7\hat i + 4\hat j + 4\hat k$ is
The angle between two vectors given by $6\hat i + 6\hat j - 3\hat k$ and $7\hat i + 4\hat j + 4\hat k$ is
${\cos ^{ - 1}}\left( {\frac{1}{{\sqrt 3 }}} \right)$
${\cos ^{ - 1}}\left( {\frac{5}{{\sqrt 3 }}} \right)$
${\sin ^{ - 1}}\left( {\frac{2}{{\sqrt 3 }}} \right)$
${\sin ^{ - 1}}\left( {\frac{{\sqrt 5 }}{3}} \right)$
The angle between two vectors given by $6\hat i + 6\hat j - 3\hat k$ and $7\hat i + 4\hat j + 4\hat k$ is
$\cos \theta = \frac{{\vec A\vec B}}{{AB}} = \frac{{42 + 24 - 12}}{{\sqrt {36 + 36 + 9} \sqrt {49 + 16 + 16} }}$$ = \frac{{56}}{{9\sqrt {71} }}$
$\cos \theta = \frac{{56}}{{9\sqrt {71} }}$
$\therefore \sin \theta = \frac{{\sqrt 5 }}{3}$ or $\theta = {\sin ^{ - 1}}\left( {\frac{{\sqrt 5 }}{3}} \right)$
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