The angle between two vectors given by 6hat i | vedclass

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Question

(d) $\cos 	heta = \frac{{\vec A\vec B}}{{AB}} = \frac{{42 + 24 - 12}}{{\sqrt {36 + 36 + 9} \sqrt {49 + 16 + 16} }}$$ = \frac{{56}}{{9\sqrt {71} }}$

Vedclass · Accepted Answer

${\sin ^{ - 1}}\left( {\frac{{\sqrt 5 }}{3}} \right)$

Vedclass · Answer

(d) $\cos 	heta = \frac{{\vec A\vec B}}{{AB}} = \frac{{42 + 24 - 12}}{{\sqrt {36 + 36 + 9} \sqrt {49 + 16 + 16} }}$$ = \frac{{56}}{{9\sqrt {71} }}$

$\cos 	heta = \frac{{56}}{{9\sqrt {71} }}$ 

$	herefore \sin 	heta = \frac{{\sqrt 5 }}{3}$ or $	heta = {\sin ^{ - 1}}\left( {\frac{{\sqrt 5 }}{3}} ight)$

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

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