If |a cdot b| = 3 and |a times b| = 4,then th | vedclass

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(B) Given that $|a \cdot b| = |a||b| \cos 	heta = 3$  $(i)$ And $|a 	imes b| = |a||b| \sin 	heta = 4$  $(ii)$ Dividing equation $(ii)$ by equation

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$\cos^{-1} \frac{3}{5}$

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(B) Given that $|a \cdot b| = |a||b| \cos 	heta = 3$  $(i)$ And $|a 	imes b| = |a||b| \sin 	heta = 4$  $(ii)$ Dividing equation $(ii)$ by equation $(i)$,we get: $\frac{|a||b| \sin 	heta}{|a||b| \cos 	heta} = \frac{4}{3}$ $	an 	heta = \frac{4}{3}$ Since $	an 	heta = \frac{	ext{opposite}}{	ext{adjacent}} = \frac{4}{3}$,the hypotenuse is $\sqrt{4^2 + 3^2} = \sqrt{16 + 9} = \sqrt{25} = 5$. Therefore,$\cos 	heta = \frac{	ext{adjacent}}{	ext{hypotenuse}} = \frac{3}{5}$. Thus,$	heta = \cos^{-1} \frac{3}{5}$.

If $|a \cdot b| = 3$ and $|a \times b| = 4$,then the angle between $a$ and $b$ is

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