If |a times b|^2 + |a cdot b|^2 = 144 and |a| | vedclass

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(A) Given that $|a 	imes b|^2 + |a \cdot b|^2 = 144$ and $|a| = 4$.We know that $|a 	imes b| = |a||b| \sin 	heta$ and $|a \cdot b| = |a||b| \cos

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$3$

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(A) Given that $|a 	imes b|^2 + |a \cdot b|^2 = 144$ and $|a| = 4$.We know that $|a 	imes b| = |a||b| \sin 	heta$ and $|a \cdot b| = |a||b| \cos 	heta$.Substituting these into the given equation:$|a|^2|b|^2 \sin^2 	heta + |a|^2|b|^2 \cos^2 	heta = 144$$|a|^2|b|^2 (\sin^2 	heta + \cos^2 	heta) = 144$Since $\sin^2 	heta + \cos^2 	heta = 1$,we have:$|a|^2|b|^2 = 144$Given $|a| = 4$,so $|a|^2 = 16$.$16 	imes |b|^2 = 144$$|b|^2 = \frac{144}{16} = 9$$|b| = \sqrt{9} = 3$.

If $|a \times b|^2 + |a \cdot b|^2 = 144$ and $|a| = 4$,then $|b|$ is equal to

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