If |vec{a} times vec{b}|^{2}+|vec{a} cdot vec | vedclass

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(C) Given that,$|\vec{a} 	imes \vec{b}|^{2}+|\vec{a} \cdot \vec{b}|^{2}=144$ and $|\vec{a}|=4$. We know the Lagrange's identity for vectors: $|\vec{a

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$3$

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(C) Given that,$|\vec{a} 	imes \vec{b}|^{2}+|\vec{a} \cdot \vec{b}|^{2}=144$ and $|\vec{a}|=4$. We know the Lagrange's identity for vectors: $|\vec{a} 	imes \vec{b}|^{2}+|\vec{a} \cdot \vec{b}|^{2}=|\vec{a}|^{2} |\vec{b}|^{2} \sin^{2} 	heta + |\vec{a}|^{2} |\vec{b}|^{2} \cos^{2} 	heta$. Factoring out $|\vec{a}|^{2} |\vec{b}|^{2}$,we get: $|\vec{a}|^{2} |\vec{b}|^{2} (\sin^{2} 	heta + \cos^{2} 	heta) = |\vec{a}|^{2} |\vec{b}|^{2}$. Therefore,$|\vec{a} 	imes \vec{b}|^{2}+|\vec{a} \cdot \vec{b}|^{2}=|\vec{a}|^{2} |\vec{b}|^{2}$. Substituting the given values: $144 = (4)^{2} |\vec{b}|^{2}$. $144 = 16 |\vec{b}|^{2}$. $|\vec{b}|^{2} = \frac{144}{16} = 9$. Taking the square root,we get $|\vec{b}| = 3$.

If $|\vec{a} \times \vec{b}|^{2}+|\vec{a} \cdot \vec{b}|^{2}=144$ and $|\vec{a}|=4$,then the value of $|\vec{b}|$ is

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