If (a times b)^{2} + (a cdot b)^{2} = 144 and | vedclass

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(C) We know that for any two vectors $a$ and $b$,$(a 	imes b)^{2} + (a \cdot b)^{2} = |a|^{2} |b|^{2} \sin^{2} 	heta + |a|^{2} |b|^{2} \cos^{2} 	he

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$3$

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(C) We know that for any two vectors $a$ and $b$,$(a 	imes b)^{2} + (a \cdot b)^{2} = |a|^{2} |b|^{2} \sin^{2} 	heta + |a|^{2} |b|^{2} \cos^{2} 	heta$. Since $\sin^{2} 	heta + \cos^{2} 	heta = 1$,this simplifies to $|a|^{2} |b|^{2}$. Given the equation $(a 	imes b)^{2} + (a \cdot b)^{2} = 144$,we have $|a|^{2} |b|^{2} = 144$. Given $|a| = 4$,we substitute $|a|^{2} = 16$ into the equation: $16 |b|^{2} = 144$. Dividing both sides by $16$,we get $|b|^{2} = 9$. Taking the square root,we find $|b| = 3$.

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

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