If |a+b|^{2}+|a cdot b|^{2}=144 and |a|=6,the | vedclass

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(C) Given that $|a+b|^{2}+|a \cdot b|^{2}=144$ and $|a|=6$. We know the identity $|a+b|^{2} + |a-b|^{2} = 2(|a|^{2} + |b|^{2})$,but here we use the pr

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$2$

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(C) Given that $|a+b|^{2}+|a \cdot b|^{2}=144$ and $|a|=6$. We know the identity $|a+b|^{2} + |a-b|^{2} = 2(|a|^{2} + |b|^{2})$,but here we use the property related to vectors. Actually,the expression $|a+b|^{2} + |a \cdot b|^{2}$ is not a standard identity. Assuming the intended expression is $|a+b|^{2} + |a-b|^{2} = 2(|a|^{2} + |b|^{2})$ or similar,let us re-evaluate the given equation $|a+b|^{2} + |a \cdot b|^{2} = 144$. If we assume the question implies $|a|^{2}|b|^{2} = 144$,then $|6|^{2}|b|^{2} = 144$. $36|b|^{2} = 144$. $|b|^{2} = \frac{144}{36} = 4$. Therefore,$|b| = 2$.

If $|a+b|^{2}+|a \cdot b|^{2}=144$ and $|a|=6$,then $|b|$ is equal to

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