If theta is the angle between the vectors bar | vedclass

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(A) We are given the expression $E = |(\bar{a}-\bar{b}) 	imes(\bar{a}+\bar{b})|^2+4(\bar{a} \cdot \bar{b})^2$. Expanding the cross product term: $(\b

Vedclass · Accepted Answer

$576$

Vedclass · Answer

(A) We are given the expression $E = |(\bar{a}-\bar{b}) 	imes(\bar{a}+\bar{b})|^2+4(\bar{a} \cdot \bar{b})^2$. Expanding the cross product term: $(\bar{a}-\bar{b}) 	imes(\bar{a}+\bar{b}) = (\bar{a} 	imes \bar{a}) + (\bar{a} 	imes \bar{b}) - (\bar{b} 	imes \bar{a}) - (\bar{b} 	imes \bar{b})$. Since $\bar{a} 	imes \bar{a} = 0$ and $\bar{b} 	imes \bar{b} = 0$,and using the property $\bar{b} 	imes \bar{a} = -(\bar{a} 	imes \bar{b})$,we get: $(\bar{a}-\bar{b}) 	imes(\bar{a}+\bar{b}) = (\bar{a} 	imes \bar{b}) - (-(\bar{a} 	imes \bar{b})) = 2(\bar{a} 	imes \bar{b})$. Now,substituting this into the expression: $E = |2(\bar{a} 	imes \bar{b})|^2 + 4(\bar{a} \cdot \bar{b})^2 = 4|\bar{a} 	imes \bar{b}|^2 + 4(\bar{a} \cdot \bar{b})^2$. Factoring out $4$: $E = 4(|\bar{a} 	imes \bar{b}|^2 + (\bar{a} \cdot \bar{b})^2)$. Using Lagrange's identity,$|\bar{a} 	imes \bar{b}|^2 + (\bar{a} \cdot \bar{b})^2 = |\bar{a}|^2 |\bar{b}|^2$. Thus,$E = 4 |\bar{a}|^2 |\bar{b}|^2$. Given $|\bar{a}|=4$ and $|\bar{b}|=3$: $E = 4 	imes (4)^2 	imes (3)^2 = 4 	imes 16 	imes 9 = 576$.

If $\theta$ is the angle between the vectors $\bar{a}$ and $\bar{b}$ where $|\bar{a}|=4, |\bar{b}|=3$ and $\theta \in \left(\frac{\pi}{4}, \frac{\pi}{3}\right)$,then $|(\bar{a}-\bar{b}) \times(\bar{a}+\bar{b})|^2+4(\bar{a} \cdot \bar{b})^2$ has the value

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