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(A) Given $|\vec{a}|=4$ and $|\vec{b}|=3$. We need to evaluate $|(\vec{a}-\vec{b}) 	imes (\vec{a}+\vec{b})|^{2} + 4(\vec{a} \cdot \vec{b})^{2}$. Expa

Vedclass · Accepted Answer

$576$

Vedclass · Answer

(A) Given $|\vec{a}|=4$ and $|\vec{b}|=3$. We need to evaluate $|(\vec{a}-\vec{b}) 	imes (\vec{a}+\vec{b})|^{2} + 4(\vec{a} \cdot \vec{b})^{2}$. Expanding the cross product: $(\vec{a}-\vec{b}) 	imes (\vec{a}+\vec{b}) = \vec{a} 	imes \vec{a} + \vec{a} 	imes \vec{b} - \vec{b} 	imes \vec{a} - \vec{b} 	imes \vec{b}$. Since $\vec{a} 	imes \vec{a} = 0$ and $\vec{b} 	imes \vec{b} = 0$,and $\vec{b} 	imes \vec{a} = -(\vec{a} 	imes \vec{b})$,we get: $(\vec{a}-\vec{b}) 	imes (\vec{a}+\vec{b}) = 0 + \vec{a} 	imes \vec{b} + \vec{a} 	imes \vec{b} - 0 = 2(\vec{a} 	imes \vec{b})$. Now,$|2(\vec{a} 	imes \vec{b})|^{2} = 4|\vec{a} 	imes \vec{b}|^{2} = 4|\vec{a}|^{2}|\vec{b}|^{2} \sin^{2} 	heta$. Also,$4(\vec{a} \cdot \vec{b})^{2} = 4(|\vec{a}||\vec{b}| \cos 	heta)^{2} = 4|\vec{a}|^{2}|\vec{b}|^{2} \cos^{2} 	heta$. Adding these: $4|\vec{a}|^{2}|\vec{b}|^{2} \sin^{2} 	heta + 4|\vec{a}|^{2}|\vec{b}|^{2} \cos^{2} 	heta = 4|\vec{a}|^{2}|\vec{b}|^{2} (\sin^{2} 	heta + \cos^{2} 	heta) = 4|\vec{a}|^{2}|\vec{b}|^{2}$. Substituting the values $|\vec{a}|=4$ and $|\vec{b}|=3$: $4 	imes (4)^{2} 	imes (3)^{2} = 4 	imes 16 	imes 9 = 576$.

Let $\theta$ be the angle between the vectors $\vec{a}$ and $\vec{b}$,where $|\vec{a}|=4, |\vec{b}|=3$ and $\theta \in \left(\frac{\pi}{4}, \frac{\pi}{3}\right)$. Then $|(\vec{a}-\vec{b}) \times (\vec{a}+\vec{b})|^{2} + 4(\vec{a} \cdot \vec{b})^{2}$ is equal to

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