If vec{a}, vec{b}, vec{c} are three vectors o | vedclass

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(A) Given $\vec{a} 	imes (\vec{a} 	imes \vec{c}) + 3\vec{b} = \vec{0}$.Using the vector triple product formula $\vec{a} 	imes (\vec{b} 	imes \vec{

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$3/4$

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(A) Given $\vec{a} 	imes (\vec{a} 	imes \vec{c}) + 3\vec{b} = \vec{0}$.Using the vector triple product formula $\vec{a} 	imes (\vec{b} 	imes \vec{c}) = (\vec{a} \cdot \vec{c})\vec{b} - (\vec{a} \cdot \vec{b})\vec{c}$,we have:$(\vec{a} \cdot \vec{c})\vec{a} - (\vec{a} \cdot \vec{a})\vec{c} + 3\vec{b} = \vec{0}$.Given magnitudes $|\vec{a}| = \sqrt{3}$,$|\vec{b}| = 1$,$|\vec{c}| = 2$,and $\vec{a} \cdot \vec{c} = |\vec{a}||\vec{c}|\cos 	heta = \sqrt{3} 	imes 2 	imes \cos 	heta = 2\sqrt{3} \cos 	heta$.Substituting these values:$(2\sqrt{3} \cos 	heta)\vec{a} - (\sqrt{3})^2 \vec{c} + 3\vec{b} = \vec{0}$.$(2\sqrt{3} \cos 	heta)\vec{a} - 3\vec{c} + 3\vec{b} = \vec{0}$.Rearranging: $3\vec{b} = 3\vec{c} - (2\sqrt{3} \cos 	heta)\vec{a}$.Taking the square of the magnitude on both sides:$|3\vec{b}|^2 = |3\vec{c} - (2\sqrt{3} \cos 	heta)\vec{a}|^2$.$9|\vec{b}|^2 = 9|\vec{c}|^2 + 12 \cos^2 	heta |\vec{a}|^2 - 2(3)(2\sqrt{3} \cos 	heta)(\vec{a} \cdot \vec{c})$.$9(1)^2 = 9(2)^2 + 12 \cos^2 	heta (\sqrt{3})^2 - 12\sqrt{3} \cos 	heta (2\sqrt{3} \cos 	heta)$.$9 = 36 + 36 \cos^2 	heta - 72 \cos^2 	heta$.$9 = 36 - 36 \cos^2 	heta$.$36 \cos^2 	heta = 27$.$\cos^2 	heta = \frac{27}{36} = \frac{3}{4}$.

If $\vec{a}, \vec{b}, \vec{c}$ are three vectors of magnitudes $\sqrt{3}, 1, 2$ respectively,such that $\vec{a} \times (\vec{a} \times \vec{c}) + 3\vec{b} = \vec{0}$. If $\theta$ is the angle between $\vec{a}$ and $\vec{c}$,then $\cos^2 \theta = $

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