Let vec{a}, vec{b}, vec{c} be three vectors s | vedclass

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(D) Given $2(\vec{a} 	imes \vec{b}) = 3(\vec{c} 	imes \vec{a})$.This can be written as $\vec{a} 	imes (2\vec{b} + 3\vec{c}) = 0$.This implies $\vec

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$3$

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(D) Given $2(\vec{a} 	imes \vec{b}) = 3(\vec{c} 	imes \vec{a})$.This can be written as $\vec{a} 	imes (2\vec{b} + 3\vec{c}) = 0$.This implies $\vec{a} = \lambda(2\vec{b} + 3\vec{c})$ for some scalar $\lambda$.Given $|\vec{a}| = \sqrt{31}$,$|\vec{b}| = 1/2$,and $|\vec{c}| = 2$.The angle between $\vec{b}$ and $\vec{c}$ is $	heta = \frac{2\pi}{3}$,so $\vec{b} \cdot \vec{c} = |\vec{b}||\vec{c}| \cos(2\pi/3) = (1/2)(2)(-1/2) = -1/2$.Now,$|\vec{a}|^2 = \lambda^2 |2\vec{b} + 3\vec{c}|^2 = \lambda^2 (4|\vec{b}|^2 + 9|\vec{c}|^2 + 12\vec{b} \cdot \vec{c})$.$31 = \lambda^2 (4(1/4) + 9(4) + 12(-1/2)) = \lambda^2 (1 + 36 - 6) = 31\lambda^2$.Thus,$\lambda^2 = 1$,so $\lambda = \pm 1$.Then $\vec{a} = \pm(2\vec{b} + 3\vec{c})$.We need to evaluate $\left(\frac{\vec{a} 	imes \vec{c}}{\vec{a} \cdot \vec{b}}ight)^2 = \frac{|\vec{a} 	imes \vec{c}|^2}{(\vec{a} \cdot \vec{b})^2}$.$|\vec{a} 	imes \vec{c}|^2 = |\pm(2\vec{b} + 3\vec{c}) 	imes \vec{c}|^2 = |2(\vec{b} 	imes \vec{c})|^2 = 4|\vec{b} 	imes \vec{c}|^2 = 4(|\vec{b}|^2|\vec{c}|^2 - (\vec{b} \cdot \vec{c})^2) = 4(1/4 \cdot 4 - (-1/2)^2) = 4(1 - 1/4) = 3$.$\vec{a} \cdot \vec{b} = \pm(2\vec{b} + 3\vec{c}) \cdot \vec{b} = \pm(2|\vec{b}|^2 + 3\vec{b} \cdot \vec{c}) = \pm(2(1/4) + 3(-1/2)) = \pm(1/2 - 3/2) = \pm(-1) = \mp 1$.So,$(\vec{a} \cdot \vec{b})^2 = 1$.Therefore,$\left(\frac{\vec{a} 	imes \vec{c}}{\vec{a} \cdot \vec{b}}ight)^2 = \frac{3}{1} = 3$.

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