Let vec{a}, vec{b}, vec{c} be three non-copla | vedclass

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(D) Given $\vec{a} 	imes \vec{b} = 4\vec{c}$,$\vec{b} 	imes \vec{c} = 9\vec{a}$,and $\vec{c} 	imes \vec{a} = \alpha\vec{b}$.Taking the magnitude of

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$36$

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(D) Given $\vec{a} 	imes \vec{b} = 4\vec{c}$,$\vec{b} 	imes \vec{c} = 9\vec{a}$,and $\vec{c} 	imes \vec{a} = \alpha\vec{b}$.Taking the magnitude of the first equation: $|\vec{a} 	imes \vec{b}| = |4\vec{c}| \implies |\vec{a}||\vec{b}| \sin 	heta_1 = 4|\vec{c}|$. Since $\vec{a}, \vec{b}, \vec{c}$ are mutually orthogonal (as $\vec{a} 	imes \vec{b}$ is parallel to $\vec{c}$),we have $|\vec{a}||\vec{b}| = 4|\vec{c}|$.Similarly,$|\vec{b}||\vec{c}| = 9|\vec{a}|$ and $|\vec{c}||\vec{a}| = \alpha|\vec{b}|$.Multiplying these three equations: $(|\vec{a}||\vec{b}||\vec{c}|)^2 = 4 	imes 9 	imes \alpha 	imes |\vec{a}||\vec{b}||\vec{c}|$.Thus,$|\vec{a}||\vec{b}||\vec{c}| = 36\alpha$.From $|\vec{a}||\vec{b}| = 4|\vec{c}|$,we get $|\vec{a}||\vec{b}||\vec{c}| = 4|\vec{c}|^2 = 36\alpha \implies |\vec{c}|^2 = 9\alpha \implies |\vec{c}| = 3\sqrt{\alpha}$.Similarly,$|\vec{a}| = 2\sqrt{\alpha}$ and $|\vec{b}| = 6$ (since $|\vec{b}|^2 = 36 \implies |\vec{b}| = 6$ is not correct,let's re-evaluate).Using $|\vec{a}| = 2\sqrt{\alpha}, |\vec{b}| = 3\sqrt{\alpha}, |\vec{c}| = 6$ is wrong. Let's use $|\vec{a}||\vec{b}| = 4|\vec{c}|, |\vec{b}||\vec{c}| = 9|\vec{a}|, |\vec{c}||\vec{a}| = \alpha|\vec{b}|$.Dividing equations: $\frac{|\vec{a}|}{|\vec{c}|} = \frac{4|\vec{c}|}{9|\vec{a}|} \implies 9|\vec{a}|^2 = 4|\vec{c}|^2 \implies 3|\vec{a}| = 2|\vec{c}|$.Also $\frac{|\vec{b}|}{|\vec{a}|} = \frac{9|\vec{a}|}{\alpha|\vec{b}|} \implies \alpha|\vec{b}|^2 = 9|\vec{a}|^2$.Given $|\vec{a}| + |\vec{b}| + |\vec{c}| = 36$. Let $|\vec{a}| = 2k, |\vec{c}| = 3k$. Then $4|\vec{c}| = 12k = |\vec{a}||\vec{b}| = 2k|\vec{b}| \implies |\vec{b}| = 6$.Then $9|\vec{a}|^2 = \alpha|\vec{b}|^2 \implies 9(4k^2) = \alpha(36) \implies 36k^2 = 36\alpha \implies k^2 = \alpha \implies k = \sqrt{\alpha}$.Substituting into sum: $2\sqrt{\alpha} + 6 + 3\sqrt{\alpha} = 36 \implies 5\sqrt{\alpha} = 30 \implies \sqrt{\alpha} = 6 \implies \alpha = 36$.

Let $\vec{a}, \vec{b}, \vec{c}$ be three non-coplanar vectors such that $\vec{a} \times \vec{b} = 4\vec{c}$,$\vec{b} \times \vec{c} = 9\vec{a}$,and $\vec{c} \times \vec{a} = \alpha\vec{b}$,where $\alpha > 0$. If $|\vec{a}| + |\vec{b}| + |\vec{c}| = 36$,then $\alpha$ is equal to:

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