Let vec{a}=6 hat{i}+hat{j}-hat{k} and vec{b}= | vedclass

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(D) Given $\vec{a} = 6\hat{i} + \hat{j} - \hat{k}$,so $|\vec{a}|^2 = 6^2 + 1^2 + (-1)^2 = 36 + 1 + 1 = 38$.Given $|\vec{c} - \vec{a}| = 2\sqrt{2}$,squ

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$\frac{9}{2}(6+\sqrt{6})$

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(D) Given $\vec{a} = 6\hat{i} + \hat{j} - \hat{k}$,so $|\vec{a}|^2 = 6^2 + 1^2 + (-1)^2 = 36 + 1 + 1 = 38$.Given $|\vec{c} - \vec{a}| = 2\sqrt{2}$,squaring both sides gives $|\vec{c}|^2 + |\vec{a}|^2 - 2(\vec{c} \cdot \vec{a}) = 8$.Substituting $\vec{a} \cdot \vec{c} = 6|\vec{c}|$ and $|\vec{a}|^2 = 38$,we get $|\vec{c}|^2 + 38 - 2(6|\vec{c}|) = 8$.$|\vec{c}|^2 - 12|\vec{c}| + 30 = 0$.Solving for $|\vec{c}|$ using the quadratic formula: $|\vec{c}| = \frac{12 \pm \sqrt{144 - 120}}{2} = \frac{12 \pm \sqrt{24}}{2} = 6 \pm \sqrt{6}$.Since $|\vec{c}| \geq 6$,we take $|\vec{c}| = 6 + \sqrt{6}$.Now,$\vec{a} 	imes \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 6 & 1 & -1 \ 1 & 1 & 0 \end{vmatrix} = \hat{i}(0 - (-1)) - \hat{j}(0 - (-1)) + \hat{k}(6 - 1) = \hat{i} - \hat{j} + 5\hat{k}$.$|\vec{a} 	imes \vec{b}| = \sqrt{1^2 + (-1)^2 + 5^2} = \sqrt{1 + 1 + 25} = \sqrt{27} = 3\sqrt{3}$.The magnitude of the cross product is $|(\vec{a} 	imes \vec{b}) 	imes \vec{c}| = |\vec{a} 	imes \vec{b}| |\vec{c}| \sin(60^{\circ})$.$|(\vec{a} 	imes \vec{b}) 	imes \vec{c}| = (3\sqrt{3}) (6 + \sqrt{6}) \frac{\sqrt{3}}{2} = \frac{3 	imes 3}{2} (6 + \sqrt{6}) = \frac{9}{2}(6 + \sqrt{6})$.

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