Let overline{a}=2 hat{i}+hat{j}-2 hat{k} and | vedclass

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(A) Given $\overline{a}=2 \hat{i}+\hat{j}-2 \hat{k}$,so $|\overline{a}|^2 = 2^2 + 1^2 + (-2)^2 = 4+1+4 = 9$,which implies $|\overline{a}|=3$. Given $|

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$\frac{3}{2}$

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(A) Given $\overline{a}=2 \hat{i}+\hat{j}-2 \hat{k}$,so $|\overline{a}|^2 = 2^2 + 1^2 + (-2)^2 = 4+1+4 = 9$,which implies $|\overline{a}|=3$. Given $|\overline{c}-\overline{a}|=2 \sqrt{2}$,squaring both sides gives $|\overline{c}-\overline{a}|^2 = (2 \sqrt{2})^2 = 8$. Expanding this,we get $|\overline{c}|^2 + |\overline{a}|^2 - 2(\overline{a} \cdot \overline{c}) = 8$. Substituting $\overline{a} \cdot \overline{c} = |\overline{c}|$ and $|\overline{a}|^2 = 9$,we have $|\overline{c}|^2 + 9 - 2|\overline{c}| = 8$,which simplifies to $|\overline{c}|^2 - 2|\overline{c}| + 1 = 0$. This is $(|\overline{c}|-1)^2 = 0$,so $|\overline{c}| = 1$. Next,calculate $\overline{a} 	imes \overline{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 2 & 1 & -2 \ 1 & 1 & 0 \end{vmatrix} = \hat{i}(0 - (-2)) - \hat{j}(0 - (-2)) + \hat{k}(2 - 1) = 2 \hat{i} - 2 \hat{j} + \hat{k}$. Thus,$|\overline{a} 	imes \overline{b}| = \sqrt{2^2 + (-2)^2 + 1^2} = \sqrt{4+4+1} = \sqrt{9} = 3$. Finally,$|(\overline{a} 	imes \overline{b}) 	imes \overline{c}| = |\overline{a} 	imes \overline{b}| |\overline{c}| \sin(\frac{\pi}{6}) = 3 	imes 1 	imes \frac{1}{2} = \frac{3}{2}$.

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