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

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(A) Given $\bar{a} = \hat{i} + 2\hat{j} - 2\hat{k}$,so $|\bar{a}|^2 = 1^2 + 2^2 + (-2)^2 = 9$,hence $|\bar{a}| = 3$.Given $|\bar{c} - \bar{a}| = 2\sqr

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$\frac{3\sqrt{6}}{2}$

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(A) Given $\bar{a} = \hat{i} + 2\hat{j} - 2\hat{k}$,so $|\bar{a}|^2 = 1^2 + 2^2 + (-2)^2 = 9$,hence $|\bar{a}| = 3$.Given $|\bar{c} - \bar{a}| = 2\sqrt{2}$,squaring both sides gives $|\bar{c}|^2 + |\bar{a}|^2 - 2(\bar{a} \cdot \bar{c}) = 8$.Since $\bar{a} \cdot \bar{c} = |\bar{c}|$,let $|\bar{c}| = x$. Then $x^2 + 9 - 2x = 8$,which simplifies to $x^2 - 2x + 1 = 0$,so $(x-1)^2 = 0$,implying $|\bar{c}| = 1$.Now,$\bar{a} 	imes \bar{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & 2 & -2 \ 1 & -1 & 1 \end{vmatrix} = \hat{i}(2-2) - \hat{j}(1+2) + \hat{k}(-1-2) = 0\hat{i} - 3\hat{j} - 3\hat{k}$.Thus,$|\bar{a} 	imes \bar{b}| = \sqrt{0^2 + (-3)^2 + (-3)^2} = \sqrt{18} = 3\sqrt{2}$.The angle between $\bar{a} 	imes \bar{b}$ and $\bar{c}$ is $	heta = 60^{\circ}$.We need to find $|(\bar{a} 	imes \bar{b}) 	imes \bar{c}| = |\bar{a} 	imes \bar{b}| |\bar{c}| \sin(60^{\circ})$.Substituting the values: $(3\sqrt{2}) 	imes (1) 	imes \frac{\sqrt{3}}{2} = \frac{3\sqrt{6}}{2}$.

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