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

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(A) First,calculate the cross product $\bar{a} 	imes \bar{b}$:$\bar{a} 	imes \bar{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 2 & 1 & -2 \

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

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(A) First,calculate the cross product $\bar{a} 	imes \bar{b}$:$\bar{a} 	imes \bar{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 2 & 1 & -2 \ 1 & 1 & 0 \end{vmatrix} = \hat{i}(0 - (-2)) - \hat{j}(0 - (-2)) + \hat{k}(2 - 1) = 2\hat{i} - 2\hat{j} + \hat{k}$.The magnitude is $|\bar{a} 	imes \bar{b}| = \sqrt{2^2 + (-2)^2 + 1^2} = \sqrt{4 + 4 + 1} = 3$.Also,$|\bar{a}| = \sqrt{2^2 + 1^2 + (-2)^2} = 3$.Given $|\bar{c}-\bar{a}| = 2\sqrt{2}$,squaring both sides gives $|\bar{c}|^2 + |\bar{a}|^2 - 2(\bar{c} \cdot \bar{a}) = 8$.Substituting $|\bar{a}|^2 = 9$ and $\bar{c} \cdot \bar{a} = |\bar{c}|$,we get $|\bar{c}|^2 + 9 - 2|\bar{c}| = 8$.This simplifies to $|\bar{c}|^2 - 2|\bar{c}| + 1 = 0$,which is $(|\bar{c}| - 1)^2 = 0$,so $|\bar{c}| = 1$.The magnitude of the cross product is $|(\bar{a} 	imes \bar{b}) 	imes \bar{c}| = |\bar{a} 	imes \bar{b}| |\bar{c}| \sin(60^{\circ})$.Substituting the values: $(3)(1)(\frac{\sqrt{3}}{2}) = \frac{3\sqrt{3}}{2}$.

Let $\bar{a}=2 \hat{i}+\hat{j}-2 \hat{k}$ and $\bar{b}=\hat{i}+\hat{j}$. If $\bar{c}$ is a vector such that $\bar{a} \cdot \bar{c}=|\bar{c}|$,$|\bar{c}-\bar{a}|=2 \sqrt{2}$ and the angle between $\bar{a} \times \bar{b}$ and $\bar{c}$ is $60^{\circ}$. Then $|(\bar{a} \times \bar{b}) \times \bar{c}|=$

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