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

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

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$\frac{3}{2}$

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(D) Given $\vec{a} = 2\hat{i} + \hat{j} - 2\hat{k}$,so $|\vec{a}|^2 = 2^2 + 1^2 + (-2)^2 = 4 + 1 + 4 = 9$,which implies $|\vec{a}| = 3$.Given $|\vec{c} - \vec{a}| = 2\sqrt{2}$. Squaring both sides,we get $|\vec{c}|^2 + |\vec{a}|^2 - 2(\vec{c} \cdot \vec{a}) = (2\sqrt{2})^2 = 8$.Since $\vec{a} \cdot \vec{c} = |\vec{c}|$,let $|\vec{c}| = c$. Then $c^2 + 9 - 2c = 8$.$c^2 - 2c + 1 = 0 \Rightarrow (c - 1)^2 = 0 \Rightarrow c = 1$. Thus,$|\vec{c}| = 1$.Now,calculate $\vec{a} 	imes \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 2 & 1 & -2 \ 1 & 1 & 0 \end{vmatrix} = \hat{i}(0 - (-2)) - \hat{j}(0 - (-2)) + \hat{k}(2 - 1) = 2\hat{i} - 2\hat{j} + \hat{k}$.The magnitude $|\vec{a} 	imes \vec{b}| = \sqrt{2^2 + (-2)^2 + 1^2} = \sqrt{4 + 4 + 1} = \sqrt{9} = 3$.The magnitude of the cross product is $|(\vec{a} 	imes \vec{b}) 	imes \vec{c}| = |\vec{a} 	imes \vec{b}| |\vec{c}| \sin(	heta)$,where $	heta = \frac{\pi}{6}$.$|(\vec{a} 	imes \vec{b}) 	imes \vec{c}| = (3)(1) \sin(\frac{\pi}{6}) = 3 	imes \frac{1}{2} = \frac{3}{2}$.

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