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

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(C) Given, $a=2 \hat{i}-2 \hat{j}+\hat{k}$, then $|a|=\sqrt{2^2+(-2)^2+1^2}=\sqrt{4+4+1}=3$.Given $|c-a|=2 \sqrt{2}$. Squaring both sides, we get $|c-

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

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(C) Given, $a=2 \hat{i}-2 \hat{j}+\hat{k}$, then $|a|=\sqrt{2^2+(-2)^2+1^2}=\sqrt{4+4+1}=3$.Given $|c-a|=2 \sqrt{2}$. Squaring both sides, we get $|c-a|^2=8$.$|c|^2+|a|^2-2(a \cdot c)=8$.Substituting $|a|=3$ and $a \cdot c=|c|$, we have $|c|^2+9-2|c|=8$.$|c|^2-2|c|+1=0 \Rightarrow (|c|-1)^2=0 \Rightarrow |c|=1$.Now, calculate $a 	imes b$:$a 	imes b = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 2 & -2 & 1 \ 0 & -1 & 1 \end{vmatrix} = \hat{i}(-2+1) - \hat{j}(2-0) + \hat{k}(-2-0) = -\hat{i}-2 \hat{j}-2 \hat{k}$.Then $|a 	imes b| = \sqrt{(-1)^2+(-2)^2+(-2)^2} = \sqrt{1+4+4} = 3$.The magnitude of the cross product is given by $|(a 	imes b) 	imes c| = |a 	imes b| |c| \sin 	heta$, where $	heta = \frac{\pi}{3}$.$|(a 	imes b) 	imes c| = 3 	imes 1 	imes \sin \frac{\pi}{3} = 3 	imes \frac{\sqrt{3}}{2} = \frac{3 \sqrt{3}}{2}$.

Let $a=2 \hat{i}-2 \hat{j}+\hat{k}$ and $b=-\hat{j}+\hat{k}$. If $c$ is a vector such that $a \cdot c=|c|$, $|c-a|=2 \sqrt{2}$, and the angle between $a \times b$ and $c$ is $\frac{\pi}{3}$, then $|(a \times b) \times c|=$

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