Let a = 2i + j - 2k and b = i + j. If c is a | vedclass

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(B) First,calculate the cross product $a 	imes b$: $a 	imes b = \begin{vmatrix} i & j & k \ 2 & 1 & -2 \ 1 & 1 & 0 \end{vmatrix} = i(0 - (-2)) - j

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$\frac{3}{2}$

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(B) First,calculate the cross product $a 	imes b$: $a 	imes b = \begin{vmatrix} i & j & k \ 2 & 1 & -2 \ 1 & 1 & 0 \end{vmatrix} = i(0 - (-2)) - j(0 - (-2)) + k(2 - 1) = 2i - 2j + k$. Next,find the magnitude of $a 	imes b$: $|a 	imes b| = \sqrt{2^2 + (-2)^2 + 1^2} = \sqrt{4 + 4 + 1} = \sqrt{9} = 3$. Given $|c - a| = 2\sqrt{2}$,square both sides: $|c - a|^2 = (2\sqrt{2})^2 = 8$. $|c|^2 - 2(a \cdot c) + |a|^2 = 8$. Since $a \cdot c = |c|$ and $|a|^2 = 2^2 + 1^2 + (-2)^2 = 9$: $|c|^2 - 2|c| + 9 = 8 \Rightarrow |c|^2 - 2|c| + 1 = 0 \Rightarrow (|c| - 1)^2 = 0 \Rightarrow |c| = 1$. Finally,calculate the magnitude of the cross product $|(a 	imes b) 	imes c|$ using the formula $|u 	imes v| = |u||v|\sin	heta$: $|(a 	imes b) 	imes c| = |a 	imes b| |c| \sin 30^\circ = 3 	imes 1 	imes \frac{1}{2} = \frac{3}{2}$.

Let $a = 2i + j - 2k$ and $b = i + j$. 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 $30^\circ$,then $|(a \times b) \times c| = $

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