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

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(C) Given $\vec a = 2\hat i + \hat j - 2\hat k$ and $\vec b = \hat i + \hat j$.First,calculate the magnitude of $\vec a$: $|\vec a| = \sqrt{2^2 + 1^2

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$2$

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(C) Given $\vec a = 2\hat i + \hat j - 2\hat k$ and $\vec b = \hat i + \hat j$.First,calculate the magnitude of $\vec a$: $|\vec a| = \sqrt{2^2 + 1^2 + (-2)^2} = \sqrt{4+1+4} = 3$.Next,calculate the cross product $\vec a 	imes \vec b$:$\vec a 	imes \vec b = \begin{vmatrix} \hat i & \hat j & \hat k \ 2 & 1 & -2 \ 1 & 1 & 0 \end{vmatrix} = \hat i(0 - (-2)) - \hat j(0 - (-2)) + \hat k(2 - 1) = 2\hat i - 2\hat j + \hat k$.The magnitude is $|\vec a 	imes \vec b| = \sqrt{2^2 + (-2)^2 + 1^2} = \sqrt{4+4+1} = 3$.Given $|(\vec a 	imes \vec b) 	imes \vec c| = 3$ and the angle between $\vec c$ and $\vec a 	imes \vec b$ is $30^\circ$,we use the formula $|\vec u 	imes \vec v| = |\vec u||\vec v| \sin 	heta$:$|\vec a 	imes \vec b||\vec c| \sin 30^\circ = 3 \implies 3 \cdot |\vec c| \cdot \frac{1}{2} = 3 \implies |\vec c| = 2$.Now,use the condition $|\vec c - \vec a| = 3$. Squaring both sides:$|\vec c - \vec a|^2 = 3^2 \implies |\vec c|^2 + |\vec a|^2 - 2(\vec a \cdot \vec c) = 9$.Substitute the known values $|\vec c| = 2$ and $|\vec a| = 3$:$2^2 + 3^2 - 2(\vec a \cdot \vec c) = 9 \implies 4 + 9 - 2(\vec a \cdot \vec c) = 9 \implies 13 - 2(\vec a \cdot \vec c) = 9$.$2(\vec a \cdot \vec c) = 4 \implies \vec a \cdot \vec c = 2$.

Let $\vec a = 2\hat i + \hat j - 2\hat k$ and $\vec b = \hat i + \hat j$. Let $\vec c$ be a vector such that $|\vec c - \vec a| = 3$,$|(\vec a \times \vec b) \times \vec c| = 3$,and the angle between $\vec c$ and $\vec a \times \vec b$ is $30^\circ$. Then $\vec a \cdot \vec c$ is equal to:

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