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

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(A) Given $|\vec c - \vec a| = \sqrt{14}$. Squaring both sides,we get $|\vec c|^2 + |\vec a|^2 - 2(\vec c \cdot \vec a) = 14$. Since $\vec a = 2\hat i

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$\frac{3}{2}$

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(A) Given $|\vec c - \vec a| = \sqrt{14}$. Squaring both sides,we get $|\vec c|^2 + |\vec a|^2 - 2(\vec c \cdot \vec a) = 14$. Since $\vec a = 2\hat i + \hat j - 2\hat k$,$|\vec a| = \sqrt{2^2 + 1^2 + (-2)^2} = \sqrt{9} = 3$. Thus,$|\vec c|^2 + 9 - 2(\vec c \cdot \vec a) = 14$,which implies $|\vec c|^2 - 2(\vec c \cdot \vec a) = 5$  ........$(1)$. Given $\vec a \cdot \vec c + 2|\vec c| = 0$,so $\vec a \cdot \vec c = -2|\vec c|$. Substituting this into $(1)$,we get $|\vec c|^2 - 2(-2|\vec c|) = 5$,which simplifies to $|\vec c|^2 + 4|\vec c| - 5 = 0$. Factoring gives $(|\vec c| + 5)(|\vec c| - 1) = 0$. Since $|\vec c| > 0$,we have $|\vec c| = 1$. Now,$\vec a 	imes \vec b = \begin{vmatrix} \hat i & \hat j & \hat k \ 2 & 1 & -2 \ 1 & 1 & 0 \end{vmatrix} = \hat i(0 - (-2)) - \hat j(0 - (-2)) + \hat k(2 - 1) = 2\hat i - 2\hat j + \hat k$. The magnitude $|\vec a 	imes \vec b| = \sqrt{2^2 + (-2)^2 + 1^2} = \sqrt{9} = 3$. The magnitude of the cross product is $|(\vec a 	imes \vec b) 	imes \vec c| = |\vec a 	imes \vec b| |\vec c| \sin(30^o) = 3 	imes 1 	imes \frac{1}{2} = \frac{3}{2}$.

Let $\vec a = 2\hat i + \hat j - 2\hat k$ and $\vec b = \hat i + \hat j$. If $\vec c$ is a vector such that $\vec a \cdot \vec c + 2|\vec c| = 0$ and $|\vec c - \vec a| = \sqrt{14}$,and the angle between $\vec a \times \vec b$ and $\vec c$ is $30^o$,then $|(\vec a \times \vec b) \times \vec c|$ is:

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