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

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(D) 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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$\frac{3}{2}$

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(D) 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 $|\vec c - \vec a| = 2\sqrt 2$,squaring both sides gives $|\vec c - \vec a|^2 = 8$.Expanding this,we get $|\vec c|^2 + |\vec a|^2 - 2(\vec c \cdot \vec a) = 8$.Substituting $|\vec a| = 3$ and $\vec a \cdot \vec c = |\vec c|$,we have $|\vec c|^2 + 9 - 2|\vec c| = 8$.This simplifies to $|\vec c|^2 - 2|\vec c| + 1 = 0$,which is $(|\vec c| - 1)^2 = 0$,so $|\vec c| = 1$.Finally,the magnitude of the cross product is $|(\vec a 	imes \vec b) 	imes \vec c| = |\vec a 	imes \vec b| |\vec c| \sin 30^o$.Substituting the values: $3 	imes 1 	imes \frac{1}{2} = \frac{3}{2}$.

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 $\vec a \cdot \vec c = |\vec c|$,$|\vec c - \vec a| = 2\sqrt 2$,and the angle between $\vec a \times \vec b$ and $\vec c$ is $30^o$,then $|(\vec a \times \vec b) \times \vec c|$ equals:

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