Let bar{a}=2 hat{i}+hat{j}-2 hat{k},bar{b}=ha | vedclass

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

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$\frac{-3}{2}$

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(D) Given $\bar{a}=2 \hat{i}+\hat{j}-2 \hat{k}$ and $\bar{b}=\hat{i}+\hat{j}$.First,calculate the magnitude of $\bar{a}$:$|\bar{a}|=\sqrt{2^2+1^2+(-2)^2}=\sqrt{4+1+4}=3$.Next,calculate the cross product $\bar{a} 	imes \bar{b}$:$\bar{a} 	imes \bar{b}=\begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 2 & 1 & -2 \ 1 & 1 & 0 \end{vmatrix} = \hat{i}(0 - (-2)) - \hat{j}(0 - (-2)) + \hat{k}(2 - 1) = 2 \hat{i}-2 \hat{j}+\hat{k}$.The magnitude is $|\bar{a} 	imes \bar{b}|=\sqrt{2^2+(-2)^2+1^2}=\sqrt{4+4+1}=3$.Given $|(\bar{a} 	imes \bar{b}) 	imes \bar{c}|=3$ and the angle $	heta$ between $\bar{c}$ and $\bar{a} 	imes \bar{b}$ is $\frac{\pi}{6}$.Using the formula $|\bar{u} 	imes \bar{v}| = |\bar{u}||\bar{v}| \sin 	heta$:$|(\bar{a} 	imes \bar{b}) 	imes \bar{c}| = |\bar{a} 	imes \bar{b}| |\bar{c}| \sin \frac{\pi}{6} = 3$.$3 \cdot |\bar{c}| \cdot \frac{1}{2} = 3 \Rightarrow |\bar{c}|=2$.Now,use the given $|\bar{c}-\bar{a}|=4$:$|\bar{c}-\bar{a}|^2 = |\bar{c}|^2 + |\bar{a}|^2 - 2(\bar{a} \cdot \bar{c}) = 4^2$.$2^2 + 3^2 - 2(\bar{a} \cdot \bar{c}) = 16$.$4 + 9 - 2(\bar{a} \cdot \bar{c}) = 16$.$13 - 2(\bar{a} \cdot \bar{c}) = 16$.$-2(\bar{a} \cdot \bar{c}) = 3$.$\bar{a} \cdot \bar{c} = -\frac{3}{2}$.

Let $\bar{a}=2 \hat{i}+\hat{j}-2 \hat{k}$,$\bar{b}=\hat{i}+\hat{j}$ and $\bar{c}$ be a vector such that $|\bar{c}-\bar{a}|=4$,$|(\bar{a} \times \bar{b}) \times \bar{c}|=3$ and the angle between $\bar{c}$ and $\bar{a} \times \bar{b}$ is $\frac{\pi}{6}$,then $\bar{a} \cdot \bar{c}$ is equal to

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