Let overline{a}=2 hat{i}+hat{j}-2 hat{k} and | vedclass

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

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$2$

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

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