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

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(B) First,calculate the cross product $\overline{a} 	imes \overline{b}$:$\overline{a} 	imes \overline{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{

Vedclass · Accepted Answer

$\frac{3 \sqrt{3}}{2}$

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(B) First,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 $|\overline{a} 	imes \overline{b}| = \sqrt{2^2 + (-2)^2 + 1^2} = \sqrt{4+4+1} = 3$ ...$(i)$.Given $|\overline{c} - \overline{a}| = 2 \sqrt{2}$,squaring both sides gives $|\overline{c}|^2 + |\overline{a}|^2 - 2(\overline{c} \cdot \overline{a}) = 8$.Since $|\overline{a}| = \sqrt{2^2 + 1^2 + (-2)^2} = 3$,we have $|\overline{a}|^2 = 9$.Given $\overline{a} \cdot \overline{c} = |\overline{c}|$,the equation becomes $|\overline{c}|^2 + 9 - 2|\overline{c}| = 8$.This simplifies to $|\overline{c}|^2 - 2|\overline{c}| + 1 = 0$,which is $(|\overline{c}| - 1)^2 = 0$,so $|\overline{c}| = 1$ ...$(ii)$.The magnitude of the cross product is $|(\overline{a} 	imes \overline{b}) 	imes \overline{c}| = |\overline{a} 	imes \overline{b}| |\overline{c}| \sin(60^{\circ})$.Substituting the values: $(3)(1)(\frac{\sqrt{3}}{2}) = \frac{3 \sqrt{3}}{2}$.

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