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

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(B) Given $|\overline{c}-\overline{a}|=2 \sqrt{2}$. Squaring both sides,we get $|\overline{c}|^2+|\overline{a}|^2-2(\overline{a} \cdot \overline{c})=8

Vedclass · Accepted Answer

$\frac{3}{2}$

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(B) Given $|\overline{c}-\overline{a}|=2 \sqrt{2}$. Squaring both sides,we get $|\overline{c}|^2+|\overline{a}|^2-2(\overline{a} \cdot \overline{c})=8$. Since $|\overline{a}|^2 = 2^2+1^2+(-2)^2 = 9$ and $\overline{a} \cdot \overline{c}=|\overline{c}|$,the equation becomes $|\overline{c}|^2+9-2|\overline{c}|=8$. This simplifies to $(|\overline{c}|-1)^2=0$,so $|\overline{c}|=1$. Next,calculate $\overline{a} 	imes \overline{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 2 & 1 & -2 \ 1 & 1 & 0 \end{vmatrix} = \hat{i}(0 - (-2)) - \hat{j}(0 - (-2)) + \hat{k}(2 - 1) = 2\hat{i}-2\hat{j}+\hat{k}$. The magnitude is $|\overline{a} 	imes \overline{b}| = \sqrt{2^2+(-2)^2+1^2} = \sqrt{4+4+1} = 3$. Finally,$|(\overline{a} 	imes \overline{b}) 	imes \overline{c}| = |\overline{a} 	imes \overline{b}| |\overline{c}| \sin(30^{\circ}) = 3 	imes 1 	imes \frac{1}{2} = \frac{3}{2}$.

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