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

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(A) 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{

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$2$

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(A) 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 that the angle $	heta$ between $\overline{c}$ and $\overline{a} 	imes \overline{b}$ is $30^{\circ}$,we use the formula for the magnitude of the cross product: $|(\overline{a} 	imes \overline{b}) 	imes \overline{c}| = |\overline{a} 	imes \overline{b}| |\overline{c}| \sin(30^{\circ})$. Substituting the known values: $3 = 3 	imes |\overline{c}| 	imes \frac{1}{2}$. This gives $|\overline{c}| = 2$. Now,use the condition $|\overline{c}-\overline{a}|=3$. Squaring both sides: $|\overline{c}|^2 + |\overline{a}|^2 - 2(\overline{a} \cdot \overline{c}) = 3^2$. Substituting $|\overline{c}|=2$ 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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