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

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(C) Given $\vec{a}=2 \hat{i}+\hat{j}-2 \hat{k}$,$\vec{b}=\hat{i}+\hat{j}$,and $|\vec{c}-\vec{a}|=3$. First,calculate $\vec{p} = \vec{a} 	imes \vec{b}

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(C) Given $\vec{a}=2 \hat{i}+\hat{j}-2 \hat{k}$,$\vec{b}=\hat{i}+\hat{j}$,and $|\vec{c}-\vec{a}|=3$. First,calculate $\vec{p} = \vec{a} 	imes \vec{b}$: $\vec{p} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 2 & 1 & -2 \ 1 & 1 & 0 \end{vmatrix} = \hat{i}(0 - (-2)) - \hat{j}(0 - (-2)) + \hat{k}(2 - 1) = 2\hat{i} - 2\hat{j} + \hat{k}$. The magnitude $|\vec{p}| = \sqrt{2^2 + (-2)^2 + 1^2} = \sqrt{4+4+1} = 3$. Given $|\vec{p} 	imes \vec{c}| = 3$ and the angle between $\vec{p}$ and $\vec{c}$ is $\frac{\pi}{6}$,we have: $|\vec{p}||\vec{c}| \sin(\frac{\pi}{6}) = 3$ $3 \cdot |\vec{c}| \cdot \frac{1}{2} = 3 \implies |\vec{c}| = 2$. Now,use the condition $|\vec{c}-\vec{a}|=3$: $|\vec{c}-\vec{a}|^2 = 3^2 = 9$ $|\vec{c}|^2 + |\vec{a}|^2 - 2(\vec{a} \cdot \vec{c}) = 9$. We know $|\vec{a}| = \sqrt{2^2 + 1^2 + (-2)^2} = \sqrt{4+1+4} = 3$,so $|\vec{a}|^2 = 9$. Substituting the values: $4 + 9 - 2(\vec{a} \cdot \vec{c}) = 9$ $13 - 2(\vec{a} \cdot \vec{c}) = 9$ $2(\vec{a} \cdot \vec{c}) = 4$ $\vec{a} \cdot \vec{c} = 2$.

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