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(A) Let $\vec{n}_1$ and $\vec{n}_2$ be the normal vectors to the planes defined by $(\hat{i}, \hat{i} + \hat{j})$ and $(\hat{i} - \hat{j}, \hat{i} + \

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$\frac{\pi}{4}$

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(A) Let $\vec{n}_1$ and $\vec{n}_2$ be the normal vectors to the planes defined by $(\hat{i}, \hat{i} + \hat{j})$ and $(\hat{i} - \hat{j}, \hat{i} + \hat{k})$ respectively.$\vec{n}_1 = \hat{i} 	imes (\hat{i} + \hat{j}) = \hat{k}$.$\vec{n}_2 = (\hat{i} - \hat{j}) 	imes (\hat{i} + \hat{k}) = \hat{i} 	imes \hat{k} - \hat{j} 	imes \hat{i} - \hat{j} 	imes \hat{k} = -\hat{j} + \hat{k} - \hat{i} = -\hat{i} - \hat{j} + \hat{k}$.Since $\vec{a}$ is parallel to the line of intersection,$\vec{a}$ is parallel to $\vec{n}_1 	imes \vec{n}_2$.$\vec{a} = \lambda (\vec{n}_1 	imes \vec{n}_2) = \lambda \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 0 & 0 & 1 \ -1 & -1 & 1 \end{vmatrix} = \lambda (\hat{i} - \hat{j})$.Let $	heta$ be the angle between $\vec{a} = \lambda(\hat{i} - \hat{j})$ and $\vec{b} = \hat{i} - 2\hat{j} + 2\hat{k}$.$\cos 	heta = \frac{|\vec{a} \cdot \vec{b}|}{|\vec{a}| |\vec{b}|} = \frac{|\lambda(1 + 2)|}{\sqrt{\lambda^2 + \lambda^2} \sqrt{1^2 + (-2)^2 + 2^2}} = \frac{3|\lambda|}{\sqrt{2}|\lambda| \sqrt{9}} = \frac{3}{3\sqrt{2}} = \frac{1}{\sqrt{2}}$.Thus,$	heta = \frac{\pi}{4}$.

$A$ non-zero vector $\vec{a}$ is parallel to the line of intersection of the plane defined by $\hat{i}$ and $\hat{i} + \hat{j}$,and the plane defined by $\hat{i} - \hat{j}$ and $\hat{i} + \hat{k}$. Find the angle between $\vec{a}$ and $\hat{i} - 2\hat{j} + 2\hat{k}$.

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