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(B) The equation of the plane containing vectors $\hat{i}$ and $\hat{i}+\hat{j}$ is given by the normal vector $\vec{n}_1 = \hat{i} 	imes (\hat{i}+\h

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$\frac{\pi}{4}$

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(B) The equation of the plane containing vectors $\hat{i}$ and $\hat{i}+\hat{j}$ is given by the normal vector $\vec{n}_1 = \hat{i} 	imes (\hat{i}+\hat{j}) = \hat{k}$. Thus, the plane equation is $z = 0$.The equation of the plane containing vectors $\hat{i}-\hat{j}$ and $\hat{i}+\hat{k}$ is given by the normal vector $\vec{n}_2 = (\hat{i}-\hat{j}) 	imes (\hat{i}+\hat{k}) = \hat{i} 	imes \hat{k} - \hat{j} 	imes \hat{i} - \hat{j} 	imes \hat{k} = -\hat{j} + \hat{k} - \hat{i} = -\hat{i} - \hat{j} + \hat{k}$. Thus, the plane equation is $x + y - z = 0$.Since $\vec{a}$ is parallel to the line of intersection of these two planes, $\vec{a}$ must be parallel to the cross product of their normals: $\vec{v} = \vec{n}_1 	imes \vec{n}_2 = \hat{k} 	imes (-\hat{i} - \hat{j} + \hat{k}) = -(\hat{k} 	imes \hat{i}) - (\hat{k} 	imes \hat{j}) = -\hat{j} + \hat{i} = \hat{i} - \hat{j}$.Let $\vec{b} = \hat{i} - \hat{j}$. The angle $	heta$ between $\vec{b}$ and $\vec{c} = \hat{i} - 2\hat{j} + 2\hat{k}$ is given by $\cos 	heta = \frac{|\vec{b} \cdot \vec{c}|}{|\vec{b}| |\vec{c}|}$.$\vec{b} \cdot \vec{c} = (1)(1) + (-1)(-2) + (0)(2) = 1 + 2 = 3$.$|\vec{b}| = \sqrt{1^2 + (-1)^2} = \sqrt{2}$.$|\vec{c}| = \sqrt{1^2 + (-2)^2 + 2^2} = \sqrt{1+4+4} = 3$.$\cos 	heta = \frac{3}{\sqrt{2} 	imes 3} = \frac{1}{\sqrt{2}}$.Therefore, $	heta = \frac{\pi}{4}$.

$A$ non-zero vector $\vec{a}$ is parallel to the line of intersection of the plane determined by the vectors $\hat{i}$ and $\hat{i}+\hat{j}$ and the plane determined by vectors $\hat{i}-\hat{j}$ and $\hat{i}+\hat{k}$. The angle between $\vec{a}$ and $(\hat{i}-2\hat{j}+2\hat{k})$ is

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