A vector $$\vec{a}$$ is parallel to the line of intersection of the plane determined by the vectors $$\hat{i}$$, $$\hat{i} + \hat{j}$$ and the plane determined by the vectors $$\hat{i} - \hat{j}$$, $$\hat{i} + \hat{k}$$. The obtuse angle between $$\vec{a}$$ and the vector $$\vec{b} = \hat{i} - 2\hat{j} + 2\hat{k}$$ is

We need to find the obtuse angle between $$\vec{a}$$ (parallel to the line of intersection of two planes) and $$\vec{b} = \hat{i} - 2\hat{j} + 2\hat{k}$$.

The first plane is determined by vectors $$\hat{i}$$ and $$\hat{i} + \hat{j}$$.

Normal to the first plane: $$\vec{n_1} = \hat{i} \times (\hat{i} + \hat{j}) = \hat{i} \times \hat{i} + \hat{i} \times \hat{j} = \vec{0} + \hat{k} = \hat{k}$$

So the first plane has normal $$\hat{k}$$, i.e., it is the $$xy$$-plane ($$z = 0$$).

The second plane is determined by vectors $$\hat{i} - \hat{j}$$ and $$\hat{i} + \hat{k}$$.

Normal to the second plane:

$$\vec{n_2} = (\hat{i} - \hat{j}) \times (\hat{i} + \hat{k}) = \begin{vmatrix}\hat{i} & \hat{j} & \hat{k} \\ 1 & -1 & 0 \\ 1 & 0 & 1\end{vmatrix}$$

$$= \hat{i}(-1-0) - \hat{j}(1-0) + \hat{k}(0+1) = -\hat{i} - \hat{j} + \hat{k}$$

$$\vec{a}$$ is parallel to $$\vec{n_1} \times \vec{n_2}$$:

$$\vec{n_1} \times \vec{n_2} = \hat{k} \times (-\hat{i} - \hat{j} + \hat{k}) = -(\hat{k} \times \hat{i}) - (\hat{k} \times \hat{j}) + (\hat{k} \times \hat{k})$$

$$= -\hat{j} - (-\hat{i}) + \vec{0} = \hat{i} - \hat{j}$$

So $$\vec{a}$$ is parallel to $$\hat{i} - \hat{j}$$.

$$\cos\alpha = \dfrac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|} = \dfrac{(1)(1) + (-1)(-2) + (0)(2)}{\sqrt{1+1}\cdot\sqrt{1+4+4}} = \dfrac{1+2}{(\sqrt{2})(3)} = \dfrac{3}{3\sqrt{2}} = \dfrac{1}{\sqrt{2}}$$

The acute angle is $$\alpha = \dfrac{\pi}{4}$$.

The obtuse angle is $$\pi - \dfrac{\pi}{4} = \dfrac{3\pi}{4}$$.

The correct answer is Option A: $$\dfrac{3\pi}{4}$$.