Let $$\vec{a}$$ be a non-zero vector parallel to the line of intersection of the two planes described by $$\hat{i} + \hat{j}, \hat{i} + \hat{k}$$ and $$\hat{i} - \hat{j}, \hat{j} - \hat{k}$$. If $$\theta$$ is the angle between the vector $$\vec{a}$$ and the vector $$\vec{b} = 2\hat{i} - 2\hat{j} + \hat{k}$$ and $$\vec{a} \cdot \vec{b} = 6$$, then the ordered pair $$(\theta, |\vec{a} \times \vec{b}|)$$ is equal to

Find the normals to the two planes.

Plane 1 is described by vectors $$\hat{i}+\hat{j}$$ and $$\hat{i}+\hat{k}$$. Its normal:

$$\vec{n_1} = (\hat{i}+\hat{j}) \times (\hat{i}+\hat{k}) = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 0 \\ 1 & 0 & 1 \end{vmatrix} = \hat{i} - \hat{j} - \hat{k}$$

Plane 2 is described by vectors $$\hat{i}-\hat{j}$$ and $$\hat{j}-\hat{k}$$. Its normal:

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

Direction of line of intersection.

$$\vec{a} \parallel \vec{n_1} \times \vec{n_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -1 & -1 \\ 1 & 1 & 1 \end{vmatrix} = (0)\hat{i} - (2)\hat{j} + (2)\hat{k}$$

So $$\vec{a} = t(0, -2, 2)$$ for some scalar $$t$$.

Use $$\vec{a} \cdot \vec{b} = 6$$.

$$\vec{a} \cdot \vec{b} = t(0 \cdot 2 + (-2)(-2) + 2 \cdot 1) = t(0 + 4 + 2) = 6t = 6$$

So $$t = 1$$ and $$\vec{a} = (0, -2, 2)$$.

Find $$\theta$$ and $$|\vec{a} \times \vec{b}|$$.

$$|\vec{a}| = \sqrt{0 + 4 + 4} = 2\sqrt{2}$$, $$|\vec{b}| = \sqrt{4 + 4 + 1} = 3$$

$$\cos\theta = \frac{6}{2\sqrt{2} \cdot 3} = \frac{6}{6\sqrt{2}} = \frac{1}{\sqrt{2}}$$

$$\theta = \frac{\pi}{4}$$

$$|\vec{a} \times \vec{b}| = |\vec{a}||\vec{b}|\sin\theta = 2\sqrt{2} \cdot 3 \cdot \frac{1}{\sqrt{2}} = 6$$

The ordered pair is $$\left(\frac{\pi}{4}, 6\right)$$.

The correct answer is Option 4.