We are given two vectors: \[ \vec{a} = 2\vec{i} + 4\vec{j} + 7\vec{k} \quad {and} \quad \vec{b} = 4\vec{i} + 7\vec{j} + 2\vec{k}. \] We are asked to find the angle between the vectors \( \vec{a} + \vec{b} \) and \( \vec{a} - \vec{b} \).
Step 1: Calculate \( \vec{a} + \vec{b} \) and \( \vec{a} - \vec{b} \) First, compute the sum \( \vec{a} + \vec{b} \) and the difference \( \vec{a} - \vec{b} \): \[ \vec{a} + \vec{b} = (2\vec{i} + 4\vec{j} + 7\vec{k}) + (4\vec{i} + 7\vec{j} + 2\vec{k}) = 6\vec{i} + 11\vec{j} + 9\vec{k}, \] \[ \vec{a} - \vec{b} = (2\vec{i} + 4\vec{j} + 7\vec{k}) - (4\vec{i} + 7\vec{j} + 2\vec{k}) = -2\vec{i} - 3\vec{j} + 5\vec{k}. \]
Step 2: Use the Dot Product Formula The cosine of the angle \( \theta \) between two vectors \( \vec{u} \) and \( \vec{v} \) is given by the formula: \[ \cos \theta = \frac{\vec{u} \cdot \vec{v}}{|\vec{u}| |\vec{v}|}. \] Let \( \vec{u} = \vec{a} + \vec{b} \) and \( \vec{v} = \vec{a} - \vec{b} \). To find the angle between them, we need to compute their dot product and magnitudes.
Step 3: Compute the Dot Product \( \vec{u} \cdot \vec{v} \) \[ \vec{u} \cdot \vec{v} = (6\vec{i} + 11\vec{j} + 9\vec{k}) \cdot (-2\vec{i} - 3\vec{j} + 5\vec{k}). \] Using the distributive property of the dot product: \[ \vec{u} \cdot \vec{v} = 6(-2) + 11(-3) + 9(5) = -12 - 33 + 45 = 0. \]
Step 4: Conclude the Angle Since the dot product \( \vec{u} \cdot \vec{v} = 0 \), this means the vectors \( \vec{u} = \vec{a} + \vec{b} \) and \( \vec{v} = \vec{a} - \vec{b} \) are perpendicular to each other. The angle between two perpendicular vectors is \( \frac{\pi}{2} \).
Thus, the angle between \( \vec{a} + \vec{b} \) and \( \vec{a} - \vec{b} \) is \( \frac{\pi}{2} \).
Thus, the correct answer is \( \boxed{\frac{\pi}{2}} \), corresponding to option (C).
If $ X = A \times B $, $ A = \begin{bmatrix} 1 & 2 \\-1 & 1 \end{bmatrix} $, $ B = \begin{bmatrix} 3 & 6 \\5 & 7 \end{bmatrix} $, find $ x_1 + x_2 $.