Step-by-step Solution:

We are given the vectors: \[ \vec{a} = \left(\vec{i} + 2\vec{j} - 3\vec{k}\right) \] \[ \vec{b} = \left(3\vec{i} - \vec{j} + 2\vec{k}\right) \] We need to find the angle between \( \vec{a} + \vec{b} \) and \( \vec{a} - \vec{b} \). \[ \vec{a} + \vec{b} = (1 + 3)\vec{i} + (2 - 1)\vec{j} + (-3 + 2)\vec{k} \] \[ = 4\vec{i} + \vec{j} - \vec{k} \] Compute \( \vec{a} - \vec{b} \) \[ \vec{a} - \vec{b} = (1 - 3)\vec{i} + (2 - (-1))\vec{j} + (-3 - 2)\vec{k} \] \[ = -2\vec{i} + 3\vec{j} - 5\vec{k} \] Compute the dot product \( (\vec{a} + \vec{b}) \cdot (\vec{a} - \vec{b}) \) \[ (4\vec{i} + \vec{j} - \vec{k}) \cdot (-2\vec{i} + 3\vec{j} - 5\vec{k}) \] Using the dot product formula: \[ (4 \times -2) + (1 \times 3) + (-1 \times -5) \] \[ = -8 + 3 + 5 = 0 \] Find the magnitudes \[ |\vec{a} + \vec{b}| = \sqrt{4^2 + 1^2 + (-1)^2} = \sqrt{16 + 1 + 1} = \sqrt{18} = 3\sqrt{2} \] \[ |\vec{a} - \vec{b}| = \sqrt{(-2)^2 + 3^2 + (-5)^2} = \sqrt{4 + 9 + 25} = \sqrt{38} \] Compute the angle
Using the formula: \[ \cos\theta = \frac{(\vec{a} + \vec{b}) \cdot (\vec{a} - \vec{b})}{|\vec{a} + \vec{b}| |\vec{a} - \vec{b}|} \] \[ \cos\theta = \frac{0}{(3\sqrt{2} \times \sqrt{38})} = 0 \] Since \( \cos\theta = 0 \), we get: \[ \theta = \frac{\pi}{2} \] The angle between \( \vec{a} + \vec{b} \) and \( \vec{a} - \vec{b} \) is \( \frac{\pi}{2} \) (90°).