First, calculate the vectors $ 2\bar{a} + \bar{b} $ and $ \bar{a} + 2\bar{b} $.

Given:

$ \bar{a} = \hat{i} - 2\hat{j} + 3\hat{k} $

$ \bar{b} = 2\hat{i} + 3\hat{j} - \hat{k} $

Calculate $ 2\bar{a} + \bar{b} $:

$ 2\bar{a} = 2(\hat{i} - 2\hat{j} + 3\hat{k}) = 2\hat{i} - 4\hat{j} + 6\hat{k} $

$ 2\bar{a} + \bar{b} = (2\hat{i} - 4\hat{j} + 6\hat{k}) + (2\hat{i} + 3\hat{j} - \hat{k}) = 4\hat{i} - \hat{j} + 5\hat{k} $

Calculate $ \bar{a} + 2\bar{b} $:

$ 2\bar{b} = 2(2\hat{i} + 3\hat{j} - \hat{k}) = 4\hat{i} + 6\hat{j} - 2\hat{k} $

$ \bar{a} + 2\bar{b} = (\hat{i} - 2\hat{j} + 3\hat{k}) + (4\hat{i} + 6\hat{j} - 2\hat{k}) = 5\hat{i} + 4\hat{j} + \hat{k} $

Now, find the dot product $ (2\bar{a} + \bar{b}) \cdot (\bar{a} + 2\bar{b}) $:

$ (4\hat{i} - \hat{j} + 5\hat{k}) \cdot (5\hat{i} + 4\hat{j} + \hat{k}) = 4 \cdot 5 + (-1) \cdot 4 + 5 \cdot 1 = 20 - 4 + 5 = 21 $

Calculate the magnitudes of the vectors:

$ \|2\bar{a} + \bar{b}\| = \sqrt{4^2 + (-1)^2 + 5^2} = \sqrt{16 + 1 + 25} = \sqrt{42} $

$ \|\bar{a} + 2\bar{b}\| = \sqrt{5^2 + 4^2 + 1^2} = \sqrt{25 + 16 + 1} = \sqrt{42} $

Using the dot product and magnitudes to find $ \cos \theta $:

$ \cos \theta = \frac{21}{\sqrt{42} \times \sqrt{42}} = \frac{21}{42} = \frac{1}{2} $

The angle $ \theta $ for which $ \cos \theta = \frac{1}{2} $ is $ \frac{\pi}{3} $.

Therefore, the angle between the vectors is $ \frac{\pi}{3} $.

Option B: $ \frac{\pi}{3} $

Vector Algebra

Continue this paper