We are given three vectors: $a = (\sqrt{3}/3, \sqrt{3}/3, \sqrt{3}/3)$, $b = (1, -1, 0)$, and $c = (-2, -2, 1)$. We need to find the angle between each pair of vectors: (a and b), (a and c), and (b and c).

1. Problem Description

We are given three vectors: $a = (\sqrt{3}/3, \sqrt{3}/3, \sqrt{3}/3)$, $b = (1, -1, 0)$, and $c = (-2, -2, 1)$. We need to find the angle between each pair of vectors: (a and b), (a and c), and (b and c).

2. Solution Steps

First, recall the formula for the dot product of two vectors $u$ and $v$:

$u \cdot v = ||u|| \cdot ||v|| \cdot \cos{\theta}$,

where $\theta$ is the angle between the two vectors.

Thus, we can find the angle $\theta$ using:

$\cos{\theta} = \frac{u \cdot v}{||u|| \cdot ||v||}$

$\theta = \arccos{\frac{u \cdot v}{||u|| \cdot ||v||}}$

(1) Angle between a and b:

$a \cdot b = (\sqrt{3}/3)(1) + (\sqrt{3}/3)(-1) + (\sqrt{3}/3)(0) = \sqrt{3}/3 - \sqrt{3}/3 + 0 = 0$

$||a|| = \sqrt{(\sqrt{3}/3)^2 + (\sqrt{3}/3)^2 + (\sqrt{3}/3)^2} = \sqrt{3/9 + 3/9 + 3/9} = \sqrt{9/9} = 1$

$||b|| = \sqrt{1^2 + (-1)^2 + 0^2} = \sqrt{1 + 1 + 0} = \sqrt{2}$

$\cos{\theta_{ab}} = \frac{0}{1 \cdot \sqrt{2}} = 0$

$\theta_{ab} = \arccos{0} = \frac{\pi}{2}$

(2) Angle between a and c:

$a \cdot c = (\sqrt{3}/3)(-2) + (\sqrt{3}/3)(-2) + (\sqrt{3}/3)(1) = -2\sqrt{3}/3 - 2\sqrt{3}/3 + \sqrt{3}/3 = -3\sqrt{3}/3 = -\sqrt{3}$

$||a|| = 1$ (calculated above)

$||c|| = \sqrt{(-2)^2 + (-2)^2 + 1^2} = \sqrt{4 + 4 + 1} = \sqrt{9} = 3$

$\cos{\theta_{ac}} = \frac{-\sqrt{3}}{1 \cdot 3} = -\frac{\sqrt{3}}{3}$

$\theta_{ac} = \arccos{(-\frac{\sqrt{3}}{3})} \approx 2.22$ radians, or $127.35$ degrees. We can also write this as $\arccos{(-\frac{1}{\sqrt{3}})}$.

(3) Angle between b and c:

$b \cdot c = (1)(-2) + (-1)(-2) + (0)(1) = -2 + 2 + 0 = 0$

$||b|| = \sqrt{2}$ (calculated above)

$||c|| = 3$ (calculated above)

$\cos{\theta_{bc}} = \frac{0}{\sqrt{2} \cdot 3} = 0$

$\theta_{bc} = \arccos{0} = \frac{\pi}{2}$

3. Final Answer

The angle between vectors a and b is $\frac{\pi}{2}$.

The angle between vectors a and c is $\arccos{(-\frac{\sqrt{3}}{3})}$.

The angle between vectors b and c is $\frac{\pi}{2}$.