If $\bar{a}, \bar{b}, \bar{c}$ are three non zero and non-coplanar vectors such that $\bar{a} \times (\bar{b} \times \bar{c}) = \dfrac{\bar{b}}{2}$, then the angle between $\bar{a}$ and $\bar{b}$ is ...

Given $\bar{a} \times (\bar{b} \times \bar{c}) = \dfrac{\bar{b}}{2}$

Using the vector triple product expansion, we have:

$(\bar{a} \cdot \bar{c})\bar{b} - (\bar{a} \cdot \bar{b})\bar{c} = \dfrac{1}{2}\bar{b}$

Rearranging the terms, we get:

$\left(\bar{a} \cdot \bar{c} - \dfrac{1}{2}\right)\bar{b} - (\bar{a} \cdot \bar{b})\bar{c} = \bar{0}$

Since $\bar{a}, \bar{b}, \bar{c}$ are non-coplanar vectors, $\bar{b}$ and $\bar{c}$ must be non-collinear. Therefore, their coefficients must be independently zero:

$\bar{a} \cdot \bar{c} - \dfrac{1}{2} = 0 \Rightarrow \bar{a} \cdot \bar{c} = \dfrac{1}{2}$

$\bar{a} \cdot \bar{b} = 0$

Since $\bar{a}$ and $\bar{b}$ are non-zero vectors, $\bar{a} \cdot \bar{b} = 0$ implies that $\bar{a}$ is perpendicular to $\bar{b}$.

Thus, the angle between $\bar{a}$ and $\bar{b}$ is $\dfrac{\pi}{2}$.

Answer: $\dfrac{\pi}{2}$