Step 1: Understanding the Concept:
We are dealing with a vector triple product identity in a disguised form.
The expression $[\bar{a} \bar{b} \bar{d}] \bar{c} - [\bar{a} \bar{b} \bar{c}] \bar{d}$ represents the expansion of a cross product of two cross products.
Step 2: Key Formula or Approach:
Scalar triple product notation: $[\vec{u}\vec{v}\vec{w}] = (\vec{u} \times \vec{v}) \cdot \vec{w}$.
Vector quadruple product identity: $(\vec{u} \times \vec{v}) \times (\vec{w} \times \vec{x}) = [\vec{u}\vec{v}\vec{x}]\vec{w} - [\vec{u}\vec{v}\vec{w}]\vec{x}$.
Magnitude of cross product: $|\vec{A} \times \vec{B}| = |\vec{A}| |\vec{B}| \sin \theta$.
Step 3: Detailed Explanation:
Let's analyze the given expression inside the magnitude:
$E = [\bar{a} \bar{b} \bar{d}] \bar{c} - [\bar{a} \bar{b} \bar{c}] \bar{d}$
We know the identity for $(\bar{a} \times \bar{b}) \times (\bar{c} \times \bar{d})$.
Let $\vec{u} = \bar{a} \times \bar{b}$. Then:
$\vec{u} \times (\bar{c} \times \bar{d}) = (\vec{u} \cdot \bar{d})\bar{c} - (\vec{u} \cdot \bar{c})\bar{d}$
$\vec{u} \times (\bar{c} \times \bar{d}) = ((\bar{a} \times \bar{b}) \cdot \bar{d})\bar{c} - ((\bar{a} \times \bar{b}) \cdot \bar{c})\bar{d}$
$\vec{u} \times (\bar{c} \times \bar{d}) = [\bar{a} \bar{b} \bar{d}]\bar{c} - [\bar{a} \bar{b} \bar{c}]\bar{d}$
So, the given expression is exactly $(\bar{a} \times \bar{b}) \times (\bar{c} \times \bar{d})$.
We need to find its magnitude: $|(\bar{a} \times \bar{b}) \times (\bar{c} \times \bar{d})|$.
Let $\vec{v}_1 = \bar{a} \times \bar{b}$ and $\vec{v}_2 = \bar{c} \times \bar{d}$.
The magnitude is $|\vec{v}_1 \times \vec{v}_2| = |\vec{v}_1| |\vec{v}_2| \sin(\text{angle between } \vec{v}_1 \text{ and } \vec{v}_2)$.
We are given the angle between $\bar{a} \times \bar{b}$ and $\bar{c} \times \bar{d}$ is $\frac{\pi}{6}$.
So, magnitude $= |\bar{a} \times \bar{b}| |\bar{c} \times \bar{d}| \sin\left(\frac{\pi}{6}\right)$.
Now, let's find $|\bar{a} \times \bar{b}|$ and $|\bar{c} \times \bar{d}|$.
Given $\bar{a}, \bar{b}$ are unit vectors, and $\bar{a} \cdot \bar{b} = \frac{1}{2}$.
$|\bar{a}||\bar{b}|\cos \theta_{ab} = \frac{1}{2} \Rightarrow (1)(1)\cos \theta_{ab} = \frac{1}{2} \Rightarrow \cos \theta_{ab} = \frac{1}{2}$.
Thus, $\sin \theta_{ab} = \sqrt{1 - \cos^2 \theta_{ab}} = \sqrt{1 - \frac{1}{4}} = \frac{\sqrt{3}}{2}$.
So, $|\bar{a} \times \bar{b}| = |\bar{a}||\bar{b}|\sin \theta_{ab} = (1)(1)\left(\frac{\sqrt{3}}{2}\right) = \frac{\sqrt{3}}{2}$.
Similarly, for unit vectors $\bar{c}, \bar{d}$ with $\bar{c} \cdot \bar{d} = \frac{1}{2}$, we have $|\bar{c} \times \bar{d}| = \frac{\sqrt{3}}{2}$.
Now, substitute these back into the magnitude expression:
Magnitude $= \left(\frac{\sqrt{3}}{2}\right) \left(\frac{\sqrt{3}}{2}\right) \sin\left(\frac{\pi}{6}\right)$
Magnitude $= \left(\frac{3}{4}\right) \left(\frac{1}{2}\right) = \frac{3}{8}$.
Step 4: Final Answer:
The value is $\frac{3}{8}$.