If `veca,vecb,vecc,vecd` are unit vectors such that `veca.vecb=1/2`, `vecc.vecd=1/2` and angle between `veca xx vecb` and `vecc xx vecd` is `pi/6` then the value of `|[veca \ vecb \ vecd]vecc-[veca \ vecb \ vecc]vecd|=`

Options

A. `3//2`
B. `3//4`
C. `3//8`
D. 2

Explanation

To solve the problem step by step, we will analyze the given conditions and use vector properties to find the required value.

### Given:
- \( \vec{a}, \vec{b}, \vec{c}, \vec{d} \) are unit vectors.
- \( \vec{a} \cdot \vec{b} = \frac{1}{2} \)
- \( \vec{c} \cdot \vec{d} = \frac{1}{2} \)
- The angle between \( \vec{a} \times \vec{b} \) and \( \vec{c} \times \vec{d} \) is \( \frac{\pi}{6} \).

### Required:
Find the value of \( |\left[\vec{a} \ \vec{b} \ \vec{d}\right] \vec{c} - \left[\vec{a} \ \vec{b} \ \vec{c}\right] \vec{d}| \).

### Step 1: Find the angles between the vectors
From the dot product:
- Since \( \vec{a} \cdot \vec{b} = \frac{1}{2} \), and both are unit vectors, we have:
\[ |\vec{a}| |\vec{b}| \cos(\theta_1) = \frac{1}{2} \implies 1 \cdot 1 \cdot \cos(\theta_1) = \frac{1}{2} \implies \cos(\theta_1) = \frac{1}{2} \]
Thus, \( \theta_1 = \frac{\pi}{3} \).

- Similarly, for \( \vec{c} \cdot \vec{d} = \frac{1}{2} \):
\[ |\vec{c}| |\vec{d}| \cos(\theta_2) = \frac{1}{2} \implies 1 \cdot 1 \cdot \cos(\theta_2) = \frac{1}{2} \implies \cos(\theta_2) = \frac{1}{2} \]
Thus, \( \theta_2 = \frac{\pi}{3} \).

### Step 2: Find the magnitude of the cross products
Using the angle between the cross products:
- The magnitude of \( \vec{a} \times \vec{b} \):
\[ |\vec{a} \times \vec{b}| = |\vec{a}| |\vec{b}| \sin(\theta_1) = 1 \cdot 1 \cdot \sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2} \]

- The magnitude of \( \vec{c} \times \vec{d} \):
\[ |\vec{c} \times \vec{d}| = |\vec{c}| |\vec{d}| \sin(\theta_2) = 1 \cdot 1 \cdot \sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2} \]

### Step 3: Calculate the magnitude of the scalar triple products
The angle between \( \vec{a} \times \vec{b} \) and \( \vec{c} \times \vec{d} \) is \( \frac{\pi}{6} \):
\[ |\vec{a} \times \vec{b} \cdot \vec{c} \times \vec{d}| = |\vec{a} \times \vec{b}| |\vec{c} \times \vec{d}| \cos\left(\frac{\pi}{6}\right) \]
Substituting the values:
\[ = \frac{\sqrt{3}}{2} \cdot \frac{\sqrt{3}}{2} \cdot \frac{\sqrt{3}}{2} = \frac{3\sqrt{3}}{8} \]

### Step 4: Use the vector triple product identity
Using the vector triple product identity:
\[ \vec{a} \times (\vec{b} \times \vec{c}) = (\vec{a} \cdot \vec{c}) \vec{b} - (\vec{a} \cdot \vec{b}) \vec{c} \]
We can express:
\[ |\left[\vec{a} \ \vec{b} \ \vec{d}\right] \vec{c} - \left[\vec{a} \ \vec{b} \ \vec{c}\right] \vec{d}| = |(\vec{c} \cdot \vec{a} \times \vec{b}) \vec{d} - (\vec{d} \cdot \vec{a} \times \vec{b}) \vec{c}| \]

### Step 5: Substitute known values
Let \( S = \vec{a} \times \vec{b} \):
\[ = |S \cdot \vec{c} \cdot \vec{d} - S \cdot \vec{d} \cdot \vec{c}| = |S| \cdot |S| \cdot \sin\left(\frac{\pi}{6}\right) \]
Substituting the values:
\[ = \frac{3\sqrt{3}}{8} \cdot \frac{1}{2} = \frac{3\sqrt{3}}{16} \]

### Final Result
Thus, the value is:
\[ |\left[\vec{a} \ \vec{b} \ \vec{d}\right] \vec{c} - \left[\vec{a} \ \vec{b} \ \vec{c}\right] \vec{d}| = \frac{3\sqrt{3}}{16} \]