Given unit vectors m, n and p such that angle between m and n. Angle between p and `(mtimesn)=(pi)/(6)`, then [n p m] is equal to

Options

A. `(sqrt(3))/(4)`
B. `(3)/(4)`
C. `(1)/(4)`
D. None of these

Answer

To solve the problem step by step, we will analyze the given information and apply the properties of vectors, particularly focusing on the scalar triple product and the relationships between the angles.

### Step 1: Understand the Given Information
We have three unit vectors \( \mathbf{m}, \mathbf{n}, \) and \( \mathbf{p} \). The angles between \( \mathbf{m} \) and \( \mathbf{n} \) and between \( \mathbf{p} \) and \( \mathbf{m} \times \mathbf{n} \) are both \( \frac{\pi}{6} \) (or 30 degrees).

### Step 2: Define the Scalar Triple Product
The expression we need to evaluate is \( [\mathbf{n} \, \mathbf{p} \, \mathbf{m}] \), which represents the scalar triple product \( \mathbf{n} \cdot (\mathbf{p} \times \mathbf{m}) \).

### Step 3: Use the Property of Scalar Triple Product
Using the property of the scalar triple product, we can rearrange the vectors:
\[ [\mathbf{n} \, \mathbf{p} \, \mathbf{m}] = \mathbf{n} \cdot (\mathbf{p} \times \mathbf{m}) = -\mathbf{p} \cdot (\mathbf{n} \times \mathbf{m}). \]

### Step 4: Calculate the Magnitude of \( \mathbf{m} \times \mathbf{n} \)
The magnitude of the cross product \( \mathbf{m} \times \mathbf{n} \) can be calculated using the formula:
\[ |\mathbf{m} \times \mathbf{n}| = |\mathbf{m}| |\mathbf{n}| \sin(\theta), \]
where \( \theta \) is the angle between \( \mathbf{m} \) and \( \mathbf{n} \). Since both \( \mathbf{m} \) and \( \mathbf{n} \) are unit vectors and the angle \( \theta = \frac{\pi}{6} \):
\[ |\mathbf{m} \times \mathbf{n}| = 1 \cdot 1 \cdot \sin\left(\frac{\pi}{6}\right) = 1 \cdot 1 \cdot \frac{1}{2} = \frac{1}{2}. \]

### Step 5: Calculate \( \mathbf{p} \cdot (\mathbf{m} \times \mathbf{n}) \)
Now we need to find \( \mathbf{p} \cdot (\mathbf{m} \times \mathbf{n}) \). We can write:
\[ \mathbf{p} \cdot (\mathbf{m} \times \mathbf{n}) = |\mathbf{p}| |\mathbf{m} \times \mathbf{n}| \cos(\phi), \]
where \( \phi \) is the angle between \( \mathbf{p} \) and \( \mathbf{m} \times \mathbf{n} \). Given that \( |\mathbf{p}| = 1 \) and \( \phi = \frac{\pi}{6} \):
\[ \mathbf{p} \cdot (\mathbf{m} \times \mathbf{n}) = 1 \cdot \frac{1}{2} \cdot \cos\left(\frac{\pi}{6}\right) = \frac{1}{2} \cdot \frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{4}. \]

### Step 6: Final Result
Thus, we have:
\[ [\mathbf{n} \, \mathbf{p} \, \mathbf{m}] = -\mathbf{p} \cdot (\mathbf{n} \times \mathbf{m}) = \frac{\sqrt{3}}{4}. \]

### Conclusion
The value of \( [\mathbf{n} \, \mathbf{p} \, \mathbf{m}] \) is \( \frac{\sqrt{3}}{4} \).

---