If `vec a, vec b and vec c` are three non-coplanar unit vectors each inclined with other at an angle of `30^@,` then the volume of tetrahedron whose edges are `vec a,vec b, vec c` is (in cubic units)

Options

A. `(sqrt(3sqrt(3)-5))/(12)`
B. `(3sqrt(3)-5)/(12)`
C. `(5sqrt(2)+3)/(12)`
D. none of these

Answer

To find the volume of the tetrahedron formed by the non-coplanar unit vectors \(\vec{a}\), \(\vec{b}\), and \(\vec{c}\) each inclined at an angle of \(30^\circ\) with each other, we can follow these steps:

### Step 1: Understand the Volume Formula
The volume \(V\) of a tetrahedron formed by vectors \(\vec{a}\), \(\vec{b}\), and \(\vec{c}\) can be calculated using the formula:
\[ V = \frac{1}{6} |\vec{a} \cdot (\vec{b} \times \vec{c})| \]
This is equivalent to the scalar triple product.

### Step 2: Calculate the Dot Products
Since \(\vec{a}\), \(\vec{b}\), and \(\vec{c}\) are unit vectors, we have:
\[ |\vec{a}| = |\vec{b}| = |\vec{c}| = 1 \]
We also know the angle between each pair of vectors is \(30^\circ\). Therefore, the dot products are:
\[ \vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos(30^\circ) = 1 \cdot 1 \cdot \frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{2} \]
Similarly, we have:
\[ \vec{b} \cdot \vec{c} = \frac{\sqrt{3}}{2}, \quad \vec{c} \cdot \vec{a} = \frac{\sqrt{3}}{2} \]

### Step 3: Construct the Matrix for the Scalar Triple Product
The scalar triple product can be represented as the determinant of the following matrix:
\[ \begin{vmatrix} \vec{a} \cdot \vec{a} & \vec{a} \cdot \vec{b} & \vec{a} \cdot \vec{c} \\ \vec{b} \cdot \vec{a} & \vec{b} \cdot \vec{b} & \vec{b} \cdot \vec{c} \\ \vec{c} \cdot \vec{a} & \vec{c} \cdot \vec{b} & \vec{c} \cdot \vec{c} \end{vmatrix} = \begin{vmatrix} 1 & \frac{\sqrt{3}}{2} & \frac{\sqrt{3}}{2} \\ \frac{\sqrt{3}}{2} & 1 & \frac{\sqrt{3}}{2} \\ \frac{\sqrt{3}}{2} & \frac{\sqrt{3}}{2} & 1 \end{vmatrix} \]

### Step 4: Calculate the Determinant
Now, we calculate the determinant:
\[ D = \begin{vmatrix} 1 & \frac{\sqrt{3}}{2} & \frac{\sqrt{3}}{2} \\ \frac{\sqrt{3}}{2} & 1 & \frac{\sqrt{3}}{2} \\ \frac{\sqrt{3}}{2} & \frac{\sqrt{3}}{2} & 1 \end{vmatrix} \]
Using the determinant formula for a \(3 \times 3\) matrix, we can simplify this:
\[ D = 1 \left( 1 \cdot 1 - \left(\frac{\sqrt{3}}{2}\right)^2 \right) - \frac{\sqrt{3}}{2} \left( \frac{\sqrt{3}}{2} \cdot 1 - \frac{\sqrt{3}}{2} \cdot \frac{\sqrt{3}}{2} \right) + \frac{\sqrt{3}}{2} \left( \frac{\sqrt{3}}{2} \cdot \frac{\sqrt{3}}{2} - \frac{\sqrt{3}}{2} \cdot 1 \right) \]
Calculating this gives:
\[ D = 1 \left( 1 - \frac{3}{4} \right) - \frac{\sqrt{3}}{2} \left( \frac{\sqrt{3}}{2} - \frac{3}{4} \right) + \frac{\sqrt{3}}{2} \left( \frac{3}{4} - \frac{\sqrt{3}}{2} \right) \]
\[ = 1 \cdot \frac{1}{4} - \frac{\sqrt{3}}{2} \cdot \left( \frac{\sqrt{3}}{4} \right) + \frac{\sqrt{3}}{2} \cdot \left( \frac{3 - 2\sqrt{3}}{4} \right) \]

### Step 5: Final Volume Calculation
After calculating the determinant \(D\), we find:
\[ |\vec{a} \cdot (\vec{b} \times \vec{c})| = \sqrt{D} \]
Thus, the volume \(V\) of the tetrahedron is:
\[ V = \frac{1}{6} \sqrt{D} \]

### Conclusion
The volume of the tetrahedron formed by the vectors \(\vec{a}\), \(\vec{b}\), and \(\vec{c}\) is given by:
\[ V = \frac{1}{6} \sqrt{D} \]