Comprehesion-I
Let k be the length of any edge of a regular tetrahedron (all edges are equal in length). The angle between a line and a plane is equal to the complement of the angle between the line and the normal to the plane whereas the angle between two plane is equal to the angle between the normals. Let O be the origin and A,B and C vertices with position vectors `veca,vecb` and `vecc` respectively of the regular tetrahedron.
The value of `[vecavecbvecc]^(2)` is

Options

A. (a)`k^(6)`
B. (b)`(1)/(2)k^(6)`
C. (c)`(1)/(3)k^(6)`
D. (d)`(1)/(4)k^(6)`

Answer

To solve the problem, we need to find the value of \([ \vec{a} \cdot \vec{b} \cdot \vec{c} ]^2\) for a regular tetrahedron with edge length \(k\). Here are the steps to derive the solution:

### Step 1: Understanding the Geometry of the Tetrahedron
A regular tetrahedron has four vertices, and all edges are of equal length \(k\). We can denote the vertices as \(O\) (the origin), \(A\), \(B\), and \(C\) with position vectors \(\vec{a}\), \(\vec{b}\), and \(\vec{c}\) respectively.

### Step 2: Position Vectors of the Vertices
We can place the tetrahedron in a coordinate system. For simplicity, we can assume:
- \(\vec{a} = k \hat{i}\)
- \(\vec{b} = k \left( -\frac{1}{2} \hat{i} + \frac{\sqrt{3}}{2} \hat{j} \right)\)
- \(\vec{c} = k \left( -\frac{1}{2} \hat{i} - \frac{\sqrt{3}}{2} \hat{j} \right)\)

### Step 3: Calculate the Dot Products
We need to calculate the dot products \(\vec{a} \cdot \vec{b}\), \(\vec{a} \cdot \vec{c}\), and \(\vec{b} \cdot \vec{c}\).

1. **Calculating \(\vec{a} \cdot \vec{b}\)**:
\[ \vec{a} \cdot \vec{b} = k \hat{i} \cdot k \left( -\frac{1}{2} \hat{i} + \frac{\sqrt{3}}{2} \hat{j} \right) = k^2 \left( -\frac{1}{2} \right) = -\frac{k^2}{2} \]

2. **Calculating \(\vec{a} \cdot \vec{c}\)**:
\[ \vec{a} \cdot \vec{c} = k \hat{i} \cdot k \left( -\frac{1}{2} \hat{i} - \frac{\sqrt{3}}{2} \hat{j} \right) = k^2 \left( -\frac{1}{2} \right) = -\frac{k^2}{2} \]

3. **Calculating \(\vec{b} \cdot \vec{c}\)**:
\[ \vec{b} \cdot \vec{c} = k^2 \left( -\frac{1}{2} \hat{i} + \frac{\sqrt{3}}{2} \hat{j} \right) \cdot k^2 \left( -\frac{1}{2} \hat{i} - \frac{\sqrt{3}}{2} \hat{j} \right) \]
\[ = k^2 \left( \frac{1}{4} - \frac{3}{4} \right) = k^2 \left( -\frac{1}{2} \right) = -\frac{k^2}{2} \]

### Step 4: Formulating the Determinant
Now, we can form the determinant \(D\) using these dot products:
\[ D = \begin{vmatrix} \vec{a} \cdot \vec{a} & \vec{a} \cdot \vec{b} & \vec{a} \cdot \vec{c} \\ \vec{b} \cdot \vec{a} & \vec{b} \cdot \vec{b} & \vec{b} \cdot \vec{c} \\ \vec{c} \cdot \vec{a} & \vec{c} \cdot \vec{b} & \vec{c} \cdot \vec{c} \end{vmatrix} \]
Substituting the values:
\[ D = \begin{vmatrix} k^2 & -\frac{k^2}{2} & -\frac{k^2}{2} \\ -\frac{k^2}{2} & k^2 & -\frac{k^2}{2} \\ -\frac{k^2}{2} & -\frac{k^2}{2} & k^2 \end{vmatrix} \]

### Step 5: Simplifying the Determinant
Factoring out \(k^2\) from each row:
\[ D = k^6 \begin{vmatrix} 1 & -\frac{1}{2} & -\frac{1}{2} \\ -\frac{1}{2} & 1 & -\frac{1}{2} \\ -\frac{1}{2} & -\frac{1}{2} & 1 \end{vmatrix} \]

### Step 6: Evaluating the Determinant
Calculating the determinant:
\[ = 1 \left( 1 - \frac{1}{4} - \frac{1}{4} + \frac{1}{4} \right) = 1 - \frac{1}{2} = \frac{1}{2} \]
Thus,
\[ D = k^6 \cdot \frac{1}{2} \]

### Step 7: Final Result
The value of \([ \vec{a} \cdot \vec{b} \cdot \vec{c} ]^2\) is:
\[ \boxed{\frac{k^6}{2}} \]