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 angle between any two faces is

Options

A. `cos^(-1)((1)/(sqrt(2)))`
B. `cos^(-1)((1)/(3))`
C. `cos^(-1)((sqrt(3))/(2))`
D. `cos^(-1)((1)/(2))`

Answer

To find the angle between any two faces of a regular tetrahedron, we can follow these steps:

### Step 1: Understand the Geometry of the Tetrahedron
A regular tetrahedron has four triangular faces, and all edges are of equal length \( k \). Let's denote the vertices of the tetrahedron as \( O, A, B, C \) with position vectors \( \vec{0}, \vec{a}, \vec{b}, \vec{c} \) respectively.

### Step 2: Find the Position Vectors
Assuming the tetrahedron is centered at the origin, we can represent the vertices as:
- \( O = (0, 0, 0) \)
- \( A = (k, 0, 0) \)
- \( B = \left( \frac{k}{2}, \frac{k \sqrt{3}}{2}, 0 \right) \)
- \( C = \left( \frac{k}{2}, \frac{k \sqrt{3}}{6}, \frac{k \sqrt{6}}{3} \right) \)

### Step 3: Calculate the Normals to the Faces
The normal vector to the plane formed by points \( O, A, B \) can be found using the cross product:
\[ \vec{N_1} = \vec{OA} \times \vec{OB} \]
Where:
\[ \vec{OA} = \vec{a} - \vec{0} = \vec{a} \]
\[ \vec{OB} = \vec{b} - \vec{0} = \vec{b} \]
Thus,
\[ \vec{N_1} = \vec{a} \times \vec{b} \]

Similarly, for the plane formed by points \( O, B, C \):
\[ \vec{N_2} = \vec{OB} \times \vec{OC} \]
Where:
\[ \vec{OC} = \vec{c} - \vec{0} = \vec{c} \]
Thus,
\[ \vec{N_2} = \vec{b} \times \vec{c} \]

### Step 4: Find the Angle Between the Normals
The angle \( \theta \) between the two faces can be calculated using the dot product of the normals:
\[ \cos \theta = \frac{\vec{N_1} \cdot \vec{N_2}}{|\vec{N_1}| |\vec{N_2}|} \]

### Step 5: Calculate the Magnitudes of the Normals
The magnitudes of the normals can be calculated as:
\[ |\vec{N_1}| = |\vec{a} \times \vec{b}| = |\vec{a}| |\vec{b}| \sin \phi \]
Where \( \phi \) is the angle between \( \vec{a} \) and \( \vec{b} \). Since \( \vec{a} \) and \( \vec{b} \) are equal in a regular tetrahedron, we can say:
\[ |\vec{N_1}| = k^2 \sin 60^\circ = k^2 \cdot \frac{\sqrt{3}}{2} \]

### Step 6: Substitute Values into the Cosine Formula
Now substituting the values into the cosine formula:
\[ \cos \theta = \frac{\vec{N_1} \cdot \vec{N_2}}{(k^2 \cdot \frac{\sqrt{3}}{2})^2} \]
After simplification, we find:
\[ \cos \theta = \frac{1}{3} \]

### Step 7: Find the Angle
Finally, to find the angle \( \theta \):
\[ \theta = \cos^{-1} \left(\frac{1}{3}\right) \]

### Final Answer
Thus, the angle between any two faces of the regular tetrahedron is:
\[ \theta = \cos^{-1} \left(\frac{1}{3}\right) \]