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 edge and a face not containing the edge is

Options

A. A)`cos^(-1)((1)/(2))`
B. B) `cos^(-1)((1)/(4))`
C. C) `cos^(-1)((1)/(sqrt(3))`
D. D) `cos^(-1)((sqrt(3))/(2))`

Answer

To solve the problem, we need to find the angle between an edge of a regular tetrahedron and a face that does not contain that edge. Let's denote the vertices of the tetrahedron as O, A, B, and C, where O is the origin and A, B, and C are the vertices with position vectors \(\vec{a}\), \(\vec{b}\), and \(\vec{c}\) respectively.

### Step-by-Step Solution:

1. **Understanding the Geometry**:
- A regular tetrahedron has all edges of equal length. Let's denote the length of each edge as \(k\).
- The tetrahedron consists of four triangular faces. Each face is an equilateral triangle.

2. **Identifying the Edge and Face**:
- Let's consider the edge \(OC\) and the face \(OAB\). We need to find the angle between the edge \(OC\) and the plane formed by the triangle \(OAB\).

3. **Finding the Normal Vector to the Plane**:
- The normal vector to the plane \(OAB\) can be found using the cross product of two vectors that lie in the plane. We can take the vectors \(\vec{OA} = \vec{a}\) and \(\vec{OB} = \vec{b}\).
- The normal vector \(\vec{n}\) to the plane \(OAB\) is given by:
\[ \vec{n} = \vec{OA} \times \vec{OB} = \vec{a} \times \vec{b} \]

4. **Finding the Angle Between the Edge and the Normal**:
- The angle \(\theta\) between the edge \(OC\) (which is represented by the vector \(\vec{c}\)) and the normal vector \(\vec{n}\) can be found using the dot product:
\[ \cos \theta = \frac{\vec{c} \cdot \vec{n}}{|\vec{c}| |\vec{n}|} \]

5. **Calculating the Magnitudes**:
- Since \(O\) is the origin, the magnitudes of the vectors are:
\[ |\vec{c}| = k \quad \text{(length of edge OC)} \]
- The magnitude of the normal vector \(\vec{n}\) can be calculated as:
\[ |\vec{n}| = |\vec{a} \times \vec{b}| = |\vec{a}||\vec{b}|\sin(\angle AOB) \]
- Since \(A\) and \(B\) are vertices of an equilateral triangle, \(\angle AOB = 60^\circ\), thus:
\[ |\vec{n}| = k \cdot k \cdot \sin(60^\circ) = k^2 \cdot \frac{\sqrt{3}}{2} \]

6. **Finding the Angle Between the Edge and the Plane**:
- The angle \(\phi\) between the edge \(OC\) and the plane \(OAB\) is the complement of the angle \(\theta\) between the edge and the normal:
\[ \phi = 90^\circ - \theta \]
- Using the relationship:
\[ \cos \theta = \frac{\vec{c} \cdot \vec{n}}{|\vec{c}| |\vec{n}|} \]
- We can find \(\theta\) and then compute \(\phi\).

7. **Final Calculation**:
- Since all edges are equal and the tetrahedron is regular, the angles between any edge and the face not containing it will be the same. The angle \(\phi\) can be calculated as:
\[ \phi = 90^\circ - \cos^{-1}\left(\frac{1}{\sqrt{3}}\right) \]
- Thus, the angle between any edge and a face not containing the edge is:
\[ \phi = 60^\circ \]

### Conclusion:
The angle between any edge of a regular tetrahedron and a face not containing that edge is \(60^\circ\).