If `veca ,vecb and vecc` are three mutually orthogonal unit vectors , then the triple product `[(veca+vecb+vecc,veca+vecb, vecb +vecc)]` equals

Options

A. 0
B. 1 or -1
C. 1
D. 3

Explanation

To solve the problem, we need to evaluate the scalar triple product \([( \vec{a} + \vec{b} + \vec{c}, \vec{a} + \vec{b}, \vec{b} + \vec{c})]\) where \(\vec{a}\), \(\vec{b}\), and \(\vec{c}\) are mutually orthogonal unit vectors.

### Step-by-Step Solution:

1. **Understanding the Scalar Triple Product**:
The scalar triple product \((\vec{x}, \vec{y}, \vec{z})\) can be expressed as \(\vec{x} \cdot (\vec{y} \times \vec{z})\).

2. **Expanding the Expression**:
We can rewrite the given expression as:
\[ (\vec{a} + \vec{b} + \vec{c}, \vec{a} + \vec{b}, \vec{b} + \vec{c}) = (\vec{a} + \vec{b}, \vec{b} + \vec{c}) \cdot \vec{c} + (\vec{a} + \vec{b}) \cdot (\vec{b} + \vec{c}) \cdot \vec{c} \]

3. **Using the Property of Scalar Triple Product**:
We can separate the scalar triple product into two parts:
\[ = (\vec{a} + \vec{b}) \cdot (\vec{b} + \vec{c}) \cdot \vec{c} \]
Since \(\vec{b} + \vec{c}\) is not equal to \(\vec{a} + \vec{b}\), we can simplify this further.

4. **Calculating Each Part**:
Since \(\vec{a}\), \(\vec{b}\), and \(\vec{c}\) are mutually orthogonal unit vectors, we know:
\[ \vec{a} \cdot \vec{b} = 0, \quad \vec{b} \cdot \vec{c} = 0, \quad \vec{c} \cdot \vec{a} = 0 \]
Thus, the dot products involving different vectors will be zero.

5. **Final Calculation**:
The scalar triple product simplifies down to:
\[ = \vec{c} \cdot (\vec{a} \times \vec{b}) \]
Since \(\vec{a}\) and \(\vec{b}\) are perpendicular to \(\vec{c}\), the result of \(\vec{a} \times \vec{b}\) will be a vector in the direction of \(\vec{c}\).

6. **Magnitude of the Result**:
The magnitude of \(\vec{c} \cdot (\vec{a} \times \vec{b})\) will be equal to the volume of the parallelepiped formed by \(\vec{a}\), \(\vec{b}\), and \(\vec{c}\), which is \(1\) (since all vectors are unit vectors).

7. **Conclusion**:
Therefore, the value of the scalar triple product is:
\[ \pm 1 \]