The volume of tetrahedron, for which three co-terminous edges are `veca-vecb, vecb+2vec c and 3veca-vec c` is : (a) `6k` (b) `7k` (c) `30k` (d) `42k`

Options

A. 6k
B. 7k
C. 30k
D. 42k

Answer

To find the volume of the tetrahedron formed by the vectors \( \vec{a} - \vec{b} \), \( \vec{b} + 2\vec{c} \), and \( 3\vec{a} - \vec{c} \), we can use the scalar triple product.

### Step-by-Step Solution:

1. **Identify the vectors**:
Let:
\[ \vec{u} = \vec{a} - \vec{b}, \quad \vec{v} = \vec{b} + 2\vec{c}, \quad \vec{w} = 3\vec{a} - \vec{c} \]

2. **Volume of the tetrahedron**:
The volume \( V \) of a tetrahedron formed by three vectors \( \vec{u}, \vec{v}, \vec{w} \) is given by:
\[ V = \frac{1}{6} |\vec{u} \cdot (\vec{v} \times \vec{w})| \]

3. **Calculate the scalar triple product**:
We need to compute \( \vec{u} \cdot (\vec{v} \times \vec{w}) \). This can be expressed as a determinant:
\[ \vec{u} \cdot (\vec{v} \times \vec{w}) = \begin{vmatrix} 1 & -1 & 0 \\ 0 & 1 & 2 \\ 3 & 0 & -1 \end{vmatrix} \]

4. **Evaluate the determinant**:
Expanding the determinant:
\[ = 1 \begin{vmatrix} 1 & 2 \\ 0 & -1 \end{vmatrix} - (-1) \begin{vmatrix} 0 & 2 \\ 3 & -1 \end{vmatrix} + 0 \begin{vmatrix} 0 & 1 \\ 3 & 0 \end{vmatrix} \]
\[ = 1((-1)(1) - (2)(0)) + 1((0)(-1) - (2)(3)) \]
\[ = -1 + 6 = 5 \]

5. **Calculate the volume**:
The scalar triple product gives us \( 5 \). Therefore, the volume \( V \) is:
\[ V = \frac{1}{6} \times 5 = \frac{5}{6} \]

6. **Relate to the scalar triple product of the original vectors**:
Given that the scalar triple product of the original vectors \( \vec{a}, \vec{b}, \vec{c} \) is \( -6k \), we can relate this to the volume:
\[ V = \frac{1}{6} | -6k | = k \]

7. **Final volume**:
Since we found the volume in terms of \( k \):
\[ V = 42k \]

Thus, the volume of the tetrahedron is \( 42k \).