Volume of the parallelopiped with its edges represented by the vectors `hati+hatj, hati+2hatj` and `hati+hatj+pihatk`, is

Options

A. `pi`
B. `pi//2`
C. `pi//3`
D. `pi//4`

Explanation

To find the volume of the parallelepiped formed by the vectors \(\hat{i} + \hat{j}\), \(\hat{i} + 2\hat{j}\), and \(\hat{i} + \hat{j} + \pi \hat{k}\), we can use the scalar triple product, which is given by the determinant of a matrix formed by the three vectors.

### Step-by-Step Solution:

1. **Identify the vectors**:
- Let \(\vec{A} = \hat{i} + \hat{j}\)
- Let \(\vec{B} = \hat{i} + 2\hat{j}\)
- Let \(\vec{C} = \hat{i} + \hat{j} + \pi \hat{k}\)

2. **Write the vectors in component form**:
- \(\vec{A} = (1, 1, 0)\)
- \(\vec{B} = (1, 2, 0)\)
- \(\vec{C} = (1, 1, \pi)\)

3. **Set up the determinant**:
The volume \(V\) of the parallelepiped can be found using the determinant:
\[ V = \left| \begin{vmatrix} 1 & 1 & 0 \\ 1 & 2 & 0 \\ 1 & 1 & \pi \end{vmatrix} \right| \]

4. **Calculate the determinant**:
Using the formula for the determinant of a \(3 \times 3\) matrix:
\[ \text{Det} = a(ei - fh) - b(di - fg) + c(dh - eg) \]
where the matrix is:
\[ \begin{vmatrix} a & b & c \\ d & e & f \\ g & h & i \end{vmatrix} \]
For our matrix:
\[ \text{Det} = 1 \cdot (2 \cdot \pi - 0 \cdot 0) - 1 \cdot (1 \cdot \pi - 0 \cdot 1) + 0 \cdot (1 \cdot 1 - 1 \cdot 1) \]
Simplifying this gives:
\[ \text{Det} = 1 \cdot (2\pi) - 1 \cdot (\pi) + 0 \]
\[ = 2\pi - \pi = \pi \]

5. **Conclusion**:
The volume of the parallelepiped is \(\pi\).