The volume of the parallelepiped whose edges are
`veca=2hati-3hatj+4hatk, vecb=hati+2hatj-hatk` and `vecc=2hati-hatj+2hatk` is

Options

A. `-2` cubic unit
B. 2 cubic unit
C. 1 cubic unit
D. 4 cubic unit

Answer

To find the volume of the parallelepiped formed by the vectors **a**, **b**, and **c**, we can use the formula for the volume, which is given by the absolute value of the scalar triple product of the vectors. The scalar triple product can be calculated using the determinant of a matrix formed by the vectors.

### Step-by-Step Solution:

1. **Identify the Vectors:**
Given vectors are:
\[ \vec{a} = 2\hat{i} - 3\hat{j} + 4\hat{k} \]
\[ \vec{b} = \hat{i} + 2\hat{j} - \hat{k} \]
\[ \vec{c} = 2\hat{i} - \hat{j} + 2\hat{k} \]

2. **Set Up the Determinant:**
The volume \( V \) of the parallelepiped can be calculated using the determinant:
\[ V = |\det(\vec{a}, \vec{b}, \vec{c})| \]
This determinant can be set up as follows:
\[ \det\begin{pmatrix} 2 & -3 & 4 \\ 1 & 2 & -1 \\ 2 & -1 & 2 \end{pmatrix} \]

3. **Calculate the Determinant:**
Using the formula for the determinant of a 3x3 matrix:
\[ \det(A) = a(ei - fh) - b(di - fg) + c(dh - eg) \]
where the matrix is:
\[ \begin{pmatrix} a & b & c \\ d & e & f \\ g & h & i \end{pmatrix} \]
For our matrix:
\[ \det = 2 \begin{vmatrix} 2 & -1 \\ -1 & 2 \end{vmatrix} - (-3) \begin{vmatrix} 1 & -1 \\ 2 & 2 \end{vmatrix} + 4 \begin{vmatrix} 1 & 2 \\ 2 & -1 \end{vmatrix} \]

Now calculating the 2x2 determinants:
\[ \begin{vmatrix} 2 & -1 \\ -1 & 2 \end{vmatrix} = (2)(2) - (-1)(-1) = 4 - 1 = 3 \]
\[ \begin{vmatrix} 1 & -1 \\ 2 & 2 \end{vmatrix} = (1)(2) - (-1)(2) = 2 + 2 = 4 \]
\[ \begin{vmatrix} 1 & 2 \\ 2 & -1 \end{vmatrix} = (1)(-1) - (2)(2) = -1 - 4 = -5 \]

Plugging these values back into the determinant:
\[ \det = 2(3) + 3(4) + 4(-5) = 6 + 12 - 20 = -2 \]

4. **Calculate the Volume:**
Since volume cannot be negative, we take the absolute value:
\[ V = |\det| = |-2| = 2 \]

5. **Final Answer:**
The volume of the parallelepiped is \( 2 \) cubic units.