For a body, angular velocity `vecomega= hati-2hatj+3hatk` and radius vector `vecr=hati+hatj+hatk`, then its velocity `(vecv= vecomega xx vecr)` is :

Options

A. `-5hati+2hatj+3 hatk`
B. `-5hati+2hatj-3 hatk`
C. `-5hati-2hatj+3 hatk`
D. `-5hati-2hatj-3 hatk`

Explanation

To solve the problem of finding the velocity vector \(\vec{v}\) given the angular velocity vector \(\vec{\omega}\) and the radius vector \(\vec{r}\), we will use the cross product formula:

\[ \vec{v} = \vec{\omega} \times \vec{r} \]

### Step-by-Step Solution:

1. **Identify the vectors**:
- Angular velocity vector:
\[ \vec{\omega} = \hat{i} - 2\hat{j} + 3\hat{k} \]
- Radius vector:
\[ \vec{r} = \hat{i} + \hat{j} + \hat{k} \]

2. **Set up the cross product**:
The cross product \(\vec{v} = \vec{\omega} \times \vec{r}\) can be calculated using the determinant of a matrix formed by the unit vectors and the components of \(\vec{\omega}\) and \(\vec{r}\):
\[ \vec{v} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -2 & 3 \\ 1 & 1 & 1 \end{vmatrix} \]

3. **Calculate the determinant**:
To compute the determinant, we can expand it as follows:
\[ \vec{v} = \hat{i} \begin{vmatrix} -2 & 3 \\ 1 & 1 \end{vmatrix} - \hat{j} \begin{vmatrix} 1 & 3 \\ 1 & 1 \end{vmatrix} + \hat{k} \begin{vmatrix} 1 & -2 \\ 1 & 1 \end{vmatrix} \]

4. **Calculate each of the 2x2 determinants**:
- For \(\hat{i}\):
\[ \begin{vmatrix} -2 & 3 \\ 1 & 1 \end{vmatrix} = (-2)(1) - (3)(1) = -2 - 3 = -5 \]
- For \(\hat{j}\):
\[ \begin{vmatrix} 1 & 3 \\ 1 & 1 \end{vmatrix} = (1)(1) - (3)(1) = 1 - 3 = -2 \]
- For \(\hat{k}\):
\[ \begin{vmatrix} 1 & -2 \\ 1 & 1 \end{vmatrix} = (1)(1) - (-2)(1) = 1 + 2 = 3 \]

5. **Combine the results**:
Now substituting back into the expression for \(\vec{v}\):
\[ \vec{v} = -5\hat{i} + 2\hat{j} + 3\hat{k} \]

6. **Final Result**:
Therefore, the velocity vector is:
\[ \vec{v} = -5\hat{i} + 2\hat{j} + 3\hat{k} \]