For a body, angular velocity `(vec(omega)) = hat(i) - 2hat(j) + 3hat(k)` and radius vector `(vec(r )) = hat(i) + hat(j) + vec(k)`, then its velocity is :

Options

A. `-5hat(i) + 2hat(j) + 3hat(k)`
B. `-5hat(i) + 2hat(j) - 3hat(k)`
C. `-5hat(i) - 2hat(j) + 3hat(k)`
D. `-5hat(i) - 2hat(j) - 3hat(k)`

Answer

To find the velocity vector \( \vec{V} \) of a body given its angular velocity \( \vec{\omega} \) and radius vector \( \vec{r} \), we can use the relation:

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

### Step 1: Identify the vectors
Given:
- Angular velocity vector \( \vec{\omega} = \hat{i} - 2\hat{j} + 3\hat{k} \)
- Radius vector \( \vec{r} = \hat{i} + \hat{j} + \hat{k} \)

### Step 2: Set up the cross product
The cross product can be computed using the determinant of a matrix formed by the unit vectors and the components of the vectors:

\[ \vec{V} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -2 & 3 \\ 1 & 1 & 1 \end{vmatrix} \]

### Step 3: Calculate the determinant
To compute the determinant, we can expand it as follows:

\[ \vec{V} = \hat{i} \begin{vmatrix} -2 & 3 \\ 1 & 1 \end{vmatrix} - \hat{j} \begin{vmatrix} 1 & 3 \\ 1 & 1 \end{vmatrix} + \hat{k} \begin{vmatrix} 1 & -2 \\ 1 & 1 \end{vmatrix} \]

Calculating each of these 2x2 determinants:

1. For \( \hat{i} \):
\[ \begin{vmatrix} -2 & 3 \\ 1 & 1 \end{vmatrix} = (-2)(1) - (3)(1) = -2 - 3 = -5 \]

2. For \( \hat{j} \):
\[ \begin{vmatrix} 1 & 3 \\ 1 & 1 \end{vmatrix} = (1)(1) - (3)(1) = 1 - 3 = -2 \]

3. For \( \hat{k} \):
\[ \begin{vmatrix} 1 & -2 \\ 1 & 1 \end{vmatrix} = (1)(1) - (-2)(1) = 1 + 2 = 3 \]

### Step 4: Combine the results
Now substituting back into the expression for \( \vec{V} \):

\[ \vec{V} = -5\hat{i} - (-2)\hat{j} + 3\hat{k} \]
\[ \vec{V} = -5\hat{i} + 2\hat{j} + 3\hat{k} \]

### Final Result
Thus, the velocity vector is:

\[ \vec{V} = -5\hat{i} + 2\hat{j} + 3\hat{k} \]