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 of a body given its angular velocity \(\vec{\omega}\) and radius vector \(\vec{r}\), we use the formula for linear velocity \(\vec{v}\) which is given by the cross product of the angular velocity vector and the radius vector:

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

### Step 1: Write down the vectors
Given:
\[ \vec{\omega} = \hat{i} - 2\hat{j} + 3\hat{k} \]
\[ \vec{r} = \hat{i} + \hat{j} + \hat{k} \]

### Step 2: Set up the cross product
We can represent the cross product using a determinant:

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

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

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

### Step 4: Calculate each of the 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 5: Substitute back into the equation
Now substituting these results back into the equation 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 Answer
Thus, the velocity vector is:
\[ \vec{v} = -5\hat{i} + 2\hat{j} + 3\hat{k} \]