The position vector of a particle is `vec( r ) = ( 3 hat( i ) + 4 hat( j ))` metre and its angular velocity ` vec( omega) =(hat( j)+ 2hat( k )) rad s^(-1)` then its linear velocity is ( in `ms^(-1)`)

Options

A. (a)`- ( 8hat( i ) - 6 hat( j ) + 3 hat( k ))`
B. (b)` ( 3hat( i ) + 6 hat( j ) + 8 hat( k ))`
C. (c)` -( 3hat( i ) + 6 hat( j ) + 6 hat( k ))`
D. (d)` ( 6hat( i ) + 8 hat( j ) + 3 hat( k ))`

Answer

To find the linear velocity of a particle given its position vector and angular velocity, we can use the formula:

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

Where:
- \(\vec{v}\) is the linear velocity,
- \(\vec{\omega}\) is the angular velocity,
- \(\vec{r}\) is the position vector.

### Step 1: Identify the vectors
The position vector \(\vec{r}\) is given as:
\[ \vec{r} = 3 \hat{i} + 4 \hat{j} \text{ (in meters)} \]

The angular velocity \(\vec{\omega}\) is given as:
\[ \vec{\omega} = \hat{j} + 2 \hat{k} \text{ (in rad/s)} \]

### Step 2: Set up the cross product
To find the linear velocity, we need to compute the cross product \(\vec{\omega} \times \vec{r}\). We can set this up using the determinant of a matrix:

\[ \vec{v} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 0 & 1 & 2 \\ 3 & 4 & 0 \end{vmatrix} \]

### Step 3: Calculate the determinant
Now we will calculate the determinant:

\[ \vec{v} = \hat{i} \begin{vmatrix} 1 & 2 \\ 4 & 0 \end{vmatrix} - \hat{j} \begin{vmatrix} 0 & 2 \\ 3 & 0 \end{vmatrix} + \hat{k} \begin{vmatrix} 0 & 1 \\ 3 & 4 \end{vmatrix} \]

Calculating each of the 2x2 determinants:

1. For \(\hat{i}\):
\[ \begin{vmatrix} 1 & 2 \\ 4 & 0 \end{vmatrix} = (1 \cdot 0) - (2 \cdot 4) = 0 - 8 = -8 \]

2. For \(-\hat{j}\):
\[ \begin{vmatrix} 0 & 2 \\ 3 & 0 \end{vmatrix} = (0 \cdot 0) - (2 \cdot 3) = 0 - 6 = -6 \implies -(-6) = 6 \]

3. For \(\hat{k}\):
\[ \begin{vmatrix} 0 & 1 \\ 3 & 4 \end{vmatrix} = (0 \cdot 4) - (1 \cdot 3) = 0 - 3 = -3 \]

### Step 4: Combine the results
Putting it all together, we have:

\[ \vec{v} = -8 \hat{i} + 6 \hat{j} - 3 \hat{k} \]

### Final answer
Thus, the linear velocity vector is:

\[ \vec{v} = -8 \hat{i} + 6 \hat{j} - 3 \hat{k} \text{ m/s} \]