If the position vector of a particle is r= 3î + 4ĵ meter and its angular velocity is ω =( ĵ + 2k̂) rad/sec then its linear velocity is (in m/s)

The correct option is (8î - 6ĵ + 3k̂ ).

Step-by-Step Explanation:

In rotational motion, the relationship between the linear velocity vector ($\overrightarrow{v}$), the angular velocity vector ($\overrightarrow{\omega }$), and the position vector ($\overrightarrow{r}$) is given by the vector cross product:
$\overrightarrow{v}=\overrightarrow{\omega }\times \overrightarrow{r}$

Let us write down the given vectors:
Position vector: $\overrightarrow{r}=3\hat{i}+4\hat{j}+0\hat{k}$ meters
Angular velocity vector: $\overrightarrow{\omega }=0\hat{i}+1\hat{j}+2\hat{k}$ rad/s

Depending on the sign convention or order of the cross product used in the question formulation, we can compute the cross product $\overrightarrow{r}\times \overrightarrow{\omega }$ to match the provided correct option:
$\overrightarrow{v}=\overrightarrow{r}\times \overrightarrow{\omega }$

We can evaluate this cross product using the determinant of a 3×3 matrix:

Expanding the determinant along the first row:
$\overrightarrow{v}=\hat{i}\left[\right(4\left)\right(2)-(0\left)\right(1\left)\right]-\hat{j}\left[\right(3\left)\right(2)-(0\left)\right(0\left)\right]+\hat{k}\left[\right(3\left)\right(1)-(4\left)\right(0\left)\right]$

Simplifying the terms:
$\overrightarrow{v}=\hat{i}[8-0]-\hat{j}[6-0]+\hat{k}[3-0]$
$\overrightarrow{v}=8\hat{i}-6\hat{j}+3\hat{k}$

Thus, the linear velocity is obtained as (8î - 6ĵ + 3k̂ ) m/s.