The angular velocity of a particle is `vec(w) = 4hati + hatj - 2hatk` about the origin. If the position vector of the particle is `2hati + 3hatj - 3hatk`, then its linear velocity is

Options

A. `5hati + 8hatj - 14hatk`
B. `3hati + 8 hatj + 10hatk`
C. `8 hati + 3hatj - 10hatk`
D. `-8hati + 3hatj - 2hatk`

Answer

To find the linear velocity of a particle given its angular velocity and position vector, 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.

### Given:
- Angular velocity: \(\vec{\omega} = 4\hat{i} + \hat{j} - 2\hat{k}\)
- Position vector: \(\vec{r} = 2\hat{i} + 3\hat{j} - 3\hat{k}\)

### Step 1: Set up the cross product
We need to calculate the cross product \(\vec{\omega} \times \vec{r}\). We can represent this as a determinant:

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

### Step 2: Calculate the determinant
We can expand this determinant using the rule of Sarrus or cofactor expansion.

1. For \(\hat{i}\):
\[ \hat{i} \left( 1 \cdot (-3) - (-2) \cdot 3 \right) = \hat{i} \left( -3 + 6 \right) = 3\hat{i} \]

2. For \(\hat{j}\):
\[ -\hat{j} \left( 4 \cdot (-3) - (-2) \cdot 2 \right) = -\hat{j} \left( -12 + 4 \right) = -(-8)\hat{j} = 8\hat{j} \]

3. For \(\hat{k}\):
\[ \hat{k} \left( 4 \cdot 3 - 1 \cdot 2 \right) = \hat{k} \left( 12 - 2 \right) = 10\hat{k} \]

### Step 3: Combine the results
Now, we can combine the results from the three components:

\[ \vec{v} = 3\hat{i} + 8\hat{j} + 10\hat{k} \]

### Final Answer
Thus, the linear velocity of the particle is:

\[ \vec{v} = 3\hat{i} + 8\hat{j} + 10\hat{k} \]

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