The linear velocity of a rotating body is given by `vec(v)=vec(omega)xxvec(r)`, where `vec(omega)` is the angular velocity and `vec(r)` is the radius vector. The angular velocity of a body is `vec(omega)=hat(i)-2hat(j)+2hat(k)` and the radius vector `vec(r)=4hat(j)-3hat(k)`, then `|vec(v)|` is-

Options

A. `sqrt(29)` units
B. `sqrt(31)` units
C. `sqrt(37)` units
D. `sqrt(41)` units

Explanation

To solve the problem, we need to calculate the linear velocity vector \(\vec{v}\) using the cross product of the angular velocity vector \(\vec{\omega}\) and the radius vector \(\vec{r}\). Then, we will find the magnitude of the velocity vector.

### Step-by-Step Solution:

1. **Identify the vectors**:
- The angular velocity vector is given by:
\[ \vec{\omega} = \hat{i} - 2\hat{j} + 2\hat{k} \]
- The radius vector is given by:
\[ \vec{r} = 4\hat{j} - 3\hat{k} \]

2. **Set up the cross product**:
We need to calculate \(\vec{v} = \vec{\omega} \times \vec{r}\).

3. **Write the vectors in determinant form**:
To compute the cross product, we can use the determinant of a matrix:
\[ \vec{v} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -2 & 2 \\ 0 & 4 & -3 \end{vmatrix} \]

4. **Calculate the determinant**:
Using the determinant formula for a \(3 \times 3\) matrix:
\[ \vec{v} = \hat{i} \begin{vmatrix} -2 & 2 \\ 4 & -3 \end{vmatrix} - \hat{j} \begin{vmatrix} 1 & 2 \\ 0 & -3 \end{vmatrix} + \hat{k} \begin{vmatrix} 1 & -2 \\ 0 & 4 \end{vmatrix} \]

- Calculate each of the \(2 \times 2\) determinants:
- For \(\hat{i}\):
\[ \begin{vmatrix} -2 & 2 \\ 4 & -3 \end{vmatrix} = (-2)(-3) - (2)(4) = 6 - 8 = -2 \]
- For \(\hat{j}\):
\[ \begin{vmatrix} 1 & 2 \\ 0 & -3 \end{vmatrix} = (1)(-3) - (2)(0) = -3 \]
- For \(\hat{k}\):
\[ \begin{vmatrix} 1 & -2 \\ 0 & 4 \end{vmatrix} = (1)(4) - (-2)(0) = 4 \]

5. **Combine the results**:
Now substituting back into the expression for \(\vec{v}\):
\[ \vec{v} = -2\hat{i} + 3\hat{j} + 4\hat{k} \]

6. **Find the magnitude of \(\vec{v}\)**:
The magnitude of the vector \(\vec{v}\) is given by:
\[ |\vec{v}| = \sqrt{(-2)^2 + 3^2 + 4^2} = \sqrt{4 + 9 + 16} = \sqrt{29} \]

### Final Answer:
\[ |\vec{v}| = \sqrt{29} \]