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 find the magnitude of the linear velocity vector \(\vec{v}\) given by the equation \(\vec{v} = \vec{\omega} \times \vec{r}\), we will follow these steps:

### Step 1: Write down the vectors
The angular velocity vector is given as:
\[ \vec{\omega} = \hat{i} - 2\hat{j} + 2\hat{k} \]
The radius vector is given as:
\[ \vec{r} = 4\hat{j} - 3\hat{k} \]

### Step 2: Set up the cross product
The linear velocity vector \(\vec{v}\) is calculated using the cross product of \(\vec{\omega}\) and \(\vec{r}\):
\[ \vec{v} = \vec{\omega} \times \vec{r} \]

We can set up the determinant for the cross product as follows:
\[ \vec{v} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -2 & 2 \\ 0 & 4 & -3 \end{vmatrix} \]

### Step 3: Calculate the determinant
Expanding the determinant, we have:
\[ \vec{v} = \hat{i} \begin{vmatrix} -2 & 2 \\ 4 & -3 \end{vmatrix} - \hat{j} \begin{vmatrix} 1 & 2 \\ 0 & -3 \end{vmatrix} + \hat{k} \begin{vmatrix} 1 & -2 \\ 0 & 4 \end{vmatrix} \]

Calculating each of the 2x2 determinants:
1. For \(\hat{i}\):
\[ (-2)(-3) - (2)(4) = 6 - 8 = -2 \]
2. For \(\hat{j}\):
\[ (1)(-3) - (2)(0) = -3 - 0 = -3 \]
3. For \(\hat{k}\):
\[ (1)(4) - (-2)(0) = 4 - 0 = 4 \]

Putting it all together:
\[ \vec{v} = -2\hat{i} + 3\hat{j} + 4\hat{k} \]

### Step 4: Find the magnitude of \(\vec{v}\)
The magnitude of the velocity vector \(\vec{v}\) is given by:
\[ |\vec{v}| = \sqrt{(-2)^2 + 3^2 + 4^2} \]
Calculating this:
\[ |\vec{v}| = \sqrt{4 + 9 + 16} = \sqrt{29} \]

### Final Answer
Thus, the magnitude of the linear velocity vector \(|\vec{v}|\) is:
\[ |\vec{v}| = \sqrt{29} \text{ units} \]

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