Find the value of `|vecV|` if the linear velocity of a rotating body is given by `vecv = vecomega xx vecr ,` where `vecomega` is the angular velocity and `vecr` is the radius vector. The angular velocity of a body is
`vecomega = hati - 2 hatj ` and the radius vector
`vecr = 2 hatj - 3 hatk.`

Explanation

To find the value of \(|\vec{V}|\), we will use the formula for linear velocity of a rotating body, which is given by:

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

where \(\vec{\omega}\) is the angular velocity vector and \(\vec{r}\) is the radius vector.

### Step 1: Identify the vectors

Given:
\[ \vec{\omega} = \hat{i} - 2\hat{j} \]
\[ \vec{r} = 2\hat{j} - 3\hat{k} \]

### Step 2: Compute the cross product \(\vec{v} = \vec{\omega} \times \vec{r}\)

To compute the cross product, we can use the determinant of a matrix formed by the unit vectors and the components of the vectors:

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

### Step 3: Calculate the determinant

Expanding the determinant, we have:

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

Calculating each of the 2x2 determinants:

1. \(\begin{vmatrix} -2 & 0 \\ 2 & -3 \end{vmatrix} = (-2)(-3) - (0)(2) = 6\)

2. \(\begin{vmatrix} 1 & 0 \\ 0 & -3 \end{vmatrix} = (1)(-3) - (0)(0) = -3\)

3. \(\begin{vmatrix} 1 & -2 \\ 0 & 2 \end{vmatrix} = (1)(2) - (-2)(0) = 2\)

Putting it all together:

\[ \vec{v} = 6\hat{i} - (-3)\hat{j} + 2\hat{k} = 6\hat{i} + 3\hat{j} + 2\hat{k} \]

### Step 4: Find the magnitude of \(\vec{v}\)

The magnitude of \(\vec{v}\) is given by:

\[ |\vec{v}| = \sqrt{(6)^2 + (3)^2 + (2)^2} \]

Calculating:

\[ |\vec{v}| = \sqrt{36 + 9 + 4} = \sqrt{49} = 7 \]

### Final Answer

Thus, the value of \(|\vec{V}|\) is:

\[ \boxed{7} \]