What is the value of linear velocity, if `vecomega=3hati-4hatj+hatk` and `vecr=5hati-6hatj+6hatk`?

Options

A. `6hati+2hatj-3hatk`
B. `18hati+3hatj-2hatk`
C. `-18hati-13hatj+2hatk`
D. `6hati-2hatj+8hatk`

Explanation

To find the linear velocity vector \( \vec{V} \) given the angular velocity vector \( \vec{\omega} \) and the position vector \( \vec{r} \), we use the formula:

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

Given:
\[ \vec{\omega} = 3\hat{i} - 4\hat{j} + \hat{k} \]
\[ \vec{r} = 5\hat{i} - 6\hat{j} + 6\hat{k} \]

### Step 1: Set up the cross product

We will calculate the cross product \( \vec{V} = \vec{\omega} \times \vec{r} \). The cross product in determinant form is given by:

\[ \vec{V} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & -4 & 1 \\ 5 & -6 & 6 \end{vmatrix} \]

### Step 2: Calculate the determinant

To compute the determinant, we expand it as follows:

\[ \vec{V} = \hat{i} \begin{vmatrix} -4 & 1 \\ -6 & 6 \end{vmatrix} - \hat{j} \begin{vmatrix} 3 & 1 \\ 5 & 6 \end{vmatrix} + \hat{k} \begin{vmatrix} 3 & -4 \\ 5 & -6 \end{vmatrix} \]

### Step 3: Calculate each of the 2x2 determinants

1. For \( \hat{i} \):
\[ \begin{vmatrix} -4 & 1 \\ -6 & 6 \end{vmatrix} = (-4)(6) - (1)(-6) = -24 + 6 = -18 \]

2. For \( \hat{j} \):
\[ \begin{vmatrix} 3 & 1 \\ 5 & 6 \end{vmatrix} = (3)(6) - (1)(5) = 18 - 5 = 13 \]

3. For \( \hat{k} \):
\[ \begin{vmatrix} 3 & -4 \\ 5 & -6 \end{vmatrix} = (3)(-6) - (-4)(5) = -18 + 20 = 2 \]

### Step 4: Substitute back into the equation

Now substituting these values back into the equation for \( \vec{V} \):

\[ \vec{V} = -18\hat{i} - 13\hat{j} + 2\hat{k} \]

### Final Result

Thus, the linear velocity vector is:

\[ \vec{V} = -18\hat{i} - 13\hat{j} + 2\hat{k} \]

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