What is the value of linear velocity. If `vec(omega)=3hat(i)-4hat(j)+hat(k)and vec(r)=5hat(i)-6hat(j)+6hat(k)` ?

Options

A. `6hat(i)-2hat(j)+3hat(k)`
B. `6hat(i)-2hat(j)+8hat(k)`
C. `6hat(i)-13hat(j)+6hat(k)`
D. `18hat(i)+13hat(j)-2hat(k)`

Answer

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

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

### Step 1: Write down the vectors

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

### Step 2: Set up the cross product

To compute the cross product \( \vec{\omega} \times \vec{r} \), we can set it up in a determinant form:

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

### Step 3: Calculate the determinant

Using the determinant formula, we can 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 4: 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 5: Combine the results

Now substituting back into the expression for \( \vec{v} \):

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

### Step 6: Final result

Thus, the linear velocity \( \vec{v} \) is:

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