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

Options

A. `6hat(i)+2hat(j)-3hat(k)`
B. `-18 hat(i)-13hat(j)+2hat(k)`
C. `4hat(i)-13hat(j)+6hat(k)`
D. `6hat(i)-2hat(j)+8hat(k)`

Answer

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

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

### Step 1: Write down the vectors
Given:
- \( \vec{\omega} = 3 \hat{i} - 4 \hat{j} + 1 \hat{k} \)
- \( \vec{r} = 5 \hat{i} - 6 \hat{j} + 6 \hat{k} \)

### Step 2: Set up the cross product
The cross product \( \vec{v} = \vec{\omega} \times \vec{r} \) can be calculated using the determinant of a matrix:

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

### Step 3: Calculate the determinant
Using the determinant formula for a 3x3 matrix, 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 the minors
Now, we 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 we can substitute these values back into our expression for \( \vec{v} \):

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

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

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