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

Explanation

To find the linear velocity \( \vec{v} \) given \( \vec{\omega} = 3\hat{i} - 4\hat{j} + \hat{k} \) and \( \vec{r} = 5\hat{i} - 6\hat{j} + 6\hat{k} \), we will use the formula:

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

### Step 1: Set up the determinant for the cross product
We can express the cross product using the determinant of a matrix formed by the unit vectors \( \hat{i}, \hat{j}, \hat{k} \) and the components of \( \vec{\omega} \) and \( \vec{r} \).

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

### Step 2: Calculate the determinant
To calculate 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: Combine the results
Now substituting back into the expression for \( \vec{v} \):

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

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

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

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