What is the value of linear veloctity, if `vecomega =hati-hatj+hatk` and `vecr=hati+2hatj+3hatk`

Options

A. `hat(i)+hat(j)+hat(k)`
B. `-hat(i)-hat(j)-hat(k)`
C. `vec(0)`
D. `-3hat(i)-3hat(j)-3hat(k)`

Explanation

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

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

### Step 1: Identify the vectors
Given:
- \( \vec{\omega} = \hat{i} - \hat{j} + \hat{k} \)
- \( \vec{r} = \hat{i} + 2\hat{j} + 3\hat{k} \)

### Step 2: Set up the cross product using the determinant method
We will set up a determinant to compute the cross product \( \vec{\omega} \times \vec{r} \):

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

### Step 3: Calculate the determinant
We can expand this determinant as follows:

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

Calculating each of the 2x2 determinants:

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

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

3. For \( \hat{k} \):
\[ \begin{vmatrix} 1 & -1 \\ 1 & 2 \end{vmatrix} = (1)(2) - (-1)(1) = 2 + 1 = 3 \]

### Step 4: Combine the results
Now substituting back into the expression for \( \vec{v} \):

\[ \vec{v} = -5\hat{i} - 2\hat{j} + 3\hat{k} \]

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

\[ \vec{v} = -5\hat{i} - 2\hat{j} + 3\hat{k} \]

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