If angular velocity of a point object is `vecomega=(hati+hat2j-hatk)` rad/s and its position vector `vecr=(hati+hatj-5hatk)`m then linear velocity of the object will be

Options

A. `(-9hati+4hatj-hatk) m/s`
B. `(9hati+4hatj+hatk) m/s`
C. `(9hati-4hatj-hatk) m/s`
D. Zero

Answer

To find the linear velocity of the object given its angular velocity and position vector, we can use the formula:

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

where:
- \(\vec{v}\) is the linear velocity,
- \(\vec{\omega}\) is the angular velocity,
- \(\vec{r}\) is the position vector.

### Step 1: Identify the vectors
Given:
\[ \vec{\omega} = \hat{i} + 2\hat{j} - \hat{k} \quad \text{(angular velocity)} \]
\[ \vec{r} = \hat{i} + \hat{j} - 5\hat{k} \quad \text{(position vector)} \]

### Step 2: Set up the cross product
We need to compute the cross product \(\vec{\omega} \times \vec{r}\). We can set this up using the determinant of a matrix:

\[ \vec{v} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & -1 \\ 1 & 1 & -5 \end{vmatrix} \]

### Step 3: Calculate the determinant
To calculate the determinant, we expand it as follows:

\[ \vec{v} = \hat{i} \begin{vmatrix} 2 & -1 \\ 1 & -5 \end{vmatrix} - \hat{j} \begin{vmatrix} 1 & -1 \\ 1 & -5 \end{vmatrix} + \hat{k} \begin{vmatrix} 1 & 2 \\ 1 & 1 \end{vmatrix} \]

Calculating each of the 2x2 determinants:

1. For \(\hat{i}\):
\[ \begin{vmatrix} 2 & -1 \\ 1 & -5 \end{vmatrix} = (2)(-5) - (-1)(1) = -10 + 1 = -9 \]

2. For \(-\hat{j}\):
\[ \begin{vmatrix} 1 & -1 \\ 1 & -5 \end{vmatrix} = (1)(-5) - (-1)(1) = -5 + 1 = -4 \]

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

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

\[ \vec{v} = -9\hat{i} + 4\hat{j} - (-1)\hat{k} \]

This simplifies to:

\[ \vec{v} = -9\hat{i} + 4\hat{j} + 1\hat{k} \]

### Final Result
Thus, the linear velocity of the object is:

\[ \vec{v} = -9\hat{i} + 4\hat{j} + 1\hat{k} \, \text{m/s} \]

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