At a given instant of time the position vector of a particle moving in a circle with a velocity `3hati-4hatj+5hatk is hati+9hatj-8hatk`.Its anglular velocity at that time is:

Options

A. `((13hati-29hatj-31hatk))/sqrt(146)`
B. `((13hati-29hatj-31hatk))/(146)`
C. `((13hati+29hatj-31hatk))/sqrt(146)`
D. `((13hati+29hatj-31hatk))/(146)`

Explanation

To find the angular velocity of a particle moving in a circle, we can use the relationship between linear velocity (V), position vector (r), and angular velocity (ω). The formula is given by:

\[ \omega = \frac{V \times r}{|r|^2} \]

Where:
- \( V \) is the linear velocity vector.
- \( r \) is the position vector.
- \( |r| \) is the magnitude of the position vector.

### Step 1: Identify the vectors

Given:
- Velocity vector \( V = 3\hat{i} - 4\hat{j} + 5\hat{k} \)
- Position vector \( r = \hat{i} + 9\hat{j} - 8\hat{k} \)

### Step 2: Calculate the magnitude of the position vector \( |r| \)

The magnitude of the position vector \( r \) is calculated as follows:

\[ |r| = \sqrt{(\hat{i})^2 + (9\hat{j})^2 + (-8\hat{k})^2} = \sqrt{1^2 + 9^2 + (-8)^2} = \sqrt{1 + 81 + 64} = \sqrt{146} \]

### Step 3: Calculate \( V \times r \)

Now we need to compute the cross product \( V \times r \):

\[ V \times r = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & -4 & 5 \\ 1 & 9 & -8 \end{vmatrix} \]

Calculating the determinant:

\[ = \hat{i} \begin{vmatrix} -4 & 5 \\ 9 & -8 \end{vmatrix} - \hat{j} \begin{vmatrix} 3 & 5 \\ 1 & -8 \end{vmatrix} + \hat{k} \begin{vmatrix} 3 & -4 \\ 1 & 9 \end{vmatrix} \]

Calculating each of the 2x2 determinants:

1. \( \begin{vmatrix} -4 & 5 \\ 9 & -8 \end{vmatrix} = (-4)(-8) - (5)(9) = 32 - 45 = -13 \)

2. \( \begin{vmatrix} 3 & 5 \\ 1 & -8 \end{vmatrix} = (3)(-8) - (5)(1) = -24 - 5 = -29 \)

3. \( \begin{vmatrix} 3 & -4 \\ 1 & 9 \end{vmatrix} = (3)(9) - (-4)(1) = 27 + 4 = 31 \)

Putting it all together:

\[ V \times r = -13\hat{i} + 29\hat{j} + 31\hat{k} \]

### Step 4: Calculate \( \omega \)

Now we can calculate \( \omega \):

\[ \omega = \frac{V \times r}{|r|^2} \]

First, calculate \( |r|^2 \):

\[ |r|^2 = 146 \]

Now substitute back into the equation for \( \omega \):

\[ \omega = \frac{-13\hat{i} + 29\hat{j} + 31\hat{k}}{146} \]

### Final Result

Thus, the angular velocity \( \omega \) is:

\[ \omega = -\frac{13}{146}\hat{i} + \frac{29}{146}\hat{j} + \frac{31}{146}\hat{k} \]