A rigid body rotates with constant angular velocity `omaga` about the line whose vector equation is, `r=lambda(hat(i)+2hat(j)+2hat(k))`. The speed of the particle at the instant it passes through the point with position vector `(2hat(i)+3hat(j)+5hat(k))` is equal to

Options

A. `omegasqrt(2)`
B. `2omega`
C. `(omega)/(sqrt(2))`
D. None of these

Answer

To solve the problem, we need to find the speed of a particle at the instant it passes through the point with position vector \( \mathbf{r} = 2\hat{i} + 3\hat{j} + 5\hat{k} \) while the rigid body rotates with a constant angular velocity \( \omega \) about the line defined by the vector equation \( \mathbf{r} = \lambda(\hat{i} + 2\hat{j} + 2\hat{k}) \).

### Step-by-Step Solution:

1. **Identify the Direction of Rotation:**
The line of rotation is given by the vector \( \hat{i} + 2\hat{j} + 2\hat{k} \). We need to find the unit vector along this line.
\[ \mathbf{n} = \hat{i} + 2\hat{j} + 2\hat{k} \]
The magnitude of \( \mathbf{n} \) is:
\[ |\mathbf{n}| = \sqrt{1^2 + 2^2 + 2^2} = \sqrt{1 + 4 + 4} = \sqrt{9} = 3 \]
Therefore, the unit vector \( \hat{n} \) is:
\[ \hat{n} = \frac{\hat{i} + 2\hat{j} + 2\hat{k}}{3} \]

2. **Express Angular Velocity Vector:**
The angular velocity vector \( \mathbf{\omega} \) can be expressed as:
\[ \mathbf{\omega} = \omega \hat{n} = \frac{\omega}{3} (\hat{i} + 2\hat{j} + 2\hat{k}) \]

3. **Position Vector of the Particle:**
The position vector of the particle is given as:
\[ \mathbf{r} = 2\hat{i} + 3\hat{j} + 5\hat{k} \]

4. **Calculate the Velocity of the Particle:**
The velocity \( \mathbf{v} \) of the particle is given by the cross product of the angular velocity vector \( \mathbf{\omega} \) and the position vector \( \mathbf{r} \):
\[ \mathbf{v} = \mathbf{\omega} \times \mathbf{r} \]
Substituting the values:
\[ \mathbf{v} = \frac{\omega}{3} (\hat{i} + 2\hat{j} + 2\hat{k}) \times (2\hat{i} + 3\hat{j} + 5\hat{k}) \]

5. **Calculate the Cross Product:**
We can set up the determinant for the cross product:
\[ \mathbf{v} = \frac{\omega}{3} \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 2 \\ 2 & 3 & 5 \end{vmatrix} \]
Expanding this determinant:
\[ = \frac{\omega}{3} \left( \hat{i}(2 \cdot 5 - 2 \cdot 3) - \hat{j}(1 \cdot 5 - 2 \cdot 2) + \hat{k}(1 \cdot 3 - 2 \cdot 2) \right) \]
\[ = \frac{\omega}{3} \left( \hat{i}(10 - 6) - \hat{j}(5 - 4) + \hat{k}(3 - 4) \right) \]
\[ = \frac{\omega}{3} \left( 4\hat{i} - 1\hat{j} - 1\hat{k} \right) \]
Thus,
\[ \mathbf{v} = \frac{\omega}{3} (4\hat{i} - \hat{j} - \hat{k}) \]

6. **Magnitude of the Velocity:**
Now, we find the magnitude of the velocity:
\[ |\mathbf{v}| = \left| \frac{\omega}{3} (4\hat{i} - \hat{j} - \hat{k}) \right| = \frac{\omega}{3} \sqrt{4^2 + (-1)^2 + (-1)^2} \]
\[ = \frac{\omega}{3} \sqrt{16 + 1 + 1} = \frac{\omega}{3} \sqrt{18} = \frac{\omega}{3} \cdot 3\sqrt{2} = \omega \sqrt{2} \]

### Final Answer:
The speed of the particle at the instant it passes through the point is:
\[ \text{Speed} = \omega \sqrt{2} \]