A spy plane is being tacked by a radar. At `t`=0, its position is reported as (100`m` ,200`m`, 1000`m`). 130`s` later, its position is reported to be (2500`m`,1200`m` ,1000 `m`). Find a unit vector in the direction of plane velocity and the magnitude of its average velocity.

Explanation

To solve the problem step by step, we will follow these procedures:

### Step 1: Identify Initial and Final Positions
The initial position of the plane at \( t = 0 \) is given as:
\[ \mathbf{R}_{\text{initial}} = (100 \, \text{m}, 200 \, \text{m}, 1000 \, \text{m}) = 100 \hat{i} + 200 \hat{j} + 1000 \hat{k} \]
The final position of the plane at \( t = 130 \, \text{s} \) is given as:
\[ \mathbf{R}_{\text{final}} = (2500 \, \text{m}, 1200 \, \text{m}, 1000 \, \text{m}) = 2500 \hat{i} + 1200 \hat{j} + 1000 \hat{k} \]

### Step 2: Calculate Displacement
The displacement \( \Delta \mathbf{R} \) is calculated as:
\[ \Delta \mathbf{R} = \mathbf{R}_{\text{final}} - \mathbf{R}_{\text{initial}} = (2500 \hat{i} + 1200 \hat{j} + 1000 \hat{k}) - (100 \hat{i} + 200 \hat{j} + 1000 \hat{k}) \]
Calculating this gives:
\[ \Delta \mathbf{R} = (2500 - 100) \hat{i} + (1200 - 200) \hat{j} + (1000 - 1000) \hat{k} = 2400 \hat{i} + 1000 \hat{j} + 0 \hat{k} \]

### Step 3: Calculate Average Velocity
The average velocity \( \mathbf{V}_{\text{avg}} \) is given by the formula:
\[ \mathbf{V}_{\text{avg}} = \frac{\Delta \mathbf{R}}{\Delta t} \]
Where \( \Delta t = 130 \, \text{s} \):
\[ \mathbf{V}_{\text{avg}} = \frac{2400 \hat{i} + 1000 \hat{j}}{130} \]
Calculating this gives:
\[ \mathbf{V}_{\text{avg}} = \left( \frac{2400}{130} \hat{i} + \frac{1000}{130} \hat{j} \right) = \left( \frac{240}{13} \hat{i} + \frac{100}{13} \hat{j} \right) \]

### Step 4: Calculate Magnitude of Average Velocity
The magnitude of the average velocity \( |\mathbf{V}_{\text{avg}}| \) is calculated as:
\[ |\mathbf{V}_{\text{avg}}| = \sqrt{\left(\frac{240}{13}\right)^2 + \left(\frac{100}{13}\right)^2} \]
Calculating this gives:
\[ |\mathbf{V}_{\text{avg}}| = \sqrt{\frac{240^2 + 100^2}{13^2}} = \frac{\sqrt{57600 + 10000}}{13} = \frac{\sqrt{67600}}{13} = \frac{260}{13} = 20 \, \text{m/s} \]

### Step 5: Calculate the Unit Vector in the Direction of Velocity
The unit vector \( \hat{u} \) in the direction of the velocity is given by:
\[ \hat{u} = \frac{\mathbf{V}_{\text{avg}}}{|\mathbf{V}_{\text{avg}}|} \]
Substituting the values:
\[ \hat{u} = \frac{\left( \frac{240}{13} \hat{i} + \frac{100}{13} \hat{j} \right)}{20} = \frac{240}{260} \hat{i} + \frac{100}{260} \hat{j} = \frac{12}{13} \hat{i} + \frac{5}{13} \hat{j} \]

### Final Answers
- The unit vector in the direction of the plane's velocity is:
\[ \hat{u} = \frac{12}{13} \hat{i} + \frac{5}{13} \hat{j} \]
- The magnitude of the average velocity is:
\[ |\mathbf{V}_{\text{avg}}| = 20 \, \text{m/s} \]