An aeroplane moves `400m` towards north, `300m` towards west and then `1200m` vertically upward. Then its displacement from the initial position is

Options

A. 1300m
B. 1400m
C. 1500m
D. 1600m

Answer

To find the displacement of the aeroplane from its initial position after moving 400 m north, 300 m west, and then 1200 m vertically upward, we can break down the problem into steps.

### Step-by-Step Solution:

1. **Define the Coordinate System:**
- Let the north direction be along the positive y-axis.
- Let the west direction be along the negative x-axis.
- Let the upward direction be along the positive z-axis.

2. **Determine the Displacement Vectors:**
- The displacement towards north (400 m) can be represented as:
\[ \vec{S_y} = 400 \hat{j} \, \text{m} \]
- The displacement towards west (300 m) can be represented as:
\[ \vec{S_x} = -300 \hat{i} \, \text{m} \]
- The displacement vertically upward (1200 m) can be represented as:
\[ \vec{S_z} = 1200 \hat{k} \, \text{m} \]

3. **Calculate the Total Displacement Vector:**
- The total displacement vector \(\vec{S}\) can be expressed as:
\[ \vec{S} = \vec{S_x} + \vec{S_y} + \vec{S_z} \]
- Therefore,
\[ \vec{S} = -300 \hat{i} + 400 \hat{j} + 1200 \hat{k} \]

4. **Calculate the Magnitude of the Displacement:**
- The magnitude of the displacement vector \(\vec{S}\) is given by the formula:
\[ S = \sqrt{S_x^2 + S_y^2 + S_z^2} \]
- Substituting the values:
\[ S = \sqrt{(-300)^2 + (400)^2 + (1200)^2} \]
- Calculate each term:
\[ (-300)^2 = 90000 \]
\[ (400)^2 = 160000 \]
\[ (1200)^2 = 1440000 \]
- Now sum these values:
\[ S = \sqrt{90000 + 160000 + 1440000} \]
\[ S = \sqrt{1690000} \]

5. **Final Calculation:**
- Simplifying further:
\[ S = \sqrt{169 \times 10000} = 1300 \, \text{m} \]

### Final Answer:
The displacement of the aeroplane from its initial position is **1300 m**.