A man walks `8 m` towards east and `6 m` towards north. The magnitude of displacement is :

Options

A. 14 m
B. 2 m
C. 10 m
D. `(8^2 + 6^2) m`

Explanation

To find the magnitude of displacement when a man walks 8 meters towards the east and then 6 meters towards the north, we can follow these steps:

### Step 1: Understand the movement
The man walks in two perpendicular directions:
- First, he walks 8 meters to the east.
- Then, he walks 6 meters to the north.

### Step 2: Visualize the path
We can visualize this movement on a coordinate system:
- The east direction can be represented along the x-axis.
- The north direction can be represented along the y-axis.

### Step 3: Represent the movements as vectors
We can represent the movements as vectors:
- The movement towards the east (x-direction) can be represented as \( \vec{A} = 8 \hat{i} \) meters.
- The movement towards the north (y-direction) can be represented as \( \vec{B} = 6 \hat{j} \) meters.

### Step 4: Calculate the resultant displacement vector
The resultant displacement vector \( \vec{R} \) can be found by adding the two vectors:
\[ \vec{R} = \vec{A} + \vec{B} = 8 \hat{i} + 6 \hat{j} \]

### Step 5: Find the magnitude of the displacement vector
The magnitude of the displacement vector \( |\vec{R}| \) can be calculated using the Pythagorean theorem:
\[ |\vec{R}| = \sqrt{(8)^2 + (6)^2} \]
Calculating this gives:
\[ |\vec{R}| = \sqrt{64 + 36} = \sqrt{100} = 10 \text{ meters} \]

### Step 6: Conclusion
Thus, the magnitude of the displacement is **10 meters**.

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