A cubical room has dimensions ` 4 ft xx 4 ft xx 4 ft` . An insect starts from one corner O and reaches a corner on the opposite side of the body diagonal.
In previous problem, the magnitude of displacement is :

Options

A. Zero
B. ` sqrt ( 80 ) ft`
C. ` sqrt ( 48 ) ft`
D. None of these

Explanation

To find the magnitude of displacement of the insect moving from one corner of a cubical room to the opposite corner along the body diagonal, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Coordinates of the Corners**:
- Let the starting corner \( O \) be at coordinates \( (0, 0, 0) \).
- The opposite corner \( P \) will then be at coordinates \( (4, 4, 4) \) since the dimensions of the cube are \( 4 \, \text{ft} \times 4 \, \text{ft} \times 4 \, \text{ft} \).

2. **Determine the Displacement Vector**:
- The displacement vector \( \vec{d} \) from point \( O \) to point \( P \) can be calculated as:
\[ \vec{d} = P - O = (4, 4, 4) - (0, 0, 0) = (4, 4, 4) \]

3. **Calculate the Magnitude of the Displacement**:
- The magnitude of the displacement vector \( |\vec{d}| \) is given by the formula:
\[ |\vec{d}| = \sqrt{x^2 + y^2 + z^2} \]
- Substituting the coordinates of the displacement vector:
\[ |\vec{d}| = \sqrt{4^2 + 4^2 + 4^2} = \sqrt{16 + 16 + 16} = \sqrt{48} \]

4. **Final Result**:
- Therefore, the magnitude of the displacement is:
\[ |\vec{d}| = \sqrt{48} \, \text{ft} \]

### Conclusion:
The magnitude of the displacement from corner \( O \) to corner \( P \) is \( \sqrt{48} \, \text{ft} \).

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