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:**
- The insect starts at corner O, which we can denote as point \( O(0, 0, 0) \).
- The opposite corner of the cube can be denoted as point \( A(4, 4, 4) \) since the dimensions of the cube are 4 ft x 4 ft x 4 ft.

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

3. **Calculate the Magnitude of the Displacement:**
- The magnitude of the displacement vector \( |\vec{d}| \) is calculated using the formula:
\[ |\vec{d}| = \sqrt{(d_x)^2 + (d_y)^2 + (d_z)^2} \]
- Substituting the components of the displacement vector:
\[ |\vec{d}| = \sqrt{(4)^2 + (4)^2 + (4)^2} = \sqrt{16 + 16 + 16} = \sqrt{48} \]

4. **Simplify the Expression:**
- We can simplify \( \sqrt{48} \):
\[ \sqrt{48} = \sqrt{16 \times 3} = 4\sqrt{3} \]

5. **Final Answer:**
- Therefore, the magnitude of the displacement is:
\[ |\vec{d}| = 4\sqrt{3} \text{ ft} \]