A bird flies from (-3m, 4m, -3 m) to (7m, -2m, -3m) in the xyz-coordinates. The bird's displacement vector is given by

Options

A. `(4hati + 2hatj - 6hatk)`
B. `(10 hati - 6hatj)`
C. `(4 hati - 2hatj)`
D. `(10 hati + 6 hatj - 6hatk)`

Explanation

To find the displacement vector of the bird flying from point A to point B in xyz-coordinates, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Coordinates**:
- Point A: (-3 m, 4 m, -3 m)
- Point B: (7 m, -2 m, -3 m)

2. **Define the Position Vectors**:
- The position vector of point A, \( \vec{R_A} \):
\[ \vec{R_A} = -3 \hat{i} + 4 \hat{j} - 3 \hat{k} \]
- The position vector of point B, \( \vec{R_B} \):
\[ \vec{R_B} = 7 \hat{i} - 2 \hat{j} - 3 \hat{k} \]

3. **Calculate the Displacement Vector**:
- The displacement vector \( \vec{D} \) is given by:
\[ \vec{D} = \vec{R_B} - \vec{R_A} \]
- Substituting the values:
\[ \vec{D} = (7 \hat{i} - 2 \hat{j} - 3 \hat{k}) - (-3 \hat{i} + 4 \hat{j} - 3 \hat{k}) \]
- Distributing the negative sign:
\[ \vec{D} = 7 \hat{i} - 2 \hat{j} - 3 \hat{k} + 3 \hat{i} - 4 \hat{j} + 3 \hat{k} \]

4. **Combine Like Terms**:
- Combine the \( \hat{i} \) components:
\[ 7 \hat{i} + 3 \hat{i} = 10 \hat{i} \]
- Combine the \( \hat{j} \) components:
\[ -2 \hat{j} - 4 \hat{j} = -6 \hat{j} \]
- Combine the \( \hat{k} \) components:
\[ -3 \hat{k} + 3 \hat{k} = 0 \hat{k} \]

5. **Final Displacement Vector**:
- Thus, the displacement vector \( \vec{D} \) is:
\[ \vec{D} = 10 \hat{i} - 6 \hat{j} + 0 \hat{k} \]
- This can be simplified to:
\[ \vec{D} = 10 \hat{i} - 6 \hat{j} \]

### Final Answer:
The bird's displacement vector is:
\[ \vec{D} = 10 \hat{i} - 6 \hat{j} \]