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 bird's displacement vector from the initial position to the final position in the xyz-coordinates, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Initial and Final Position Vectors**:
- The initial position vector \( \mathbf{r_1} \) is given as \( (-3 \, \text{m}, 4 \, \text{m}, -3 \, \text{m}) \).
- The final position vector \( \mathbf{r_2} \) is given as \( (7 \, \text{m}, -2 \, \text{m}, -3 \, \text{m}) \).

2. **Express the Position Vectors in Unit Vector Notation**:
- The initial position vector can be expressed as:
\[ \mathbf{r_1} = -3 \hat{i} + 4 \hat{j} - 3 \hat{k} \]
- The final position vector can be expressed as:
\[ \mathbf{r_2} = 7 \hat{i} - 2 \hat{j} - 3 \hat{k} \]

3. **Calculate the Displacement Vector**:
- The displacement vector \( \mathbf{d} \) is given by the difference between the final and initial position vectors:
\[ \mathbf{d} = \mathbf{r_2} - \mathbf{r_1} \]
- Substituting the values:
\[ \mathbf{d} = (7 \hat{i} - 2 \hat{j} - 3 \hat{k}) - (-3 \hat{i} + 4 \hat{j} - 3 \hat{k}) \]

4. **Simplify the Displacement Vector**:
- Distributing the negative sign:
\[ \mathbf{d} = 7 \hat{i} - 2 \hat{j} - 3 \hat{k} + 3 \hat{i} - 4 \hat{j} + 3 \hat{k} \]
- Combine like terms:
- For \( \hat{i} \):
\[ 7 + 3 = 10 \hat{i} \]
- For \( \hat{j} \):
\[ -2 - 4 = -6 \hat{j} \]
- For \( \hat{k} \):
\[ -3 + 3 = 0 \hat{k} \]
- Thus, the displacement vector is:
\[ \mathbf{d} = 10 \hat{i} - 6 \hat{j} + 0 \hat{k} \]

5. **Final Result**:
- The displacement vector can be written as:
\[ \mathbf{d} = 10 \hat{i} - 6 \hat{j} \]