If a has magnitude 5 and points North-East and vector b has magnitude 5 and point North-West, then |a-b| is equal to

Options

A. `25`
B. `5`
C. `7sqrt(3)`
D. `5sqrt(2)`

Answer

To solve the problem, we need to find the magnitude of the vector \( | \mathbf{a} - \mathbf{b} | \), where vector \( \mathbf{a} \) has a magnitude of 5 and points North-East, and vector \( \mathbf{b} \) has a magnitude of 5 and points North-West.

### Step-by-Step Solution:

1. **Understanding Directions**:
- North-East means the vector is at a 45-degree angle from both the North and East axes.
- North-West means the vector is at a 45-degree angle from both the North and West axes.

2. **Components of Vector \( \mathbf{a} \)**:
- The components of vector \( \mathbf{a} \) can be calculated using trigonometry:
\[ \mathbf{a} = 5 \cos(45^\circ) \hat{i} + 5 \sin(45^\circ) \hat{j} \]
- Since \( \cos(45^\circ) = \sin(45^\circ) = \frac{1}{\sqrt{2}} \):
\[ \mathbf{a} = 5 \left( \frac{1}{\sqrt{2}} \hat{i} + \frac{1}{\sqrt{2}} \hat{j} \right) = \frac{5}{\sqrt{2}} \hat{i} + \frac{5}{\sqrt{2}} \hat{j} \]

3. **Components of Vector \( \mathbf{b} \)**:
- The components of vector \( \mathbf{b} \) are:
\[ \mathbf{b} = 5 \cos(135^\circ) \hat{i} + 5 \sin(135^\circ) \hat{j} \]
- Here, \( \cos(135^\circ) = -\frac{1}{\sqrt{2}} \) and \( \sin(135^\circ) = \frac{1}{\sqrt{2}} \):
\[ \mathbf{b} = 5 \left( -\frac{1}{\sqrt{2}} \hat{i} + \frac{1}{\sqrt{2}} \hat{j} \right) = -\frac{5}{\sqrt{2}} \hat{i} + \frac{5}{\sqrt{2}} \hat{j} \]

4. **Calculating \( \mathbf{a} - \mathbf{b} \)**:
- Now we subtract \( \mathbf{b} \) from \( \mathbf{a} \):
\[ \mathbf{a} - \mathbf{b} = \left( \frac{5}{\sqrt{2}} \hat{i} + \frac{5}{\sqrt{2}} \hat{j} \right) - \left( -\frac{5}{\sqrt{2}} \hat{i} + \frac{5}{\sqrt{2}} \hat{j} \right) \]
- This simplifies to:
\[ \mathbf{a} - \mathbf{b} = \left( \frac{5}{\sqrt{2}} + \frac{5}{\sqrt{2}} \right) \hat{i} + \left( \frac{5}{\sqrt{2}} - \frac{5}{\sqrt{2}} \right) \hat{j} = \frac{10}{\sqrt{2}} \hat{i} + 0 \hat{j} \]
- Therefore:
\[ \mathbf{a} - \mathbf{b} = \frac{10}{\sqrt{2}} \hat{i} \]

5. **Magnitude of \( \mathbf{a} - \mathbf{b} \)**:
- The magnitude is given by:
\[ | \mathbf{a} - \mathbf{b} | = \left| \frac{10}{\sqrt{2}} \hat{i} \right| = \frac{10}{\sqrt{2}} = 5\sqrt{2} \]

### Final Answer:
\[ | \mathbf{a} - \mathbf{b} | = 5\sqrt{2} \]