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 the Vectors**:
- Vector \( \mathbf{a} \) points North-East, which corresponds to an angle of \( 45^\circ \) from the positive x-axis (East).
- Vector \( \mathbf{b} \) points North-West, which corresponds to an angle of \( 135^\circ \) from the positive x-axis.

2. **Expressing the Vectors in Component Form**:
- The components of vector \( \mathbf{a} \) can be expressed as:
\[ \mathbf{a} = 5 \left( \cos 45^\circ, \sin 45^\circ \right) = 5 \left( \frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2} \right) = \left( \frac{5\sqrt{2}}{2}, \frac{5\sqrt{2}}{2} \right) \]
- The components of vector \( \mathbf{b} \) can be expressed as:
\[ \mathbf{b} = 5 \left( \cos 135^\circ, \sin 135^\circ \right) = 5 \left( -\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2} \right) = \left( -\frac{5\sqrt{2}}{2}, \frac{5\sqrt{2}}{2} \right) \]

3. **Finding \( \mathbf{a} - \mathbf{b} \)**:
- Now, we calculate \( \mathbf{a} - \mathbf{b} \):
\[ \mathbf{a} - \mathbf{b} = \left( \frac{5\sqrt{2}}{2} - \left(-\frac{5\sqrt{2}}{2}\right), \frac{5\sqrt{2}}{2} - \frac{5\sqrt{2}}{2} \right) \]
\[ = \left( \frac{5\sqrt{2}}{2} + \frac{5\sqrt{2}}{2}, 0 \right) = \left( 5\sqrt{2}, 0 \right) \]

4. **Calculating the Magnitude**:
- The magnitude of \( \mathbf{a} - \mathbf{b} \) is:
\[ | \mathbf{a} - \mathbf{b} | = \sqrt{(5\sqrt{2})^2 + 0^2} = \sqrt{50} = 5\sqrt{2} \]

### Final Answer:
Thus, the magnitude \( | \mathbf{a} - \mathbf{b} | \) is equal to \( 5\sqrt{2} \).

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