To determine the location of the school relative to the student's home, we will calculate the net displacement using vectors.

Step 1: Break each movement into components

The movements are:

1. 2 miles N45°W:

   • North-West at a 45° angle implies equal components in the north and west directions.
   • North component: $2 \cdot \cos(45^\circ) = 2 \cdot \frac{\sqrt{2}}{2} = \sqrt{2}$ miles.
   • West component: $2 \cdot \sin(45^\circ) = 2 \cdot \frac{\sqrt{2}}{2} = \sqrt{2}$ miles.

2. 2 miles N45°E:

   • North-East at a 45° angle implies equal components in the north and east directions.
   • North component: $2 \cdot \cos(45^\circ) = \sqrt{2}$ miles.
   • East component: $2 \cdot \sin(45^\circ) = \sqrt{2}$ miles.

3. 3 miles West:

   • North component: $0$ miles.
   • West component: $3$ miles.

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Step 2: Add components

North components:

$\text{Total North} = \sqrt{2} + \sqrt{2} = 2\sqrt{2} \, \text{miles}.$

East-West components:

$\text{Total West} = \sqrt{2} + 3 - \sqrt{2} = 3 \, \text{miles}.$

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Step 3: Calculate net displacement

The net displacement is a vector with:

• North component: $2\sqrt{2}$.
• West component: $3$.

The magnitude of the displacement is: $\text{Distance} = \sqrt{(2\sqrt{2})^2 + 3^2} = \sqrt{8 + 9} = \sqrt{17} \, \text{miles}.$

The direction (relative to west) is: $\theta = \tan^{-1}\left(\frac{\text{North}}{\text{West}}\right) = \tan^{-1}\left(\frac{2\sqrt{2}}{3}\right).$

Approximating: $\theta = \tan^{-1}(0.943) \approx 43^\circ.$

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Final Answer:

The school is approximately $\sqrt{17} \approx 4.12$ miles at $43^\circ$ north of west relative to the student's home.

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Do you need further clarification or details? Here are 5 follow-up questions:

1. How are angles in navigation typically measured, and why is this important for such problems?
2. Could you confirm the direction if the student reversed their steps?
3. What would be the net displacement if the third leg of the journey was east instead of west?
4. How would you compute displacement if the path included altitude changes?
5. Can this problem be solved graphically instead of mathematically?

Tip: Always break vectors into components to simplify complex directional problems.