Vectors — Displacement Problems (Class 10 & 11)

Distance: Total path length travelled (scalar — no direction) Displacement: Straight-line distance from start point to end point (vector — has direction)

Example: A car travels 17 km north, then 17 km east.

• Distance = 17 + 17 = 34 km (total path)
• Displacement = √(17² + 17²) = √(289 + 289) = √578 = 24.04 km (NE direction)

Type 1: Perpendicular movements A car moves 17 km north from point A, then 17 km east to point B. Find displacement AB.

Solution: Since north and east are perpendicular: AB² = 17² + 17² AB² = 289 + 289 = 578 AB = √578 = 17√2 ≈ 24.04 km

Direction: tan θ = 17/17 = 1, θ = 45° → Northeast (NE)

Type 2: Opposite directions A person walks 5 km east, then 3 km west. Displacement = 5 − 3 = 2 km east (same line, subtract opposite)

Type 3: Using the formula for any angle Resultant R = √(A² + B² + 2AB cosθ) Where A, B are magnitudes and θ is angle between them.

1. Same direction: Add magnitudes, same direction
2. Opposite direction: Subtract smaller from larger, take larger's direction
3. Perpendicular (90°): Use Pythagoras: R = √(A² + B²)
4. Any angle: R = √(A² + B² + 2AB cosθ)

Direction of resultant (for perpendicular): tanθ = (vertical component)/(horizontal component)