From A to B: 3 km East. From B, turn left and go 3× that distance to C. From C, turn left and go 5× the A→B distance to reach D. What is the shortest distance from A (start) to D (destination)?

Introduction / Context:Multi-leg vector problems require careful scaling of segments and consistent turn directions. We will compute coordinates after each segment and then the straight-line distance from start to finish.

Given Data / Assumptions:

• A→B: 3 km East.
• B→C: left turn, distance = 3 × 3 = 9 km (East→left = North).
• C→D: left turn again, distance = 5 × 3 = 15 km (North→left = West).

Concept / Approach:Use coordinate accumulation: East +x, West −x, North +y, South −y. Then apply the distance formula.

Step-by-Step Solution:Start A at (0,0).After A→B: (3, 0).After B→C (North 9): (3, 9).After C→D (West 15): (−12, 9).Distance A→D = sqrt((−12)^2 + 9^2) = sqrt(144 + 81) = sqrt(225) = 15 km.

Verification / Alternative check:The 9-12-15 triple is a scaled 3-4-5 triangle (×3). Diagram confirms displacement of 12 West and 9 North.

Why Other Options Are Wrong:12, 16, 18 km do not match the exact Pythagorean distance computed.

Common Pitfalls:Applying “left” relative to the fixed East–West axis instead of the current heading.

Final Answer:15 KM