Umesh directly went from P, to Q which is 9 feet distant. Then he turns to the…

Concept: In a direction-sense problem, plot the walker's path on a coordinate plane. Assign the starting point as the origin and a starting heading. Each 'turn right' rotates the current heading by 90 degrees clockwise (East to South to West to North to East). The straight-line distance from the start to the final point is found from the vector sum of every leg: horizontal displacements in opposite directions cancel each other, and vertical displacements in opposite directions cancel each other, leaving a net horizontal and a net vertical offset that combine via the Pythagorean theorem.

1. Place P at the origin and let the direction from P to Q be East. Since PQ = 9 ft, Q is 9 ft east of P.

2. At Q, turning right changes the heading from East to South. Walking 4 ft south reaches a point 9 ft east and 4 ft south of P.

3. Turning right again changes the heading from South to West. The next leg is a distance equal to PQ, i.e. 9 ft west, which exactly cancels the earlier 9 ft eastward leg, bringing the horizontal position back in line with P.

4. Turning right once more changes the heading from West to North. Walking 3 ft north from a point 4 ft south of P leaves the final point 1 ft south of P.

5. Since the two horizontal legs are equal and opposite, they cancel completely, so the straight-line distance from P to the final point is simply the net vertical offset.

Cross-check: Because both horizontal legs are exactly 9 ft in opposite directions, the horizontal component of the final position is 0, so the distance formula (square root of horizontal squared plus vertical squared) reduces to just the vertical difference, confirming the same result without needing the full calculation.

Therefore, Umesh is 1 foot from P.