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Difficulty: HardDirection and Distance Test

One morning after sunrise, Rohan starts walking from point P directly towards the shadow cast by a tall vertical flagpole. After walking 18 m18\text{ m}, he turns 135∘135^\circ clockwise and walks 102 m10\sqrt{2}\text{ m}. Next, he turns 45∘45^\circ anti-clockwise and walks 5 m5\text{ m}. Finally, he turns to his right and walks 16 m16\text{ m} to reach point Q. What is the shortest straight-line distance between point P and point Q, and in which direction is point Q with respect to point P?

  1. 17 meters17\text{ meters}, North-EastAnswer
  2. B
    17 meters17\text{ meters}, South-East
  3. C
    17 meters17\text{ meters}, South-West
  4. D
    25 meters25\text{ meters}, North-East

Answer

The shortest distance between point P and point Q is 17 meters17\text{ meters}, and point Q is in the North-East direction relative to point P.
The correct option correctly accounts for the shadow facing West at sunrise, breaks down the 135∘135^\circ clockwise angular movement into +10 m+10\text{ m} East and +10 m+10\text{ m} North, adds 5 m5\text{ m} North, and finishes with 16 m16\text{ m} East to reach coordinates (8,15)(8, 15). Applying the Pythagorean theorem yields 82+152=17 m\sqrt{8^2 + 15^2} = 17\text{ m} in the North-East quadrant.

Step-by-Step Solution

1
Determine initial direction from morning shadow context.
At sunrise, the sun is in the East, so shadows fall toward the West. Walking towards the shadow means walking West for 18 m18\text{ m}. Position relative to P(0,0)P(0,0) is (−18,0)(-18, 0), facing West (180∘180^\circ).
Morning sun casts shadows due West.
2
Calculate displacement after a 135∘135^\circ clockwise turn.
From West (180∘180^\circ), turning 135∘135^\circ clockwise faces North-East (45∘45^\circ). Walking 102 m10\sqrt{2}\text{ m} North-East adds +102cos⁡(45∘)=+10 m+10\sqrt{2}\cos(45^\circ) = +10\text{ m} East and +102sin⁡(45∘)=+10 m+10\sqrt{2}\sin(45^\circ) = +10\text{ m} North. New position: (−18+10,0+10)=(−8,10)(-18 + 10, 0 + 10) = (-8, 10).
Decompose diagonal vector into Cartesian components.
3
Calculate displacement after a 45∘45^\circ anti-clockwise turn.
From North-East (45∘45^\circ), turning 45∘45^\circ anti-clockwise faces North (90∘90^\circ). Walking 5 m5\text{ m} North updates position to (−8,10+5)=(−8,15)(-8, 10 + 5) = (-8, 15).
Anti-clockwise rotation shifts facing direction 45∘45^\circ to the left.
4
Calculate displacement after a right turn.
Facing North, a right turn faces East (0∘0^\circ). Walking 16 m16\text{ m} East updates position to (−8+16,15)=(8,15)(-8 + 16, 15) = (8, 15). Point Q is at (8,15)(8, 15).
A right turn relative to North points East.
5
Calculate straight-line distance and final direction relative to origin P.
Distance PQ=82+152=64+225=289=17 mPQ = \sqrt{8^2 + 15^2} = \sqrt{64 + 225} = \sqrt{289} = 17\text{ m}. Since x=+8x = +8 (East) and y=+15y = +15 (North), Q is in the North-East direction relative to P.
Apply the Pythagorean theorem x2+y2\sqrt{x^2 + y^2} for displacement magnitude.

Key Concept

Vector displacement combining cardinal direction shadows, angular rotations, and Pythagorean distance calculation.
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Topics

Direction and Distance Test

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