Two walls and a ceiling of a room meet at right angles at a point P. A fly is in the air 1 m from one wall, 8 m from the other wall and 9 m from the point P. How many meters is the fly from the ceiling?
- (a) 4 ✓ UPSC's answer
- (b) 6
- (c) 12
- (d) 15
Why the answer is (a)
• Let the point P be the origin (0,0,0) where the two walls and the ceiling meet at right angles.
• The fly's position can be represented by coordinates (x, y, z), where x, y, and z are the perpendicular distances from the respective planes.
• The problem states the fly is 1 m from one wall and 8 m from the other, so let x = 1 and y = 8.
• The distance from the point P (origin) to the fly is given as 9 m, which corresponds to the magnitude of the position vector: $\sqrt{x^2 + y^2 + z^2} = 9$.
• Squaring both sides gives $x^2 + y^2 + z^2 = 81$.
• Substituting the known values: $1^2 + 8^2 + z^2 = 81 \Rightarrow 1 + 64 + z^2 = 81 \Rightarrow 65 + z^2 = 81$.
• Solving for $z^2$: $z^2 = 81 - 65 = 16$, so $z = \sqrt{16} = 4$.
• Therefore, the fly is 4 meters from the ceiling.
Why the other options are wrong
- (b) 6
- Option (b) 6 is incorrect because $1^2 + 8^2 + 6^2 = 1 + 64 + 36 = 101$, which is not equal to $9^2 = 81$.
- (c) 12
- Option (c) 12 is incorrect because $1^2 + 8^2 + 12^2 = 1 + 64 + 144 = 209$, which is not equal to $9^2 = 81$.
- (d) 15
- Option (d) 15 is incorrect because $1^2 + 8^2 + 15^2 = 1 + 64 + 225 = 290$, which is not equal to $9^2 = 81$.
Asked in the CSAT Paper II of the UPSC Civil Services Preliminary Examination 2017, held on 18 June 2017. Question and answer key: Union Public Service Commission. Explanation: UPSC Answer Check.