Distance Between Parallel Planes
Evaluate the perpendicular distance between the two parallel planes 2x - y + 2z = 5 and 2x - y + 2z = 11 in space.
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About This Question
- Subject
- mathematics
- Chapter
- three dimensional geometry
- Topic
- distance between parallel planes
- Difficulty
- Easy
- Year
- 2025
Solution
Correct Answer:
2
The applicable result is the distance between parallel planes ax + by + cz = d_1 and ax + by + cz = d_2, given by |d_1 - d_2| / \sqrt{a^2 + b^2 + c^2}. This compact formula is a routine JEE Advanced fact, valid only after the normal coefficients are made identical, which they already are here. With a = 2, b = -1, c = 2, the normal magnitude is \sqrt{4 + 1 + 4} = 3. The constants are d_1 = 5 and d_2 = 11, so |d_1 - d_2| = 6, and the distance is 6/3 = 2. Option 3 mistakenly divides by 2 instead of 3. Option 6 forgets to divide by the normal magnitude entirely. Option 1 halves the constant difference incorrectly. This relies on the parallel-plane distance theorem, which is the difference of constant terms projected onto the unit normal. Plausibility check: the planes share the same normal direction so they are genuinely parallel, and the positive value 2 represents a consistent uniform gap between them everywhere in space.
This easy difficulty mathematics question is from the chapter three dimensional geometry, covering the topic of distance between parallel planes. It appeared in the 2025 exam.
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