Plane Through Line Of Intersection
A family of planes passes through the line of intersection of x + y + z = 1 and 2x + 3y - z = 4 in space.
Select the correct option:
Solution
(x+y+z−1)+λ(2x+3y−z−4)=0
The governing concept is the pencil-of-planes idea: any plane through the line of intersection of P_1 = 0 and P_2 = 0 can be written as P_1 + \lambda P_2 = 0 for some scalar \lambda. This linear-combination representation is a powerful and frequently examined JEE Advanced construction. Writing P_1 = x + y + z - 1 and P_2 = 2x + 3y - z - 4, the family becomes (x + y + z - 1) + \lambda(2x + 3y - z - 4) = 0. Any point on both original planes makes P_1 and P_2 zero, hence satisfies this combination for every \lambda, so the whole line of intersection lies in each member. Option two drops the constant terms, so its members no longer pass through the correct line. Option three is a product representing two separate planes, not a single plane. Option four mismatches the structure and the constant. This applies the family-of-planes theorem. Plausibility check: substituting any point common to P_1 = 0 and P_2 = 0 yields 0 + \lambda \cdot 0 = 0, confirming every such plane contains the intersection line.
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About This Question
- Subject
- mathematics
- Chapter
- three dimensional geometry
- Topic
- plane through line of intersection
- Difficulty
- Medium
- Year
- 2025
Solution
Correct Answer:
(x+y+z−1)+λ(2x+3y−z−4)=0
The governing concept is the pencil-of-planes idea: any plane through the line of intersection of P_1 = 0 and P_2 = 0 can be written as P_1 + \lambda P_2 = 0 for some scalar \lambda. This linear-combination representation is a powerful and frequently examined JEE Advanced construction. Writing P_1 = x + y + z - 1 and P_2 = 2x + 3y - z - 4, the family becomes (x + y + z - 1) + \lambda(2x + 3y - z - 4) = 0. Any point on both original planes makes P_1 and P_2 zero, hence satisfies this combination for every \lambda, so the whole line of intersection lies in each member. Option two drops the constant terms, so its members no longer pass through the correct line. Option three is a product representing two separate planes, not a single plane. Option four mismatches the structure and the constant. This applies the family-of-planes theorem. Plausibility check: substituting any point common to P_1 = 0 and P_2 = 0 yields 0 + \lambda \cdot 0 = 0, confirming every such plane contains the intersection line.
This medium difficulty mathematics question is from the chapter three dimensional geometry, covering the topic of plane through line of intersection. It appeared in the 2025 exam.
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