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Question #41664
The Question
Three planes and intersect in a common line . Let and be the unit normal vectors to and respectively, such that their components are strictly positive. It is given that . The position vectors of the foot of the perpendiculars from the origin to and are and respectively. If is the sum of the intercepts made by the plane on the coordinate axes, then the value of is:
Options
- A.
- B.
- C.
- D.
Check Your Answer
Correct Answer
Step-by-Step Solution
Explanation
How the Examiner Thinks
Specialist
Targeted practice for niche, high-difficulty concepts that frequently appear in JEE Advanced exams.
Concepts Behind This Question
Variable plane intercepts|Linear Independence vs Dependence|Dot Product Projection Sign|Collinearity vs Coplanarity
Why This Question Is Tricky
Common Mistakes
This question tests the student's understanding of the vector equation of a plane and the geometric interpretation of the dot product projection.
The primary trap lies in the sign of the perpendicular distance from the origin.
Students often incorrectly assume that the distance in must be strictly positive, taking the absolute value .
This 'Dot Product Projection Sign' error flips the orientation of the plane, resulting in inverted intercepts and a sum of instead of .
Additionally, the problem requires recognizing that for planes to be concurrent in a line, the linear dependence of their coplanar normal vectors must perfectly mirror the linear combination of their scalar distances, penalizing those who confuse collinearity with coplanarity.
Quick Facts About This Question
Question Information
| Subject | Mathematics |
| Chapter | Three-Dimensional Geometry - Class 12 |
| Concept | Variable plane intercepts |
| Difficulty | 8 |
| Cognitive Load | 7 |
| Target Solve Time | 160 sec |
| Access | Free |