Question bank/JEE Main/Mathematics
JEE MainMathematics
Let the line L1: [IMAGE] and L2:
[IMAGE] intersect at P. The least distance of P from the
plane 3x - 4y - 12z + 4 = 0, equals
Show answer and solution
Any point of L1 and L2 can be
(2λ + 1, 3λ + 2, 4λ + 3) and (µ + 2, 2µ + 3, 4µ + k) respectively.
For point of intersection P
2λ + 1 = µ + 2....(i)
3λ + 2 = 2µ + 3....(ii)
4λ + 3 = 4µ + k....(iii)
Solving (i) and (ii) we have λ = 1 and µ =
(2λ + 1, 3λ + 2, 4λ + 3) and (µ + 2, 2µ + 3, 4µ + k) respectively.
For point of intersection P
2λ + 1 = µ + 2....(i)
3λ + 2 = 2µ + 3....(ii)
4λ + 3 = 4µ + k....(iii)
Solving (i) and (ii) we have λ = 1 and µ =
Source: Grass-G / M-01 Mathematics Paper-1.docx
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