For L1, x=λy+(λ−1)⇒y=λx−(λ−1)z=(λ−1)y+λ⇒y=λ−1z−λ From (i) and (ii) λx−(λ−1)=1y−0=λ−1z−λ The equation (A) is the equation of line L1. Similarly equation of line L2 is μx−(1−μ)=1y−0=1−μz−μ Since L1⊥L2, therefore λμ+1×1+(λ−1)(1−μ)=0⇒λ+μ=0⇒λ=−μ⇒λ=μ
If two lines L_1 and L_2 in space, are defined by … z=(1-√μ) y+√μ then L_1 is perpendicular to L_2, for all nonnegative reals λ and μ, such that :
JEE Main 2013 — Mathematics Coordinate Geometry
If two lines L1 and L2 in space, are defined by L1={x=λy+(λ−1),z=(λ−1)y+λ} and L2={x=μy+(1−μ), z=(1−μ)y+μ} then L1 is perpendicular to L2, for all nonnegative reals λ and μ, such that :
Held on 23 Apr 2013 · Verified 6 Jul 2026.
Options
- A
λ+μ=1
- B
λ=μ
- C
λ+μ=0
- D
λ=μ
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