If two lines L_1 and L_2 in space, are defined by aligned & L_1=\ x= λ y+( λ -1), & z=( λ -1) y+ λ and & L_2=\ x= μ y+(1- μ ), aligned z=(1- μ ) y+ μ then L_1 is perpendicular to L_2 , for all nonnegative reals λ and μ

Correct answer: D. λ=μ For L_1 , aligned & x= λ y+( λ -1) y= x-( λ -1) λ & z=( λ -1) y+ λ = z- λ λ -1 aligned From (i) and (ii) x-( λ -1) λ =(y-0)/(1)= z- λ λ -1 The equation (A) is the equation of line L_1 . Similarly equation of line L_2 is x-(1- μ ) μ =(y-0)/(1)= z- μ 1- μ Since L_1 L_2 , therefore aligned & λ μ +1 ×₁⁺( λ -1)(1- μ )=0 & λ + μ =0 λ =- μ & λ=μ aligned

JEE Main 201323 AprilOnline

Exam: JEE Main
Year: 2013
Date: 23 April
Mode: Online
Subject: Mathematics
Chapter: Straight Lines

JEE Main 2013 · 23 April · Online · Mathematics · Straight Lines

If two lines L_1 and L_2 in space, are defined by aligned & L_1=\...

If two lines

L_1 and L_2

in space, are defined by

\begin{aligned}& L_1=\{x=\sqrt{\lambda}y+(\sqrt{\lambda}-1), \\ & \quad z=(\sqrt{\lambda}-1) y+\sqrt{\lambda} \text{ and }\\ & L_2=\{x=\sqrt{\mu}y+(1-\sqrt{\mu}), \end{aligned}

z=(1-

\sqrt{\mu})

\sqrt{\mu} then

L_1

is perpendicular to

L_2

, for all nonnegative reals

\lambda and \mu$, such that :

Options

A $\sqrt{\lambda}+\sqrt{\mu}=1$
B $\lambda \neq \mu$
C $\lambda+\mu=0$
D $\lambda=\mu$

Correct answer

D. $\lambda=\mu$

Step-by-step solution

For

\mathrm{L}_1

\begin{aligned}& x=\sqrt{\lambda}y+(\sqrt{\lambda}-1) \Rightarrow y=\frac{x-(\sqrt{\lambda}-1)}{\sqrt{\lambda}}\\ & z=(\sqrt{\lambda}-1) y+\sqrt{\lambda} \Rightarrowy=\frac{z-\sqrt{\lambda}}{\sqrt{\lambda}-1} \end{aligned}

From (i) and (ii)

\frac{x-(\sqrt{\lambda}-1)}{\sqrt{\lambda}}=\frac{y-0}{1}=\frac{z-\sqrt{\lambda}}{\sqrt{\lambda}-1}

The equation (A) is the equation of line

\mathrm{L}_1

. Similarly equation of line

\mathrm{L}_2

\frac{x-(1-\sqrt{\mu})}{\sqrt{\mu}}=\frac{y-0}{1}=\frac{z-\sqrt{\mu}}{1-\sqrt{\mu}}

Since

\mathrm{L}_1 \perp \mathrm{L}_2

, therefore

\begin{aligned}& \sqrt{\lambda} \sqrt{\mu}+1 \times₁⁺(\sqrt{\lambda}-1)(1-\sqrt{\mu})=0 \\ & \Rightarrow \sqrt{\lambda}+\sqrt{\mu}=0 \Rightarrow \sqrt{\lambda}=-\sqrt{\mu}\\ & \Rightarrow \lambda=\mu \end{aligned}

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