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  • If two lines L1 and L2 inspace, are defined by  <img alt="L_{1}= \left \{ x=\sqrt{\lambda } y+(\sqrt{\lambda }-1) , z= (\sqrt{\lambda }-1)y+\sqrt{\lam

If two lines L1 and L2 inspace, are defined by 

L_{1}= \left \{ x=\sqrt{\lambda } y+(\sqrt{\lambda }-1) , z= (\sqrt{\lambda }-1)y+\sqrt{\lambda }\right \} and  L_{2}= \left \{ x=\sqrt{\mu } y+(1-\sqrt{\mu }) , z= (1-\sqrt{\mu })y+\sqrt{\mu }\right \} then L1 is perpendicular to L2 for all non-negative reals \lambda and \mu, such that

Option: 1

\sqrt{\lambda} +\sqrt{\mu}=1


Option: 2

\lambda \neq\mu


Option: 3

\lambda +\mu=0


Option: 4

\lambda= \mu


Answers (1)

best_answer

For L1

\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} \Rightarrow y=\frac{z-\sqrt{\lambda}}{\sqrt{\lambda}-1} \end{aligned}

From above,

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

This is the equation for L1

Similarly, for L2

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

Since, L1 and L2 are perpendicular

\\\sqrt{\lambda} \sqrt{\mu}+1 \times 1+(\sqrt{\lambda}-1)(1-\sqrt{\mu})=0 \\ \Rightarrow \sqrt{\lambda}+\sqrt{\mu}=0 \Rightarrow \sqrt{\lambda}=-\sqrt{\mu} \\ \Rightarrow \lambda=\mu

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