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JEE Mains · Maths · STD 12 - 11. three dimension geometry
If two lines \(L_1\) and \(L_2\) in space, are defined by \({L_1} = \{ x = \sqrt \lambda y + \left( {\sqrt \lambda - 1} \right),z = \left( {\sqrt \lambda - 1} \right)y + \sqrt \lambda \} \) and \({L_2} = \{ x = \sqrt \mu y + \left( {1 - \sqrt \mu } \right),z = \left( {1 - \sqrt \mu } \right)y + \sqrt \mu \} \) then \(L_1\) is perpendicular to \(L_2\), for all non-negative reals \(\lambda \) and \( \mu \), such that
- A \(\sqrt \lambda + \sqrt \mu = 1\)
- B \(\lambda \ne \mu \)
- C \(\lambda + \mu = 0\)
- D \(\lambda = \mu \)
Answer & Solution
Correct Answer
(D) \(\lambda = \mu \)
Step-by-step Solution
Detailed explanation
For \(L_{1}\) \(x=\sqrt{\lambda} y+(\sqrt{\lambda}-1) \Rightarrow y=\frac{x-(\sqrt{\lambda}-1)}{\sqrt{\lambda}}\) .....\((i)\) \(z=(\sqrt{\lambda}-1) y+\sqrt{\lambda} \Rightarrow y=\frac{z-\sqrt{\lambda}}{\sqrt{\lambda}-1}\) ......\((ii)\) From \((i)\) and \((ii)\)…
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