Question

If two lines \(L_1\) and \(L_2\) in space, are defined by
\(L_1 = \{x = \sqrt{\lambda}\,y + (\sqrt{\lambda} - 1),\ z = (\sqrt{\lambda} - 1)\,y + \sqrt{\lambda}\}\) and
\(L_2 = \{x = \sqrt{\mu}\,y + (1 - \sqrt{\mu}),\ z = (1 - \sqrt{\mu})\,y + \sqrt{\mu}\}\)
then \(L_1\) is perpendicular to \(L_2\), for all nonnegative reals \(\lambda\) and \(\mu\), such that :

• A. \(\sqrt{\lambda} + \sqrt{\mu} = 1\)
• B. \(\lambda \ne \mu\)
• C. \(\lambda + \mu = 0\)
• D. \(\lambda = \mu\)

Hint:

Write each line with \(y\) as parameter, read the direction ratios, and set their dot product to zero.

Answer 1

The Correct Option is C

Take \(y\) as the parameter.
\(L_1\): \(x = \sqrt{\lambda}\,y + (\sqrt{\lambda}-1)\), \(z = (\sqrt{\lambda}-1)\,y + \sqrt{\lambda}\), so direction ratios are \((\sqrt{\lambda},\ 1,\ \sqrt{\lambda}-1)\).
\(L_2\): direction ratios are \((\sqrt{\mu},\ 1,\ 1-\sqrt{\mu})\).
Perpendicular: \(\sqrt{\lambda}\sqrt{\mu} + 1 + (\sqrt{\lambda}-1)(1-\sqrt{\mu}) = 0\).
Expand: \(\sqrt{\lambda\mu} + 1 + \sqrt{\lambda} - \sqrt{\lambda\mu} - 1 + \sqrt{\mu} = 0\), i.e. \(\sqrt{\lambda} + \sqrt{\mu} = 0\).
With \(\lambda, \mu \ge 0\), both square roots are non-negative, so both must be zero: \(\lambda = \mu = 0\), equivalently \(\lambda + \mu = 0\).
Answer: (3) \(\lambda + \mu = 0\)