If lines \(\displaystyle\frac{x+5}{5\lambda+2}=\frac{4-2y}{10}=\frac{1-3z}{-3}\) and \(\displaystyle\frac{x-2}{1}=\frac{1+2y}{4\lambda}=\frac{2+z}{3}\) are perpendicular, then value of '\(\lambda\)' is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 1
The denominators give the direction ratios only when x, y and z each have coefficient 1 in the numerators, so the terms are rewritten first.
First line:
\(\displaystyle \frac{4 - 2y}{10} = \frac{-2(y - 2)}{10} = \frac{y - 2}{-5}\)
\(\displaystyle \frac{1 - 3z}{-3} = \frac{-3\left(z - \frac{1}{3}\right)}{-3} = \frac{z - \frac{1}{3}}{1}\)
So its direction ratios are \((5\lambda + 2,\ -5,\ 1)\).
Second line:
\(\displaystyle \frac{1 + 2y}{4\lambda} = \frac{2\left(y + \frac{1}{2}\right)}{4\lambda} = \frac{y + \frac{1}{2}}{2\lambda}\)
So its direction ratios are \((1,\ 2\lambda,\ 3)\).
For perpendicular lines, \(a_1a_2 + b_1b_2 + c_1c_2 = 0\):
\((5\lambda + 2)(1) + (-5)(2\lambda) + (1)(3) = 0\)
\(5 - 5\lambda = 0\)
\(\lambda = 1\)