The value of \(\lambda\) for which the lines \(\displaystyle\frac{2-x}{3}=\frac{3-4y}{5}=\frac{z-2}{3}\) and \(\displaystyle\frac{x-2}{-3}=\frac{2y-4}{3}=\frac{2-z}{\lambda}\) are perpendicular is:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \(\displaystyle\frac{19}{8}\)
Direction ratios can be read off only from the standard form \(\frac{x-x_1}{a}=\frac{y-y_1}{b}=\frac{z-z_1}{c}\), where x, y and z each have coefficient +1.
First line:
\( \displaystyle \frac{2-x}{3} = \frac{x-2}{-3},\quad \frac{3-4y}{5} = \frac{y-\frac{3}{4}}{-\frac{5}{4}} \)
\( \displaystyle \text{direction ratios } \left(-3,\ -\frac{5}{4},\ 3\right) \)
Second line:
\( \displaystyle \frac{2y-4}{3} = \frac{y-2}{\frac{3}{2}},\quad \frac{2-z}{\lambda} = \frac{z-2}{-\lambda} \)
\( \displaystyle \text{direction ratios } \left(-3,\ \frac{3}{2},\ -\lambda\right) \)
The lines are perpendicular when \(a_1a_2 + b_1b_2 + c_1c_2 = 0\):
\( \displaystyle (-3)(-3) + \left(-\frac{5}{4}\right)\left(\frac{3}{2}\right) + (3)(-\lambda) = 0 \)
\( \displaystyle 9 - \frac{15}{8} - 3\lambda = 0 \)
\( \displaystyle 3\lambda = \frac{57}{8} \)
\( \displaystyle \lambda = \frac{19}{8} \)
Tip: Reading the denominators as printed, (3, 5, 3) and (−3, 3, λ), gives −9 + 15 + 3λ = 0, that is λ = −2, Option (1). The coefficients of x, y, z must be made +1 before the direction ratios are read.