3D Geometry
Three Dimensional Geometry
star_batch_jee_advanced_2025
Grade 12
Question:
The straight lines whose direction cosines are given by the relations $al + bm + cn = 0$ and $fnn + gnl + hlm = 0$ are perpendicular if:
\frac{f}{a} + \frac{g}{b} + \frac{h}{c} = 0
\frac{a}{f} + \frac{b}{g} + \frac{c}{h}
\frac{h}{a} + \frac{g}{b} + \frac{f}{c}
None of these
Step-by-Step Solution
Key Concept: Product of slopes of two lines from a quadratic equation equals ratio of constant to leading coefficient, and perpendicularity requires sum of products of direction cosines equals zero.
Given $I = \left(\frac{-bm-cn}{a}\right) = f\mn + (gn + hm)\left(\frac{-bm-cn}{a}\right) = 0$, we derive $-\frac{hb}{a}\left(\frac{m}{n}\right)^2 + \left(f - \frac{gb}{a} - \frac{ch}{a}\right)\frac{m}{n} - \frac{gc}{a} = 0$. If $\frac{m_1}{n_1}$ and $\frac{m_2}{n_2}$ are roots, then $\frac{m_1}{n_1} \cdot \frac{m_2}{n_2} = \frac{gc}{hb}$. Two lines are perpendicular when $l_1l_2 + m_1m_2 + n_1n_2 = 0$, which simplifies to $\frac{gc}{hb} + \frac{f}{a} + \frac{g}{b} + \frac{h}{c} = 0$.
Correct Answer: 1