If the lines \( x=ay+b,z=cy+d \) and \( x=a'y+b',z=c'y+d' \) are perpendicular, then
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( a\,a'+c\,c'=-1 \)
With y as the parameter, the lines are
\( \displaystyle \frac{x-b}{a}=\frac{y}{1}=\frac{z-d}{c} \)
\( \displaystyle \frac{x-b'}{a'}=\frac{y}{1}=\frac{z-d'}{c'} \)
with direction ratios (a, 1, c) and (a′, 1, c′). For perpendicular lines,
\( aa'+1\cdot 1+cc'=0 \)
\( aa'+cc'=-1 \)
Why the other options are incorrect
Option (1) \( a\,a'+c\,c'=1 \) — Incorrect. The sum is −1, not 1.
Option (2) \( a\,a'+c\,c'=0 \) — Incorrect. This forgets the product of the middle direction ratios, 1 × 1.
Option (4) \( a\,a'=c\,c' \) — Incorrect. The condition is \( aa'+cc'=-1 \), not \( aa'=cc' \).