The lines x = ay + b, z = cy + d and x = a’y + b’, z = c’y + d’ will be mutually perpendicular provided :
Answer & explanation
Correct answer: option 2
Give lines are
$\frac{x-b}{a}=y=\frac{z-d}{c}$ and $\frac{x-b'}{a'}=y=\frac{z-d'}{c'}$
These lines will be mutually perpendicular, provided
a.a' + 1.1' + c.c' =0
⇒ a.a' + c.c' + 1 = 0