Question:
If the lines $x=a y+b, z=c y+d$ and $x=a^{\prime} z+b^{\prime}$, $\mathrm{y}=\mathrm{c}^{\prime} \mathrm{z}+\mathrm{d}^{\prime}$ are perpendicular, then:
Correct Option: , 2
Solution:
Line $x=a y+b, z=c y+d \Rightarrow \frac{x-b}{a}=\frac{y}{1}=\frac{z-d}{c}$
Line $x=a^{\prime} z+b^{\prime}, y=c^{\prime} z+d^{\prime}$
$\Rightarrow \frac{\mathrm{x}-\mathrm{b}^{\prime}}{\mathrm{a}^{\prime}}=\frac{\mathrm{y}-\mathrm{d}^{\prime}}{\mathrm{c}^{\prime}}=\frac{\mathrm{z}}{1}$
Given both the lines are perpendicular $\Rightarrow \mathrm{aa}^{\prime}+\mathrm{c}^{\prime}+\mathrm{c}=0$